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= — j‘ dx dx' --1: 6(x'
= (c)
—
f dx
p(x, x')
x)
(2.38)
p(x, x')] x
120:
Finding the probability that a Probability that a system is in state system is found in the state Ix> is equivalent to finding the expectation value of the operator Ix> <xl. To see this for the case of a pure state, notice that if the system is in a state 10>, the probability of experimentally finding it in state Ix> is
i<X101 2
= .
Therefore, the probability that the system is found in state Ix> can be written
P = probability = Tr pixXxl = Tr
E
w11I 2 = <X1PiX>.
=
(2.39)
The equation
P
=
E wiKilx>i 2
is in accord with our interpretation of the w i as the probability that the system actually is in state ii>. The probability density at value x0 for a coordinate is the probability that the system is in state ['c o >, which by Eq. (2.39) is p(x o, x0). For a pure state it/i>, p(x o , x 0) = 10001 2. 2.3 DENSITY MATRIX IN STATISTICAL MECHANICS
When 19i > is an eigenket (eigenfunction), and Ei is the corresponding eigenvalue of the Hamiltonian H of the system, the probability that the system is in the state 19 i> is (2.40) (11Q)e - PE i.
Density matrix in statistical mechanics
2.3
47
Thus the density matrix is
P =7-
E wniqin> = — + 2 coth f (2.87) 2 1 - e -hcilkT = 2 1 - e -2f — Equation (2.86) is exactly half of Eq. (2.87). Therefore we know that --= = (h0)/4) coth f.
(2.88)
The partition function is derived from Eq. (2.84):
e - " = f p(x, x) dx . \I
1 mw \I nii 2rch sinh 2f ma) tanh f 2 sinh f
(2.89)
which leads to
1 F = - in (2 sinh f) = kT In (e'l 2kT _ e-h0)12kT)
fi
hco --=-- —
2
+ kT ln (1 - e -hallkT).
(2.90)
This is the free energy, as already derived. We may examine the limit of high temperature, or small f in Eq. (2.84). In this limit V(x) p(x, x; i3) _, exp (- inco2x2 /2) = exp (--(2.91) , kT kT ) except for the factor in front of the exponential. This result, Eq. (2.91), agrees with classical mechanics. The low-temperature limit is found by inspection of Eq. (2.83):
PI"
p(x, x') -> \
(_ ma)2x,2) (-mwx2 ) 2 exp exp (- h(D p) ex .
(2.9 )
E e -/LE'9i(x)4(x') -> e -ftE1) (P0(x)VC•(x'),
(2.93)
nh This limit is calculated as P(x, x') ----
2kT
2h
2h
Anharmonic oscillator
2.6
53
because when fl .- oo, only the ground state has an effective contribution. Equation (2.92) agrees with Eq. (2.93) because we know that for the ground state /MC° \ 4 (-mo.v2h)x2 90(X) .= (-)1/ e nh
(2.94)
2.6 ANHARMONIC OSCILLATOR
Consider the potential energy shown by the curve in Fig. 2.1, given by the following function: V(X) =
„2 2 3 Muj .X ± kx .
2
(2.95)
We will consider the area near x --= 0 only, so that the region V(x) < 0 does not come into the calculation. Because of the anharmonicity, when the temperature increases the mean position of the oscillation moves out, as shown in Fig. 2.2. Bearing this in mind,
Fig. 2.1 Anharmonic oscillator (k < 0).
Fig. 2.2 Mean position of the oscillation is displaced.
54
Density matrices
we will treat the problem using the minimum principle discussed previously. The principle is written as (2.96) F < F0 + nr = O. For example, in Eq. (2.118) we have the case for which 2
S = [— 2
y -P-i— + E T 2m
if j 1, let ( m \3v/2 f
h„ == -
dx i dx 2 • - • dx,
27Th 213) X
111 exp -- -- Rx 1 — x2) 2 + (x2 — x3) 2 -I- • • • 2h213
+ (x ,,_ 1 — x„) 2 + (x,, — x 1 ) 2]
(2.149)
in
hl = ( )3/2 f dx i -, V ( in ) 3/2 \27ch2fij \27rh2fli Then Eq. (2.148) can be written
e — fiFs =
1
N!
E.
( 11
( h)
cv) .
(2.150)
ab 2 a + b (x — z)1.
(2.151)
P v
It is not difficult to show that
J'
dy exp [
—
a(x — y) 2] exp [ .-_-
—
b(y — z) 2 ]
+ )3/2 b ( a7r
exp [ —
Symmetrized density matrix for n particles
2.8
63
Using Eq. (2.151) we obtain )3/2 hv = V
(2.152)
27rh 2 fly
We must now count the possible permutations. number of permutations that have
Let M(C i ,
. Cq)
the
C1 cycles of length 1 C2 cycles of length 2
Cq cycles of length q Then e —V S
1 -
N!
Cq)
H y
h
(2.153)
v
CO, consider the case of As a preliminary to the determination of M(C i , N 6, Ci = 3, C3 = 1, C2 = C4 = C5 = C6 =O. An example is the permutation P ---- (5)(3)(2)(461). Notice that the sequence Q = 5, 3, 2, 4, 6, 1 corresponds to P. In other words, from the sequence Q, we can uniquely construct the permutation P with C 1 = 3, C3 = 1. But from Q' = 2, 5, 3, 4, 6,1, 5,3,2,1, 4, 6, we can also construct in the same way the same P. or from Q" In fact, it can be easily seen that there are 3! times 3 = 18 different sequences that can be used in the way described above to construct the permutation P = (5)(3)(2)(461). Generalizing to any set of C's, we see that there are two ways we can get a new Q that leads to the same P: a) we can interchange cycles of the same length, b) we can make cyclic permutation within a given cycle. There are FL, Cs,! ways of interchanging cycles of the same length, and fly vcv cyclic permutations within cycles. Thus for each P there are 11, v' differentQ's that lead to the same P. Because there are N! possible Q's,
ni,
M(C o , CI) =
N!
fly
cy! v c v •
It follows that — 13 Fs
where E v vCv = N.
E
hC" ,v
vc- '
(2.154)
64
Density matrices
Instead of trying to do the sum in Eq. (2.154), we will find the free energy of the grand canonical ensemble (that is, we will let N vary). co
-13F = e
(2.155) N=1
As usual, let a = e + Pfl. Then, e
— flF
h cv v
--c„E... c,II,, cv ! vcv
ce rC, =
E ... H ci, ...c, „
where now each Cq runs from 0 to oo. Interchanging
[h,(aviv)rv
'
e — PF
y c„—o
n
Cv !
[h,(av/v)] c v '
Cy !
H and E, we have
oc v
ay
exp (it, --) --= exp (E h,
y
y
y
v
Then, fiF = —
c°
,,
h av
E -12- _ ( v= 1 V
3/2
27r /i213 -
v
czy
(2.156)
vv
This result is identical to the one we obtained in a much simpler way earlier. When the density of the gas is not large, only the identity permutation is important for high temperatures or for the limit of h — ■ 0, since we have such factors as e - E(inkT/2h2)(x - '02]. This means that the quantum effect appears only for low temperatures. For the antisymmetric case, pA (x 1 - • - x k • - • xN ., xl, • - - x;, • • • fiv)
1
E (- 1)PpD(x 1
•••
X;
Px
- • Px k' • • - Px).
P
N!
where for even permutation
(-1) p = 1 —1 for odd permutation.
Problem: Modify the argument leading to Eq. (2.156) so as to obtain the corresponding result for a noninteracting Fermi gas. 2.9 DENSITY SUBMATRIX
For N particles, the density matrix can be written p(x 1 , x 2 , . . . , x N ; x 1 . , x'2 , ... x'N) =
E i
wip i(x l , . .. , xN )4(./cri , ... , xtiv).
(2.157)
Density submatrix
2.9
65
Notice that we are not now specializing to the case of statistical mechanics. To compute the expectation value of PtI2m, we proceed as follows: 1
Tr p
\2m/
—
Tr p
1 h2
2m c a2
, xN ; x 1 .. , 4)
p(x l , Tr p 2m j .0)4 2 h2 0 =— dx -, p i (x i , 2m Ox,2
dx, • • • dx N
,
p2
= Tr p i xi =x,
1,
(2.158)
2m
where we define p i(x, x') =
p(x,
X2, . • .
XN; X', X2, ...
XN) dX2 •
dXN
Tr [p]
(2.159)
Pi is the one-particle density submatrix. Using this matrix we can calculate, for example,
P k
2
2m/ = —N
r h2 0 2 p (x x')] dx L 2m ax12 1 x'=x
n2 N Tr 0[r p i 2m
(2.16 )
(assuming that the particles are essentially identical). Similarly,
V(x k)) N f (V x)p,(x, x) dx = N Tr [Vp i ].
(2.161)
The two-particle density submatrix is defined as p2 (x i x 2 ;
= p 2(x i x2x3 • • xN ; x'l x'2 x3 • • xN) dx 3 • - • dx N
Tr [p]
(2.162)
From Eq. (2.162), the two-particle potential energy V(x 1x2) has expectation
E
V(x i, x i)
N(N — 1) 2
V(x 1 x2 )p 2(x 1 x 2 ; x 1 x2) dx, dx 2 .
(2.163)
Special cases of density submatrices are the distribution functions. The singleparticle distribution function is defined as
p 1 (x) = p1 (x;
(2.164)
The two-particle distribution function is
P2(x,
= P2(x, Y; Y).
(2.165)
In a uniform substance, such as a gas or a liquid, p 1 (x) is independent of x, and P2(x, y) depends only on the difference (x —
Density matrices
66
2.10 PERTURBATION EXPANSION OF THE DENSITY MATRIX
Recall that the density matrix in statistical mechanics satisfies 813 _-= —Hp.
(2.166)
011
There are very few Hamiltonians for which we can solve Eq. (2.166) exactly, but it may be that H is close to one such Hamiltonian, Ho . H = H o + H1, apo
(2.167) (2.168)
— HoPo•
We would like to use p o to obtain an approximation to p, which we expect to be close to po = Cfill°. Because p is close to e fill °, we expect ell°ap to vary slowly with f3. 0 011
(e"p) = Ho eH°Pp + ewe aP
018 —
eHofiH o p — ell°fi HP = — e"HiP. (2.169)
Integrating Eq. (2.169) from zero to fl, and remembering that if f3 = 0, we find efi°Iip = 1, eu°fip(fl )
—
1 Jo
efi'H 14') Of -
(2.170)
Therefore,
P(13 ) = po(fi) —
J oi3
Po(fl
iP(6') cbq'
(2.171)
The last term in Eq. (2.171) is small if H 1 is small, and it is a correction term to the approximate equation p p o . If the correction term is small, we can use an approximate p(13) to obtain a much more accurate approximation to p(#). For example, if we make the approximation p(r) po (fl'), we have
PO) ''Z'd
Pa)
fi
f Po(fl —
1P0(P1 ) dfl'.
(2.172)
Proof that F 5_ F0 + 0
2.11
67
In using Eq. (2.172) as a new approximation to p(f3), we can find a still better approximation to p. Continuing in this manner we get /3
p (fl) =
—f 0
p0(6 )
+
fi
I Jo
dir
d flip 0 (fl
fir o
—
fl')H I P 0 (r).]
dfl"[Po(fl — fi i)11 iPar — fl")11 tPo06")] ('fi"
— f fl dr f ir dfl" o Jo
dfr[
] + • ••
(2.173)
We can easily rewrite Eq. (2.173) in coordinate representation. For example, P(x, x'; 10 = < xiP(MIx'› Po(x, x';
p 11) —{ . <x1P0U3 — fil ) (fx"><x"Idx ")ix'> HiPogn dr,
o
(2.174) where we use _ .0 ix"›<x"I dx" --= 1. J'
If H1 = V(x), then
= v(x")P0(x", x'; r),
<x"iilip0(fl')Ix'> and Eq. (2.174) becomes
r
P(x, x'; 16) = Po(x, x', 16)
.
i
- f Pax, f -.0 o
;
fl
-
fr)v(x1P0(x", x'; fl') dr dx" + • • • (2.175)
2.11 PROOF THAT F _-_ Fo + o
Now that we have a perturbation theory, we are in a position to prove as promised in Section 2.3, that F < Fo + 0 . Let V . H — Ho . Since F is determined by the equation 61- 13F •= Tr e - fl (1113÷v), the most obvious approach is to find an approximation to F by using the perturbation expansion of e -fl(hbo+v) ,__
p.
C fl(11°±11) = e fill° .
+ f
fi0fl e -(") "°Ve - "H° du
13uf i
o 0
du i du 2
e-(13-u1)11°Ve-(u1-u2)H°Ve-u2H°
(2.176)
Density matrices
Taking the trace of Eq. (2.176) we obtain Tr e- P (It o 4-1"
)
e
-PF0
p
Tr [e
H°Ve - '1
du
.0 ±
JO
- u n ove -( ui -112)u oVe - u2H1 du i du 2 Tr [e-oo
o
(2.177)
Using the fact that Tr AB e -pF
PF0
fi
= Tr BA, we find that
du Tr e - "°V
O
du i du 2 Tr [e x
Ii 1 —
u2 )}10 Ve — ul — u 2.)11oV]
(2.178)
(
It is possible to simplify the second-order term. Let w = U1 — u2 and let ui . Then the integral becomes Tr [e- Psc'ewa°Ve - wuoil
(2.179)
where A is the shaded region in Fig. 2.4. Now let w' = Then the integral becomes
w and let x' = fi x.
A
Tr [e- Putiew".°Ve - wH°V].
(2.180)
A'
Averaging Eqs. (2.179) and (2.180), we find, from Eq. (2.178),
e - PF = e - PF°
fi Tr [e - PH°V] + 132
fi
dw Tr [e
ewlIQVe - w H° V
]
+
•
• .
O
Fig. 2.4 The shaded area A is the region over which Eq. (2.179) is integrated.
Proof that F Fc, + 0
2,11
Let 1tn> and In) be eigenstates of Ho, and let En, and E eigenvalues. Then
their respective
E
Tr [e- PH°V] =
E
Tr [e-0e*H0 Ve - wH0 V]
69
(2.182)
50 e-wo. (3.49) Now we suppose that S and So are real and use the inequality
> e- .
(3.50)
3.4
Variational principle for the path integral
87
e-
Fig. 3.5 Geometrical interpretation of (e - f ) > e- .
The geometrical interpretation of this relation is as shown in Fig. 3.5. In Fig. 3.1, <e- f> is always above the curve Cf. Note that Eq. (3.50) does not depend on how thef's are distributed. When we apply Eq. (3.50) to Eq. (3.49) we can write e - PF e - <s -so> e (3.51) where
<S
(S — So)e -s°
SO >
fJe
(3.52)
-so gx
Usually Eq. (3.52) is simpler to calculate than the first factor in Eq. (3.47). We thus have a theorem:
F < Fo + 1<S So>50 Exercise: When
s
J
So =
.
(3.53)
.:
L 13
2
rrni2(4)
L
2
v(x(u))1 du, (3.54) Vo(x(u)ddu,
show that _
< s — so >0 o.
(3.55)
Thus Eq. (3.53) contains Eq. (3.45) as a special case. (Hint: from Eq. (3.54)), S
So =
(V — Vo) du.
(3.56)
Jo The numerator of the right-hand side of Eq. (3.52) becomes an integral over the integrand of which is independent of u.
u,
88
Path integrals
3.5 AN APPLICATION OF THE VARIATION THEOREM
Finally in this chapter let us explore an application of our variation theorem, Eq. (3.53). In a one-dimensional, one-particle case, we have
r u [m(u)2 S= 1 + V (x(u))idu, h o 2
(3.57)
where
(3.58)
flh.
U
To avoid unnecessary clutter, we will set h equal to unity in most of the following work. Every once in a while we will restore the h's, so that the reader can see where they belong. The partition function may be written as exp {_sou r mi(u)2 + lqx(u))1 du} 2‘x(u) dx(0). (3.59) = L 2 Here, we first fix x(0) and x(U) = x(0), and then do the path integral over all paths; then we vary x(0). The zero in indicates that we integrate over x(0). In the classical limit of high temperature, or small h, U is small, so that the path does not deviate much from the initial point x(0). Thus the first approximation in Eq. (3.59) is to replace V(x(u)) by the initial value V(x(0)) to obtain
mkT 227rti f
dx.
(3.60)
This is the well-known classical result. How can we improve on this approximation to take the quantum effect into account? First we observe that, because x(0) and x(U) are equal, it appears more natural to use an average 1 x — (3.61) x(u) du U0 in place of x(0). Second, the deviation of the path from the classical straight line may be taken into account by some average of V(x) over the path rather than by a constant V(x(0)). These considerations lead to the following trial So : So
=f
U
0
2
du + Uw(x).
(3.62)
Here w(x) is a still undetermined function, which is to be varied later to minimize Eq. (3.53). In the first approximation one would expect to choose w(x) = V(x) but we can do much better with our variational principle.
An application of the variation theorem
15 1'1
U
gx exp [— ..
• [ MX 2
2
fo
./
all closed paths
+ UW(X)1
89
du]
tv-74
= f dx
exp all closed paths x with fixed average x
— f u [ 11/12 du + Uw(x)]du]gx. 2 o [
Now let y =-- x — x(0). Since .p(u) = ,e(u), we obtain = I dx all closed paths
u m V• 2 exp [— d gy. fo 2
y with y(0) = 0
In the path integral above, there is no restriction on y because y = x — x(0), and we integrate over x(0). The reasons for integrating over the paths chosen can be made clearer by drawing diagrams; the reader should do this if he needs clarification. m e -uwo ) dy. e - PF° = dx e -uw(x) Pfree(0, 0 ; CI) = (3.63) 27rhU It should be noted here that in the classical case Eq. (3.60), w(y) is replaced by V(y). Next, we calculate <S — So >. Using Eqs. (3.57) and (3.62), we see by definition
f
-1 <S — So> fi ... 11 11/3(S — S 0) exp [—f mi 2/2 du] exp [— Uw(x)] gx fi exp [—f mi 212 du] exp [— Uw(x)] gx V(x(u')) du' — w(x)} exp [—f m5c 2I2 du] exp [— Uw(x)] gx . jj exp [—f m. 2/2 du] exp [— Uw(x)] gx (3.64) Here x is defined in Eq. (3.61) and is functional of x(u). The first term in the numerator is simplified as follows: =
g WU fg,
SU V(x(u')) du' exp [—
U
So
m(u) 2
' 2
dui exp [— Uw(x)] gx
'
u du' s'" u mi(u) 2 d u — V(x(u , )) exp [ ] exp — So U „ —f 2 0 (-: tro.c 2 (u) du' f , dx V(x') exp [— ' ' du o x(o)=u x(u')=,, , -
J
u
f
O
]
•2 )
MX (ii
2
du
x exp [ — Uw(x)] gx(u).
90
Path integrals
14 =
u=0
Fig. 3.6a Paths of integration.
Fig. 3.6b
We are integrating over paths like those in Fig. 3.6a. By making the change of variables v = tu + U u' 0 < u < u' u u' u' < u < U —
—
we integrate over paths like those in Fig. 3.6b. Because of the condition that x(u = 0) = x(u = U), we do not get a discontinuity in x at V = U — u'; therefore
Su m5c2(0u m±2(V) " du = J dV. 2
= fo
2
Also,
x ,---
1 fu 1 u x(u) du = f x(V) dV. U 0 U 0
3.5
An application of the variation theorem
91
The first term in the numerator of Eq. (3.64) becomes
ru du' r dx' V (x')
Jo U j
x(o)=x(u)=x,
exp [— roil m5c2(V) dV] exp j 2
0.-7.
u ,,,,, ;,. (u)
= x(0)x(U)
V(x(0)) exp [—fo ...2 "" du] exp [— Uw(x)] gx(u). 2
(3.65)
Thus Eq. (3.64) may be written as
1 — <S — So >so
13
= < V (x(0))) — <w(x)>.
(3.66)
To transform further we use the following result (proved previously):
exp [—f u
m$(u) 2 2
lik—Av
=
2
u
dui exp [— if f(u)x(u) du] gx(u)
o
7M1
u u u 1 o( S f(u) du exp {— [f f lu — u'lf(u)f(u') du du' 0 4m o o )
u
2
+— 2 (f u uf(u) du)
u
0
1
(3.67)
In evaluating Eq. (3.65) we first do the path integral keeping x fixed; then Eq. (3.65) is written I.-m±(u)2 Ilx(0)) exp f[ du] exp [— Uwx)] 2'x 2 .1.... =
f K(y) exp [— Uw(y)]
dy
(3.68)
where
K(y)
=
V(x(0)) exp [— u ni5c(u)2 dtd3x 2 So .—., such that x = y. Equation (3.69) may be written ,o—rt U mit ul 2 K(y) = 1/(x(0)) exp [—fo 2,` ' du] (5(x
(3.69)
—
y)
gx(u).
(3.70)
al—io
Por this we use the Fourier transform: V(x) -=.- f v(q)e lqx dq
(3.71)
92
Path integrals
to write Eq. (3.70) as
exp [iqx(0)] exp [ik(x
K(y) = f v(q) dq f dk u mi. x exp [ —So 2
—
y)]
du] x.
(3.72)
Here 6(x y) of Eq. (3.70) was also Fourier transformed, and we are ignoring factors such as 2n. Use Eq. (3.61) in Eq. (3.72) to write —
K(y) = f(q) v dq f dk exp [ — iky] '1-'1 exp [iqx(0)] exp [i — ku f x(u) U0
du]
el—al
U
exp [—
•
171X 2 X
fo 2
du] gx.
(3.73)
Now we see that the path-integral part of this expression can be brought into the form of Eq. (3.67) if we define f(u) --=---.: qb(u — 0) + --I-U
(3.74)
because f(u)x(u) du = qx(0) + — k ru x(u)
u Jo
f ou
du.
(3.75)
When we use Eq. (3.74) in Eq. (3.67) we see that u
(3.76)
f(u) du = q + k,
jo u
JO
uf(u) du =
2
(u — u 1)f(u') duf(u)
du 0
0 U
=2
du jo
(3.77)
,
u
u
lu — u'l f(u)f(u 1) du du' = 2
kU
[fu (u - Li')
o
k _ du' + udf(u) U
= 2 r u durku2 + uqlf(u) = k2 U + kqU. 3 Jo [2 u (3.78)
An application of the variation theorem
3.5
93
Use Eqs. (3.76), (3.77), and (3.78) in Eq. (3.67) and also in the path-integral part Eq. (3.73) to write the latter as
a
k 2 + kq + 3 2
K(y) = f dqv(q) dk exp [— iky]6(q + k) exp [4m
= dqv(q) dk exp [—ik.YAq + k) exp [
k2U ] 24m
= f dkv(— k) exp [ — iky] exp [ — k2 1 .
(3.79)
24m
Using the inverse Fourier transform, we find that Eq. (3.79) becomes K(y) = f dk exp [— iky] f V(z) exp [ikz] dz exp [
24m
k21
6m
f dz V(z) exp [ — u (y
z) 2 ].
(3.80)
When the correct factors are taken into account, the final form is K(y) =
6mkT
nh 2
dzV(z) exp [—
6mkT(y — z)2 ].
(3.81)
It should be noticed that K(y) is V(z) averaged over a Gaussian. The root mean square of the Gaussian spread is
h
(3.82)
N/12 mkT This spread is about 1 A for a helium atom at 2°K, and is a very narrow spread at room temperatures. At the limit of infinite temperature, the Gaussian becomes a delta function, and K(y) V(y). Summarizing the results obtained so far, we have
F Fo + e - PF° = f exp
1 fi
Ks
— s o >so =
1
7,
(3.83a)
-
dy]
fin(K w) -
ke - fiw dy ($e'' dy) 2
e - fiw(K w)dy
cflw dy
(3.85b) Thus 1 6 (F0 + - <S
(3.86)
S 0 >s) --=
leads to
w(Y) = MY).
(3.87)
This is the best choice of w(y). In this case Eq. (3.83c) yields
‹S
So >s. = 0,
(3.88)
so that, from Eq. (3.83a), F < Fe1K
(3.89)
where Fc, K means the classical free energy with the potential V(y) replaced by K(y). Alternatively, e
-pFc, K
m
e -PIC(Y) dy
27rhU
(3.90)
compare Eq. (3.60) . K(y) is defined in Eq. (3.81). f'd K is a better approxima tion than the ordinary classical form Fc, in Eq. (3.60). Example: To find out how good Fci K is, let us consider the harmonic oscillator. In this case the potential is V(x) 7-= rii
(3.91)
2
From Eq. (3.81)
K(y) = \16rnkT nic°2 f z 2 exp [ - 6mkT h2 nh2 2
(y - z)21dz =
172(0 2
2
(y 2
± h2fl ) . 12m (3.92)
An application of the variation theorem
33
95
Using this result in Eq. (3.90), we obtain e -13F6 K -
.1 m
f
27ch2#
exp [ #
ma) 2
(y 2 ±
2
h21 dy
12m
= (11 liflo))e- h211202/ 24 .
Thus, we have the following results: F true =
1
In 2 sinh
(3.93)
Cal 2
)6
(3.94a)
,
(3.94b)
Fe lc = -1 [In (hcofl) + , h2)6242w1 )6
1
(3.94c)
F1 = - in (hab6))6 When we write G
2F —
ha)
and
ho)fl 2 .
f :7_
(3.95)
Eq. (3.94) becomes Gtrue
1
in (2 sinh f),
(3.95a)
f
1 Gc i K = —
f2
In (2f) ± .1-- ,
(3.95b)
6
f
1 Gd = .-f- In (2f).
(3.95c)
Table 3.1 is a numerical comparison of G, rue, Gcl K and Ga. This table shows the remarkable improvement Ga K has over Ga . Remember that f = 1 corresponds to the temperature kT = ha) where the quantum effect is large. Table 3.1
Gtrue Goic
Gel
f= 1/2
f = 1
f = 2
0.08263 0.08333 0
0.8544 0.8598 0.6931
0.9908 1.0264 0.6931
96
Path integrals
The expression Fe! K in Eq. (3.90) combined with Eq. (3.81) is better than the series expansion of K(y) in terms of V(y), d 2 VIdy 2 , . . . . Nevertheless, Fc1 K is not as useful as Table 3.1 might indicate. First, it cannot be used when quantum-mechanical exchange effects exist. Second, it fails in its present form when the potential V(y) has a very large derivative as in the case of hard-sphere interatomic potential.
CHAPTER 4
CLASSICAL SYSTEM OF N PARTICLES
4.1 INTRODUCTION The classical partition function of N interacting particles is written as e -(3F
) 3N/2 in 1 ( , ZN -- —
(4.1)
N ! 27h 2 13
where e-fiv(R,R2• • •
ZN =
RN) d
3R1 d 3R2 • d 3R N .
(4.2)
J These equations can be derived from the quantum-mechanical results we have obtained previously. From Chapter 2, Section 8, we have C al's
=
I -
E i. PDa 1,... , XN; PX1,... , PXN)
dX 1 ' ' • dXN
N! p
for Bose statistics. For Fermi-Dirac statistics, BF
e ' =
1 X" —
N! P
(
—
OPPD(Xil••• , XN; PX 1 , . . . , PX N) dX i • • • dX N .
In either case, only the identity permutation is important, because at sufficiently high temperatures, factors such as e-mkTI2h2(X-PX)2 kill the other
terms. It follows that: e
1 ,=:., — D(X 1 , . . . , X N ; X1,..., X N) dX, • • - dX N . N! p
We have already estimated this integral for the case N = 1 by the use of path integrals (see Chapter 3, Eq. (3.60)). Generalizing the method used to the case of N particles in three dimensions, we arrive at Eqs. (4.1) and (4.2). When a gas particle is a polyatomic molecule, the internal motion of a molecule and the motion of the center of gravity can be separated; then the specific heat is a sum of the two contributions. In Eq. (4.2), R i is regarded as the center of gravity of the ith particle; the partition function for the internal motion is not included here. The latter can be calculated using information 97
98
Classical system of N particles
about the energy levels obtained from the infrared spectra, or calculated using quantum mechanics. For a dense system such as a liquid the internal motion of polyatomic molecules and the motion of center of gravity are entangled with each other, and the two are hard to separate. The following discussion excludes the internal motion of a particle, and the system is regarded as a gas or a liquid of inert gas particles. The free energy is written from Eq. (4.1) as F=
1N- In fl +
3N 27rh 2. 1 N N in In —, ln Z, ± 2$ m fl fle
2)6
(4.3)
The e in the last term of Eq. (4.3) is the base of natural logarithms, rather than the electronic charge. This e appears because we used the Stirling formula:
N! — (— N
)N
e
N ). -s127rN =. In N! r.-., N (ln — e
(4.4)
From Eq. (4.3), we have for the internal energy U U
= — T2
a(FIT) OT
= iRT +
RT 2 azN NZ N aT
'
(4.5)
where
(4.6)
R --.,-- Nk.
The equation of state is found to be
V OZN PV RT NZ N al/ '
(4.7)
from Eq. (1.6). Example: For an ideal gas, V(R i , . . . , RN) . 0; therefore ZN = V N and Eq. (4.7) becomes PV = RT. (4.8) When we assume that the potential energy can be written as a sum of pairwise potentials: V(R i R 2 • ' • RN) --- EV(r ii),
ii
(4.9)
pair
Eq. (4.2) can be written ZN
= f exp [ — )6 E V(r)1l d 3R 1 d 3 R 2 • • • d 3 RNpair
(4.10)
Introduction
41
99
V(r)
0
Fig. 4.1 An unrealistic potential; yo is the volume of a cluster of N particles.
Note that this expression contains all the information about the phase transitions of the system. In other words, the nature of the phase transitions can, in principle, be discussed by examining the behavior of the integral in Eq. (4.10) purely mathematically, without physical knowledge. Equation (4.10) suggests that when the number of particles Nis finite there is no discontinuity in physical qualities. Only when N co do we expect discontinuities, and this is the case we are interested in. The assumption of a central force potential V(r ii) in Eq. (4.9) is not exactly justified. The necessity of a noncentral force is illustrated by the following observation: In the solid state, if we assume the nearest-neighbor interaction only, the face-centered cubic and the hexagonal close-packed structures are not distinguishable. When the second nearest neighbors are taken into account the central-force assumption leads to the conclusion that hexagonal close packing has the lower energy, whereas face-centered cubic is actually more stable for solid argon for example, which is inert in its gaseous phase. The free energy F of Eq. (4.3) is not necessarily a function of VIN only. If we take an unrealistic potential like the one in Fig. 4.1, the energy of the System is roughly of the form CN 2 , where C is a constant and Nis the number of particles forming a cluster of volume yo . Thus, we can think of two extreme forms of ZN: 41) = e— fiN2c vlov, ZW ) = uN
(4.11)
When the temperature is low, for large N, the contribution of 41 ) can be larger than 42) ; then F is not a function of VIN only.
100
Classical system of N particles
4.2 THE SECOND VIRIAL COEFFICIENT
When the gas is not dense, the data of the equation of state are summarized i] practice as B C PV --= 1 + — + (4.12 +••• RT
V V 2
where B is called the second vinai coefficient, and C the third vinai coefficieni In this section we will derive an expression for B. We assume a pairwise potentiz as in Section 4.1. Consider a system of N + 1 particles. In this case, Eq. (4.10) of Section 4. becomes ZN+ 1 — f
exp [— )6
E voc - R id exp [— )6 E
V(r ii)]d3X d 3N R.
pair
When we integrate over X first, all the It i's are fixed. If the gas is not dense the volume for which V(X R i) is appreciably different from zero is small; s we may approximate (4.1 ZN + IZZN. Equation (4.14) will lead to the ideal-gas equation of state. To obtain a bette estimate of ZN + 1 , we write Eq. (4.13) as ZN +1
exp [
E, v(x —
R i)] exp [— f3 pair V(r ii)] d 3X d 3NR
Lair V (r ii)]d3NR
J exp [ x exp [p
Vfr iidd 3N R.
(4.15)
pair
The ratio part of this equation has the form of the weighted average of
exp [—fl
v(x
-
R 1)] d 3X.
We will evaluate this average by neglecting the three-body and higher-order collisions. When Xis close to R i, this pair is assumed to be far away from any of the rest of the R. This assumption is equivalent to replacing the weight factor -16
exp [
E
V(r ii)1
pair
by unity (see Fig. 4.2). When X approaches R i in Fig. 4.2(b), we have a threebody collision among X, R i, and R. Therefore, we neglect the configuration (b).
The second vinai coe ffi cient
4.2
•
Ri •
•
•
•
•
•
•
R.
R.
•
• •
•
•
•
•
•
Weight: 1
101
•
•
•
•
•
Weight:
(a)
(b)
Fig. 4.2 Weighting factors in the derivation of the ideal gas equation of state.
Thus, when we neglect the three-body and higher-order collisions, we can approximate Eq. (4.15) as
exp [—fl L
ZN+1
v(x - R i)] d 3X d 3N R f d 3 NR
ZN
d 3X [f exp [— V(X — R)] d 3Rr 7 VN
(4.16a) (4.16b)
Now, we make the transformation: e —V(X—R1 d 3 R.
e 1317(X R)d3R -
-
(4.17)
When we write the second term as a, a
fo°
[1
—
s v( ]47rr2 dr.
(4.18)
Using Eqs, (4.17) and (4.18), we write Eq. (4.16b) as V(V — a)N
Z N+1
(4.19)
ZN.
VN
This is a recurrence relation for follows:
ZN.
ZN = V (1
We can solve for ZN from Eq. (4.19) as a ) N-1 V
Z N _ i = V (1
Z1
V.
a
V
y-2
ZN_1,
ZN-2
9
(4.20)
102
Classical system of N particles
Multiplying all the equations in Eq. (4.20), we have
zN
avyN-1)+(N-2)+ • • +1 = V N
= vN
N(N a v) ——
1)/2 .
(4.21)
When N is very large, it can be approximated as
zN
vN e N2al2V -
(4.22)
This exponential part is interpreted physically to mean that each pair contributes a factor Coin. The equation of state is derived from Eqs. (4.22), and (4.7):
PV ValnZ = —
RT
eV
V N
=—--
(3v
"
(DI
N
T7 N 2 a ) in v 2V
(4.23)
or PVIRT =-- 1 ± Na/2V
(4.24)
Comparing this with Eq. (4.12), we obtain the expression for the second vinai coefficient: B =-- MO. (4.25) This is an exact result, in the sense that the three-body and higher-order collisions that we have neglected have effect only for the third vinai coefficient and higher-order terms of Eq. (4.12).
Example: Let us consider a potential as shown in Fig. 4.3. Then Eq. (4.18) is evaluated as co
a
=
%
JO
0)4nr 2 dr ± f [1 — efl 9(14nr 2 dr.
Fig. 4.3 A hypothetical potential.
(4.27)
The second vina! coefficient
4.2
When the temperature is high we may approximate Eq. (4.27) as 00 4gb 3 1 a = (p(r)4nr 2 dr, 3 kT SI,
103
(4.28)
which we may write in the abbreviated form:
a
D
vb — —.
(4.29)
When we use this in Eq. (4.24), PV RT
Nv b ND 2V 2VT
(4.30)
We can find out if this expression fits experimental data well by comparing it with the Van der Waals equation of state, which is a good fit to the data. The Van der Waals equation: (P + awIV 2)(V — bw) = RT,
(4.31)
can be expanded as PV a IRT — b2 = 1 b' + +• V V2 RT
(4.32)
If we set NVI,J2 = b„, and ND/2 awIR, we arrive at Eq. (4.30) up to the second vinai coefficient. If the Van der Waals equation is valid, it gives a second vinai coefficient of (4.33) B„ = b,, aw IRT. A special feature of this expression is that, because aw is positive, Bw becomes negative for low temperatures. Experimental data for helium are drawn in Fig. 4.4. The slight decrease for high temperatures does not follow
Fig. 4.4 Experimental data for helium gas.
104
Classical system of N particles
from Eqs. (4.33) or (4.30). It must be remembered that Eq. (4.30) is based on the hard-core assumption of Fig. 4.3. In actuality, the potential V(r) in Eq. (4.18), which is accurate, is slightly soft inside r = b, so that for high temperatures, when particles are hitting each other hard, they can come slightly coser than b. Thus there is a decrease in the "hard-core" radius, and hence a decrease in B. Numerical results: When we use the Slater potential V(r) = Cc ir
c21r6 ,
(4.34)
B'/r 6 ,
(4.35)
or the Lennard Jones potential —
V(r) = A' Jr"
—
in Eq. (4.18) we can calculate a. Incidentally, in Eqs. (4.34) and (4.35), the r 6 term comes from the Van der Waals attraction. Table 4.1 lists the numerical results for helium gas (in units of cc/mop. Table 4.1 T°K
350 250 100 35 20 15
* BCai, (CL.)
Bobs
tBc a l e (Q.M.)
11.60 11.95 10.95 4.80 –4.00 –14.0
10.80 11.34 10.75 —6.95 –20.0
10.82 11.16 4.44 –5.14 –15.1
* The Slater potential is used. J. G. Kirkwood, Phys. Rev. 37, 832 (1931). t J. G. Kirkwood, Phys. Zs. 33, 39 (1932).
The quantum-exchange effect on the second virial coefficient was neglected in this section. This effect becomes important only for temperatures below about 1°K. The general references on the subject of the second vinai coefficient are: J. de Boer, Report on Progress in Physics, Vol. 2, p. 305 (1948-1949). MayerMayer, Statistical Mechanics (John Wiley & Sons, Inc. New York, 1940). Problem: For a mixture of two gases with atomic fractions X1 and X2, such that
+ X2 = 1, show that the second vinai coefficient is B = B 11 X + 213 12 X 1 X 2 B22X1.
Mayer cluster expansion
4.3
105
43 MAYER CLUSTER EXPANSION
From Eq. (4.2), we know that the classical partition function for an N-particle interacting gas is
1 7 Mk
e - liF .
N! ZN =
f
e -I3V(RI ' R2 ' • • •
Tr /2 ZN
(4.36)
27r/i2)
'RN) d 3R i d 3R 2 - • • d 3RN .
(4.37)
The equation of state can be written, as in Eq. (4.12), as PV
B C
RT
V V 2
(4.38)
+•
This is the vinai expansion; as before, B is the second vinai coefficient, C the third, and so on. This formal expansion can, in principle, give all the vinai coefficients. It was originally hoped that this expansion would yield the triple point, and related phenomena. This was not to be. For reasons to be discussed later, the formal program, which we now review, has not proved practical. In the following work, it is implicitly recognized that the /th vinai coefficient is due to /-fold clusters. TAR i , R 2 , ... , RN) can be written (we assume) as ZN = f exp [—fl
E V(r if ).
E V (r ii)] dNR.
(4.39)
Let WN(Ri, R 2
,. • • RN)
= eXP E — 16
E
V(ri3)1
(4.40)
If there is one particle, then W1 (R 1 ) -- e - I") = 1.
(4.41)
For two particles, W2(R1, R 2) = e - fly (ri2) W3(R 1 , R 2 , R3) = exp {— fi[V(r 12) + V(r 23) + V(r13)]}, etc.
(4.42)
Let W1 (R 1 ) --= U1(R1).
(4.43)
If there are N points characterized by WN(Ri, • . . , R N) and if points 1 to M are far from points M + 1 to N, W N (12 1 ,. . . , R N) ,-zi, Wm (R 1 , . . . , RIO x lirri-m(R m+i , ... , RN). For example, if N = 2 and R 1 is far from R2,
106
Classical system of N particles
W2(R1, R2)';'"'d U1(R1)U1(R2). If the points are close to each other, this approximate equation must be corrected by the addition of a quantity, called U 2 (R 1 , R2), that is large only when R, and R2 are close. Thus W2 (R 1 , R2) can be written as W2(R 1 , R2) --= U1 (R 1)U1 (R 2) + U2(R 1 , R 2). Similarly,
W3(R1, R2, R3)
W3(R1, R2, R3) =
(4.44)
can be written as U(R 1)U(R 2)(J(R3) ▪ [U2(R1, R2)UI(R3)
U2(R2, R3)U1(R1) U2(R1, R3)U1(R2)]
(4.45)
U3(R1, R2, R3).
▪
is the contribution due to all three particles close to one another; U3 (R 1 , R2, R3) 0 unless the particles are "clustered." U3(R1, R2, R3)
H/N (R,, R2 ,. , RN) = U(R,) . . . U(RN) + • - • + UN(R i , R 2 ,. , R N).
(4.46)
There is another way of expressing the above equations. If we set U2(R i, R3) = then e -1"") = I + f and, from Eq. (4.40),
WN(R,, R 2 , . . . , RN) = fi (1 +
(4.47)
For example, if N = 3, f 4W3(R1, R2, R3) =l+f 12 +f • J 1.3 + • f23 •+f J12.123 • Jf3./f23
+ fi2f13 + fi2f23f13. We can express each individual term in this sum as a graph consisting of points and lines. If N = 3, then 3•
1 4-4 • •
12
3
3
..f12 4-4 sii• f12123 4—+
12
3 112f23f13
12
.
12
If G represents a graph, and W(G) is the integral over all space of its corresponding function, then (4.48) Z N = E W(G), where the sum is taken over all possible graphs with N points. For disconnected graphs, such as 4 12
Mayer cluster expansion
43
107
W(G) is a product of its disconnected parts. A cluster of N points is represented by a sum of all possible graphs in which those points are connected. For example: 3
3
U3(R1, R2, R3) = ip--% ±
12
A
12
3
+
L.
12
3
±
A.
12
The UN can also be expressed in terms of the W's. From Eq. (4.43) U1 (R 1 ) W1 (R 1 ). From Eq. (4.43) and (4.44), U2 (R 1 , R2) = W2(R1, R2) — W1(R1)W1(R2).
(4.49)
From Eqs. (4.45) and (4.49), U3(R 1 , R2, R3) = W3(R1, R 2 7 R3) - E w2(R1, R2)w1(R3) (4.50)
± 2 WAROW2(R2)W3(R3),
where E w2(R1, R2)w1(R3) -. w2(R1, R2)w1(R3) + w2(R 2 , R3)w1(R1) + w2(RI, R3)w1(R2)•
In general, Ui(k) = Ui(Ri, - • • , Rz)Wi(Ri, . . . , R/) — E w2(Ri, R3)Wz - 2(Rk, • • • , Ri) + • • ' ± ( - 1)1- '(1 — 1)! Wi(Ri) • ' ' Wi(Ri), (4.51) Here the coefficient for a term with k groups is (-1)k-1 (k — 1)!. Thus, from Eq. (4.40), all the WN are known. From Eq. (4.51), U(R) in terms of the Wi is given; thus the U1(k) can be found. (4.52) In any particular partition of the N particles, there will be m i clusters of size 1; and of course E, 1m i = N. For example (as is evident from Eq. (4.45)), a three-particle gas can be partitioned into 1. 1 cluster of 3, m 3 = 1, m 2 = m 1 = 0. 2. 3 clusters of 1, m i = 3, m 2 = m 3 = 0. 3. 1 cluster of 2 and 1 cluster of 1 (this can happen 3 ways), in3 = 0.
m1 =
m2 = 1,
From the above discussion, with Eqs. (4.45) and (4.46), WAR N) = Sono
{I us(k)
E hr, = N.
(4.53)
Sono means the sum over all possible partitions of N, that is, all divisions of
Particles into groups.
108
Classical system of N particles
From Eqs. (4.37), (4.39), and (4.40), XN
N! J N!
d NR.
(4.54)
Now construct the expression bi =
1 Vi!
Ui(k) dR i. • • • dR i .
(4.55)
This equation is known as Mayer's cluster integral. From Eqs. (4.51), (4.53), and the statement labeled (4.52), XN
1
=
s,„„,,
n
U i(R 1)
1 son , (Vi! H N!
=
br.
But if N = 5, for example, U(R 1 R 2)U(R 3 , R 4)U(R 5 ) gives the same contribution as U(R 1 R 3)U(R 2R 5) U(R 4), which is a different partition with the same set of m i. For each possible set of m 1 there are several terms, all of which are equal. To see how many such terms there are, note that there are N! permutations of the N coordinates, and each such permutation corresponds to a terni in Sono with the same set of m i . But we are counting each term more than once. We should not count different permutations of the coordinates within a given U, for example, U(R i , R2) a U(R 2 , R 1). We also do not count changes of order of the U's as separate contributions. For example, U(R i , R 2)U(R 3 , R 41)U(R 5)---U(R 3 , R 4)U(R 1 , R 2)U(R 5) are not counted separately. Thus for a given set of m i, there are Nvn, (p )mt in t terms. So we can write !
X N = ZN N!
= 1
= {m,) E ZH
N!
ail possible sets of m i : Elmi = N
(V b ir mi !
H (VI! V' (1rm i !
•
(4.56)
It is important to note that (for large volume), 1) 1 is independent of volume. That this is so can be seen as follows: Ii(R i) vanishes unless the particles are clustered. Holding dR i fixed and varying dR 2 , . . , dR i then gives some number completely independent of volume, because unless R 2 , . ,R 1 are near R 1 . U1(R ) = 0. The integral over dR i then gives a quantity proportional to the volume, V. and as b i was defined with a V, in the denominator, b i is independent of V. We must now evaluate XN = Sono fi t (Vb i)mlim i ! subject to E m il = N.
Mayer cluster expansion
109
Recalling a similar summation encountered earlier (for the quantum statistics of a many-particle system), we note that if the restriction E m ii = N could be removed, XN can be easily evaluated. Using the same methods as before we define 3N/2
and e - fig
ZN
h 2fi) N1 ! (27rm
—
= E cfl( Fz.,--,2N ) = E cr NxN N
(4.57)
N
where ( in )312 a =
e".
27ch 2f3
But first taking the sum E on° with E in,/ = Nand then summing over all N is clearly equivalent to summing over all ni l with no restriction. e - fig =
(Vb irlah E {I nl m
I
(no restrictions)
MI!
= m,,,,." ,E
(Vb i riamt (Vb 2 yn 2a2 m2
m,!
rn2!
= exp [I7 alb]
g = —kTV Ealbi. -
(4.58)
1
This is known as Mayer's cluster expansion. From Eq. (1.51) Oft
= +kTV
E
Oa locl-1 — b i = V Op
E
Wk.
1
Therefore the equation p = 11/ = E laibi
determines a if we know the density. From Eq. (1.53) the pressure is given by .-- kT A
E A i. 1
(4.59)
Classical system of N particles
110
Therefore we get the equation of state PV
PV
RT
NkT
_
1 b1& Ei lbill.
(4.60)
For small density, a is small—as can be seen from Eq. (4.59). Then a1 + 1 KK al ; so that expansion (Eq. (4.60)) need only be carried out a few terms. Fo r small 1, b 1 can be calculated easily. Inverting Eq. (4.59) to get a as a power series in 1/ V, and writing PV
RT
=1+
B C + +-••, V V 2
we have
b2 N B=—— N=— b? 2 C
= 4q, — 2b i b 3 = 4B2 bt
[1 — e - "(r121 dR 2
(4.61)
N2 J1(2 — e - "(" 2) — Cflv(r23) — e -"(r13) 3
+ e— fl[V(r 1 2) + V(r2 3)+ 17 ( r13)]) dR 2
d R 3.
(4.62)
When the density becomes large, the Mayer cluster expansion is of little use. This is because at very high density (liquid), the most important term in the expansion becomes the Nth term. That is, almost every particle becomes a member of a very large cluster— and in a liquid all the particles are in an N-fold cluster. Hard Sphere Gas The question may be asked: What is the partition function of a hard-sphere gas—that is, a gas composed of small inpenetrable spheres that exert no forces on one another when not touching? Unfortunately, this question is still unanswered. The question may be stated in a different way.
,. 1 d NR.
ZN = .1
no two closer than 2a. a= radius of sphere
But V N --=
f
d NR.
no restriction
Thus j. dN R/1 dN R is essentially the probability that if N points are "dropped' . a volume V, no two particles will be closer than 2a. into
Radial distribution function
44
111
44 RADIAL DISTRIBUTION FUNCTION
Consider a system of N particles interacting with pair-wise forces. The partition function ZN is
E
f exp [—fl
ZN =
(4.63)
pair
The probability density for particle 1 being at R 1 , particle 2 at R2, etc., is exp [—fiV(R i ,
, RN)],
where the factor of 1/ZN provides normalization. The probability that a given particle, say particle 1, is at R 1 and the rest of the particles are anywhere is (
1
N)
f
, RNA d 3 R2 d 3R 3 , . . . , d 3R N .
exp [— /5V(R 1 ,
Because there are N particles, the probability density for one of these particles being at R is N ni(R) = — f exP [ N N =—
exp [—
ZN
fiV(Ri, R2 , . . . , R N)] d 3R2 • • d 3RN
E
Z
V(rii )]d 3 R 2 — • d 3RN ;
(4.64)
pair
nt is the one-particle density, or distribution function. The two-particle distribution function n 2 (R 1 , R2) is defined as n 2 (R I , R2) =
N(N — 1) f ZN
exp [
E
1/]d 3R 3 • - • d3R N ;
(4.65)
pair
n2 is the probability density for finding a particle at R 1 and another particle at R2. For a liquid or a gas, n 2 (R 1 , R 2 ) is a function of the distance between the two Particles. n 2 (R 1 , R2 ) = n2 (r 12). co, n 2 (r) The _general shape of n 2 (r) is as shown in Fig. 4.5. When r approaches n? where n 1 is the one-particle density; this is because when the distance becomes large, the existence of one atom does not influence the distribution of the other atom. The radial distribution function n2 (r) is related to x-ray or neutron diffraction as shown in Fig. 4.6. The x-ray is coming in along the z direction and is scattered by atoms. The scattered x-ray is observed at P somewhere far from the
112
Classical system of N particles
n2(r)
n 21
r
Fig. 4.5 Two-particle distribution function.
Z.Incoming x-ray
R i\
wave vector Kin
unit vector
C Outgoing wave wave vector K„„t =
C, reference center
somewhere in the liquid . We can define Z --= 0 at C.
k ip
P
Fig. 4.6 X-ray diffractions.
liquid. The amplitude observed at P is the sum of amplitudes scattered by each atom, the lager being of the form a(0)(el" I R).
Here R is the distance from the scattering atom and a(0) indicates the possible angular dependence of scattering, which is not important here. Taking into account the phase of the incoming wave, e z , we write Amplitude = a(0)E(eikR jpe ikZ j 1f 1pp,‘./P\) ' j
P is the point of observation. Then, approximately, Rip = Rcp — Ri • ip
and we may write kZ i = Kin • R.
kip = Kout .
(4.66)
Thermodynamic functions
45
113
So Eq. (4.66) becomes Amplitude = a(0)
e ikRcp
E C iR"t•Ri e iRin.R i
Rcp j
or Amplitude = a(0)
e ikRcp
(4.67)
G,
R CF
with G .=
E exp [i(Kin • R.
— Kout • Ri)] = E exp (iq • R1),
where
q
Kin — Kont•
The intensity of the x rays is Intensity = 1Amplitude1 2 = (1a(19)1 2/RM G1 2 .
(4.68)
To compute the intensity, we need to know the expectation value of 1G1 2 . Now IGI 2
=
E
1
For a liquid, neglecting the i = j terms in the double sum, we have
=
N(N
1) eig . (R ,_ R2)
[exp — 13 Ep ir a
2
V(Riil dR 3N
(4.69)
ZN
because the term in brackets, [ ], is the probability for the configuration R I R2, . . . RN Now, using Eq. (4.65), we write
,
.
-,--- 1
f
eig .(R l -R2) n2(R 1 , R 2) d 3R 1 d 3R 2 .
(4.70)
When n2 (R 1 , R2) depends on 1R 1 — R2 1 only, V = the volume of the liquid, and q= lq I,
= fV e'q•x n2 (1X1) d 3X =
27rV
rn2(r) sin qr dr.
(4.71)
4.5 THERMODYNAMIC FUNCTIONS
In the classical case, if we know
n 2 (r),
we can derive all the thermodynamic
functions. The internal energy is
U=RT-
1 OZN ZN
fl
(4.72)
114
Classical system of N particles
where
ZN
is the partition function defined in Eq. (4.2), now 1 zN
azN 43a
fV(r12) exP [ — /3
E 11 d 3NR
zN
—
I V(r23 ) exP [ — #
E
II] d 3NR
...
zN
=_
N(N — 1) f V(r12) exP [—fl E if] d mR. j 2 ZN Here, we use Eq. (4.65) to arrive at 1 OZ N . — ZN
ap
1 .11 V(r 1 2)n2(r12) d 3 R 1 d 3 R2 2
or 3
Vol f
V(r)n 2(r)4nr 2 dr. (4.73) 2 This equation could have been expected. The first term of Eq. (4.73) is the kinetic energy, and the second term is the potential energy. The latter comes from the pair potential V(r), and the factor 1/2 appears because each pair is counted twice. The pressure is expressed as follows: U=
pV
-
2
=
RT +
RT
J la
Vol —
6
or
R • VV(R)n 2 (R) d 3R
(4.74a)
Vol
(4.74b) rr(r)n 2(r)4nr 2 dr. 6 These are classical formulae, which do not hold for liquid helium. Equation (4.74b) is derived as follows. From Eq. (2.121) pV = RT —
3pV = 2 — K
E
rii r(r ii)) .
(4.75)
iJ i<j
As in Eq. (4.73), 0 0 for all n. Then by repeated application of Eq. (6.6), an la> is an eigenvector of a+ a with eigenvalue a — n. This contradict Eq. (6.3), because a — n < 0 for sufficiently large n. Therefore we must have allot) 0
0
but
an +1 1a> = 0
(6.9)
for some nonnegative integer n. Let la — n> = (1101 I(fin> II, so that la — n> is a normalized eigenvector with eigenvalue a — n. Then from Eqs. (6.7) and (6.9), -■/a
— n .---- lala — n> I = 0,
and therefore a = n. This shows that the eigenvalues of a+ a must be nonnegative integers, and that there is a "ground state" 10> such that
(210> = 0.
(6.10)
By repeatedly applying a+ to the ground state we see that (a + )"10> has the eigenvalue n and, because of Eq. (6.8), it is never zero. Thus the eigenvalues of a+ a are 0, 1, 2, 3, . ... If in> is a normalized eigenvector with eigenvalue n, then, from Eq. (6.8), In — 1> = (11NI n)aln> is a normalized eigenvector with eigenvalue n — 1. Also
ak in — 1> = (11-\,1 Oa + aln> = NI nin>. So applying a + to in — 1> gives us back In> (within a factor), rather than some other state with eigenvalue n. We may then construct the eigenstates of a+ a as follows: First we find a state 10> such that (6.11 ) 00> = 0. (10> may be unique; if not, we find 'other operators commuting with a and a - and classify the 10>'s according to their eigenvalues.) Then we define Il> = a + 10>; 12> = 1, all> = 1_ (a + )2 10>; • • V2 V2
A simple mathematical problem
6.1
153
and in general 1 (a + ) n 10) . In> = ,_ V n!
(6.12)
(Note that we could have included arbitrary phase factors in the definition of In> ; our convention here is to make them unity.) With this definition, the In> are orthonormal* and satisfy a+ In> =.-- ■ /n + 1 In + 1> 02> , - N/nn In — 1>
(6.13) (6.14)
a+aln> — nin> .
(6.15)
Equations (6.11) through (6.15) form the answer to the problem posed at the beginning of this section. The operators a+ and a are called "raising" and "lowering" operators, respectively, because they raise and lower the eigenvalue of a+ a. In later applications a+ a will be interpreted as the observable representing the number of particles of a certain kind, in which case a+ and a are called "creation" and "annihilation" (destruction operators, or "emission" and "absorption" operators. Equations (6.13) and (6.14) may be alternatively expressed in terms of matrix elements: -,-- \/ fl + 1 5,,,,,, <mialn> — \I n 6m, - 1 .
(6.14')
* For, by (6.12) we have
0-00
I
= (Olan(a+)"10>(11\ n!m!).
From Eq. (6.1) we easily obtain [a,(a411 = so that (Olan(a + )70) ---= (0k11-1 (a+)ma 0) + (Olnan-1 (a + )m-1 10)
--- n ,-- n(n — 1) • • - (n — m + 1)(qa" - 70> -_, n! 6,,„,
and the orthonormality follows.
(6.13')
154
Creation and annihilation operators
6.2 THE LINEAR HARMONIC OSCILLATOR
Our first application of the results of Section 6.1 will be to the one-dimensional harmonic oscillator, which has a Hamiltonian of the form
H= 1
2
2 MW 2 X,
+
2m
2
(6.16)
where x and p are the position and momentum operators for the particle and satisfy [x, p] = ih. (6.17) Our task is to find the eigenvalues and eigenstates of H. Note that N,I (mcolh)x and (11Vmoh)p are dimensionless. Let us define
a
(\II") x -F
= 1
i , 1 p) . NI mcoh
h
Ni 2
(6.18)
Because x and p are Hermitian it follows that a
+ = 1 (\Imo Ni-2
. 1
— h x - 1 NI fnco -h P) .
(6.19)
From Eq. (6.17) we obtain
[a, a+] = 1. Expressing x and p in terms of a and x=
we have
\I h a ma)
I
P
a+ ,
,,
(6.20)
a
+
a+
V2 -
,
a+
(6.21)
(6.22)
i\72
We get, for the Hamiltonian, hw 2
+
H = — (a a + aa+) = hco(a+ a + 4-).
(6.23)
Thus, the eigenstates of H are those of a+ a. Now we can apply the results of Section 6.1, obtaining the eigenstates 10>, II>, 12>, .. . that satisfy
Illn> = (n + Dholn). The energy levels are thus
En =
(n + Dhco.
(6.24)
The linear harmonic oscillator
6.2
155
The eigenstates themselves are given by Eqs. (6.11) and (6.12). We can easily obtain the wave functions cp(x) = <xln> as follows: from Eqs. (6.18) and (6.11),
0 = a10> =
'no) ( x+ p) 1 0>. ma) 2h
Applying <xi and noticing that <xlpi(P>
—ih(d<xi(P>Idx), we get
h d)
0= mco x 2h
(6.25)
(6.26)
mco dx <x[0>
(where x is now a number, rather than an operator.) Equation (6.26) is merely in coordinate representation, in which it takes the form of a differEq. ential equation. Solving it, we get
(6.11)
<x10> =Ae-(nw)I2h)x2
where A is a constant. Normalization requires that 1 = =
= 1Al 2
fc°. < 01xXx10> dx
ma)
1Al 2
e -(woh),2 dx
,
SO
A = eie
ith
The phase O of A is arbitrary, and we set it equal to zero. Then m(0 )
A =
1/ 4
7rh (
so e -on.120x2
(6.27)
We have thus found the wave function for the ground state. For the other states we apply a+ according to Eq. (1.12): (6.28)
156
Creation and annihilation operators
Since \irno)
<xla +
mio)
<xl
h
2h we have
1 ( mcoy2
<xln>
inco dx
x
h
2h 7rh
mo) -c-14
1 (mo)) 114 (incoy/ 2
N/n!
d \n
\ 2hJ
)
x
d rnw dx h
e
_ (m0)120 ,2
(6.29)
The matrix elements of observables between harmonic oscillator states can be found without having to express the states in coordinate representation and integrating over x. We simply express the observable in terms of the raising and lowering operators. An example of this procedure is given in the following section.* 6.3 AN ANHARIVIONIC OSCILLATOR
Suppose a system has the Hamiltonian 2
mc02 2 X + 2
H= P 2m
(6.30)
Assume that A is small enough ( = e -1iIthi/(0)> = WO» — itHitfr(0)) + 0(t 2), we will soon start finding different numbers of particles. The quartic terms similarly change the number of particles, except for terms like
+ +aya6, a: apay+ aa, .. a„ao which conserve the number of particles but act as a mutual interaction between them; that is, the particles are no longer independent. The description of a mutual interaction will be considered in more detail later. Exercise: Verify that the number-of-phonons operator N, defined by Eq. (6.62), commutes with a product of creation and destruction operators if and only if the number of a's equals the number of a's in the product. 6.6 FIELD QUANTIZATION
A notable example of a system with infinitely many degrees of freedom is a field. Examples are the amplitude of sound waves, drumhead vibrations, light, and so on. Consider a real scalar* field cp(x) whose motiont is described by the
Lagrangiant L(ço, (p) = 1 f d 3 x0(x)0(x) — 1
where K(x — x') = K(x' — x). varying p(x), are
0=
f f d'x
d'x'K(x — x')(p(x)(p(x),
(6.69)
The classical equations of motion, found by
0 SI, at ST(x)
= 0(x) +
Sid ST(x)
f
d 'x'K(x —
(6.70)
* The following procedure can be generalized for a multicomponent field by putting indices on everything. t Classically v(x) depends on t, but (as with qt in Section 7.4) we will not show it explicitly. Besides, in the Schrtidinger picture, the operator v(x) is time-independent. 1 We assume that the system is invariant under translations, so that K is only a function of x — x'.
Field quantization
6.6
163
Note how Eqs. (6.69) and (6,70) resemble the corresponding equations for the system of harmonic oscillators described in Section 6.4:
O
=4., + E
u,,q
Thus, we are justified in treating the field as a system of harmonic oscillators (at least formally): cp corresponds to the symbol "q" and x corresponds to i. (p(x) can be thought of as a separate coordinate of the system for each x. As an example, suppose that (6.71)
K(x — x') = —c 2V 2 b 3 (x — x').
Then Eq. (6.69) becomes, after a few integrations by parts, L= 4
d 3 40(X)0(X)
C 2 VV(X)
•
Wp(X)],
(6.72)
and Eq. (6.70) becomes V2(p(X)
1 .. cp(x) e2
= 0,
(6.73)
which is the usual wave equation. If we assume that (p(x) is a coordinate of the system for each x, the conjugate momentum to (p(x) is 6L (6.74) 11(x) =0( = (p(x).
(5x)
The Hamiltonian is then
H =f d'xFI(x)cp(x) L
f
d 3x11(x)11(x) + 4 f d 3 x f d 3x' K(x
—
x')(09(x)(p(x 1).
(6.75)
To quantize the system we let (p(x) and II(x) be Hermitian operators satisfying
Pp(x), cp(x')] = [11(x), 11(x')] = 0,
(6.76)
Po(x), Mx)] = ih(5 3 (x — x'),
(6.77)
and assume that the Hamiltonian is given by Eq. (6.75), except for a scalar term to make the ground-state energy zero. We next express everything in terms of "normal modes." The situation turns out to be slightly different from that of Section 6.4 because it is convenient here to use "complex" (that is, non-hermitian) normal coordinates.
164
Creation and annihilation operators
Because the system is translationally invariant, we expect that it might help to express the fields in "momentum" representation. Therefore we define Ci6(k) =
d 3x(p(x)e -ik
(6.7 8 )
fl(k) = f d 3xf -1(x)e -lk.
(6.79)
The inverse transformation is* (p(x) =
d3k -cXk)e ik..x
(6.80)
(270 3
with a similar expression for fl(x). Since 9(x) andfl(x) are Hermitian, we ha‘ e (16 + (k)
(k) = (-k).
(6.81)
From Eqs. (6.76) and (6.77) we obtain
Mk), 0(k' - )] = [ n(k),ticio] =
0,
(6.82)
[(To(k), 11(k')] = iii(270 3 6 3 (k + k').
(6.83)
w 2(k) = f d 3 xK(x)e -ik. x.
(6.84)
Now let
From K(x) = K( x) = K*(x) it follows that (0 2 (k)
_ w2(k)*. _
(6.85)
Rewriting the Hamiltonian of Eq. (6.75), we get H = 1 f dk
2 (270 3
-
[El( - k)11(k) + co 2(k)A- kg)(k)]
1
d 3k [11 + (k)I1(k) + co2 (k)(f) +(k)(p(k)]. 2 (270 3 We assume that w 2 (k) > 0, so that the Hamiltonian is positive definite. [(1-6.18-1161') w(k) is real, and we take co(k) > 0.] In the example described in Eqs. (6.71), (6.72), and (6.73), (o(k) = elk. * Throughout all of this the following integrals will be useful:
fd 3Xe- ik
C d 3k
(27) 36 3(k)„ j (20 3
eik . x
6 3 (x).
Field quantization
165
Next we will define annihilation and creation operators (compare Eqs. (6.47) through (6.50)):
L
a(k) (k)
\I co - (k) (79(k) + ,
fl(k)1,
\ 12h
L.
\,1 2h
co(k) ço(- k) - -, ri(- k)1, w(k) L
v w(k)
(6.87)
(6.88)
or ç7(k) = n(k)
h [a (k) + a + (kg
(6.89)
2w(k)
hco(k) [a(k)
i
(6.90)
2
/.,The commutation relations are, from Eqs. (6.82) and (6.83),
[a(k), a(k')] = [a+ (k), a + (ki)] = 0, [a(k), a+(k')] = (27) 3 6 3 (k
(6.91)
k').
(6.92)
If we write H in terms of a and a + , making the change of variable k when necessary, we obtain
k
d3 k hco(k)[a + (k)a(k) + a(k)a + (k)]
H= 1
2 j (27) 3
plus a correction term to make the vacuum energy zero.
The corrected
Hamiltonian is evidently H =
ci3k -
- hco(k)a + (k)a(k).
(27) 3
(6.93)
[Note that the correction term is the infinite quantity - 12- S d 3khco(k)6 3 (0)]. Finally, we express the original field variables in terms of the creation and destruction operators, using Eqs. (6.80), (6.89), and (6.90):
co(x) = 11(x)
(13 k - -
h
- [a(k)
(27) 3 2co(k)
d 3k
+ a + (k)e -
hw(k) [- ia(k)e'k X + ia+(k)e - jk. x]. (2m)\/ 2 J
(6.94)
(6.95)
Equations (6.91) through (6.95) are the important results of the quantization procedure. The commutation relation of Eq. (6.92) may appear strange, in that
166
Creation and annihilation operators
[a(k), a + (k)] is infinite (instead of unity), so that the analysis of Section 6. does not apply directly to a(k). Suppose, however, that we use a less singular representation. Choose a complete orthonormal set of functions IP OE(k), where 7 is a discrete index: k
tfr:( 1c)/i
(6.96)
= 60,
iff 2(k)1/4(k) = (270 3,53(k —
k')
(6.97)
and define
d3 k =— (270 3
(k)a(k).
(6.98)
Then
[ao„
= Sas ,
(6.99)
so that we can apply previous results and construct In i • • • ricc • >. But these states may not be eigenstates of H. The states 1k> = a +(k)10>,
1k,
= a + (k)a + (V)10),
and so forth, though unnormalizable, are eigenstates of H. What kind of phonons do that unnormalizable states represent? The state 1k> is a phonon of energy ha)(k), and we may also say that it has momentum hk. To discover the reason for this, consider the operator
P
=
d3k hka + (k)a(k), (270 3
(6.100)
which satisfies
[P, a + (k)]
hka + (k),
[P, ct(k)] = — hka(k)
(6.101)
so that
k 2 1 . . . > = (hk
hk 2 + • - )1k 1 , k 2 ,.
,>.
(6.102)
Now, from Eqs. (6.94) and (6.101) we obtain
[P, (p(x)] = ihV(p(x).
(6.103)
One can then show that ea ' Pl ihp(x)e - a • Pilh = p(x + a),
(6.104)
so that P generates translations and is therefore the momentum operator. In Sections 6.7 and 6.8 we will describe further the relation between the operators and the states they create and destroy, as well as how other operators.
Systems of indistinguishable particles
6.1
167
such as the Hamiltonian, can be written in terms of the creation and destruction Operators using any basis of states. Note: In relativistic quantum mechanics, when one quantizes a free-particle k 2 + m2 (h = c 1), a different normalization and field with co(k) summation convention is commonly used for momentum states. Everything is k'), which happen written in terms of S d 3kI(270 3 2co(k) and (27) 3 2(4)6 3 (k to be relativistically invariant; to accomplish this change of normalization one uses a "relativistic" a(k) equal to v f 2co(k) times our a(k). Equations (6.92), (6.93), and (6.94) then become [a(k), a -4- (k)] = (27r)3 2co(k).3 3 (k — k');
H =
1*
d 3k (27) 32w(k)
d 3k
(p(x) =
(27) 3 2co(k)
co(k)a +(k)a(k); [a(k)e'' + a + (k)e'
In some texts the (270 3 is also treated differently. We will not use the relativistic normalization here, but it is mentioned in case the reader finds it elsewhere and wants to reconcile it with our notation.
6.7 SYSTEMS OF INDISTINGUISHABLE PARTICLES In the preceding sections we considered the quantum states of an oscillator system as being states of various numbers of a "particle" called a phonon. We identified certain states as one-phonon states, and others as states containing more than one phonon. In this section we shall follow a different line of reasoning. We will start with a space of states describing a single particle, either Bose or Fermi, and Construct the multiparticle states according to standard methods. For the Bose ease we will arrive at a system of states and operators that is mathematically the same as that found previously for a system of oscillators, thereby showing that the interpretation of oscillator states as many-phonon states is consistent with the usual description of many-particle systems. In the meantime we will have also developed a formalism for dealing with Fermi particles, for which the states do not resemble those of a harmonic-oscillator system. We will treat the Bose and Fermi cases simultaneously, distinguishing them by the number C;
{ +1 if the particles —1
are Bose if the particles are Fermi.
(6.105)
We will use the symbol CP (where P is a permutation) to denote I for the Bose ease and (-1) P for the Fermi case.
168
Creation and annihilation operators
Consider first the case of distinguishable particles. one-particle states, then = 00102>
If IV/ i >,
IhI1 n>
On> are (6.106)
describes the n-particle state with the ith particle in state ItP i >. 191)1492> • • • I9.>, then we write
If l(p>
= (• • •
(6.107)
= (91101> Ix 2 > Ix,>. As each x i runs over all space, the resulting states form a complete orthonormal set for the n-particle space (ignoring spin and other variables):
((xil - • • x„1)(iY
• ' iYn>) = 63 (x 1
Yi) • • • 6 3 (x. — Yn),
• ixn))((x 1 1 • • • (x„I) = I.
d 3 x 1 - • • f d 3 x,,(1x1)
(6.108) (6.109)
Using this basis we can express the n-particle states in coordinate representation: = ((xil• •
<x,D1 0 .
(6.110)
For the particle Ii/i> of Eq. (6.106), we have, using Eq. (6.107), Oxi, •
(6.111)
xn) = 01(x1)On • • • (xn),
where /i(x 1) = Next, let us consider indistinguishable particles. We assume that the particles obey Bose or Fermi statistics, which means that we must symmetrize or antisymmetrize, respectively, the states obtained in Eq. (6.106). We therefore define 101> X
102>
x •••
On) = ( 1 h/n
E kit P(1)>I0
P(2))
kJ/No>,
(6.112)
where P runs through all permutations of n objects. It will often be convenient to write Itfr i , ifi 2 , , On > for 0 1 > x It 1 2 > x • • • x The space of n-particle states is that spanned by all "products" of the form is totally symmetric in the Bose of Eq. (6.112). Note that 10 1 > x • • • x case and totally antisymmetric in the Fermi case, as it should be.
Systems of indistinguishable particles
0-1
Example: Let la) and lb> be two single-particle states. particles) , NI
= +1 (Bose
If
jb)la>),
b) = 1/ (la>lb>
la> x lb> =
169
2
la) x la> = la, a) = /2[a )[a>.
—1 (Fermi particles) 1
la, b> =
N./ 2
(la)lb> — lb>la»
la, a> = O. Thus, we have the expected result that two Fermi particles cannot be in the same state. What is the inner product of two of these n-particle states? The answer is given by the following theorem* • • •
(
E E )(: Q (9 P(I )10 Q( 1 )) •
n! p
Q
,
Li
ifQ(n)>)
• < (13 tii;
QP 1 (1)>
(where we have permuted the factors by P) 1 —
v 7-QP - I
n! P Q
Vi" 1
QP -1 (1)> • • • ( q) ni0QP - 1 (n))
* Compare this theorem with the well-known formula for vectors in 3-space, (a x h) (c X d)
a •
cad
b•c-d
170
Creation and annihilation operators
(since CP = CP-1 and CQC13-1 = CQP-1 )
n!
E E V'OpiloRti>> • • • ( (NitP Rcio) PR
(letting R = QP - 1 ) CR • • •
=
(9„10R (0 >
= 1( 10 )
which is the desired result. Now let {11>, I2>... be a complete orthornormal set of states:
< 2 1/3 > = 60;
(6.115 )
E ia>, where a, < • A complete set of n-particle states consists of la i , a 2 , < an in the Bose case, and a l < • < an in the Fermi case. These states are orthogonal to one another, but not always normalized. The reader can sho‘N,. using Eq. (6.113), that for a complete orthonormal set of states we may take
i at, • • • , — an>
(a l
• • • < an) for bosons
N/n 1 ! n2 ! • an>(a t < • • • < an >
la i ,
for fermions,
, at, (for where nc, is the number of times that a occurs in the sequence a t , Fermi particles, n„ = 0 or 1). For either case the completeness relation can be written in the following convenient form: n
E • • • E
al
an
an> +...
If {fa>} is a complete orthonormal set of states, so that Eq. (6.115) holds, then using Eqs. (6.113), (6.115), and (6.118) we may summarize orthogonality by < 2 1 / • " / 2irlfii
9 • " 5
13m>
= (5 nm
6 a113 1
•
• •
(5 an#1
66( On 61 nfin
(6.119)
;
From Eq. (6.116) we also have the completeness relation
. 1
E ---;
n=0 n:
E
(6.120)
iai , • - - , cc,r)<X1, • • • , X n 1
=
,
.1.
(6.121) (6.122)
We may expand an arbitrary multiparticle state 10) as follows, using Eq. (6.122):
E -- f cc
it» =
1
ii =o n!
d 3 x, • • -
i' d 3 x„Ix i , ... , xn )0 (")(x 1 , ... , x„).
(6.123)
Here
11(n) (X 11 - • • , x n) --= < X 11 • • • , X nkii>
1
(6.124)
172
Creation and annihilation operators
is (if Itii) is normalized) the amplitude for the state 10 to have n particles, o ne at each xi. Note that 1 1/ (n )(x 1 , , xn) is symmetric or antisymmetric accordi ng the statistics. Note also that if 10> is an n-particle state and is of the forrn to ItPi, • • • , 0>, where each 10 i ) is a one-particle state, then 0 x.) =
unless
m = n;
1(xi) .(xi)
(6.125)
i(xn) 9
where iji i(x) = <xi ltfr i>. Equation (6.125) follows from Eqs. (6.124) and (6.113). In the Fermi case the determinant is called the "Slater determinant." We are now ready to define creation and destruction operators. These operators are fundamental for two reasons; first, we constructed the multiparticle states so that we could describe changing numbers of particles, and v+e need some operators that can effect this change, and second, other operators, such as the total energy, will turn out to be simply expressible in term,s of the creation and destruction operators. Let 10 be any one-particle state. We define a(q) to be that linear operator which satisfies On> =
••-
(6.126)
'lia>
on>.
for any n-particle state For n = 0 this is understood to mean a + (p)Ivac> 10. We call a+(p) the creation operator for the state kp>, and its adjoint a(p) the destruction operator. A creation operator clearly converts an n-particle state into an (n + 1)particle state. It is easily seen that a destruction operator turns an n-particle state into an (n 1)-particle state and annihilates the vacuum state. To find the effect of a(p) on an n-particle state lip i • • • On > we multiply on the left by an arbitrary (n 1)-particle state (x t • • L.,_ 1 1. —
—
= *
Xi
• Xn- i> *
(0119> • • • > "' < 0.1(P>
ilzi) •
=
k -= 1
Ck < tPkbP>
= • •
r
Systems of indistinguishable particles
173
(an expansion by minors)
=E k=1
l
amitp i
= k=1 E
k -1 (9100101 "
(no 00 • ' • On>.
(6.127)
Thus the destruction operator removes the states 10 i >, one at a time, leaving a sum of (n — 1)-particle states. In the Bose case = 1) the terms all have a + sign, whereas in the Fermi case ( = —1) they alternate in sign. Equations (6.126) and (6.127) describe the action of creation and destruction operators on many-particle states. From Eq. (6.126) it follows that a + (90a + (92) =
or Ea + (.91), a + (92)]-
(6.128)
0,
where [A, .13]_ AB — OA; that is, [AB]_( = +1) is the commutator and [A, B] +(C = —1) is the anticommutator. Taking the adjoint of Eq. (6.128) we obtain the further result (6.129)
[a(91), a(92)]-; = O.
Thus, the creation operators commute for Bose particles and anticommute for Fermi particles, and similarly for destruction operators. Now, what is [a(9 1 ), a + (ç,9 2 )]_? Does it (or any similar expression) reduce to anything simple? We first calculate a (91)a + 03 2)101
• 0.>
4901(92, i0i • • •
• On> On ) +
E
ck192, 0i • •
k=1
(no
00
•••
n)
(6.130)
and then a + (9 2) a((,0 1) 101 • • • 0. >
k=1
k=-- 1
Ck-1 (9110k>a + (92)101 ' • (no 11/1‘)
(k - '
on>. (6.131)
174
Creation and annihilation operators
Multiplying Eq. (6.131) by
and subtracting it from Eq. (6.130) we see that
[a(pi), a+(g9 2)1- =
(
(6.132)
+9 11 (19 2)•
Equations (6.128), (6.129), and (6.132) are the fundamental "commutation" relations for creation and destruction operators. The relations we have derived are usually stated in ternis of an orthonormal basis, and we shall now do that. Let {la)} = [II>, 12>, . I be a complete orthonormal set of one-particle states. It is usual to let aŒ = a(Œ). Then, since
>=
(6.133)
N/n 1 ! n 2 ! • •
the number of times a appears in the ket on the right. Then the where n in i n2 • • • > (each n, = 0, 1, 2, 3, ... ) form an orthonormal basis for the whole multiparticle space. From Eqs. (6.133), (6.126), and (6.127) we find a: Ini n2 . • • n„• •
=
;kith • • no(' • •
= \Inc, In i n 2
+ lln i n2 • • n„ + 1 • • n, • • • >.
>,
(6.134) (6.135)
The commutation relations are [aŒ, ap] = [a:,
= 0;
[a„,
= 60 .
(6.136)
Equations (6.134), (6.135), and (6.136) are identical to Eqs. (6.60), (6.61). (6.51), and (6.52) for raising and lowering operators for a system of harmonic oscillators. The operator for the number of particles in the state la> is
N„ = a: aa. The notation of Eq. (6.133) (in terms of "occupation numbers") is generally not the most convenient. It is more natural, in fact, to continue using the notation we have been using all along in this section, that is, the notation !a l , a 2 , an >. This notation was discussed in Section 6.5 in connection with phonons. Note that Eqs. (6.126) and (6.127) of this section, when applied to states of the , c(„>, become identical to Eqs. (6.67) and (6.68). Thus, creation form loc i , and destruction operators for a system of Bose particles look just like those for what we called a "phonon" system.
Systems of indistinguishable particles
6.7
175
Fermi Case Using the notation la,,
,
7„),
we have
a,c+ IY•i, • • • , Y.> =
(6.137)
• • • , Œn>
and aŒlx,,.
••,
an>
=
E
(-1)k '62aklai, • • •
+ 1 • • •
(6.138)
k= 1
We could also use the occupation-number notation inin2: • • > = 11 1, 1 2
•>
= O. One could also derive Eqs. (6.137) and (6.138) directly from the anticommutation relations
[a„, atd, = [a:, a]+ = 0;
[a„, a]+ =
(6.139)
as was done in previous sections for the a:s of the harmonic-oscillator system. But for the Fermi case there does not appear to be any a priori reason for postulating Eq. (6.139). (Remember, for oscillators the corresponding commutation rules followed from the canonical quantization procedure.) One may rather consider Eq. (6.139) as derived from the antisymmetrization postulate for fermions. Let us return to the general case where Eqs. (6.126) through (6.132) apply. One advantage of deriving them in such a general form is that we are not tied down to a particular basis of states. Suppose we use a basis of momentum eigenstates, Ip>. Because = a + (x i ) • • • a + (xn)lvac>.
6 3 (x
x'). (6.142 ) (6.143)
If we use a basis of hydrogen-atom energy eigenstates In/ni>, then [a(nbn). = 611 , 6.m , , and so on. How do the creation and destruction operators change when we make a change of basis? This question is easily answered by noting that if
Ix> = 1 10> ± #1(P>,
(6.144)
then
a +(z) = a(X)
(0) + Pp),
= 00'00)
+ #*a((p),
(6.145)
This means that creation operators "transform" like kets, whereas destruction operators "transform" like bras (because <xi =iI + (3*
=
d 3x1xXxlp> = d 3 xix>eiP"x, f d 3p
f d 3p IP>, and we let the "wave functions" of these states be <xlcx) = tiGix). Write down the formulas for a +(x) and a(x) in terms of a,c+ and a„, and vice versa. 6.8 THE HAMILTONIAN AND OTHER OPERATORS
In the last section we developed a method for describing systems containing many Bose or Fermi particles, and we defined creation and destruction operators
The Hamiltonian and other operators
SO
177
for the particle states. We will now show that these operators have other uses an merely creating and destroying states. Suppose A (1 is an operator that acts only on one-particle states. We wish to find an operator A that represents the "sum of A" ) over all of the particles." That is, for any n-particle state • • • , q/n> =
=
(6.148)
> X• • •X I>
we want Alt» to satisfy Alt» = A (1) lt1> X
lt2> X • • • X 10„›
+•••+
> X A(1)12> X • • • X
(6.149)
> X Iti/2> X • • • X A (1) 1q/ n >,
To see what this means, suppose each lt,IJ1> is an eigenstate of A" ) with eigenvalue ai. Then Eq. (6.149) implies that
AltP> =
(al
(6.150)
+ ' • ' + anO>.
For example, if A (1) is the single-particle Hamiltonian, then A is the total energy (ignoring mutual interactions, which we shall consider later in this section). If A" ) is the momentum operator for a single particle, then A is the total momentum. If A" ) = 1") (the unit operator on one-particle states), then A = N, the "number-of-particles operator." The desired operator A is easy to find. We first find it for the special case A(11 10001. In this case Eq. (6.149) becomes AP> = <M01>la, tk2,/ ±•••
•
tri>
66 102>11,
11 , t2 , • • •
9
'•
fl>
(6.151)
O*
C
Now look back at Eq. (6.131) and notice that when 9 2 = c and (p i = 13 we have a + (ctja(fi)10 k=
1
Ck
ilx)til 1, • • • , (no ti/k),
ti/n)• (6.152)
But 1
5 • • • 7
(no
ti-1 09 • • • 9 ilf st>
= 019 • • •
9 th-19 °( 9 q/k-i-11•
••
due to the symmetry property of the n-particle state. Using this equation in Eq. (6.152) and comparing it with Eq. (6.151) we find A
=
a + ()a(13)
when
A" ) = 17><M.
(6.153)
The generalization of Eq. (6.153) for an arbitrary one-particle operator A (11 is immediate. We choose a basis any basis of one-particle state la>, and write A" ) =
E lx> - V<xl
(6.158)
i
we have for the total momentum
P =
C d 3p
i (2703
+ Pa (P)a(P)
1 = f d3xa+(x) - Va(x). i
(6.159)
(Compare the first of these expressions for P with Eq. (6.100), another expression for P.) Finally, suppose the Hamiltonian for a single particle is p2
H" ) =
2m
+ V(x),
(6.160)
The Hamiltonian and other operators
6.8
179
where x is the position operator. In coordinate representation, 1 V26 3 (x — x') + V(x)6 3(x — x), 2m
<xIH mix') = —
(6.161)
so that H =-- - it d 3 x it d 3 x' [
= i' d 3xa + (x) [-
( + x)a(x) 1 V26 -— 3(x — x') + V(x)5 3(x — x'da
2m V2
i-m- + V(x)1(0).
(6.162)
In momentum representation* 1 2 (270 363(P j*k
= E yk-, ç riik 6 3(x — x k)6 3 (y — x j)lx k , xi, . . . , (no xk, xi), . . . , xn> j*k
= E 6 3(x — x k)6 3(y — x i)ix i ,..., JO. j*k
Multiplying by -1-V(2) (x, y) and integrating over x and y, we find that V as given by Eq. (6.171) indeed satisfies Eq. (6.170). In view of Eq. (6.168) and the remarks following it, we might expect that the mutual interaction could also be described in terms of the particle density by the operator V' = 1 f d 3 x f d 3yV (2)(x, y)p(x)p(y).
(6.172)
However, V' is not quite the same as V. To see the difference we write p(x)p(y) = a + (x)a(x)a + (y)a(y) = Ca(x)a(y)a(x)a(y) + 6 3 (X — y)a + (x)a(y) = a+ (x)a+(y)a(y)a(x) + 63 (X — y)a +(x)a(x),
so that V ' = V + 1f d 3 xV (2) (x, x)p(x).
(6.173)
Thus V' contains an extra term, which may be interpreted as a self-energy; it contributes even when there is only one particle present. The true mutual interaction V is zero unless there are two or more particles. We want only the
182
Creation and annihilation operators
—q
p1 + q
Fig. 6.1 Operator V adds momentum
.7 to one particle and subtracts it from the other.
mutual interaction, because any self-energy (if it exists in nature) can be included in the Hamiltonian of Eq. (6.162). Besides, for many potentials (for example, the Coulomb potential), V' is infinite and is not what we would consider to be the true energy. If we express v in momentum representation, using Eqs. (6.171) and
(6.147), and assume V(2) (x, y) . V(x — y) ----= V(y — x),
(6.174)
we arrive at
V
.... 12 if(270 3 rid3q (2r0 d3P fd3Pi 7(q)a+(p + q)a+(p' — q)a(p')a(p). 3 (270 3 f
(6.175)
17(q) is as defined in Eq. (6.164) (in proving Eq. (6.175) we use P(q) -fk—q), because V(x) = v(— x)). Note that
Here
d3
VIP 1, P2) = it (203 (2t)
— V(q)i Pi +
q, P2 - q>7
which says that V adds momentum q to one particle and subtracts it from the other with amplitude V(q). This process is denoted by the diagram in Fig. 6.1. The total momentum is conserved, as we would expect, because Eq. (6.174) implies that the mutual interaction is invariant under translations. The Hamiltonian and mutual interactions described here all conserve the total number of particles (see the exercise at the end of Section 6.5). 6.9 GROUND STATE FOR A FERMION SYSTEM
In this section we will consider fermions, so that the results of Sections 6.7 and 6.8 apply with C = —1. Suppose we know the eigenstates and eigenvalues of the single-particle Hamiltonians, H" ) : Hwil> =- Eja>,
(a
= 1, 2, 3, .. . ).
(6.176)
Ground state for a Fermion system
49
183
Then, ignoring mutual interactions, we have for the multiparticle Hamiltonian: H =
Every state may be operators:
built
E E„a„+ a„.
(6.177)
up from the vacuum state by applying creation
lai, • .. , an> . a:, ... , aZa ivac>.
(6.178)
(Both sides vanish unless the a i are all distinct.) It is often inconvenient to refer everything to the vacuum state, as in Eq. (6.178). In practice we may be considering states that differ from some "ground" state only by the presence or absence of a few particles. Suppose there are G particles present. Assume that the energy levels are ordered such that Ei •Ç E2 LÇ, E3
'••.
Then the state of lowest energy is
Ignd> = IL ... , G> = at • • • 41vac>,
(6.179)
which we call the ground state; its energy is Egnd
-E l + • • • + E G*
(6.180)
Any other G-particle state will have some of the levels 1, ... , G unoccupied and some higher levels occupied. It is convenient to use ignd> as a reference, describing the removal of a particle from 1, ... , G as the creation of a "hole." Particles in the levels G + 1, G + 2, ... are still called "particles." If a fi), then we say that a particle is excited from the level a to the level Ma < G hole with energy —EOE has been created as well as a particle with energy E. The concept of holes may be formulated mathematically as follows: Define bc, = a:
for
a < G.
(6.181)
From the anticommutation relations in Eq. (6.139) we have, if a, a' > G and 13, 13' G,
[as, a]+ = [a„, bp] * = [bp, bp ] + = 0, [aŒ, a]+ = (5; [bp, bin + = bpp,, [a„, VI_ = 0.
(6.182)
Thus the operators a:(a > G) and b:(cx < G) behave like creation operators. We can now express the states in the form la" ... , am, P t , .. . , fin , gnd> _- ao,+, • - - a: b'pri • • • b in - ignd>
(6.183)
184
Creation and annihilation operators
(ai > G, fl i G). We call this a state with m particles and n holes. ground state acts like a vacuum state in that
az ignd> = bp Ignd> = 0.
Th e
(6.184)
The Hamiltonian of Eq. (6.177) may be written in the form
E
H =
E
Eaa:a„ +
a›G
Ebb :
ccG
Œ
E (—b:b„
+ 1)
=G+
(6.186) a›G
a.G
thus a hole counts as — 1. The number of particles and holes outside the ground state (that is, the number in + n in Eq. (6.183)) is given by the operator
N' =
E
ci:aOE +
a>G
E
(6.187)
b: bOE,
a = ilflOE),
cr2y lococ> = OP. Note that o- 1 orr2; = cr2Ja 1 i, which is to say that operators pertaining to different particles commute. We have previously considered (and will consider further in the section to follow) the operator S1 • S2 = iff 1 • a2 . By explicit calculation, we find ai ' cr2loot> = loot>, ai ' 0'2 lotfi> = 2 ifioc> — lotfl), O" - tr2Ifh> = 21OEfl> — I floc>
,
f'i '
0.
21flfl> =
Iflfl>
•
(7.23)
202
Spin waves
Writing laa> = 21Œx> the useful relation
laa> and similarly for . 0,2
11313>, we immediately come up on
2p 1,2 _ 1,
(7.24)
where p" is the operator that exchanges the spins of the particles; that is, p 1,2 1000 = Icco, p 1 . 2 143>= floc>, and so on. The symmetric states
lafl) +
11,1> = lace>, 11, 0> =
lflot> , 11,
2
—1
> --=
I! /3 >
all satisfy p 1,211 , ni ■2 = and therefore a lni> 0'21 1 , m> 1, 0, — 1). The antisymmetric state
10, 0> = 14> _ Iflot
>
( 7.2 5 )
m> (in (7.26)
N/2 — 310, 0>. satisfies p"210, 0> = — 10,0> and thus a l • a2 10, 0> the eigenvalues of a l • 62 are 1 and —3. Now the total spin is a'1 + cr2
S
Therefore (7.27)
and its square is SS =
+
• cr 2 .
(7.28)
Therefore
S • S11, m> = 211, m>,
• 510, 0> --= 0,
so that (as we anticipated in the notation) the states 11, m> have spin I and 10, 0> spin 0. The fact that the symmetric states have spin 1 and the antisymmetric ones have spin 0 may also be seen directly from the relation
S • S = p" + 1,
(7.29)
which is derivable from Eqs. (7.27) and (7.28). 7.3 SPIN WAVE IN A LATTICE
Consider a lattice, the ith member of which has spin a i . The interaction between the spins is due to the exchange effect, and the Hamiltonian is given by
H =
E
•a
(7.30)
Consider the lattice to be cubic, and let each point on the lattice be denoted by the integer vector N a(nxe„ + ney + nze,), where a is the lattice spacing. Then
H =
E A
N,N"IN
(7.31)
Spin wave in a lattice
1.3
203
It is clear that the interaction depends only on the distance N — N' = M, a nd if in addition it is assumed that only nearest-neighbor interactions are important, H =
E Ai(o'N' 6 N
N -f- aex + (T N . (7N+ae, 4- CrN . Cr N+aez)•
(7.32)
Occasionally next-nearest-neighbor interactions are important. They may be even more important than the nearest-neighbor ones, but such a situation can be analyzed in the same way that we use here. We now return to Eq. (7.32) and consider the one-dimensional case. The extension to three dimensions will be immediate. We also are interested in the ferromagnetic case (A 1 < 0), and we can neglect boundary effects by considering the system to be a closed loop. Let there be Nlattice points; o-N+1 Thus, H = —A an • 0.„ ± , (7.33)
E n
where A = — A l . The minus sign in front of A is included to emphasize the fact that in the case of ferromagnets, —A < O. That is, the energy is most negative when the spins are parallel. The (rare) instances when —A > 0 are due to a double-exchange effect. Recall that we used x to represent spin up and il spin down, and we defined the spin exchange operator p" 2 = 10 + a l • a 2 ), satisfying
r 1,2 1 „›
1fliq) = Iflfl), p"100) = lib>,
p1 '2
p 1 ,21137>
(7.34)
. 1 113) .
Then N
H = —A
E n=1
N
(2p"'" + ' — 1) = NA — 2A
E
p"'"'.
(7.35)
n=1
To solve Schriidinger's equation with the above Hamiltonian, we must find the eigenstates and eigenvalues of H. To do so, notice first that Eq. (7.34) implies that H operating on a state with a fixed number of fl's gives a state with the same number of 13's. Since all states can be represented as linear combinations of those with a fixed number of fl's, we have a method of partially diagonalizing H. Rather than trying to completely diagonalize H, let us merely find eigenstates With low energy. One suspects that the lowest state (most negative energy) occurs when all the spins are parallel. For example lxx - • • x) is such a state. More generally, When we say all spins are parallel what we really mean is that the total spin is as large in magnitude as possible, and it can be shown that the states that satisfy this condition are completely symmetric states. The completely symmetric
204
Spin waves
states are invariant under pn'n+ 1; so p" can be replaced by l in computing the energy of those states. Thus, we expect that Egnd =
= E1/i>, n
(Ea)/, n > = 0» =
H 1 10> N
= —2A n=1
= —2A i
n= 1
an E
(rm+ 1 —
m=1
an [(pn'n +1 —
1 )1 11/ n>
1) + ( PH— 1 'n — 1 )1 1/1 n>
N
= —2A Ei an[Pn+ t> + I-i> — 2 kfrn>]. n=
Because the 10> are orthogonal, we can equate coefficients of
Ean =
—
141> to get
2A(a n+ 1 + (1n _ 1 — 2an ). (7.38)
To solve this set of equations, let an = en and substitute into Eq. (7.38). We thus obtain 5 (7.39) E = —2A(e iô — 2 + e -i(1 ) ,--- 4A(1 — cos .5) = 8A sin' 2
Spin wave in a lattice
7.3
205
E
Fig. 7.1 Energy as a function of .5.
Neglecting boundary conditions as we have done is tantamount to assuming that the N lattice points are arranged in a circle, or that the N + 1 point is the first point. In other words, aN +1 = al,
or
e ib(N+1) . eib,
or
e l " -= - 1.
This means that 6 = (27/N)I; 1 an integer;
— N12 :5_ 1 :5_ N12.
(7.40)
We have, from Eqs. (7.39) and (7.40): E = 8A sin' -6 = 8A sin' 27 1 . 2 N 2
(7.41)
This function is plotted in Fig. 7.1. There is, in other words, a band of energies, the minimum energy occurring at 6 = 0. If 6 = 0 we have a completely symmetric state, which we have already concluded has the lowest possible energy, E = 0. Another way of seeing that the state with 6 -,-. 0 has the same energy as Ion • • • a> is as follows. Let each spin be tipped from its "up" position by an angle 0, where 0 is very small. Such a state is given by
>
Ix = (cos P2 a
.
0
+ sin ---6 fl)(cos --° a + sin - dq) • • • ) . 2 2 2
Because 012 is small, cos 0/2 ,':.-_, 1 and sin 0/2 ,c,-, ri = Koc + tifi)(Π+
')›. Neglecting terms in ri 2 or higher powers of ri, we have rifl • •
Ix> = Ica • • • > + ri(Ifica • • • cc> + I)(ficx ' ' • a> + • ' • ). Subtracting out the ground-state term, we obtain 17,'> = En rill/in > where II',> IS given by Eq. (7.36). Since ix> must be an eigenvector of H with the same eigenvalue (energy) as il, a, ... , isc>,* so must I-Z>, and (1/016 = E. ItPn>. 'Since in the absence of external interaction there could be no change in energy when all of the spins are rotated in a certain angle.
206
Spin waves
For 6 0 0, we see that the spins are out of phase. This suggests the concept of a wave, and we now rewrite ei6" to make the nature of the wave somewhat clearer. The one-dimensional integer "vector" is na, where a is the lattice spacin g. which describes a plane wave of wal,e We can write a, as an = 6.16n = e number k. Such waves are known as "spin waves." Thus,
6 k =- = E= 27ri
4A(1 — cos ka),
1 — k 2a212, and
For small ka, which means long waves, cos ka
E
(7.42)
a Na
2Aa 2 k2 (long wavelength).
(7.43)
The extension to three dimensions gives us no trouble: The lowest state again has all spins parallel. Subtracting out the ground energy, we obtain
H' = —
E
—
A m(aN aN , s,
1)
N ,M
and, in a manner similar to that used in the one-dimensional case,
10> = E e'NalifrN>
and
E A m(eiK . m — 1).
E=
For nearest-neighbor interactions (in a cubic crystal),
E=
—
2A{(e 1Kxa
—
1) + (C1Kx"
—
1) + (e`KY"
—
1) + - - -}
. Ka { s in Ka + sin 2 i' =+ 8A sin 2 Kza 2 2 2
(7.44)
For long waves
E
(7.45)
41,
2Aa 2 K 2
where 1/2p = 2Aa 2 . To find the next band of energies, we must consider the case when all the spins but two are parallel. But first we will digress to discuss the semiclassical interpretation of spin waves. Problems dealing only with spins have often an immediate semiclassical interpretation. Let us look at the spin wave case. 7.4 SEMICLASSICAL INTERPRETATION OF SPIN WAVES
The Hamiltonian is H = —2A E (p"'" +1 — 1) = —A E (an • an + n=1
n1
—
)
(7.46)
Two spin waves
1.5
207
The Heisenberg equation of motion for the nth spin is .
i a n = - (Han h
(7.47)
—
We transform this using the commutation relations: (7.48)
axo-y — o-yax = 2io-z etc.
so that (62 ' 6 1)aix
— 0- 1),( 0'2' al) =-- (a2xa1x + a2ya1 y + 0-2c7 1z)cr1x — ( 7 1x(''') = 2i(o- 1yo-2y — o - 1z o-2y)
(7.49)
---- 2i(ff 1 x a 2 )„.
Using Eq. (7.49) in Eq. (7.47), we can write hir, = 2A(■ an x an + 1 — an - 1 X an).
(7.50)
We regard this as the classical equation of motion for the vector a„. This is a nonlinear equation, but it can be linearized when we consider a n, to be close to 1, in which case we can approximate thus: fha„,x y t,han
&'
2A(2o- , — an+i,y — an- 1,0 , ,x — an+ 1,x — an- 1,x ')—
(7.51)
2A(2a
We expect the vector an to rotate around the z-axis, and we also expect the solution to be of the form c sin (weak,
(7.52)
c cos oneinak ,
where
a is the lattice
constant. When we use Eq. (7.52) we see that
(7.53)
hco = 4A(1 — cos ak).
7.5 TWO SPIN WAVES When the spins at positions function (pni ,n2 :
n1
and n 2 point downward we define the wave
Içon,,n 2 > = lot • • ' fic( • • •
ocfla • • •
a>,
(7.54)
Where the B's are in positions n i and n 2 . The Hamiltonian is the same as that of N. 7.46, and the Schr6dinger equation is Hiti/ (2) ) = 4Ae (2) 1(11 2 >;
Energy = 4AE(2)
(7.55)
Where
10(2)> =
a
(Pn1
,n2>'
(7.56)
208
Spin waves
When Eq. (7.56) is used in Eq. (7.55) and the coefficients of (19,40 2 are compared we can distinguish two cases. When n 2 0 n 1 + 1, in other words when the two down spins at n 1 and n 2 are nonadjacent, we have 2Eani,n2
= (an1,n2
(an 1 1 n2
a nt— 1,n2)
(an 1 ,n2
ant+ 1,n2)
an 1, n2— 1) + (an 1,n2
an1,n2+1)•
(7.57)
On the other hand, when n2 = n 1 + 1 the two down spins are next to each other and Eq. (7.57) does not hold. We then have 2ta„,„ +1
(an,n+1 — an— 1,n + 1)
+ (an,n+ 1
— an,n+ 2)•
(7.58)
Before solving Eqs. (7.57) and (7.58) rigorously let us derive some approximate results. Because the case n 2 = n 1 + 1 is only one out of N (the number of lattice points), for most of the time Eq. (7.57) holds. Therefore, the first crude approximation should correspond to two noninteracting waves: — e ikinia eik2n2a .
(7.59)
When we use Eq. (7.59) in Eq. (7.57), we see that — cos
E(2) =
k l a) + (1 — cos k 2a)
(7.60)
c(k 2),
= E(k 1 )
where 4A8(k 1 ) is the energy for one spin wave of wave number k l . At low temperatures, only the lowest energy modes are excited; thus for one spin wave, the energy might be, k2 E — 21u and for two spin waves 7
E
2
(11 =
1 4Aa 2)
k2
2p 2p
.
We can form wave packets with group velocity Owlek, where E = ha) and the phase velocity is co/k. The statements above bring out strikingly the similarity of spin waves to crystal vibrations or phonons. By analogy to the concepts of "phonon," "photon," and so on, we call a spin-wave excitation a "magnon." Since the approximate system discussed above is made up of independent Bose particles, we can use the expressions that we discussed before. The free energY Fis F = kT lin ln (1 —
e - fi E(k) ) d3k
(7.61)
(270 3 *
For low temperatures only the E(k) near the bottom are significant, so that we may approximate E(k) k 2 /2u,
Two spin waves (rigorous treatment)
7.6
Fig. 7.2 Two spin waves, or the
209
scattering of two quasi-particles.
and F = kT f
rk In (1 _ e-(k2/21,,,T)) 47- 2 dk (210 3 .o
(7.62)
The energy is
œ k2 2p. 4rrk2 dk U = f k2/2,,kbT _ 1 (270 3 . 0 e
(7.63)
This energy depends on T to the 5/2 power: U oc T 512 .
Therefore the specific heat is proportional to the 3/2 power of T: C oc T 312 . This result for spin waves was derived by Bloch. Although the system is made up of ideal Bose particles, Bose-Einstein condensation does not occur because the number of particles is not fixed.
Problem: When there is an external magnetic field, derive the energy of the spin waves. Find the magnetic susceptibility. 7.6 TWO SPIN WAVES (RIGOROUS TREATMENT)
Bethe treated the two spin waves in a linear system rigorously.* When there are two spin waves in a system, the problem can be looked upon as a scattering of two quasiparticles, as in Fig. 7.2. For the scattering process the energy and the momentum are conserved, and the eigenstate is a superposition of the incoming and outgoing waves: a ni,n2 ___ orei(kinia +k2n2a) +
fle i(k2n la 4- k in2a) .
(7.64)
From this point onward we will let a ,-- 1. That is, for convenience we are either assuming unit lattice spacing or absorbing the factor a in k. As mentioned earlier, n 1 < n2 . The ratio alf3 of the coefficients is related to the phase shift * See A. Sommerfeld and H. Bethe, Handbuch der Physik, Vol. XXIV/2, p. 604 (1933).
210
Spin waves
resulting from the scattering, and it is to be determined from the scattering equation. The left-hand side, a 12 of Eq. (7.64), must satisfy Eqs. (7.57) and (7.58). First, when ci„,, is substituted in Eq. (7.57), we see that ,
6(2) = (1 — cos k 1 ) + (1
—
cos k2).
(7.65)
This equation has the same form as Eq. (7.60), although we expect k 1 now satisfies a different relation from that found in Section 7.1. Next we can use Eq. (7.64) in Eq. (7.58). Rather than doing that immediately, notice that Eq. (7.58) is a special case of Eq. (7.57) if we define 42 (which was left unspecified before) by
2a„,n + 1 ---- an,,, + an + 1,n+ 1
(7.66)
Then, substituting Eq. (7.64) in Eq. (7.66) is equivalent to substituting Eq. (7.64) in Eq. (7.58). From Eqs. (7.66) and (7.64) we have
2(aeik2 + fleik i) = a + fi + aei(" k2) + (3e i(ki+k2) . It follows that 1 + e1(" k2) 2eik ' I + ei(" k2) — 2e2
cc
fi
e i(k1—k2)/ 2
cos (k 1 + k2)/2 k 2)12 cos (k 1
(7.67)
e i(k1 - k2)12 -
If k 1 and k 2 are real, a
z
where
z*
z = i cos
k 1 + k2 2
iei (ki-k2)12 .
Therefore 14131 = 1, and in fact (7.68)
4/3 = where 9/2 is the phase of z: cot
2
=
Re z _ Im z cos (k 1
=
2
(cot
k2 )/2 sin (k 1 k2)/2 — cos (k 1
2
— cot
k2)/2
2 '
or
k2 2 cot — cot — cot . 2 2 2
(7.69)
We will also consider cases in which k 1 and k2 are complex. Then 14131 is no longer 1, but 9 may still be defined by Eq. (7.68) (9 will now be complex).
Two spin waves (rigorous treatment)
7.6
211
Equation (7.69) holds, for such cases, as an algebraic identity or by analytic continuation from the case of real (p. For scattering problems we assume k i and k 2 to be real. Then Icx//31 ---- 1, as is expected, and 9i, which is real in this case, is the phase shift after the scattering. One special solution is = ei(p12 /3 = e -h012
(7.70)
From the substitution of Eq. (7.70) in Eq. (7.64), this becomes 00,-Fk,n2+,0/2) e ick2n.-Ekin2-(p12) =e
(7.71)
When we impose the periodicity requirement with the period of N, we see that (7.72)
an ! ,n2.
an2,111+117
(Because n i + N > n2 , we do not write ani 4 satisfied when, from Eq. (7.71),
an1,n2 .) Equation (7.72) is
Nk l = 27rm 1 + 9 Nk 2 = 27m 2 —
(7.73)
where m l and m 2 are integers. It must be noticed here that Nk i is not equal to 27rm 1 , as was the case for one spin wave, but is changed slightly by the amount of 'p. Using Eqs. (7.69) and (7.73) we can eliminate 1( 1 and k 2 , and determine 9 as a function of m l and m 2 . It can be shown that the real values of k do not exhaust the possibilities. We can examine the case of complex wave number by putting k l = u + iv, k 2 = u — iv.
(7.74)
Equations (7.70) through (7.73) still hold (although 9 is complex). Eqs. (7.74) we can rewrite Eq. (7.73) in terms of u and v: ,
With
(7.75)
u — (m + m2), Tr(m 2 — m 1 ) + iNu.
(7.76)
From Eq. (7 .7 5) it follows that u is real. Using Eqs. (7.74) in Eq. (7.67), and (7.76) in Eq. (7.68), we obtain a relation between u and v: e 1n(m 2 -m 1 )e -Nv
_cos
— e'
cos u — ev
Now, if we hold v fixed and let N large N
op, the left hand side
cos u,
0; so we have for (7.77)
212
Spin waves
so that y is approximately real. (We have assumed that Re y > 0, but the following result will be the same if Re y < O.) The energy is derived from Eqs. (7.65) and (7.74); use of Eq. (7.77) leads to
iv) — cos (u — iv) • = 2(1 — cos u cosh y)
e(2) = 2
—
cos (u
sin2 u = 1(1 — cos k),
(7.78)
where we have defined k = 2u = k1
k2.
(7.79)
To get a„,, we use Eqs. (7.74) and (7.76) in Eq. (7.71). The result is (with a change in the normalization) arr i ,n2
ei( " nzAl2 cosh
IN
a 12
e i(n1 -Fn2)kl2 Sinn
N 2
2
or y
+ n 1 — n2
(7.80) +
n1 — 12 2
according to whether m 1 + m 2 is even or odd. a„,, has a sharp maximum when n1 n 2 (or n i = n2 — N) and decreases as n2 — n 1 increases (remember we n 2). assumed n 1 Our solution represents a bound state. What is the binding energy for this quasiparticle? When there are two independent spin waves both having wave number K, the energy is e = 2(1
—
cos K) = K 2 — K 4112 + • • •
(7.81)
For a quasiparticle of wave number 2K, we find from Eq. (7.78) the energy s ----- 1(1 — cos 2K) = K2 — K4/3
••
(7.82)
Equation 7.82 gives a lower energy than Eq. (7.81) does, but the difference is very small, namely, 10/4. In other words the binding is very weak, and it depends on K in such a way that the long-wavelength spin waves do not interact much. This is the reason why the approximation of non-interacting spin waves is a good approximation for low temperatures. 7.7 SCATTERING OF TWO SPIN WAVES In this section and the one that follows we will approach the two-spin-wave system in a slightly different way. In Section 7.6 we found the eigenstates of the Hamiltonian containing two spin waves. Here we will consider the simple waves found in Section 7.5, which are not eigenstates of H but describe pairs Of independent spin waves. Then we will apply perturbation theory to calculate the amplitude for scattering of these waves.
Scattering of two spin waves
1.1
213
The general Hamiltonian for spin waves in three dimensions is written as (7.83) H = — E 2A m(p N,N-m _ 1); M,N
where A m is a constant. For the one-dimensional, nearest-neighbor exchange interaction case, we may write H = —2A
EN (p„,.+1
-
(7.84)
1).
The scattering of two spin waves will be described in this section for the onedimensional case, although a similar analysis holds for the case of three dimensions. We know that a ground state of Eq. (7.84) is Iota
(7.85)
• • • 00,
for which all spins point up. When one of the spins points down, Itiik) =
1
-
E
e
ikn kpn ) ,
(7.86)
\IN
is an eigenfunction of Eq. (7.84) with the eigenvalue Ek = 4A(1 — cos k).
(7.87)
When there are two spins pointing down, 1 '
\I N 2
E
e i kin j e ik 2 n21 91102 )
(7.88)
It in2
describes a state in which two spin waves exist. However, Eq. (7.88) is not an eigenfunction of Eq. (7.84), as we saw above, and the two spin waves scatter each other. Let us examine how the two spin waves interact. By definition, frpnin2 > . 19,,,,) is the state where spins are down at n 1 and n2 , and i9„1 „,› = O. We want to evaluate N IN
2 Hililk 1 k2)
. E e ik i nk e ik2n2 H I ( p nin2.
).
n 12
Here 1119, 12 > is calculated for four cases: i) n 1 and n2 are separated, that is n 2 0 n i ,n 2 0 n i + 1, n2 0 n 1 — 1. ii) n2 = n i + I. iii) n 2 = n 1 — 1. iv) n2 . n 1 . In this case 19) = O. For case (i), 1119,21n2 > = —2A[(19 11 + 1 ,n2> —
19,102)) + (19 n i — 1 ,n2>
191702>)
+ (19„,,„ 2 +1) — 19,,,,>) + (19n, ,n2-1> —
(7.89)
214
Spin
waves
For case (ii), =
2 4(19n – 1 ,n 1>
19 n,n + 1>) + ( 19n ,n 2> — I9n,n4-1>
)].
In order to conform with Eq. (7.89) we will write the equation above in the form I/19„,„+1 )
=
—
2A [ — 210 n,n+1, + ( 1 (0 n-1
19n,n+ 1))
— 4A 19n ,n+ 1>
I9n,n+ 1>)]
+ (19 +2>
( 7 . 90)
where the term in square brackets is Eq. (7.89) with n 2 = n 1 + 1. Similarly, for case (iii),
>
—2 A[ -2 19n,n-1> + (19..n-2> —
HI9n,n-1)
)
— 4 A19„,.- 1 >•
+
(7.91)
For case (iv), HI,>
o = — 2A[19,+ 1 jj > + 19n– 1,n> ± 19 n,n 1> + 19 n,n – 1>]
(7.92)
+ 4 419.,1+1> + 19.,.-1)].
Here the first term is Eq. (7.89) with n 1 = n2. When we use Eqs. (7.89) through (7.92) to evaluate V.N2 Hlii1k1k2>, the first terms of Eqs. (7.90), (7.91), and (7.92) combined with Eq. (7.89) lead to the sum of Eq. (7.92) over n 1 and n 2 without restriction; thus N/N2
---
—2A E e 1 (kin1+k2n2)19n02)Reik1 - 1) 4-
— I)
nin2 (e
— 4,4
E [e
ik2
1)
( e –ik2
— 1)]
k,,,,±k20+1»,
in+ k2(n– 1)) içon,n– 19+i> + et (k
— e i(kin+ k2.)( 9n,.
= 44(1 — cos
k1) + (1 —
cos k 2 )]\/ N2 tfr k 1 k 2
+ +
N 2 IX> .
(7.93)
The "remainder" (x) is written as N2
Ix> ---- —4A
E ei(kl +k2)n(eik2
eiki —
I
—
e
ick1+k2)1 9n,H+ i >
(7.94)
The first term of Eq. (7.93) is, as seen in Eq. (7.88), a double sum over n, whereas Eq. (7.94) is a single sum over n. Therefore, the second term of Eq. (7.93) can be regarded as expressing the scattering of the two spin waves that appear in the first term of Eq. (7.93). Notice that when H operates on Ith ik,› a new state emerges, but the new state is still a combination of states with two flipped spins. In other words, when two spin waves encounter, two spin waves come out.
Non-orthogonality
7. 8
215
With the help of Eq. (7.87) we may write Eq. (7.93) as
(7.95)
Hill1k 1 k 2 > = (Eki + Ek2)iikkik2> + fx>,
where (x
) = _
4A 1 e i(ki +k 2)%1
N/INT 2 n
(7.96)
9n,n + 1),
with
= eik2 + e a' _ 1 _ e i (ki + k2 ) .
7.8
(7.97)
NON-ORTHOGONALITY
The general theory of scattering tells us that the cross section a satisfies 2 (number of final states) 2n ay = --- IMfi l x ,
h
AE
(7.98)
'where y is the velocity of the colliding particles. From perturbation theory, Mfi = H H 1±
H ' 11H' 1 n +"• n*i Ei — En
E
(7.99)
Hf'i is the matrix element of the perturbation Hamiltonian between the initial
and final states. In this section the discussion will be limited to the first term of Eq. (7.99). We are interested in the scattering from the initial state 410 ,1,2 > to the final state Itk L1 ,L2 > where
(7.100)
Eki + Ek 2 = EL I + EL2 .
Then we cannot use Eq. (7.99) as such, because the states I T kik2. d/ are not orthogonal to each other. This is shown as follows. We compute the inner product, using Eq. (7.88): 1 OPLI,L21 1Pki,k2> = W2 n 02 n;ni
1E
e i( k i n i + k2n2 – L OI – L2ni •
(7.101)
From the definition of 19,,„ 2 >, we see that
(7.115)
N N
where we define the creation operator of a magnon to be = 1 E e ik • N o.N
(7.116)
NiN N
The destruction operator is then ak =
1
E e
NN
Further,
r ki,k2f
-ik-N
N+•
(7.117)
of Eq. (7.88) is then kfr kik2> = 4142190'
(7.118)
218
Spin waves
To see that this is correct, remember that = O.
Equation (7.117) may be solved for aN , as
1v
aN+ = —,=---- 1,.• VN k
ik
•N
(7 .119)
ak.
The Hamiltonian of Eq. (7.83) includes pN'N+M, which contains aN aN-hm or terms of the form 12
12 12 axax + TTy +
in which the first two terms can be rewritten in terms of products of two ais. The last term corresponds to products of four a's because of the result shown in Eq. (7.113). Now let us find out if the a's satisfy commutation relations such as
( 7.12 0 )
[ak ,ait] = (5kL.
The answer is no, and is related to the fact that the tfrkik are not orthogonal to each other. We can disprove Eq. (7.120) using Eqs. (7.116) and (7.117) together with the fact that [a + , a _] = az : ,
[ak,
an
1
E e - ik • N 1 eiL • N2 r
=—
N
,
L" N 1 - ,
aN,+]
NiN2
= j_ E e - i(k - JO • N , N
" Nz
N
1
L e- i(") . N(aNz = (5k,L + —1-7
N
— 1),
(7.12 1 )
N
because
1 E e - i(k - I.) • N NN
:=
On account of the second term in Eq. (7.121), Eq. (7.120) does not hold exactly. although the second term is numerically of the order of 1/N when there are only a few particles with spin down. 7.10 SCATTERING OF SPIN WAVES—OSCILLATOR ANALOG
Figure 7.3 shows two waves of momenta Ki and K2 interacting and "emerging . ' with momenta L 1 and L2. The scattering matrix Aff i is given by MN = - 2A
Em eit2. m(1
—
e - lm ' ( 2)(1 — eim ' ( J).
(7.122)
Scattering of spin waves—oscillator analog
1.10
219
L1= L2= K2 -
Q
K2
Fig. 7.3 Scattering of two spin waves.
For small K1 , K2, Mr' approximately by
K1 K2 - K2 . The scattering cross section a is given
kw 1 2
Ka.
(7.123)
T. From Eqs. (7.122) and At a temperature T, hK 2 12ti = kT and K 2 1; hence a 1, and the vinai (7.123), a- - T2 . For ordinary particles, Alf T 1 /2 . In the case of spin waves, however, o- - T2 and thus coefficient b 2 b2
T512 .
(7.124)
It is plausible to suppose that in a vinai expansion we must first consider independent spin waves, for the next term, two-particle collisions, then three, and so on. Such plausibility arguments can be made rigorous by the use of an "harmonic oscillator analog," In this analog, corresponding to the spin state where the ith spin is down we will take the oscillator state where the ith oscillator is excited to the first level and the others are not excited at all. Hence, 00100 corresponds to ccocihoc, and so on. Let a; be the creation operator for the oscillators. That is, cc Z creates a phonon (say) in the Nth oscillator.
a K-1-
1 E
e - iK • N
'NM N
(7.125)
•
Now the real H (for spin waves) acts as follows: H (1431a) --÷
Waal) (Ica f31) .
(7.126)
We want now to construct an analog H for the oscillators so that H(00100) -*
(01000) (00010)
Now H=
—
- E A m (am+ +N
— 2 A(y . I I I. =
E sK altaK ,
Xj+
CO]
(7.127)
220
Spin waves
where sic = —E Am(e'm — 1) At
will do the trick as long as only one /3 is present. However, consider the following (aflaficxa) H(cec43flaoc) —+ (7.128) (ocallafla) . However, the H defined by Eq. (7.127) acts differently:
H(001100) —* (010100) (001010) -\7 2 (000200) 4-N/2 (002000) 4—
(7.129)
Note that the arrowed functions in Eq. (7.129) have no analog in Eq. (7.128). Therefore we must add terms to H so that these arrowed terms will disappear: that is, we want to define a new H' = H + A such that
H'(00100) —>
(010100) (001010) '
Hence the term A that is added to H behaves like
A(001100) --+
— N/2 (000200) —V2 (002000) .
Then A -----
E 2A [oti++ 1 0Q++ loci+ loci
— °cif+ itxi+ 'xi + ioti]
(7.130)
is the new term we need. For example, ai+l a i annihilates two adjacent phonons at i and i + 1, and at++ l oci++ , creates two at i + 1. H' = H + A defined as above is not Hermitian, and terms such as oti++ l oci+ / ja i must be added. H is an exact analog for the real H if states with only one 16 are considered; H' is exact if states with one or two iTs are considered: and so forth. Note that the state (00200) may contribute an energy Ei such that cfiEl is not negligible, in which case the C flEi term must be subtracted out when forming the partition function, C flEn . Problem: Using as's and ak's which satisfy anticommutation relations will solve the problem of states with more than one "particle." What are the difficulties?
CHAPTER 8
POLARON PROBLEM
8.1 INTRODUCTION
An electron in an ionic crystal polarizes the lattice in its neighborhood. The interaction changes the energy of the electron, and furthermore, when the electron moves the polarization state must move with it. An electron moving with its accompanying distortion of the lattice has sometimes been called a polaron. It has an effective mass higher than that of the electron. We wish to compute the energy and effective mass of such an electron. Even without any vibration, there is a (periodic) potential acting on the
electron. This potential will be approximated by assuming that its presence causes the electron to behave as a free particle but with an effective mass, M. Hence the energy of the electron when there are no lattice vibrations is P2 /2M, where P is the electron's momentum. It will be recalled that the dispersion in a crystal is as shown in Fig. 8.1. We will be interested in the region described by the almost constant part of the optical branch, that is, the branch for which the positive ions move in a direction opposite to that of the neighboring negative ions. If we consider the permittivity as a function of frequency (Fig. 8.2), the portion of the curve that will interest us is that part where the lattice vibrations are important. co
Optical branch
lo K
Fig. 8.1 Dispersion in a crystal. 221
222
Polaron problem
Fig. 8.2 Permittivity as a function of frequency.
Let us take the crystal lattice as being a continuum with a unit cell at each point x. The lattice vibrations are quantized according to Section 6.10, where the transition from a discrete lattice to a continuous one is made according to Eq. (6.201) of that section. Before writing down the Hamiltonian let us consider which vibrational modes actually interact with the electron; the others may be ignored in this problem. Using the notation of Section 6.10, we assume that the electron's potential due to an undisturbed lattice is either zero or else is taken into account by reducing the mass as already described. If the lattice is displaced, the potential AVi (x) felt by the electron is due to the charge density p(x) resulting from the displacement, and this in turn is due to the polarization P(x): V 2 AV i (x) = ep(x) -=
—
eV • P(x)
(8.1)
(where the electron has charge — e). Assume that the crystal lattice has a positive ion and a negative ion in each unit cell. There are then six modes of vibration for each wave number K. In three of the modes both ions move the same distance in the same direction; as K 0 these modes approach rigid translations of the whole crystal and co(10 —* O. Such modes do not contribute much to the polarization P(x). In the other three modes the positive and negatke ions move in opposite directions in each unit cell, producing a polarization that is proportional to the amplitude of the mode. For these modes the frequenc approaches a nonzero value co as K 0, and we shall make the approximation that w(K) = co for all K. That is, we assume that the frequency is independent of the wave number. Let a + (K, a) be the creation operator for phonons. By analogy with Eqs. (6.200) and (6.202) (in which A + (k, a) is the creation operator for phonons ) must be proportional to the displacement) is westhapolrizn(wch
223
Introduction
8.1
of the form P(x) = a' f
-d 3 K 13
(20 3 a= 1
[a(K, a)ei ' • leica + a + (K, a)e" . xe* K,a]
where a' is a real constant. The charge density is then
p(x) =
—V•
= iOE
P(x)
, fd K 3 E [a(K, a)e'K 'K e- K,a — a±(K,a)e'K e,„]. (2703 3 a =1
To a very good approximation we have one elia which is in the K direction (the longitudinal optical mode) and therefore two which are perpendicular to it (transversal optical modes). Actually these differ in frequency, and have a frequency ratio of (eo le„)112 , where co and e,c are the static and high frequency dielectric constants of the crystals. Only the longitudinal mode, the one with (eca fIK) contributes to p(x); denote the creation operator for that mode simply by al+, and ignore the other modes. Thus
p(x) --= ia'
f d3K K ei' — a z e -iK • xij, (2703 K[a
(8.2)
so that, from Eq. (8.1), the electron's potential energy due to the lattice vibrations
is AVI (x) =
iea'
d 3 K 1 Ea e iK • x (2/0 3 K K
= i(N 2.7M) 1 2
d3K
3 1/4 1/4
(i
(2703 K
j
M
iK
aK e iK
[aK e - i K • x
(8.3)
where a is a dimensionless constant. It can be shown that
1) e 2 (2Mco) 112 c o hw h
= I (1 2k
,
(8.4)
where co and 8,, are again the static and high-frequency dielectric constants of the Crystal. In a typical case such as sodium chloride, a is about 5, and in general it runs from about I to 20. The Hamiltonian for the free electron and lattice is p2 H6±
H05
--=
2M
±
ha)
I d 3k
(2703
,
(8.5)
Where we are considering only one electron with momentum operator P, and We are ignoring the other lattice modes (which do not interact with the electron according to our assumptions).
224
Polaron problem
To simplify slightly the calculations of the following sections we shall assum e units to be such that h = M w = 1. Also we shall sometimes use thethe notation
C d 3K (270 3
(Slue
J
=
(270 36 3(K — K').
With the assumptions made above (due to Friihlich*), the problem is reduced to that of finding the properties of the following Hamiltonian:
H = -1132 +
[aK+ e
E aK-F aK + i(\/2 cur) 1 /2 E K
K
—
iK • X
aK eiK.x ].
(8.6)
Here, X is the vector position of the electron, P its conjugate momentum, and aK are the creation and annihilation operators of a phonon of momentum K. The polaron problem is not of special importance, but we have now reduced it to mathematics. The method we shall use to solve this mathematical problem will be applicable to different problems of a similar nature. Our method will be valid for arbitrary coupling, but we will first assume small a and will use conventional perturbation theory to get, first, an answer that we can compare with our results for arbitrary coupling, and second, a description of what happens if the electron goes too fast.
8.2 PERTURBATION TREATMENT OF THE POLARON PROBLEM
We wish to find AE0, the perturbation energy when there are originally no lattice vibrations. The Hamiltonian is H = Ho + H' where p2 =—
2
H
E
1.4K L4K
1
=
K
AE 0 = Woo
E n
uKeEK • X
).
K
H 0„H '
E g — E,?
+
(8.7)
where H' „ = <mIH'In>, lin> and In> are eigenstates of 1/0 and E° is the eigenvalue of 1/0 corresponding to In>. Because H' acting on a state changes the number of phonons, we bave H'00 = O. We are interested in the energy of an electron traveling with the momentum P. Therefore, we consider the diagram in Fig. 8.3. Here the initial * H. Friihlich, Adv. in Physics, 3, 325 (1954).
Perturbation treatment of the polaron problem
8.2
225
P—K
Fig. 8.3 Electron with momentum (P
—
K) and phonon with momentum K.
state has an electron of momentum P and no phonons. In the intermediate state the electron's momentum is (P Thus the energies Eg and E °„ are
—
K) and a phonon of momentum K exists. p2
Eg = -- , 2
E ° = (P
—
2
K)2
+ 1.
(8.8)
Here "1" in E,° is the energy of the phonon (in our present units).
Hno = i(-\,72 noc) 112 (n i
1
E - (a + e -
— cl iceiK ' x)1P; no phonons>,
lc K K
which is 0 unless In> represents a state with one phonon and an electron of momentum P'. If the phonon has momentum K, lino = 4\12 7OE) 112 (131 , no phonons j(11K)e -i'x IP, no phonons>.
(8.9)
Since e -i "IP> = IP— K>, P' must be P K, as expected, so 14 0 -.--- i(N/ 2 itc) 112 ( 1 /106r,p-K, —
and
AEo = — N72 Tra
1 1 K (P — K) 2 /2 + 1 — P 2 /2 x 1
(8.10)
The extra factor of 2 is a consequence of the spin of the electron. There are two intermediate states with momentum P — K. Generalizing Eq. (8.10) to three dimensions and writing the sum as an integral, we find 2d 3 K/(27) 3 --, - , AEo --- —2-./2 7ra f — K..(K.,.. 2P.K + 2) *
(8.11)
226
Polaron problem
In evaluating this integral, we may make use of the identity rt 1 dx o + b(1 — x)] 2 • ab j[ax Regarding b =
K 2 and
K 2 — 2P .
a=
(8.12)
K + 2, we can write Eq. (8.11) as
r d 3 K j"
j (27
) 3
dx
dx o [ X( K 2 — 2P • K + 2) + (1 — x)K 2r d 3 K1(27r) 3 f 2 (K — 2xP • K ± 2x)2
dx f
d3K1(27)3 [(K — xP) 2 + (2x — x2P 21 2
Now we use the integral: C d 3K/(270 3 _ 1 [K 2 + a ] 2 87r-\/a
to find
J
1
AEi = —47rN/2 a
dx
Jo 8m\/2x — x 2P 2
47r-\/2 a 2 . _ 1 P
8m
P
sin
.\12 •
Thus, finally, AE i = —a
N/2
P
(8.13)
P When P = 0, Eq. (8.13) gives
(8.14)
AEi = —oc.
If the perturbation expansion is carried to higher order,* Eq. (8.14) becomes
AE = —a + 1.26(a/10) 2 . This is the perturbation energy when the electron is at rest. When P is small, we may expand Eq. (8.13) in the form 1 [ p ) +..2 AE ; = —a N/2 [ P, + P -\12 6 "\,/ 2 D
2
=—a —— a
12
+•".
* E. Haga Progr. Theoret. Physics (Japan) 11, 449 (1954).
]
(8.15)
Perturbation treatment of the polaron problem
227
When we combine this perturbation with the kinetic energy, the total energy if
E=
p2
2
P2
a
12
P2 2(1 + a/6)
x+•••
I
+.••.
(8.16)
Equation (8.16) shows that the mass of the electron is increased (1 + a/6) times by the interaction with the phonons. That is,
±
(8.17)
6
ni
In Eq. (8.13), if P > \/2, AE, becomes imaginary. This implies that the energy of the electron dissipates by creating a phonon in the same sense as in terenkov radiation. Let us evaluate the smallest value of P sufficient for this decay effect. Obviously, the process that could give the smallest possible P is the one in which P is in the direction of K. Consider the case shown in Fig, 8.4. By conservation of energy, we know that
102
(P
2
p2 + I =
2
(8.18)
.
Equation 8.18 can be rewritten as K
1 + - - = P. 2 K
(8.19)
The smallest P that can satisfy Eq. (8.19) corresponds to the minimum value of the left-hand side, which is -\/2, so that P > -\/2
(8.20)
is the condition for the "terenkov" effect to occur, as we might have expected from Eq. (8,16). Let us next calculate the rate of this C' erenkov effect. This rate is given as Rate =
E
2nilif0 1 26(Ef — E i).
(8.21)
Final states
The initial state i consists of an electron of momentum P, and in the final state we have an electron of momentum (P K) and a phonon of momentum K. From Eq. (8.6), we have —
Hf
i = ( 13 ' 1
intiP) =
(-\/2
nc
)l/ 2 6
PK
Fig. 8.4 Electron dissipates energy by creating
a phonon.
228
Polaron problem
where 1 K means one phonon of momentum K. Thus 1
Rate = _
27r f N/2 Ircx 6(P K)2 ± 1 K2 2
d3K 2. 2 ) (270 3
(8.22)
The 6-function insures conservation of energy. We may write
d 3K = 47rK2 dK(d0147r) so that
K2 dK 6 (- PK cos 0 + — + 1) , 2
dfl Rate = 2\/2 a f 47r
(8.23)
where 0 is the angle between P and K. When Ko is a solution of -PK 0 cos 0 + 1(2° + 1 = 0, 2
(8.24)
we may transform dK 6(- PK cos 0 +
K2 ± 1) = 2
dK 6[10K P cos 0) 2 - (K 6 - P cos 0) 2)] 2 - P cos 01
Thus Eq. (8.23) becomes
di 4 70(o - P cos 01
Rate = 4N/2 a
(8.24)
When the momentum of the incoming electron P is given, Ko of the created phonon is a function of the angle 0, and is given by Eq. (8.24). Solving Eq. (8.24) we obtain pcose--12 N/2. sin 0 dO = 2a - cosh -1 P . (8 . 25) Rate = 2 -\/2 a P N/2 o N/P 2 cos' 0 - 2
f
It should be noticed that since 1 3/-\72 > 1, we may write sin -1
= N/2
7r 2
+ i cosh -1 N/2
Thus, cosh -1 Ph/2 in Eq. (8.25) is the imaginary part of sin -1 P/N/2, a factor we obtained in Eq. (8.13). The connection between Eqs. (8.13) and (8.25) can be seen as follows. When the rate of transition is y, the amplitude of remaining in the original state has a factor e -."12 , and the time-dependent wave function has a factor e -(y/2
)
t e - iEt
e -i(E - iy/2)t.
Perturbation treatment of the polaron problem
i.2
229
no1 2
E
111.
Eg
E°
Fig. 8.5 The summand in Eq. (9.7) thus —y/2 is the imaginary part of the energy. Equation (8.25) agrees with Eq. (8.13) because both are for the limit of small x and both are based on a perturbation treatment. Although we have found Eq. (8.25) to be consistent with Eq. (8.13), you may object; Eq. (8.13) should be valid only when P < -\/2. The situation is clarified as follows. Strictly speaking, Eq. (8.7) does not hold when E c`n Eg, or when the sum over E (,), diverges near E. Schematically drawn, the summand in Eq. (8.7) has the shape shown in Fig. 8.5. The correct way to take care of the region near Eg is to write the perturbation as AE0
=
E
Onl nO
E + ie
E
(8.7')
,
and take the limit as e > O. —
We see
1 X + iE
is X 2 ± E2
2 X ± E2
The first term has the shape shown in Fig. 8.6, and to use this term is equivalent to taking the principal value when integrating. The imaginary part approaches it times a 6-function as e --* 0, because
E 2
2± 6dx
=
7C.
Thus we may write
1
Inn ---+ ie c--4)
principal value
1
-) — in
(8.26)
230
Polaron problem
f(x)
X
—x Fig. 8.6 The function f(x) — x 2 + 8 2
Fig. 8.7 The integral
foo
X
2
+
5
2
dx = n.
Thus Eq. (8.7') may be written as
A E0 =
EH
Eg principal n
en — lit tH o I 2 6(Ef
OnH nO -
E0).
(8.27)
part
Equation (8.27) shows the consistency of Eqs. (8.7) and (8.21). Although Eq. (8.13) was originally calculated for P < ../2, the analytic continuation v, as tacitly made by raising the pole slightly.
Problem: For low-frequency sound waves co = KC.„ Cs being the velocity of sound. Calculate the critical value of P for radiation of sound. Calculate the direction of the emerging sound waves as a function of energy. Evaluate the rate of emission.
Formulation for the variational treatment
231
FORMULATION FOR THE VARIATIONAL TREATMENT
pie partition function may be written
as
= Tr [e- PH] = E e. When fi ->
CO,
(8.28)
the leading term is e - fiE°. Therefore, we claim
lim[ —1 ln (Tr e)] = Emin . fi-cc
(8.29)
Ibus, we first calculate the partition function to evaluate Emin . We are particdarly interested in the case of large [I. Using the path-integral representation, the partition function is written as Tr (e") =
(path).
The path-integral representation is used when
(8.30)
the Hamiltonian is written in
twins of the coordinates and momenta. Since H is written in terms of the creation and annihilation operators we must first undo it and write it using coordinates and momenta. If we quantize the motion of a crystal, we must choose the creation and annihilation operators so that 1 qK = _(a a _ K) \12 (8.31) i pK = ,_ (a _ K - aK). N/ 2
Rut then the interaction Hint
between the electron and the phonons is of the form
N/2 = i( J2
1
)1/2
IKI raK
= i(12 n ) 1 f 2 E K
=
.\7 2 (V2 7[0
) 112
IKI
-iK • X
— aK eiK 19
(a K - aK)e iK• X 1pKe
1K 1
X.
We do not find
the above form for Hint suitable, because in path integrals we Want potentials that are functions of position, rather than of momentum. We can exchange the roles of qK and pK either before quantization of the crystal Ilkotion (by means of a canonical transformation) or afterward, as follows. Let a ia _ K . Then if we define qK' and pK' as in Eq. (8.31) we find -
q
= pK;
lift
qK.
232
Polaron problem
The resulting Hamiltonian (dropping the primes) is H_
p2
2.\/
lqic)
E
(\/ 2 wr)/ 2
v gK elK X
K
(8.3 -) )
Thus, the Hamiltonian is written in terms of the electron coordinate X. momentum P, phonon coordinates qK and momenta pK . We write
e
Tr (e - ") =
gq 1 (u) gq 2(u) • • •
(8.33)
X(0)=X(13)
where the action integral S is
S=
.2
2
qi(u)) + -\/2 (-\/2 77) 1 / 2
E 9K( u2! 1" (11 (8.34)
The advantage of this method is that the path integral over the phonon coordinates can be performed because qK and K both appear quadratically (and linearly) in Eq. (8.34). We use the result proved previously*, and find
f
exp
[M4 2 ± co2 q + q(u)y(u)] du}
gq(u)
y(t)y(s)e
= exp {+ 40-)
t'l dt dst .
(8.35)
o o
Since y and q are complex here, we modify Eq. (8.35) to read y*(t)y(s) instead of y(t)y(s). We will need this modification after Eq. (8.36). Also, Eq. (8.35) is approximate, for e - fl`-') is neglected. 1- When we use Eq. (8.35), Eq. (8.33) is written as Tr (Col)
e -s
x(u)
= f exp (--
g 2 (u) du) exp (2-\/2
x exp (1K (X(t)
—
X(s))e - it'l dt ds) gX(u)
*Equation (3.39.) The complete form is written by replacing
c("It - s 1 1 — e - "s
But remember that fi is very large.
i fl
e - (01t - s1
1
—
e - "3
in Eq. (8.35) by
0 2K 2
)
Formulation for the variational treatment
L3
or, changing
233
EK into Ç d 3 K1(27-0 3 ,
1 2 t J0
)i.r 2 d 0 —
arc
jo
fl
f
fl
CI 3K
eiK"(x(r)-xcso
(210 3
K2
dt
So
TV du
1
_ ffl 8 0 J0 1X(t) — X(s)1
2
dt
(8.36)
CIS.
Actually, we are in error when we treat all the qii's as independent real variables as we did in Eq. (8.35) above. The 's are complex, with qt . But if we went to the trouble of doing it properly, Eq. (8.36) would still come out as It did above. Our aim in using this path-integral approach is to combine it with a variational theorem described before. We write
qic
fe
s
g(path)
Ç e)e-so g(Path) f e-so g(path) e so gr(path) -
. ,e
where
= q_ x
0)> I e - so (path).
S0 is an appropriately chosen
(8.37)
approximation of S. We use the inequality
<e- f> > e- to write Eq. (8.37) as
where
f
e -s g(path)
e - < s-s°> f e -s° g (path),
(8.38)
(S — So)e -s° g(path) e- g(path)
(8.39)
<S — So > --= -
Because for the approximate choice of the Hamiltonian the free energy F0 is given by e - PF°
, 14
We can write Eq. (8.38) as
1 F < Fo + <S — Se>. (8.40) fi Before choosing So , let us examine the meaning of S in Eq. (8.36). The lirst term is analogous to a kinetic energy, and the second term represents the Potential energy. The special feature of this potential is that the potential at t,
Polaron problem
234
where t is thought of as a "time," depends on the past, with weight e - i t- s . This is a type of retarded potential, and the significance of the interaction with the past is that the perturbation caused by the moving electron takes "time" to propagate in the crystal. When a is small, we may take only the first term of Eq. (8.36) as So :
Then, from, Eqs. (8.40) and (8.29), we obtain 1 E < Eo + hm - <S — S0 ). P-co )6 This gives the old perturbation answer
8.4 THE VARIATIONAL TREATMENT
We have seen that the use of So for the kinetic energy 1 fg g 2dt gives the perturbation result AE .__ — a (second-order perturbation) and furthermore this result is shown to be an upper bound (a result proved only by much greater effort with the usual methods). However, it is possible to imitate the physical situation much better with a more judicious choice of So and thus obtain a much better estimate of E. The next most obvious thing to try for So is the action for an electron bound in a classical potential V(X). This choice can be shown to be equivalent to the use of some trial wave function in the ordinary (Ritz) variational method. In particular, if one chooses V(X) to be a Coulomb potential, one obtains for E (at large x) the same result as that which follows from a trial wave function of the form If one chooses V(X) to be a harmonic potential, then one obtains an e_ improved estimate for E that could also be obtained with a trial wave function of the form e - '2 (at least at large a). However, it can be shown that for I less than approximately 6, no V(X) can improve on the result V = 0! This indicates that a classical potential is not a very good representation of the physical situation, except possibly at very large binding energies. There are two major reasons for this. First, the electron is not constrained to any particular part of the crystal but is free to wander. Any potential V(X) obviously tends to keep the electron near its minimum. Second, we see from the form of the exact action that the potential the electron feels at any "time" depends on its position at previous times, with a weighting factor ea It - 31. That is, the effect the electron has on the crystal propagates at a finite velocity and can make itself felt on the electron at a later .
sA
235
The variational treatment
Electron
Mass
Particle
=1
Force
=- K
Al
Fig. 8.8 An electron coupled by a "spring" to another particle of mass M. time. This is less so for tighter binding, where the reaction of the crystal occurs much faster; hence a classical potential might well be expected to be a good approximation only for tight binding. A model that would not suffer from either of the above objections would be one where, instead of the electron's being coupled to the lattice, it is coupled by some "spring" to another particle and the pair of particles are free to wander. See, for example, Fig. 8.8. The trial action for the system after the coordinates of the mass M have been eliminated is
P f 2 0 where W
di
+
C
-
1# i° [X„) — X (s) Pe -w it - sl di ds,
2 o 0
(8.41)
'
\/K/M and C' = MW 3 /4.
Of course, we could have seen mathematically that S o would overcome the objections raised above without deriving So from any physical model, because we are free to choose any form for the trial action that we please (if it is not Complex). However, the recognition of the physical nature of So greatly helps in avoiding mathematical difficulties when questions of boundary conditions, asymptotic behavior and so on, arise, and also may facilitate computation (see below). We write according to our variational principle E
1 E0 + - ‹.5 — so >,
(8.42)
[3
where E c, is the binding energy of our model and Q means "average with weight Let
I(K,t,$)
(X(t)—X(s))>
r e iK • (X(t)—X(s)) e — So .2X x
236
Polaron problem
If we can evaluate 1(K, t, s), then it is easy to find <S
0 - S 0> =
fl rp
:8
-
So>:
dt ds \
e-it-si
(f 0 Jo IX(t) - X(s)1/
{ 2 jo 0
C f(r p
p x(,) _ x (,),, e-mr-si dt ds) .
For example,
(f f ix(t)_x(s),„—wit—s,
dt ds) = fT dtdse - " - sl Ko , where hKo is V = volume and F(K) = 1 if WI the maximum momentum, that is, E0 = h2KV2m = Fermi level. E
,
K0 h2K 2 4nK 2 V' dK = V —1 h2 1 2 la 5 2m 27r j o 2m (27 ) 3
(9.7)
3 V ° 47rK 2 . N=V dK = -IRK 0 7 3 (20 7 3 So (20 K
Thus, if p o = NIV and
E =---
Ely,
3 h2 K,?) 1 h2 Po = 5 2m 5 4m7r2
B=—
(67E 2\ 5/3 _5/3 _
I
Po
5/3
— aPo .
(9.8)
It is sometimes convenient to define r o and rs so that 1/p 0 --- volume per electron = 47r/3 rci. rs = r0/a0 where ac, = h 2/Zme 2 = Bohr Radius.
(9.9)
rs is of course dimensionless, and for most metals, rs is between 2 and 6. The energy is often expressed in Rydbergs where 1 Rydberg = me4/2h 2 . The results of Eqs. (9.7) and (9.8), plus, of course, many more, can be obtained from 9 of Eq. (9.3). First the normalization factor appearing in Eq. (9.3) will be derived.t We require f (1)*(1) d 3 R i d 3 R 2 • - • d3 R„ , I. Consider a specific term in the summation of Eq. (9.3), without any normalization factor. Such a term is, for example, u 1 (R 2)u2 (R 1 ) - • • . Because the u • are orthonormal,
Also,
f
J
uT(R 1 )ul(R 2) • • • u i (R 2 )u 2(R 1 ) • • • d 3 R 1 ,d 3 R 2 • • • = O.
4(R 1 )4(R 2) • • • u 1 (R 1 )u 2 (R 2) - • • d 3 R 1 d 3 R2 • • • = 1.
* We assume periodic boundary conditions. t The factor 1/N ! did not appear in Eq. (6.125) because th enormalization there was 1/N ! .f (0* v d3Ri • - • d3R,, = L
244
Electron gas in a metal
In other words,
J
cp*u i ( R 2 )u 2 (R i ) • • • PR ' d 3 R 2 • • • = 1, etc.
There are N! terms of the type u 1 (R 1 )u2 (R 2 ) • • • and thus ço*cp d 3N R =-- N! for an unnormalized (p. When u(R) is normalized, the normalization factor is therefore 1/N/N!. The kinetic energy T can be calculated in the same manner as the normalization integral. T = ((p,
Ei
hV? 2 cp) 2m
—
(p*
E h2V i 2m
d 3N R.
(9.10)
Consider a single term of 9, which is now assumed to be normalized. Such a term is 11\ N! u 1 (R 2 )u 2 (R 1 ) • • • and (dropping the 11\ N!), —h 2n2mu i ( R 2 )u2 (R i )- • • =- u 1 (R 2 )u3 (R 3 ) • • •(—h 2 VfL I2mu 2 (R i )).
Although it is not a priori evident that (— h 2 \q/2m)u 2 (R 1 )
where A.2 is some eigenvalue, this must be so since electron number one has nowhere except u2 to go. That is, VI' operates only on u2 (R 1 ) and does not bother other electrons. The argument used to evaluate the normalization integral now may be used, and since T
2m
T=
E
Since V = 0,* the kinetic energy T equals the total energy E and the ui are plane waves. Hence AK = h2 K 2/2m and E h2K 2 12K2 d3K 3 h2K F(K) (9.11) — 5Po = E V K 2m 2m (270 3 2m Equation (9.11) is the same as Eq. (9.8). The energy per electron = 4(h 212m)K cir has been used.
2.22/rs rydbergs, where Eq. (9.9)
9.2 SOUND WAVES
It is interesting to ask what the velocity of sound co is in an electron gas. it can be found by introducing a perturbation of density p = pavg + (5p) cos kx OPlep and determining how it propagates, or more directly by calculating c * Try not to confuse V = volume with V = potential.
Sound waves
9.2
245
where P is given in Eq. (1.53) and p = the mass density (not the number density po). Then co =
(9.12)
The velocity of an excitation V g
=1
08
h OK °
hK o = Po
in
Vg > co ,
in
(9.13)
where Po is the Fermi momentum (not the pressure). Equation 9.12 is a general result; if the velocity of a gas particle is a constant, C, the velocity of sound is C/\/3. For, suppose P is the pressure of the gas, a its internal energy density and p its internal momentum density. Then, to first order in the sound perturbation, we have, using standard hydrodynamics,
= —VP = — C 2V p
(energy conservation).
But for constant velocity we also have P = obtain
Combining these equations we
c2 — V 2 8 = 121,
3
which represents a wave propagating with velocity Chti. The perhaps startling conclusion may be drawn that sound propagation in the collisionless electron gas is impossible. In other words, the energy of a density disturbance will be dissipated very quickly. This may be seen as follows: Consider a Fermi sphere (Fig. 9.1).
Fig. 9.1 A Fermi sphere.
246
Electron gas in a metal
A given excitation of momentum K can increase the total energy of the gas by as small an amount as we please (for instance when an electron is excited from A to B tangentially to the Fermi surface). Thus a great many energy level s of excitation exist for a given momentum increase. These energy levels are below the energy of a sound wave and hence the sound-wave energy will be
quickly dissipated. 9.3 CALCULATION OF P(R)
We now remove the restriction V = O. Let
V=
E V(R i, Ri). if
The reader can easily verify that p, (
E V (R i, R i)(p) = Eif f [u(R 1 )u(R 2) - up(R 2)uT(Ro] ii x V(R 1 ,R 2)[ui(R 1 )ui(R 2)]d 3R 1 d 3R 2 . (9.14)
What is the probability, P(R), of two electrons (of same spin) being a given distance R apart?
P(a, b) = (cp,
E c5(Ri —
a) 6(R ; — b)(p)
if
is the probability that there is a particular at a and one at form as Eq. (9.14), so
P(a, b) =
Eif flu i(a)1 2 lui(b)1 2
Take V = O. Then uli
—
P(a, b) = f f
e
—
a/\/V, and
-11)
This is of the same
Eu(a)ui(a)]Eul(b)u i(b)]}.
(9.14)
E i -) f cPKV/(270 3 , so that
d 3K d31, [1 _ ei(K- L) • (270 6
---- 1 f ei K V
b.
(a
F(K) d 3 K N (2703
- bl F(K)F(L)
N2 2
(9.15)
P(R)E-_-- f d 3a d 3 bP(a, b) 6(a — b — R) 47rR 2 { 1 V
F [
3 K R3
2
(sin K oR — K oR cos K oR)
.
(9.16)
Equation 9.16 is an important result. For example, it tells us that the hard-sphere model of a one-spin gas should be easy to formulate, since under most conditions the electrons are not very close anyway (P(R) = 0 at R = 0). The short-range
Calculation of P(R)
9.3
247
P(R)/R 2
II-R
Fig.
9.2
Probability P(R ) of two electrons (of same spin) being some distance R apart.
Coulomb interactions are now understandably not important—although the long-range ones are. It is also clear that Eq. (9.14) could have been written as
(p (
Eii V (R i, Ri») = f V (a, b)P(a, b) d 3a d 3 b.
(9.17)
The mean potential energy is the integral over a and b of the product of the potential energy between two particles at a and b and the probability of two particles being at a and b. Suppose that one wants to know the number n of electrons that are in some region R. n -----,
E R(R i) = E f 6(R i — a)R(a) d 3 a = JP(a)R(a) d 3a
where
R(a) = and p(a) =
inside R 0 outside R,
{1
Ei 6(R 1 — a)
is the density of electrons at point a. Fluctuations can also be handled. P(a, b), we have
(c p, n 2(p) = (9,
For example, using the definition of
Eif f 6(R i — a) 6(R i — b) R(a)R(b) d 3a d 3 b)
= f R(a)R(b)P(a, b) d 3a d 3b. PK
f p(a)e ±
'lc ' a d
(9.18)
3a
Electron gas in a metal
248
or
pK
= E
E c5(R i —
Ri
a)ei" d 3a.
(9.19)
O. But by squaring Eq. (9.19), we derive
Clearly = 0 for K
is E=
(go
0(k, q))1 f (k)I 2 +
N ( V (q) — V (k — k' f * (k) f (k
)
,
(0(k, q) = k q + q2 in 2m and let us try to minimize it with the normalization condition
E
allowed
If(k)1 2 = 1.
Obviously, this amounts to a diagonalization of a matrix A (allowed k's only), + 0 (k,
Ak,k' = ( g0
k,k' + V (q) — V (k — k') .
For large q, 0(k, q) seems to control the result so that the favorable excitations are electron-hole pairs, f(k) = 5 kk0 . But when q is small enough, the long range of the Coulomb potential suggests that V(q) controls the result, so that the minimizing f(k) is f(k) const. Then the favorable Ô is Ô
which is exactly the operator for a density fluctuation with w_a_ve_number q. We therefore conclude, by this qualitative argument, that for small q. density fluctuations should play an important role. Let us now see how we can make the microscopic picture more exact. 9.6 RANDOM PHASE APPROXIMATION
Bohm and Pines discussed the plasma oscillation in the following way. The Hamiltonian is H=
E i
_ h2 2m
vi2 +1 Ee 2. 2
(9.38)
rii
The potential energy is Fourier transformed and written as , — h 2 vr H=L— i
2m
2
E 47re E 2V
K'
Ri E
K'
(Subtracting N eliminates the self-energy term with i = j.)
.
(9.39)
Random phase approximation
9.6
253
We will proceed to write the equation of motion of density fluctuations. The Fourier components of the density operator p(x) =
E 6(R i — x)
(9.40)
are PA
E e iK • Ri .
(9.41)
Let us examine the Heisenberg equation of motion of pK . The equation of motion of pK is calculated as follows PKI-1 ) =
PA = 411 PK
E e iK.
K • (Pi
+
K12)
(9.42)
171,
where Pi is the momentum operator of the /th electron. In deriving Eq. (9.42), we used the fact that pK of Eq. (9.41) commutes with the potential energy of Eq. (9.39), and also we used the relation — e iK. R ip, = KeiA.R1 (9.43) we note that Eq. (9.42) (remember that in our units h = 1). In evaluating no longer commutes with the potential part of Eq. (9.39), so that we obtain _IK. R i
[K • (Pi + K/2)] 2
e
1 v , 47re2 eu I i e IK R ietK R iK • K'. -
in
V K' I,j
m K 12
(9.44) Now we make a crude approximation: v.- 2.14. iK. Ri [K • (PI + If/2)]2
Ee
K .2„.2f
Le' 3m t
r pK , 3m
(9.45)
where pf is the momentum at the Fermi surface. The "3" in the denominator comes from averaging over three directions. For the second term of Eq. (9.44), we separate the K' = K term from K' 0 K. The former gives
47re2 N
4,2,, ZW.
—
ni V
ft
(9.46)
PA'
where ii = NIV is the number of electrons per unit volume. For K K', we may write
E K' , j
K' # K
47e 2
_ _ -
mK' 2
ei8'
E e i(K-
81
(9.47)
Bohm and Pines showed that this term is small for the relatively high-density case. When we neglect this term we call the result the random phase approximation. This approximation is interpreted physically as the fact that the sum Ei is
254
Electron gas in a metal
small if R 1 are distributed over a wide variety of positions, so that the various components making up the term tend to cancel. It should be emphasized that Bohm and Pines proved this approximation rather than simply assuming it. After these considerations, Eq. (9.44) simplifies to
iiK
(9.48)
colPit
with col = p1( 2/3m 2 + 4ne2 n1m.
(9.49)
The last term gives the plasma frequency = 4ire2nlm.
(9.50)
9.7 VARIATIONAL APPROACH
The plasma state can be investigated using a variational method. It can be regarded as the state of free electrons modified by plasma modes. Thus, we may use as a trial function TTrial =
exP
EŒjÇ
TEree ,
(9.51)
where pK is the Fourier component of the density operator pK =
E
In Eq. (9.51) ΠK is the variational parameter. When we minimize the energy we find that Bohm and Pines's treatment corresponds to the following choice of aK :
=
for for
K
Kcritical
(9.52)
Thus, the ground state for the plasma oscillation is (i9 exp [ —
2ne2
, IPK 1 2 V''''Free•
(9.53)
k Po.if
The introduction of the last factor (involving the f(P)) enables us to dispense with the comments after Eq. (9.66) explaining the limits of integration. Now the integrations can go from — 00 to 00 The sign of each amplitude is determined as follows. For Fermi-Dirac statistics (the present case), the sign is — 1 for every closed "matter" loop. For Bose-Einstein statistics the sign is +1 for every closed matter loop. Figures 9.10(a) and (b) are each considered as one closed matter loop. Figure 9.11 is considered as two closed matter loops. Hence the sign of the first term in Fig. 9.5 is ( +) and the sign of the second term is (—). To sum over spins for the second diagram of Fig. 9.5 gives another factor of 2 in Eq. (9.67). .
(b)
Fig. 9.10 Each
of these is considered as one closed matter loop.
00
Fig. 9.11 Two closed matter loops.
262
Electron gas in a metal
9.9 HIGHER-ORDER PERTURBATION
Consider the diagram of Fig. 9.12. It should be noticed that the interaction is always wave vector q. The perturbing energy contributed from this diagram is calculated (including the sum over spins) as
(
1)323 it (27re2 h23 d 3q d 3Pi d 3 P2 d 3 P3 V4 V g 2 ) (27 ) 3 j (27)3 j (27) 3 j (27) 3 q
q,11 2
q
q P2 ) (1 q 2
2
q
q
2
q 113 )
,
(9.68)
(
(
where the integration is over the appropriate range. Because of the existence of the powers of g in the denominator, diagrams of this type diverge when integrated over g, even with the restriction on the range of integration. However, the final quantity of interest is the sum of contributions from these diagrams, and the final result is expected not to diverge. In order to arrive at the final result without getting involved in the divergence, 1/q 2 is first modified into 1/(q 2 + e 2) using a small E, the integrations are carried out including E, and at the last stage e is brought to zero. The worst divergences comes from diagrams such as those of Fig. 9.13. These are called the sausage diagrams, and the q's are the same for all interactions shown by dashed lines. Consider the integrals
d 3p F(p)0 (210 3
Fg(t) = and
1 =-
• • •
-
O e - It1( 11 2 1 2-F q• P)
f000r) d t 1 • • • dt„(5(t 1 + t 3
+ •
(9.69)
•+
.0
X Fg (tOFg(t 2) - - Fq(t).
(9.70)
The simplest case of A„ is
dt,
A2
6,„
d 3p 1 d 3 p 2 (27) 3 (27) 3 d 3p2 . f d 3p i r (2 7-
=-'
f
) 3
i
(27
)
1
31
file j
-
oodt 1 Cit,l(q2 -"2-(Pi+p2»
d 3p 1 i' d 3p 2 (27
) 3
(27
) 3
1
q2 + q
(pi + P2)
(9.71)
Higher-order perturbation
99
263
l+g
Fig. 9.12 Diagrams of this type diverge when integrated over q.
0110
o
Q
"
o
0 0%
10
°0
0
Fig. 9.13 Sausage diagrams.
where all integrals are over the appropriate ranges. Thus we see that the contribution from an n-loop diagram is
C = n
d 3 (27re2h2
f (27
)3
A n (q)(
Vq 2
—
On
(9.72)
.
)
In order to simplify A n (q), use the integral expression of the 6-function 6(t)
f
et du 2m
so that Eq. (9.70) becomes 1
An = 1 —
n
dt, • • • dtn ei(" - t2+
du
f [f
dteauF (t)] g
+ tou - g(t
) g(t2) • - F g(tn)
du 27r
(9.73)
264
Electron gas in a metal
Use Eq. (9.72) in Eq. (9.71) to write the sum from all sausage diagrams as
E Ç —nEjf (27) Y( d3q3 A„(q)( 2ge2h2 17q2 )
on
E ( — l r pne2h2 f d3q r du L f 3q f du in ri + 2ne2h2 f = fd L =
(2703 j 2n n
Vq2
n
Vq2
(2703 j 2n
n
dteituFg(t)
dteituFq(01 .
(9.74)
The integral over f dt • • • has been worked out for small q and is
F(p)[1 — F(p + q)] 2(1 '12 + q . p ) -Qq(u) = f dte"uFg(t) = fd3P ( 2n)3 (le + q • p) 2 + u2 4n (1 — u tan -1 -61) . u q For small q,
Qq(
(9.75)
/4)
"::-,, '")
4n q 2 3 u2 —
(9.76)
—
Thus Qq(u) is finite for small q. The final answer for the contributions from all the sausage diagrams is
2 4ar g = — (1 — in 2) [ln s+ n 72
f (ln R)R 2 dy f R 2 dy
11 2i
(9.77)
where
R = 1 — y tan -1- y,
(9.78)
Π= (4/90 1 /3 .
rs is the average spacing of electrons measured in units of the Bohr radius. It should be noticed that a in rs term appears in Eq. (9.76). The final result for the total energy is 2.22 0.916 (9.79) E= — + 0.0622 In rs — 0.096 + O(r). rs rs2 Here, the last constant term —0.096 is the exchange term calculated apart from the diagram sum. It should be remembered that the mathematics in this section summed only the sausage diagrams as shown in Fig. 9.13. Although it is true that each of these diagrams causes the worst divergence, the theory would not be complete if the rest of the diagrams like the one in Fig. 9.9 were neglected. Gell-Mann and Brueckner examined the contribution from these diagrams also and — 0.096 is the result; as was expected, it is smaller (for small r) than the contribution from the sausage diagrams, but it should not be neglected.
CHAPTER 10
SUPERCONDUCTIVITY
10.1 EXPERIMENTAL RESULTS AND EARLY THEORY
Just about any material can be brought into a superconducting state, in which there is no measurable electrical resistance, by cooling it below a certain critical temperature. That temperature depends on the material; it gets as high as 21°K for Nb I2 A1 3 Ge. Although the vanishing of the resistance is perhaps the most spectacular effect in superconductors, we can perhaps find clues to the cause of superconductivity by examining other effects. One such effect is the discontinuity in the specific heat at the critical temperature. The specific heat less the aT3 contribution of the lattice gives the specific heat of the electrons, Ce. Comparison of normal elements with superconducting ones gives the two curves shown in Fig. 10.1. For the normal element Ce varies as T, because the thermal energy is kT and only those electrons within an energy range kT near the Fermi surface can be excited. Thus, the thermal energy varies as T 2 . Integrating the specific-heat curve shows that the superconductor is lower in energy than the normal element.
Ce
Normal element
T,
Fig. 10.1 Specific heat of electrons as a function of temperature for a superconductbr and a normal element. 265
266
Superconductivity
Meissner Effect. A changing magnetic field produces an electric field, so a changing magnetic field. if asomethingwzracenoti in a magnetic field, the field lines will be forced supercondtighlae away from the sphere. If a ring is cooled in a magnetic field until it is a super-. conductor, the flux through the ring will remain after the external source of th e is turned off. Apparently the resistance is exactly zero, for experimental] \ field the flux remains constant indefinitely (provided the ring is not allowed Co warm up). More surprising is the Meissner effect. If a solid (simply connected) piece of superconducting material is placed in a magnetic field and then cooled belok‘ the critical temperature, the magnetic field is pushed out of the superconductor. Technically, some lines might be trapped in the object, because some parts reach the superconducting state before others. Furthermore, if the magnetic field is strong enough, it might not be pushed out at all. In such a case, the material does not become superconducting. Its resistance and specific heat are normal. Because of its magnetic domains, iron cannot be cooled into superconductivity. The Gibbs function is defined as G F + PV where P is the pressure and V is the volume. Here the pressure can be taken to be the energy per unit volume required to push the field out. From classical thermodynamics, the Gibbs function has the property that in a reversible change of phase at constant temperature and pressure, G does not change. So the critical field is characterized by pH 2
Fsupercond
"
an
V = Gsupercond = Gnormal = Fnormal-
(10.1)
At the critical temperature, Her = 0, and as the temperature decreases F„,„,„,,, increases, so that Hcr increases.
Fig. 10.2 Variation of critical field with temperature.
Experimental results and early theory
10.1
267
London observed, on the Meissner effect, that if n = density of electrons, mass = in, charge = —e and electric field = E, then acceleration is given by
—eE = mi, and the current density j is
—nei = j. In a magnetic field of vector potential A (if we use cgs units),
E=
1A c
t
gives a rate of change of current density of
dj dt
A OA
et
Hence,
— A(A — A con„) = j, where A — ne2 Imc = constant. A const is constant in time, but may vary with ,position. It is fixed by the Meissner effect, so that no arbitrary magnetic fields can be put into the superconducting region. London proposed that Aconst be taken equal to zero for superconductors. We then must satisfy the boundary condition that Loma' — 0 at the surface of the superconductor, so that A normai = 0. We do this by an appropriate choice of gauge, called "transverse gauge." We conclude then that (10.2) j = —AA. Equation (10.2) implies a modification of the statement that there is no magnetic field in a superconductor. Since v2A
4n AA c
(10.3)
there is no sudden drop in the magnetic field to zero when we enter the superconductor. In one dimension, for example, A cc exp (+
47rA 7c x),
where the sign in the exponential is chosen so that A decreases as the distance into the solid increases. The magnetic field penetrates to a depth of order 700 A (•/c/4nA). The existence of a finite penetration depth can be determined experimentally by measuring the diamagnetic susceptibility of small drops of superconducting material, or by working with thin films. The above theory suggests where the A might come from and approximately how big it should be. But remember that electrons in a metal are not free, so A
268
Superconductivity
is not exactly ne2 1mc. The theory can be made to correspond with reality better by taking ./ = — AA ', where A' is A averaged over the position using an appro.. priate function. For example, in one dimension the average might be taken as follows :
f
A' = normalization constant x
A(y)e - lx - Yll dy.
For "hard" or "type II" superconductors, is much smaller than the penetration depth. For "soft" or "type I" superconductors we have large Impurities make a superconductor hard. A further contribution by London came from consideration of the quantummechanical electric current, which is —e times the probability current: =
—he 2i in
(V0) * (P)-
( 1"1/1
In a magnetic field the momentum operator becomes p + vIcA, so that the
current is then
—e 2m 1
+ [1h
v i
eA
c
c
—he Etevo (vom e 2A teip.
mc
2im
(10.4)
With A = 0 we get -
When A
EiPt [V/i0(v00)*00] = 0.
1m 2
0, 0 0 changes into
=
0.
If 0
0 0 , then Eq. (10.4) gives
e2Ane 2 IP *0 = A = —AA. me me
(10.2)
We see then that if the wave function is "rigid" (that is, if it does not change when A is introduced) then Eq. (10.4) implies Eq. (10.2). What could cause this rigidity'? From perturbation theory, we have
= 00 + r)C1 E
E
-
E0
In>,
where E0 is the ground-state energy and En is the energy of an excited state. if there is a gap between the ground-state energy and the energy of the first excited state, then E„ — Et) is large and 0
Setting up the Hamiltonian
10.2
269
Fig. 10.3 Variation of energy gap A with temperature.
The assumption of an energy gap can also explain the anomaly in the specific heat. Instead of the energy varying at T 2, a gap would cause it to vary as C A I', where A is the size of the gap. More direct confirmation of the existence of a gap has been afforded by experiments involving microwaves. The energy needed to excite the material across the gap could even be measured as a function of temperature, and A decreases as the temperature increases. It took almost fifty years for the problem of superconductivity to be reduced to that of explaining the gap. In what follows we will explain the gap following the theory of Bardeen, Cooper, and Schreiffer. This theory is essentially correct, but I believe it needs to be made more obviously correct. As it stands now there are a few seemingly loose ends to be cleared up.
10.2 SETTING UP THE HAMILTONIAN
The energy gap is A a., kT,, n_s 10 - 3 eV/electron, which is a small quantitative effect to produce a large qualitative effect. Thus the Coulomb correlation energy is too big, and can therefore be neglected. If this reason for neglecting Coulomb effects seems odd, remember that we are not trying to explain and predict everything about the solid. We are just trying to understand superconductivity. If some effect is associated with an energy that is too large, then we know that effect is not the cause of superconductivity. Similarly, as we know that the energy associated with superconductivity is small, we can predict that certain phenomena (such as e + e annihilation) are not going to be affected much by superconductivity. If we change the isotope out of which the superconductor is made, the critical temperature changes. Spin-spin and spin-orbit interactions do not change with changes in the isotope, so they should be neglected.
270
Superconductivity
The speed of sound is definitely dependent on the mass of the atoms of the som e_ supercondt.Iilasbehnogutprcdivyhas thing to do with interactions between the electrons and the phonons. Let us try, then, a Hamiltonian of the form 8
KaZak +
(DIX:CI-
CK + E M KK'CK'-K
a l+C a K
K,K'
+ K,K' E mZiccIt_KaPialc. aZ creates an electron, CZ creates a phonon.
(10.5 )
EK is the energy of an independent electron of momentum K, and WK (h = 1) is the energy of an independent phonon. We measure the energy of the electrons from the Fermi surface, and holes are treated as electrons with negative e. The last two terms of Eq. (10,5) represent the interaction between phonons and electrons. Usually this interaction is taken to be responsible for resistance. High resistance at normal temperatures implies high AI, which in turn implies a special propensity towards superconductivity. There are many effects that cause large energy changes, but which are easily understood and have nothing to do with superconductivity. For example, the effect represented by Fig. 10.4 changes the energy of the electrons by an amount larger than the gap, but does not produce a gap. The diagram in Fig. 10.5 just corrects the properties of the phonon.
Fig. 10.4 An effect that causes large energy changes.
Fig. 10.5 An effect that corrects the properties of the phonon.
Setting up the Hamiltonian
10.2
271
Fig. 10.6 A two-electron process.
Evirtual
K2 1
(€ K2 )
Fig. 10.7 One electron distorts the lattice, the other is affected by the distortion. We see, finally, that the superconductivity must involve more than only one electron or one phonon. We must consider diagrams such as that in Fig. 10.6. Physically, this diagram can be interpreted as a consequence of one electron distorting the lattice, which in turn affects another electron. Figure 10.7 shows that two electrons K1 and K2 interact via a phonon, causing IC; and K1 to come out. For the momenta the relations are
K2
= Q.
—
eK2 in the intermediate region, and the
The initial energy is Einitial virtual energy is EK
Evirtual
h0Q•
6K2
The perturbation energy due to the mechanism of Fig. 103 is
1
— AlK2 K•2
M:11C1 virtual
Einitial
1
,
MK2E ( 8Ki
EK2)
( E ICE ±
6 1(2
± 11W Q )
M ZKI•
(10.6)
272
Superconductivity
Ki
K2
Fig. 10.8 Another two-electron process, of similar type to Fig. 10.7.
Figure 10.8 shows another mechanism of a similar kind.
The perturbation
energy coming from this mechanism is V2
1 MK1K; , V1( 1 + E1C2)
(EK 1
+ eio + hcoQ)
(10.7)
For the calculation of the perturbation energy, the initial state and the final state have the same energy: 4- EK2
or
== eici 4- 81C2
8 1(2 — EC, = — (K 1 — EICO•
The perturbation energy is the sum ViiK'2;KIK2' We are mainly concerned with electrons near the Fermi surface, so we can take all e's to be approximately
equal. Then VICIK'2;K1K2
1
A
,
,
K 2KiMliK
M KlKi M ItK2)•
(10.8)
Q
If the M's are approximately equal, the perturbation energy is negative; so the electrons near the Fermi surface attract each other. In metals, the Fermi surface is curved. If we disregard the curvature of the surface, the Hamiltonian can be treated exactly. However, when we do that Nke find no superconductivity. Therefore the curvature of the Fermi surface is essential for superconductivity. The Coulomb interaction among electrons gives the interaction energy : 47re 2 Vcoulomb
Ki'1 2
(10.9) (const) 2
which is always positive, so the interaction is repulsive. The "(const) 2 " is a consequence of shielding. In order to achieve overall attractive interaction between electrons we need V
Vcoulomb < o•
A helpful theorem
10.3
K1
273
K
-NI
K1
0
K2
-41
Fig. 10.9 A case where
1K1
—
CI is large.
Looking at Eqs. (10.8) and (10.9) we see that for overall attraction when EK1 , larger values of 11(1 — K; I are more favorable. For example, look at the case shown in Fig. 10.9. We can summarize these results by writing a new Hamiltonian for the electrons alone. The phonons have the effect of modifying the eK for the electrons and of modifying the interaction between the electrons so that the interaction is, We will be working with the under certain circumstances, attractive. Hamiltonian ,..-1- ± , (10.10) H = E excilaK + 2.44.10 cIKcilii alif E VKiKx K
1616;K1K2
Our problem will be to discover how Eq. (10.10) can lead to a ground state with especially low energy. 10.3 A HELPFUL THEOREM
Consider a Hamiltonian H = Ho + U. The eigenvalues and the eigenfunctions of Ho are written as Ei and (p i . U
(9 i, U(p).
If all the Ei's are nearly equal, and also all the U's are nearly equal, then a large amount of lowering of energy can be achieved. This is shown as follows : Try a wave function tfr — i: aiçoi and evaluate the energy expectation value for this state: =
E Ei la i l 2 ± i
ii
Following the assumption let us put E0 ,
Uii r:., —V.
The normalization of the a i's is
Then
E lai r i
. 1.
= E0 — V E ara. u
(10.11)
274
Superconductivity
Suppose there are in states of the required nature; what is the best choice of the a i's that makes of Eq. (10.11) a minimum? It is* for
ai = 11\1m, when Eq. (10.11) becomes
= Eo — m V. Thus, when V is positive, or U11 is negative, the stabilization is m-fold intensified. 10.4 GROUND STATE OF A SUPERCONDUCTOR
With the theorem of the previous section as a guide, let us proceed to find a set of such (p i's for which the Uils are nearly equal and negative, and the E,'s are all nearly equal. In the k-space, a wave function is defined by the configuration of occupanc ■ and vacancy of states. For Uo , consider
= fi Pk),
(10.13)
where
Itpk > = Uk kok(1)> + Vkl9k(0)> and
It41 2 + IVkl 2 = 1 . The actual ground state is some linear combination of states of the form given in Eq. (10.13), but we will make the simplifying assumption that with appropriate choice of Uk and 1/k , Eq. (10.13) is the ground state. By adjusting the phase of 19k(1)) and kok(0)>, we can take Uk and Vk real. Our program will first be -\/1 — UZ. Then we to find the ground-state energy by varying Uk and Vk will find the energy of the excited states, which will turn out to be a finite, macroscopic amount above the ground state. The energy of a candidate for the ground state is E = =E (0E07: ak + E-ka -± ka
E Vie k <tplak+, a+ _ k fa k a _01J) k',k
(10.14) where all sums are taken over half the k's and where Vick = E
E Ilk'k + k',k
Then 11 0'
Ground state of superconductor (continued)
10.5
277
Let
sk = t and tk =
OP kl4i a l: ki th> •
E=
E eksk + E
Then k
Vie ktktt:.
k',k
If
I tPk> = ukkpk(1)> + vk I (pk( 0)>, then sk --.---- 2U1,
E =-
tk
= Uk Vk =
E 26k ui + E k
VkrkU k , J/k , U kVk.
k'k
(10.17)
The Uk 's for the unperturbed state are Uk =
1
for
k < kFermi .
Uk =
0
for
k > kFern., i .
The best choice of the ground state wave function is obtained by minimizing Eq. (10.17) with respect to the Uk 'S, Vk being a function of Uk . In the minimization it is not necessary to fix the total number of electrons explicitly. By varying Uk and Vk, we vary the number of electrons. If we choose an appropriate zero of the energy ek , we wind up with the correct number density of electrons after minimizing the total energy. We will take the zero of the energy to be at the Fermi surface, so that &Fermi = O• 10.5 GROUND STATE OF SUPERCONDUCTOR (CONTINUED)
Following our program we must next minimize Eq. (10.17) with respect to the Uk 's. A set of equations results:
4Ek Uk + 2 E
Vkk,U k,V k , Vk —
k'
U 2) k Vk
= 0.
Because Vkk , is real and negative, we can define Ak = —
E vkieuvvk, > 0 k'
and we write Eq. (10.18) as A 1 — 2U 12, 2Ek Uk = Llk NI 1 — U il
Introduce x as U i = 4-(1 + x)
(10.18)
278
Superconductivity
so that
1 - Uri,
f( 1 - x)
= -
and
(10.22) 1 - 2U 12, =
-
x.
Square Eq. (10.20) and use Eqs. (10.21) and (10.22) to solve for x2 : 2
,2 c'k
(10.23)
E;,' where we define
(10.24)
Taking the right sign we obtain from Eq. (10.23):
-(1 - 2Ui) = x --- - ek-
(10.25)
Ek
so that for Ek < 0 (below Fermi level) x > 0 and Uk > Vi„ as it should be. Putting this back into Eq. (10.21) we have
U , = 1 (1 2\
1 V?, = - (1 + 2
Ek 8k)
(10.26)
Ek
tli is the probability that the pair k is occupied. The form of tli, in Eq. (10.26) shows qualitatively that the distribution Ul is a rounded Fermi distribution. Assuming A k is small, we see
LTi --* 1 deep inside the Fermi sphere (ek < 0) (10.27)
and
t/i -+ 0 far outside the Fermi sphere (sk > 0). These limits suggest the distribution shown in Fig. 10.13.
I F.- Unperturbed distribution
2. Uk in
o
Eq. 1 006 fk
Fermi
Fig. 10.13 The rounded Fermi distribution.
Excitations
10.6
279
Fig. 10.14 The range of nonzero Vkic, as assumed by Bardeen, Cooper, and Schrieffer.
To complete the calculation we have to solve for Ak. If we are to use Eq. (10.19), we shall find Uk Vk . Because Eq. (10.26) gives 0
1 Uk Vk = —
1—
2
2
k
=
Ak
,
2Ek
EZ
(10.28)
Eq. (10.19) can be written as Ak,
Ak = — k
21I Ei , ±
Al,
(10.29)
This is the equation to be solved for Ak. In order to obtain some insight, Bardeen, Cooper, and Schrieffer assumed Vkk , is a constant for k and k' within the range hwc, above and below the Fermi surface, and zero otherwise. See Fig. 10.14. When we make this assumption, Ak of Eq. (10.28) becomes a constant independent of k, and the equation for it is
1
1= k
2,1 el2, + A 2
Changing Ek into an integral over s, we have h(,,Ç,. m(g) h(0.,de V de 7.....:, . iviM (0) fo 1 = fwc, V e 2 + A2 I6 2 v 2 + A2
_h
M(s) is the density of states for energy e; it is close to a constant M(0) near the Fermi surface. Solving for A, we obtain A =
ho)„ sinh(1/IVIM(0)) •
(10.30)
10.6 EXCITATIONS Now we are ready to find the excitation energy. To describe excited states we will have to consider the possibility of a given pair of momenta being halfoccupied. Define (10.31) (pk(2) = aipk(0), (Pk(3) = a -Ilk VP k( 0).
280
Superconductivity
on> is some linear combination of the four following orthonormal states:
Then
Uk cpk (1) + Vo k (0) = the appropriate I'/k> for the ground state; Ith(0)) 10k( 1 )> = VOk( 1 ) — Uk9k( 0) =l'//k> for the excited state of a pair; unoccupied; hilk(2)> = (10k( 2) = ith> for a single excitation, with ith( 3)> = (A (3 ) = Pk > for a single excitation with kt unoccupied. (10.32)
If = 91(1 (2) and = UkTk(1) + VOk(0) for k o k l , then the energy that we will call E', is given by Eq. (10.16). Take Ec, as the ground-state energy. Then ski = 1 and tki = 0, so
E' — E0 =
ek i
E vkk suk yk uki vki) - E vkik(uk ykl Uk Vk)
24 1 1)
= 4,0
2 (E
2L/i 1)
Vk i kUkVk Uk i Vk i .
k
We have used the fact that Vkk i and (10.28) we see that
= Vk k •
Using Eqs. (10.19), (10.25), 2
E' — E 0 =
If Ok i
Ek i ( 8 k i )
E ki
2( — Ak,) (,..6,-,k1 ) 2tki
i
A
2
1 = Ek,= Ck iEk 1- iLlk
(10.33)
Tk1 (3), ski = 1 and tki = 0, so Eq. (10.33) still holds. If Ok i Vk1 T k1 (1) — U k1 N 1 (0), the energy becomes E". It is easy to show that ski tki = —Uk Vk1 , and =
E" — E 0 = 2Ek1 . Note that Ek i > Ak i . Suppose Oki = 9k1 (2) and tp k, = Tk2(2). count Vkik2 terms too often, we get
E
E0
Ek,
Ek 2
(10.34)
Then, if we are careful not to
2 Vk,k2Uk i Vk 1 Uk2 Vk 2 ;..." Eki
Ek 2,
because Vk i k 2 Uk i Vk i Uk 2 Vk i is an infinitesimal quantity. Thus,
Etwo excitations
— E0 == Ek i
Ek 2 .
(10.35 )
If Oki = Vk,9k,( 1 ) — Uk 1 9k,( 0), it is plausible to consider the excitation to be a double excitation, with both k 1 and —k 1 excited. We conclude that there is a gap between the ground state and the excited states, and that excitations consist of the breaking of pairs. The excitation energy is the sum of the Ek for each excited electron in each pair.
Finite temperatures
10.7
281
10.7 FINITE TEMPERATURES To find the energy of a system at a finite temperature, we will use: Expectation value of the energy = E
= E (Probability of state 0
x (energy of state i).
Describe a state by the product fjj Itpki (n.d>, or by the sequence n 1 , n2 , , where ni = 0, 1, 2, or 3. If Pk (n) is the probability that th> = 10k (n)>, then the probability of a state is fl. Pki(n). Of course, E n Pk(n) = 1. Then
_
y
E=
=
I
(
+
vk ,,,,,,tk ,(ndt,t(n„,)]
E. j*/ fl Pkini)) Pk iod8k i sk,(n1) nt
ni,.
not
ni
E
+
E
nt,
. j #1,m
Pkini))
E
Pk,(ni)Pknoovk,knitkin,Kn(n.).
nr,nm
not ni,nrn
But
E
ni,
not it/
= j*1
fl E Pk j(ni) = jj 1 =
1,
i* I
101 ni
and similarly,
E
fl
nt,jr 1,m
not
P(n1 )
=
ni,nm
Thus,
E=
E E Pk,(n)ekesk ,(n) + E E Pki(n)Pkm(n')VkA m tk,(0:(0.
(10.36)
I,m n,n'
n
Suppose we call fk the probability that a k state is excited. Then the probability that a given pair of k states is unexcited is
Pk(0) = (1
fk) 2 .
(10.37a)
The probability that one of the states (say k) is excited, and the other is not, is
Pk( 2)
Pk(3) = fiO — A).
(10.37b)
The probability that both states of a pair are excited is
Pk( 1 ) = f
(10.37c)
282
Superconductivity
From Eqs. (10.37) and (10.15),
Sk(0) = 2U 12,, Sk(1) = 2E1,
tk
(
o) = u k vk,
tk(1) = — UkVk, tk(2) — t k( 3) — 0.
Sk(2) = Sk(3) = 1, Thus, from Eqs. (10.36) and (10.37),
E—
E ak[u(t --- 2fk) + fk] + E
f
Vkk , U k , Vk , U kVal. - 2fk,)(1 - 2fk),
(10.38) Here we assume that U2k + V,2, = 1, but we do not assume that Uk is such that E is a minimum for zero temperature. Uk and fk are functions of temperature. For a given set of orthonormal states Uk and fk , define a set of Pi = probability of state i. Then
E=
E Eip, i
S = — k E Pi in Pi (k = Boltzmann constant) i
F(Pi) = E — TS.
(10.39)
If the Pi are proportional to e - k T , then F(P1) — free energy. It is easy to prove that if F(Pi) is minimized with respect to Pi (subject to the condition that E Pi .--- 1), then the Pi have the correct values for temperature T. Using the method of Lagrange multipliers, we find that
0=
OF 0 Pt
.1
0
E P•t = Ei + kT (in Pi + 1)
0131
—
A.
Then Pi ---- e( A/kT-1)e -Ei lkT . .1 must be chosen so that E Pi = 1, in which case the Pi are correctly given for equilibrium. For the case of superconductors at a finite temperature, minimizing F with respect to Pi is equivalent to minimizing with respect to Uk and A. Before we do this, note that we are really considering an F that has an undetermined number of electrons. We then minimize subject to the condition that the expectation value of the number of particles is a given number. So, for example, Pi should be written P(ni , n2 ,. . .) where n i , n 2 , .. . are the number of particles in state 1, 2, .. . respectively. In addition to the condition that
E
ni,n2,
P(n i , n 2 , ...) = 1 • • •
we have
E 1-11,n2, - - •
(n 1 + n2 + n3 + • • • )P(ni , n 2 , . . .) = fixed number.
Finite temperatures
10.7
283
The method of Lagrange multipliers then gives what amounts to the chemical potential of Chapter 1, Section 6. When working with superconductivity, we do not explicitly write the chemical potential because it is taken care of by an appropriate choice of the zero of the energy. S = entropy = —
k
—
k
E probability of state î
x in [probabilty of state
E Pk(n) ln Pk(n).
It follows that
TS = — 2)6 0=
E rf, In fk + (1 — jek ) in (1 — fk)]
TS (51' 6E = — 6fk bfk (5fk
= 2ek(1 — 2U1) — 4
Ek' v,,,,uk,vk,ukvk(l
2fk,) +
2)3- 1 ln
fk
1
fk
(10.40) Set k
Ek( 1
2
2 U1)
Ek'
Vkk'Uk'Vk'U kl /k( 1
2,10
Then 1 / (efiek
fk
+ 1).
If we now take 6F 6E 6U k (5U k
[4.EkU k ±
2E
VkktUk•Vk•
k'
U2 k Vk
2f1( 1 )]( 1
2fk),
(10.43) we have (in principle) equations that can be solved for both Uk and fk . Now, define Ak —
E VkkoU kono( 1
(10.44)
2fk').
Then Eq. (10.43) becomes identical to Eq. (10.20). All our results from Eq. (10.20) through Eq. (10.28) are true at finite temperatures. Equation (10.41) becomes 63 k —
—
2U
+ 2U kVkAk =4,k gk
Ak
2Ek
A
—k
F
(10.45)
where we have used Eq. (10.44), then Eqs. (10.25) and (10.28), and finally Eq. (10.24).
284
Superconductivity
The final result is Ak = —
, Ak ' tanh E k Ek' Vkk' 2Ek 2k T ' ,
(10.46)
'-k = ei + Al. To simplify this equation, assume Vkk' = — V
A. Thus for V < A there is no current. For V> A, it is easy to show with our simplifications that I oc I V 2 — and looks like the function in Fig. 10.19. At finite temperatures, the curve looks like the function in Fig. 10.20. This situation, with metal 1 a superconductor and metal 2 normal, can be visualized as shown in Fig. 10.21. At V = 0, E 1 is the same as EF2 . When V lowers EF2 more than A 1 below EF„ current can flow from 1 to 2.
Real test of existence of pair states and energy gap
10.8
289
I
V
o
Fig. 10.19 Current across a junction with one superconducting metal at zero temper-
ature.
I
V
Fig. 10.20 Current in the same junction at finite temperatures.
Unoccupied .[ levels .r.
Gap
-
Unoccupied levels
--I
EF,
Filled levels
1 (superconductor)
2
(normal)
Fig. 10.21 Levels in a junction of two metals, separated by an insulating layer. Metal 1 is a superconductor and metal 2 is normal.
290
Superconductivity
T=0
o
T>0
o
6'1 1- 2
1 6'1 —6' 21
z 6, 1+ ' ■ 2
Fig. 10.22 Current in a junction of the superconductors, at zero and finite temper-.
atures.
Suppose both metals 1 and 2 are superconductors. The current as a function of voltage looks like the curves in Fig. 10.22. From the second curve in Fig. 10.22, both A l and A2 can be obtained. For more details concerning the energy gap, see Progress in Low Temperature Physics, Vol. IV, page 97 and on (especially, page 140 on).
10.9 SUPERCONDUCTOR WITH CURRENT In Section 10.4, we assumed that the electron pairs had zero total momentum, so that k i + k2 = 0. The more general situation is k i + k2 = 2Q = constant for all pairs, in which case there exists a current. As before, we can describe a pair by a single momentum vector, k, but in this case,
k i ---= k + Q, k 2 =--
—
k + Q.
Formerly, the energy of a pair was ek -I- c-k = 2. But h2
h 2 k2 (k + Q)2 + h2 ( — k + Q) 2 = 2 2m 2m 2m
h 2 Q2
2m '
SO
ekt + ek2 = 2ek ± 2(h 2 Q2 /2m). In analogy to what we did before, we define ViV as Vk±Q,_ k+Q;k' +Q,-k' Q• Notice, however, that V (kQk), ,-z.-, Vut + terms of the order Q 2 or less. The change in Vide of order Q2 is similar to the doppler frequency shift in sound emitted from moving objects. Neglecting "Doppler shift " we modify Eq. (10.17) only by the replacement of 2ek by ak + 2h2 Q 2 12m. The energy of the ground state 20. is increased by ( h2Q2,20
n
Superconductor with current
P.9
291
The number density of electrons is A( = f 2Ui, and we can define V = •01M. Then excess energy = ,K(mv 2 /2). (10.56) The current is
where p is the momentum operator. It follows that J=
—eh E [(k + Q) + (—k + Q)YI 12, = —e hQ E 2Ui ,-- — 'rev. Mk Mk
(10.57) How is the wave function changed by replacing pairs +k with pairs ik + Q? The equations for Uk can be brought up to date by replacing ek with ek + h2Q2/2m. As well as h2 Q 2/2m, we must add another constant to make the zero of the energy appropriate for the correct number density of electrons. The net result is that the zero of energy is again at the Fermi surface, and the Uk come out the same as before. But now the (A describe the amplitude for +k + Q to be occupied. Suppose we describe the ground-state wave function by 00(x 1 , x2 , . . . ). We want to modify this wave function to become tk,2(x l , x2 , ...). This modification is effected by replacing each component of momentum p by a component of the same amplitude, but of momentum p + hQ. That is, eiPe'rem is multiplied by eif2. x.. Then
= exp (IQ •
E xe)(11 0 =
exp[i E
e
mu h
-
xd11/ 0 . e
(10.58)
Equations (10.56) and (10.58), it should be noted, represent simply the consequences of a Galilean transformation of Schriidinger's equation. According to our approximations, a superconductor with a current is the same as a superconductor in its ground state as seen by a moving observer. In a real metal the current can vary from place to place. Because k is large near the Fermi surface, the lowest-energy wave functions are not affected much in energy if the variation in current is slow. To see how to deal with variation of current with respect to position, assume a metal is divided into regions. Each electron is constrained to move Within a given region, and for each region there is a different current. Then the wave function becomes
m 0 = exp (i — v i h
E xe exp electrons in region 1
im = exp (- E v(re) h
e
te)
J
ipo .
m
i — v2 h
E
xe • • tir 0 )
region 2
(10.59)
292
Superconductivity
We can now let v(x) vary gradually with x e . But this is not quite right. To st why this modification of is inaccurate, remember that the equation for curre t density involves derivatives. The fact that V(Xe) is not constant modifies a current. To improve our description of a current in a superconductor, let 11S ne: try
= exp (i
0(x e)) 'P o .
The number density of electrons is given by K(R) =
Ef
6(x, — R) dx i dx 2 • • • ,
and the current density is given by .0)
—he e 2im
ji [leVetfr — (V M*111] 6(x e — R) dx i dx 2 • • • .
If we remember that the current for 1// = tk o is zero, we find
h j(R) = — jre — VB.
(10.60)
For example, if O = mv • xlh, j(R) is the same as that for Eq. (10.57). Note that we have only described currents for which V x j = 0. For the steady state, V j = 0. In a ring, the above conditions can be satisfied with nonzero current. In ds = —Afe(hlm) AO, where (AO) is the difference in O as we go this case, around the ring. AO need not be zero, but it must be a multiple of 2m, for \A e — require that the wave function be single valued. If we define y = (111m) VO, then v • ds
27rn
A
,
( 10. 6 1)
where n is an integer. If the ring's central hole shrinks to a point, the only vs, * n can be kept greater than zero is to have lines along which the superconductor breaks down. By calling v = —j/4 1e, we have implied that this quantity can be thought of as a velocity. This interpretation is confirmed by computation of the expected value of the energy.
Current versus field
10.10
—2
f 0*H0 . 10* [
E=
ii 2
: 2
293
vi + vi 0
f Vele • Vetk + ii *V O
= 2m
h2
„
= 2. j P R j (vee
• Vetk) 6 ( re - R) + 5 0*i/0 (10.62)
VI*VO = tPôVtko. Also,
vtp* • vtp = vg . voo + ive . [00vg — qto'vvio] + (v0)2400. Equation (10.62) becomes (if Eo is the energy for
E = E0 +
h2 f c1 3 RiVO(R) • 2m
h2
+
2m
fd3Rive(R)12
SE e
0 = 00):
[ioveg — 1e/'O] 6(x, - R)
E to,(0 06(x, —
R)
e
= E0 + -h Vf d 3R O(R) • fo(R) + e
h2
2m
fRd 3RIVO( A 2 jr(R).
(10.63)
But j0(R) = current for ground state = 0. Then
E = E0 + f d 3 R(imv 2) K(R),
(10.64)
as we would expect from the interpretation of v(R) as a velocity. 10.10 CURRENT VERSUS FIELD
In the equation j(r) = - AA'(r), A' is averaged over some region about the point r. In other words, we must use a sort of "spread-out" A(r). We can interpret this spreading as an effect of the size of the electron pairs, and we can calculate the spreading by means of the BCS theory. Empirically we have
A'(r 1) -,-- C
f[A(r)e - Ir'r4° • (r - r')](r - r') d3 r, Ir - r'14
(10.65)
294
Superconductivity
where C is an appropriate normalization factor (so that constant A gives A' = A). Note that Eq. (10.65) has the form of a convolution. If we write j(q)---- f i(r)e -ig" d 3 r,
di(q) = JA(r)e" d3 r, (10.65)
K(q) =
I K(r)e - iq'r d3 r,
K(r) =
Ce - r/c)rr
r4
then
(10.67)
j(q)---- A(q) • K(q).
In what follows we will find the component of j(q) that is in the direction of A(q). Suppose that A(r) is in the z direction and has magnitude A o e - iq 'X , where y is perpendicular to A (we are using a gauge in which V • A = 0). Then
A z(q') =f Az(x)e -ig" • x d 3 x =
A0 (5q of ,
where everything is normalized within a unit volume. In Section 10.1, we showed that if the wave function is "rigid," jp . -AA. In so far as the wave function is not rigid, we get a contribution from the paramagnetic current of - he
2im
[0*Vti/ - (V0) * 0]
. itotal =
i + ifr
Set h = 1. By considering tp as an operator (as in second quantization) we could get the operator for fir). But we will use a slightly different approach. For a single particle, the operator for Mr) is j1 (r) =
e
2tri
[Pi 6 (r - r1 ) + 60' - rOP ii ,
where r1 is the operator for position and r is a vector (a set of three numbers). Then ii
(
q) =
f
0 1 (r))e -ig". --=
I Odii(r)10>e
-—e f ii d 3ri d 3re(r1 )[-1 V, m
i
d 3r
c5(r - 1. 1 ) + (3(r - ri )
1 i
V 1 ] 0(r 1)e- iq • r
Current versus field
10.10
295
After some manipulation we find —
MO
e
= 2m
= 00> + E where
TÀ. (Hel) and a < for T < T,i, (Hell). This result has not been obtained theoretically and remains one of the unsolved problems of superfluidity. At very low temperatures the specific heat is proportional to T 3 , a relation that is understood. Properties of Liquid Helium: Liquid helium below the il-point, that is, Hell, or "superfluid" helium, exhibits several remarkable properties. 1. As has already been noted, He ll does not boil (although there is evaporation). This can be explained by assuming the heat conductivity of Hell to be essentially infinite. 2. More remarkable than the apparent infinite heat conductivity is the zero viscosity (under certain conditions) or superfluidity of Hell. It has been shown that below a critical velocity Vi, Hell flows through a thin capillary or "superleak" with zero resistance. Furthermore, V,. increases as the capillary diameter decreases. Observing flow through a capillary is not the only way to measure viscosity, however. If a cylinder is placed in a liquid helium II bath and rotated, there is a momentum transfer from the rotating cylinder to the helium, indicating that under the conditions of this experiment, the viscosity is not zero! These viscosity experiments can be crudely explained if it is assumed that Hell is composed of an intimate mixture of two fluids—one fluid with zero viscosity and density ps, and the other with normal viscosity and "density" pn .* Thus it is the zero-viscosity component that flows through capillaries and_ the normal component that interacts with the rotating cylinder. To explain the observed data, we must assume that NIA, is a function of the temperature. 3. A third remarkable property of Hell is the "thermomechanical" or "fountain effect," together with the related "mechanocaloric effect."
Consider two containers of Hell connected by a superleak (Fig. 11.3). Let the density, p, and temperature T be kept constant on each side. Then there will be superfluid flow through the superleak until the pressure head AP is equal to AP = ps AT. s is the specific entropy. This temperature difference giving rise to a pressure difference is known as the thermomechanical effect. If one container is a thin tube the pressure head will cause a fountain of liquid helium (Fig. 11.4). * The reason for the quotation marks about "density" will become evident later.
314
Superfluidity
A
B
PF-FAP, Ti-F-AT
He II
He II
Fig. 11.3 The thermomechanical effect.
Heat
Fig. 11.4 The fountain effect. If the containers of Fig. 11.3 are kept at a constant pressure and there is mass flow from A to B, container B will cool. This is the mechanocaloric effect. If the two-fluid concept is extended, so that now it is assumed that the superfluid component not only has zero viscosity but also carries zero entropy, the thermomechanical and mechanocaloric effects can be explained both qualitatively and quantitatively. For example, the cooling of container B in the mechanocaloric effect is exactly what one would expect if the mass transferred had zero entropy and hence zero temperature. The superfluid flow of a zero-entropy, zero-viscosity fluid also explains the abnormal heat conductivity of Hell. Most of the other "superfluid" properties of Hell can be explained in terms of the properties described above. For example, liquid helium II placed in a beaker will creep up the surface of the beaker and spill over the sides. This will continue until the beaker is empty (it being assumed throughout that the temperature of the helium is always below the A-point). This rather peculiar behavior can be explained by the ordinary physics of evaporation and the infinite thermal conductivity and zero viscosity of Hell. If any normal liquid is placed in a beaker we would expect that, due to Van der Waals attraction between liquid and beaker molecules, a layer of liquid, whose thickness decreases with height, will be formed on the beaker walls. In all liquids but Hell, however, the layer formation is inhibited by small but finite temperature, and hence vapor pressure, differences between the wall and the liquid. Thus, depending on whether the wall is warmer or colder, there
11.1
Introduction: nature of transition
315
will be either rapid evaporation from the wall layer or the formation of droplets that will fall back into the liquid. In Hell, however, the abnormal heat conductivity prevents the establishment of any temperature differences, and as a result the layer creeps up the side and over the top. Landau Two-Fluid Theory: We have seen that both the viscosity experiments and the thermomechanical effect can be explained (crudely) by a two-fluid model for liquid helium II. A phenomenological theory using the two-fluid concept was introduced in 1940 by Tisza* and in another form in 1941 by Landau. Since Landau's view is the deeper one, we shall concentrate on his formulation. Consider Hell to consist of a perfect background fluid (which has zero entropy and viscosity), and some type of excitations, which for the momeni may be regarded as phonons. This simple hypothesis explains a great many ol the superfluid properties. (The model is somewhat analogous to a solid consisting of a background lattice plus phonon excitations.) The excitations according to the Landau view are the normal component. First, the specific heat of a "phonon gas" is proportional to T 3 at low temperatures in agreement with Fig. 11.1. Second, as the perfect background fluid flows through a superleak the phonons are inhibited because they collide with the walls. Thus the perfect fluid emerges with no excitation and hence zero entropy. The two-fluid model is further strengthened by the following ingenious experiment due to Andronikashvili.t A pile of closely packed discs is rotated in a Hell bath. The superfluid component is unaffected, but the phonon: (and any other excitations) are dragged around with the discs and have ar inertial effect, which can be measured. In this way the "density" pn of the inertial or normal component can be measured. (Fig. 11.5). The two-fluid concept suggests that the two components can oscillate oui of phase in such a way that the total density of Hell at a point is essentiall3 constant but the difference or ratio of the superfluid and normal components i: not. The density of excitations, however, is a function of temperature, and hence as pn l p varies so must the temperature. This leads to a new type of wave propaga. tion known as "second sound." Second sound is a temperature wave and wil be excited by heat rather than pressure pulses. According to the Landau view, one can think of second sound as a densit) wave in a phonon gas. If the phonon velocity near T = 0 is c (c is, of course, th( * Some of Tisza's ideas can be found in his following papers: Nature, 171, 913 (1938) J. Phys. et Radium, (8) 1, 164, 350 (1940); Phys. Rev. 72, 838 (1947). t L. D. Landau, J. Phys. USSR 5, 71 (1941). See also: L. D. Landau and E. M Lifshitz, Fluid Mechanics, Addison-Wesley, Reading, Mass., 1959, ch. XVI; am L. D. Landau and E. M. Lifshitz, Statistical Physics, Addison-Wesley, Reading, Mass. 1959, ch. VI, §§66-67. E. L. Andronikashvili, J. Phys. USSR 10, 201 (1946); Zh. Exsperim. i Teor. Fiz. 18 424 (1948).
316
Superfluidity
Fig. 11.5 Andronikashvili's experiment.
c,
X I
1 1
I
------.
7.\ 7-1 2
3
.- T (° K)
Fig. 11.6 Velocity of second sound.
velocity of first sound), the ordinary theory of sound propagation predicts that the velocity of second sound cs = c/-\73 as T ---* O. (See also Section 9.2.) Tisza associated second sound with the superfluid component rather than the normal (phonons), and as a result he predicted that c --4 0 as T --* O. For a while there was a controversy on this point, but experiment soon proved Landau's supremely confident prediction that cs ---> c/-0 to be essentially correct. (Actually es > cW3 at very low temperatures, less than 0.5°K.) The actual curve of second sound vs. temperature is given in Fig. 11.6. To explain this curve between the 2-point and 1°K, it must be assumed that there are other excitations besides phonons. Landau derived the excitation curve of Fig. 11.7 empirically; it has since been developed theoretically. At the
11.1
Introduction: nature of transition
317
E(P)
Fig. 11.7 Excitation curve for phonons and rotons.
lowest temperatures, the mean free path of the phonons becomes large; so it becomes difficult to measure the speed of second sound. At low temperatures, E(P) = cP and the excitations are phonons. In the region around P o , E(P) = A + (P
P0) 2 2tt
—
where A is a constant and ti is some effective mass. Landau called these excitations "rotons." The two-fluid model of Landau can be put on a more quantitative level by considering the statistical mechanics of a phonon-roton gas. The free energy is F = kTln (1 —
e-
E( p)/ kT) d3P
v.
(11.1)
(27th) 3
It is clear that the low-energy excitations (phonons) and the excitations around Po (rotons) contribute the most to the integral. Well below 1°K, however, the contribution due to the phonons dominates. Above 1°K, the rotons dominate. The average energy and specific heat due to the phonons (E = Pc) is found as follows: The expectation value of the number of phonons of energy E is
E, nc 0 ne -13(nE) Enco 0 e —B(nE) . 9,
-_..--
11 =eBE _ 1 e Pc / kT ____ 1 -
Then c) Eph =
f
o
Pc e
P c 1 kT
4nP 2 dP 4700T 4 ____ 1 (27Th) 3 15h 3 C 3 ' -
Cph =
167r5 VT' 15h 3 C 3
.
(11.2)
318
Superfluidity
The specific heat due to the rotons (E = A + (P - P0) 2 /2p) is
kT A
2p 1 / 2PgA 2 Crot
(2703/ 2 0/ 2 T 312 h3
kTy] e-AMT 4k A
(11.3)
To understand the Andronikashvili experiment and the meaning of p„, we must consider the statistical mechanics of a moving phonon-roton gas. Perhaps the simplest way to proceed is to assume that the superfluid background component is moving with velocity V. Then the energy of an excitation, phonon or roton, is given by
E = E(P) + P • V,
(11.4)
A similar equation was derived for superconductors. An easy way to see this relation is as follows. Suppose a gremlin floating in the fluid gives it a kick and produces an excitation of momentum P. Then the velocity of the rest of the fluid is decreased by PIM, where M is the mass of the background fluid. If the fluid winds up with velocity Vs plus the excitation, it must originally have had velocity Vs + PIM, along with the energy necessary to create the excitations. The total energy must therefore be 1M( Vs + P/M) 2 + E(P) MV + P • V3 + E(P). The energy the fluid has if it moves at velocity Vs without an excitation is Therefore the energy necessary to excite the fluid without changing its velocity is E(P) + P • V. Now, applying ordinary statistical mechanics, we find that the expectation value of the number of phonons of momentum P is
N= (E(P)+P • Y.)IkT e
(11.5)
The total momentum density is evidently given by pV, + 1. At reasonably high temperatures y is small and the sum converges. Also, at such temperatures there are not too many polygons, so assumptions (a), (b), and (c) are not too unrealistic. Thus we expect our results to be qualitatively correct above the A point. Equations (11.55b) and (11.56b) resemble the results for a Bose gas (see Section 1.8) and lead to an increase in specific heat with decreasing temperature. The shape of the specific heat curve is shown in Fig. 11.37. Specific heat 1
T
Fig. 11.37
Specific heat near the lambda point.
350
Superfluidity
This analysis (which is supposed to be good for T> T2) thus explains why C increases as T approaches 1'2 from above. This treatment does not lead to the discontinuity of the kind shown in Fig. 11.34, probably because of the approximations made between Eqs. (11.52) and (11.55a). In other words, we expect that the volume of the helium atoms described by p in Eq. (11.52) is the cause of the discontinuity in the specific heat. Kikuchi and others have written papers in which it is indicated that more accurate computations starting from Eq. (11.52) would lead to a discontinuity.*
Problem: How do these polygon calculations differ from the ones we did in Chapter 5 for the Onsager problem?
* Kikuchi, R., Phys. Rev. 96, 563 (1954). Kikuchi, R., Denman, H. H., and Schreiber, C. L., Phys. Rev. 119, 1823 (1960).
INDEX
Annihilation operators, see creation and annihilation operators Bardeen-Cooper-Schrieffer (BCS) model, 269 BCS model, see Bardeen-Cooper-Schrieffer model Blackbody radiation, 10-12 Boltzmann's constant, 1, 6 Bose-Einstein condensation, 31-33 Bose-Einstein statistics, 30-33 classical limit of, 31 density matrix for, 60-64 field operator for, 173, 174-176 free energy for, 64 Chemical potential, p, 26-27 in a Bose-Einstein system, 26 in a classical system, 109 determination from number of particles, 27, 28, 30, 34, 109 in a Fermi-Dirac system, 26 for the lambda transition problem, 347 for a one-dimensional classical gas, 119, 121 for a one-dimensional order-disorder problem, 130 for a two-dimensional order-disorder problem, 131 Classical systems, 97-126 condensation of, 125-126 equation of state of, 98, 100, 102, 103, 110, 114 Mayer cluster expansion for, 105-110 of one-dimensional gas, 117-125 partition function for, 97 radial distribution function for, 111-113 second vinai coefficient for, 100-104 thermal energy of, 98, 114
Cooper pairs, 274 Correlation energy of electrons, 249, 255-264 Creation and annihilation operators, 153 for Bose-Einstein systems, 173, 174-176 expression of other operators in terms of, 176-182, 188, 189, 191, 192 for Fermi-Dirac systems, 173, 174-176 and the harmonic oscillator, 154 for phonons, 159, 162 for spin waves, 217 for a system of harmonic oscillators, 157-159 and an unharmonic oscillator, 156 Critical temperature for Bose-Einstein condensation, 31 for liquid helium, 313 for order-disorder systems, 140 for a superconductor, 284 Debye approximation, 19 Debye temperature, 20 Density matrix, 39-71, 40, 47 in the classical limit, 77 differential equation for, 48 for free particles, 48-49, 59, 73-74 for free Bose-Einstein particles, 60-64 for free Fermi-Dirac particles, 64 functions of, 45 for an harmonic oscillator, 49-53, 80-84 for an interacting harmonic oscillator, 81-84 for one-dimensional motion, 74-76 path integral formulation of, 72-96, see also path integrals perturbation expansion for, 66-67 for polarized light, 42-43 for a pure state, 41 in the x-representation, 41 351
352
Index
time derivative of, 44 of an unharmonic oscillator, 53-55 Wigner's function and, 58-59 see also density submatrix Density of states, 2 connection with temperature, 3 independence of energy, for a heat bath, 3-5, 6 Density operator, 248, 253, 326 equation of motion of, 253 Density submatrix, 64-65 see also distribution function Distribution function, 65 Kirkwood approximation for three particles, 116 for a single particle, 65, 111 for two particles, 65, 111-113, see also two particle distribution function Effective mass, 227, 241, 317, 345 Electrons in a metal, see electrons in metals and phonons, 185-190, 221, 224, 231, 270-273 and photons, 190-192 superconducting, 301 Electrons in metals, 37, 242-264 correlation energy of, 249, 255-264 energy of, 243, 244, 249, 264 Fermi energy for, 243 Fermi-Thomas model for, 250 Fermi-Thomas wavelength for, 250 phonons in, 244-246 plasma oscillations in, see plasma oscillations specific heat of, 37-38 structure factor for, 248, 330 two particle distribution function for, 246 Energy, thermal, see thermal energy Entropy, 6, 45 of Bose-Einstein particles, 29 of Fermi-Dirac particles, 29 law of increase of, 8-9 of liquid helium, 313-314 for order-disorder problems, 136 for a superconductor, 283, 299 Equation of state, 98, 100, 102, 103, 110 Equilibrium, thermal, see Thermal equilibrium
Exchange integral, 199, 200 Expectation value, of an observable, 1, 42, 59 of energy, 7, see also thermal energy of kinetic energy, 52, 57, 58, 65 of momentum, 45-46 of position, 45 of potential energy, 52, 57, 65 Fermi-Dirac statistics, 34-38 density matrix for, 64 field operator for, 173, 174-176 free energy for, 64 Fermi energy, 34, 243 Fermi temperature, 34 Fermi-Thomas model, 250 Fermi-Thomas wavelength, 250 Ferromagnetism, 129 Feynman's diagrams, 192-196 correlation energy of electron gas and, 257-264 Field, 162 Field operator, 163-167 for Bose-Einstein systems, 173, 174-176 for Fermi-Dirac systems, 173, 174-176 see also, creation operators, annihilation operators Free energy, 6 of blackbody radiation, 11 of Bose-Einstein particles, 30, 64 in the classical limit, 77, 94, 95, 97, 98, 109 of combined systems, 9 of a crystal, 17 for Fermi-Dirac particles, 64 of free particles, 49 of an harmonic oscillator, 9, 52 inequality for, 47, 54-55, 67-71, 87, 93, 233 for liquid helium, 317, 345, 347 for one-dimensional motion, 76, 119, 120 for one-dimensional order-disorder problems, 130 perturbation expansion for, 56, 67-71 of spin waves, 209 for a superconductor, 266, 282, 283, 299, 302 for two-dimensional order-disorder systems, 148, 149
Index
Gibbs function, 27-28 for Bose-Einstein systems, 30, 32 for Fermi-Dirac systems, 34 for a superconductor, 266 Graph representations for cluster expansions, 106 for the correlation energy, 257 Feynman's diagrams, 192-196, 257-264 for the lambda transition, 347 for the Onsager problem, 137 sausage diagrams, 262 in superconductivity, 270-273 Ground state for a fermion system, 182-185 for liquid helium, 321-324 for many particle systems, 152 for a superconductor, 274-279 Heat bath, 1, 2 of harmonic oscillators, 3-4 of free particles in a box, 4-5 Heat capacity of a macroscopic system, 6 Hydrodynamics, 305, 319 Junctions, with superconductors, 287-290, 306 Kirkwood approximation, 116 Lambda point of liquid helium, 33, 313, 343 Liquid helium, 312-350 critical velocity, 313, 334, 342 effective mass, 345 energy, 317 entropy of, 313-314 excitations in, 324 free energy in, 3_17, 345, 347 ground state of, 321-324 lambda point of, 313, 343 mechanocaloric effect in, 313-314 phonons in, 315, 326-327 quantization of vorticity, 336 rotons in, 317, 327-333 second sound in, 315 specific heat of, 34, 313, 315, 318, 349-350 structure factor of, 330 thermomechanical effect in, 313-314 two fluid theory for, 315
353
viscosity of, 313 vortex lines in, 339-342 Mayer cluster expansion, 105-110 Meissner effect, 266-269 Mössbauer effect, 22-25 Normal modes, of vibrations in a solid, 13-17 Onsager problem, 136-150, 350 Order-disorder problems, 127-150 critical temperature for, 140 in one dimension, 130-131 in two dimensions, 131-150 Partition function, 1-9 of Bose-Einstein particles, 25-27 for classical systems, 97 for Fermi-Dirac particles, 25-27 of an harmonic oscillator, 9 for order-disorder systems, 127 for the polaron problem, 231 for two-dimensional order-disorder systems, 136-148 Path integrals, 72-96 calculation by expansion about the classical path, 79-81 calculation by perturbation expansion, 84-85 definition of, 72-73 and the polaron problem, 232 variational principle for, 86-87 Pauli matrices, 200 spin exchange operator and, 202 Permutations, of identical particles, 62 Phonons, 12-22, 159-162 in a cubic lattice, 20-22 dispersion relations of, 16, 221 and electrons, 185-190, 221, 224, 231, 270-273 in an electron gas, 244-246 free energy of, 17 in liquid helium, 315, 326-327 thermal energy of, 18, 19-20 Planck radiation law, 12 Plasma oscillations, 249-254 dispersion relation for, 251, 254 random phase approximation for, 253 variational approach to, 254 wave function for, 254
354
Index
Polaron, 221 dissipation of energy of, 227-228 effective mass of, 227, 241 energy of, 226, 227, 235, 239, 240 partition function for, 231 path integral for, 232 variational treatment for, 234-241 Pressure, 7-8 of Bose-Einstein particles, 28 of classical particles, 98, 114 of Fermi-Dirac particles, 28 Probabilities, of a quantum mechanical state, 1, 2-3 of combined systems, 5-6 Propagator, 260 Propagator factor for electron gas, 260 Rayleigh-Jeans law, 12 Random phase approximation (RPA), 252-254, 254 Rotons, 317, 333 RPA, see random phase approximation Second sound, 315 Specific heat, 7 of blackbody radiation, 12 of Bose-Einstein particles, 33 of Fermi-Dirac particles, 37-38 of liquid helium, 34, 313, 315, 318, 349-350 of spin waves, 209 of a superconductor, 265, 285 of vibrations in a solid, 18, 20 Spin exchange operator, 202 Spin-spin interactions, 198-200 Spin waves, 204, 206 combination of two, 207-217 energy of, 205, 206, 209 free energy of, 209 scattering of, 212-220 semiclassical interpretation of, 206-207 specific heat of, 209 Stefan-Boltzmann law, 12 Structure factor for an electron gas, 248 for liquid helium, 330 Superconductors, 265-311 BCS model for, 269
critical temperature of, 284 with current, 290-311 energy of, 277, 281, 293, 298, 303 energy gap in, 269, 279, 284, 285 entropy for, 283, 299 excitations in, 279-280, 283-284 free energy of, 266, 282, 283, 299, 302 ground state of, 274-279 Hamiltonian for, 273 Hard, 268 and hydrodynamics, 305 junction of, 287-290, 306 Lagrangian formalism for, 303 and magnetic field, 266, 267-268, 293-295, 311 in rotation, 310-311 soft, 268 specific heat of, 265, 285 type I, 268 type II, 268 Superfluids, 312-350 Temperature, 1, 6 of harmonic oscillators, 4, 10 increase of energy with, 6 Thermal energy, 7 of blackbody radiation, 11 of Bose-Einstein particles, 31, 33 for classical systems, 98, 114 of a crystal, 18, 19-20 of Fermi-Dirac particles, 35-37 of an harmonic oscillator, 9, 52 of spin waves, 209 of a two-dimensional order-disorder problem, 135 Thermal equilibrium, 1 Two particle distribution function, 65, 111-113 Born-Green equation for, 115-116 for an electron gas, 246 at low densities, 115 Vibrations in solids, see phonons Vinai coefficients, 100-104, 109 Vinai theorem, 55-57 Vortex lines, 336, 338-339 Wigner's function, see density matrix
