Thermodynamics Statistical Mechanics Notes


       
 
  


   
 


                 

                          

              

  


   
  
                                                     

                         

  

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     !                 $   0 for all ai . Let amax be the largest of all the ai 's. Clearly, then

amax  Pamax  1

P X i=1 P X i=1

ai ai

Thus, we have the inequality

amax    Pamax or ln amax  ln   ln amax + ln P This gives upper and lower bounds on the value of ln . Now suppose that ln amax >> ln P . Then the above inequality implies that ln   ln amax This would be the case, for example, if amax  eP . In this case, the value of the sum is given to a very good approximation by the value of its maximum term. Why should this theorem apply to the sum expression for (E )? Consider the case of a system of free particles P N H = i=1 p2i =2mi, i.e., no potential. Then the expression for the partition function is (E ) 

Z

D(V )

dN r

Z

dN p


! N p2 X i ;E  VN 2m i=1

i

since the particle integrations are restricted only the volume of the container. Thus, the terms in the sum vary exponentially with N . But the number of terms in the sum P also varies like N since P = E= and E  N , since E is extensive. Thus, the terms in the sum under consideration obey the conditions for the application of the theorem. Let the maximum term in the sum be characterized by energies E1 and E2 = E ; E1 . Then, according to the above analysis,   S (E ) = k ln (E ) = k ln + k ln 1 (E1 )2 (E ; E1 ) + k ln P + k ln C Since P = E= , ln + ln P = ln + ln E ; ln = ln E . But E  N , while ln 1  N . Since N >> ln N , the above expression becomes, to a good approximation   S (E )  k ln 1 (E1 )2 (E ; E1 ) + O(ln N ) + const Thus, apart from constants, the entropy is approximately additive: S (E ) = k ln 1 (E1 ) + k ln 2 (E2 ) = S1 (E1 ) + S2 (E2 ) + O(ln N ) + const Finally, in order to compute the temperature of each system, we make a small variation in the energy E1 dE1 . But since E1 + E2 = E , dE1 = ;dE2 . Also, this variation is made such that the total entropy S and energy E remain constant. Thus, we obtain 0

0 = @S1 + @S2 @ E1 @ E1 0 = @S1 ; @S2 @ E1 @ E2 0= 1 ; 1

T1

T2

from which it is clear that T1 = T2 , the expected condition for thermal equilibrium. It is important to point out that the entropy S (N V E ) dened via the microcanonical partition function is not the only entropy that satises the properties of additivity and equality of temperatures at thermal equilibrium. Consider an ensemble dened by the condition that the Hamiltonian, H (x) is less than a certain energy E . This is known as the uniform ensemble and its partition function, denoted (N V E ) is dened by (N V E ) = C

Z

H (x)<E

dx = C 2

Z

dx (E ; H (x))

where (x) is the Heaviside step function. Clearly, it is related to the microcanonical partition function by

@ (N V E ) (N V E ) = @E

Although we will not prove it, the entropy S~(N V E ) dened from the uniform ensemble partition function via S~(N V E ) = k ln (N V E ) is also approximately additive and will yield the condition T1 = T2 for two systems in thermal contact. In fact, it diers from S (N V E ) by a constant of order ln N so that one can also dene the thermodynamics in terms of S~(N V E ). In particular, the temperature is given by

!

 @ ln   1 = @ S~ = k T @E N
V @E N
V

II. REVERSIBLE LAWS OF MOTION AND THE ARROW OF TIME Hamilton's equations of motion

q_i = @H @pi

p_i = ; @H @q i

are invariant under a reversal of the direction of time t ! ;t. Under such a transformation, the positions and momenta transform according to

qi ! qi i ! m dqi = ;p pi = mi dq i d(;t) i dt Thus, or

d(;pi ) = ; @H d(;t) @qi

dqi @H d(;t) = ; @pi q_i = @H @pi

p_i = ; @H @q i

The form of the equations does not change! One of the implications of time reversal symmetry is as follows: Suppose a system is evolved forward in time starting from some initial condition up to a maximum time t at t, the evolution is stopped, the sign of the velocity of each particle in the system is reversed, i.e., a time reversal transformation is performed, and the system is allowed to evolve once again for another time interval of length t the system will return to its original starting point in phase space, i.e., the system will return to its initial condition. Now from the point of view of mechanics and the microcanonical ensemble, the initial conditions (for the rst segment of the evolution) and the conditions created by reversing the signs of the velocities for initiating the second segment are equally valid and equally probably, both being points selected from the constant energy hypersurface. Therefore, from the point of view of mechanics, without a priori knowledge of which segment is the forward evolving trajectory and which is the time reversed trajectory, it should not be possible to say which is which. That is, if a movie of each trajectory were to be made and shown to an ignorant observer, that observer should not be able to tell which is the forwardevolving trajectory and which the time-reversed trajectory. Therefore, from the point of view of mechanics, which obeys time-reversal symmetry, there is not preferred direction for the ow of time. Yet our everyday experience tells us that there are plenty of situations in which a system seems to evolve in a certain direction and not in the reverse direction, suggesting that there actually is a preferred direction in time. Some common examples are a glass falling to the ground and smashing into tiny pieces or the sudden expansion of a gas into a large box. These processes would always seem to occur in the same way and never in the reverse (the glass shards never reassemble themselves and jump back onto the table forming an unbroken glass, and the gas particles never 3

suddenly all gather in one corner of the large box). This seeming inconsistency with the reversible laws of mechanics is known as Loschmidt's paradox. Indeed, the second law of thermodynamics, itself, would seem to be at odds with the reversibility of the laws of mechanics. That is, the observation that a system naturally evolves in such a way as to increase its entropy cannot obviously be rationalized starting from the microscopic reversible laws of motion. Note that a system being driven by an external agent or eld will not be in equilibrium with its surroundings and can exhibit irreversible behavior as a result of the work being done on it. The falling glass is an example of such a system. It is acted upon by gravity, which drives it uniformly toward a state of ever lower potential energy until the glass smashes to the ground. Even though it is possible to write down a Hamiltonian for this system, the treatment of the external eld is only approximate in that the true microscopic origin of the external eld is not taken into account. This also brings up the important question of how one exactly denes non-equilibrium states in general, and how do they evolve, a question which, to date, has no fully agreed upon answer and is an active area of research. However, the expanding gas example does not involve an external driving force and still seems to exhibit irreversible behavior. How can this observation be explained for such an isolated system? One possible explanation was put forth by Boltzmann, who introduced the notion of molecular chaos. Under this assumption, the momenta of two particles do not become correlated as the result of a collision. This is tantamount to the assumption that microscopic information leading to a correlation between the particles is lost. This is not inconsistent with microscopic reversibility from a probabilistic point of view, as the momenta of two particles before a collision are certainly uncorrelated with each other. The assumption of molecular chaos allows one to prove the so called Boltzmann H-theorem, a theorem that predicts an increase in entropy until equilibrium is reached. For more details see Chapters 3 and 4 of Huang. Boltzmann's assumption of molecular chaos remains unproven and may or may not be true. Another explanation due to Poincare is based on his recurrence theorem. The Poincare recurrence theorem states that a system having a nite amount of energy and conned to a nite spatial volume will, after a suciently long time, return to an arbitrarily small neighborhood of its initial state.

Proof of the recurrence theorem (taken almost directly from Huang, pg. 91):

Let a state of a system be represented by a point x in phase space. As the system evolves in time, a point in phase space traces out a trajectory that is uniquely determined by any given point on the trajectory (because of the deterministic nature of classical mechanics). Let g0 be an arbitrary volume element in the phase space in the volume !0 . After a time t all points in g0 will be in another volume element gt in a volume !t , which is uniquely determined by the choice of g0. Assuming the system is Hamiltonian, then by Liouville's theorem: !0 = !t Let ;0 denote the subspace of phase space that is the union of all gt for 0  t < 1. Let its volume be 0 . Similarly, let ; denote the subspace that is the union of all gt for  t < 1. Let its volume be  . The numbers 0 and  are nite because, since the energy of the system is nite and the spatial volume occupied is nite, a representative point is conned to a nite region in phase space. The denitions immediately imply that ;0 contains ; . We may think of ;0 and ; in a dierent way. Imagine the region ;0 to be lled uniformly with representative points. As time progresses, ;0 will evolve into some other regions that are uniquely determined. It is clear, from the denitions, that after a time , ;0 will become ; . Also, by Liouville's theorem: 0 =  Recall that ;0 contains all the future destinations of the points in g , which in turn is evolved from g0 after a time . It has been shown that ;0 has the same volume as ; since 0 =  by Liouville's theorem. Therefore, ;0 and ; must contain the same set of points (except possibly for a set of zero measure). In particular, ; contains all of g0 (except possibly for a set of zero measure). But, by denition, all points in ; are future destinations of the points in g0 . Therefore all points in g0 (except possibly for a set of zero measure) must return to g0 after a suciently long time. Since g0 can be made arbitrarily small, Poincare's theorem follows. Now consider an initial condition in which all the gas particles are initially in a corner of a large box. By Poincare's theorem, the gas particles must eventually return to their initial state in the corner of the box. How long is this recurrence time? In order to answer this, consider dividing the box up into M small cells of volume v. The total number of microstates available to the gas varies with N like V N . The number of microstates corresponding to all the particles occupying a single cell of volume v is vN . Thus, the probability of observing the system in this microstate is approximately (v=V )N . Even if v = V=2, for N  1023 , the probability is vanishingly small. The Poincare recurrence time, on the other hand, is proportional to the inverse of this probability or (V=v)N . Again, since N  1023, if v = V=2, the required time is 4

t  21023 which is many orders of magnitude longer than the current age of the universe. Thus, although the system will return arbitrarily close to its initial state, the time required for this is unphysically long and will never be observed. Over times relevant for observation, given similar such initial conditions, the system will essentially always evolve in the same way, which is to expand and ll the box and essentially never to the opposite.

5

G25.2651: Statistical Mechanics Notes for Lecture 4

I. THE CLASSICAL VIRIAL THEOREM (MICROCANONICAL DERIVATION) Consider a system with Hamiltonian H (x). Let xi and xj be specic components of the phase space vector. The classical virial theorem states that

@H i = kT
hxi @x ij j

where the average is taken with respect to a microcanonical ensemble. To prove the theorem, start with the denition of the average:

Z @H C @H
(E ; H (x)) hxi @x i = (E ) dxxi @x j j where the fact that
(x) =
(;x) has been used. Also, the N and V dependence of the partition function have been

suppressed. Note that the above average can be written as

@H i = C @ hxi @x (E ) @E j

Z

@H (E ; H (x)) dx xi @x j

@ Z @H = (CE ) @E dx xi @x j H (x)<E Z @ = (CE ) @E dx xi @ (H@x; E ) j H (x)<E

However, writing

xi @ (H@x; E ) = @x@ xi (H ; E )] ;
ij (H ; E ) j j allows the average to be expressed as


dx @x@ xi (H ; E )] +
ij (E ; H (x)) j j H (x)<E I  Z @ xi (H ; E )dSj +
ij dx (E ; H (x)) = (CE ) @E

@H i = C @ hxi @x (E ) @E

Z

H =E

H<E

The rst integral in the brackets is obtained by integrating the total derivative with respect to xj over the phase space variable xj . This leaves an integral that must be performed over all other variables at the boundary of phase space where H = E , as indicated by the surface element dSj . But the integrand involves the factor H ; E , so this integral will vanish. This leaves:

@H i = C
ij @ hxi @x (E ) @E j C
Z = (Eij)

Z

H (x)<E

H (x)<E

= (
ijE ) (E )

1

dx

dx(E ; H (x))

where (E ) is the partition function of the uniform ensemble. Recalling that we have

@ (E ) (E ) = @E @H i =
(E ) hxi @x ij @ E j

( )

@E

=
ij @ ln 1(E) @E

= k
ij @1S~

@E

= kT
ij which proves the theorem. Example: xi = pi and i = j . The virial theorem says that

hpi @H @p i = kT i

2 h mpi i = kT i 2 p h 2mi i = 12 kT

i

Thus, at equilibrium, the kinetic energy of each particle must be kT=2. By summing both sides over all the particles, we obtain a well know result 3N 3N 2 X X h 2pmi i = h 21 mi vi2 i = 23 NkT i i=1 i=1

II. LEGENDRE TRANSFORMS The microcanonical ensemble involved the thermodynamic variables N , V and E as its variables. However, it is often convenient and desirable to work with other thermodynamic variables as the control variables. Legendre transforms provide a means by which one can determine how the energy functions for dierent sets of thermodynamic variables are related. The general theory is given below for functions of a single variable. Consider a function f (x) and its derivative

df
g(x) y = f (x) = dx The equation y = g(x) denes a variable transformation from x to y. Is there a unique description of the function f (x) in terms of the variable y? That is, does there exist a function (y) that is equivalent to f (x)? Given a point x0 , can one determine the value of the function f (x0 ) given only f (x0 )? No, for the reason that the function f (x0 ) + c for any constant c will have the same value of f (x0 ) as shown in the gure below. 0

0

0

2

f(x)

x0

x

b(x 0 )

FIG. 1.

However, the value f (x0 ) can be determined uniquely if we specify the slope of the line tangent to f at x0 , i.e., f (x0 ) and the y-intercept, b(x0 ) of this line. Then, using the equation for the line, we have f (x0 ) = x0 f (x0 ) + b(x0 ) This relation must hold for any general x: f (x) = xf (x) + b(x) 1 Note that f (x) is the variable y, and x = g (y), where g 1 is the functional inverse of g, i.e., g(g 1 (x)) = x. Solving for b(x) = b(g 1 (y)) gives b(g 1(y)) = f (g 1 (y)) ; yg 1(y)
(y) where (y) is known as the Legendre transform of f (x). In shorthand notation, one writes (y) = f (x) ; xy however, it must be kept in mind that x is a function of y. 0

0

0

0

;

;

;

;

;

;

;

III. THE CANONICAL ENSEMBLE A. Basic Thermodynamics In the microcanonical ensemble, the entropy S is a natural function of N ,V and E , i.e., S = S (N V E ). This can be inverted to give the energy as a function of N ,V , and S , i.e., E = E (N V S ). Consider using Legendre transformation to change from S to T using the fact that 3

  T = @E @S NV The Legendre transform E~ of E (N V S ) is

E~ (N V T ) = E (N V S (T )) ; S @E @S = E (N V S (T )) ; TS The quantity E~ (N V T ) is called the Hemlholtz free energy and is given the symbol A(N V T ). It is the fundamental energy in the canonical ensemble. The dierential of A is

 @A   @A    dT + dV + dA = @A @T NV @V NT @N TV dN

However, from A = E ; TS , we have

dA = dE ; TdS ; SdT From the rst law, dE is given by

dE = TdS ; PdV + dN Thus,

dA = ;PdV ; SdT + dN Comparing the two expressions, we see that the thermodynamic relations are

 @A  S = ; @T  @A NV P = ; @V  @A  NT  = @N VT

B. The partition function Consider two systems (1 and 2) in thermal contact such that

N2  N1 E2  E1 N = N1 + N2 

E = E1 + E 2

dim(x1 )  dim(x2 ) and the total Hamiltonian is just H (x) = H1 (x1 ) + H2 (x2 ) Since system 2 is innitely large compared to system 1, it acts as an innite heat reservoir that keeps system 1 at a constant temperature T without gaining or losing an appreciable amount of heat, itself. Thus, system 1 is maintained at canonical conditions, N V T . The full partition function (N V E ) for the combined system is the microcanonical partition function (N V E ) =

Z

dx
(H (x) ; E ) =

Z

dx1 dx2
(H1 (x1 ) + H2 (x2 ) ; E )

Now, we dene the distribution function, f (x1 ) of the phase space variables of system 1 as 4

f (x1 ) = Taking the natural log of both sides, we have

Z

dx2
(H1 (x1 ) + H2 (x2 ) ; E )

Z

ln f (x1 ) = ln dx2
(H1 (x1 ) + H2 (x2 ) ; E ) Since E2  E1 , it follows that H2 (x2 )  H1 (x1 ), and we may expand the above expression about H1 = 0. To linear order, the expression becomes



Z

Z

ln f (x1 ) = ln dx2
(H2 (x2 ) ; E ) + H1 (x1 ) @H @(x ) ln dx2
(H1 (x1 ) + H2 (x2 ) ; E ) 1 1 H1 (x1 )=0

Z

Z @ = ln dx2
(H2 (x2 ) ; E ) ; H1 (x1 ) @E ln dx2
(H2 (x2 ) ; E ) where, in the last line, the dierentiation with respect to H1 is replaced by dierentiation with respect to E . Note

that

Z

ln dx2
(H2 (2 ) ; E) = S2k(E )

@ ln Z dx
(H (x ) ; E ) = @ S2 (E ) = 1 2 2 2 @E @E k kT where T is the common temperature of the two systems. Using these two facts, we obtain ln f (x ) = S2 (E ) ; H1 (x1 )

k

1

kT

f (x1 ) = eS2(E)=k e H1 (x1 )=kT ;

Thus, the distribution function of the canonical ensemble is

f (x) / e H (x)=kT ;

The prefactor exp(S2 (E )=k) is an irrelevant constant that can be disregarded as it will not aect any physical properties. The normalization of the distribution function is the integral:

Z

dxe H (x)=kT
Q(N V T ) ;

where Q(N V T ) is the canonical partition function. It is convenient to dene an inverse temperature  = 1=kT . Q(N V T ) is the canonical partition function. As in the microcanonical case, we add in the ad hoc quantum corrections to the classical result to give Z 1 Q(N V T ) = N !h3N dxe H (x) ;

The thermodynamic relations are thus, Hemlholtz free energy:

A(N V T ) = ; 1 ln Q(N V T ) To see that this must be the denition of A(N V T ), recall the denition of A:

A = E ; TS = hH (x)i ; TS 5

But we saw that

 @A  S = ; @T NV

Substituting this in gives

A = hH (x)i ; T @A @T

or, noting that

@A = @A @ = ; 1 @A @T @ @T kT 2 @

it follows that

A = hH (x)i +  @A @

This is a simple dierential equation that can be solved for A. We will show that the solution is A = ; 1 ln Q( )



Note that

1 ln Q( ) ; 1 @Q = A ; hH (x)i  @A = @  Q @

Substituting in gives, therefore

A = hH (x)i + A ; hH (x)i = A

so this form of A satises the dierential equation. Other thermodynamics follow: Average energy:

Z 1 E = hH (x)i = Q CN dxH (x)e H (x) ;

@ ln Q(N V T ) = ; @

Pressure:

Entropy:

 @A   @ ln Q(N V T )  P = ; @V = kT @V NT NT @A @ 1 @A S = ; @A @T =; @ @T = kT 2 @  @ ; 1 ln Q(N V T ) = ;k @ ln Q + k ln Q = k 2 @  @ E = kE + k ln Q = k ln Q + T 6

Heat capacity at constant volume:

 @E  @ = k 2 @ ln Q(N V T ) CV = @T = @E @ 2 NV @ @T

C. Relation between canonical and microcanonical ensembles We saw that the E (N V S ) and A(N V T ) could be related by a Legendre transformation. The partition functions (N V E ) and Q(N V T ) can be related by a Laplace transform. Recall that the Laplace transform f~() of a function f (x) is given by

Z

f~() =

1

dxe x f (x) ;

0

Let us compute the Laplace transform of (N V E ) with respect to E : ~ (N V ) = CN

Z

1

Z dEe E dx
(H (x) ; E ) ;

0

Using the
-function to do the integral over E : Z ~(N V ) = CN dxe H (x) ;

By identifying  =  , we see that the Laplace transform of the microcanonical partition function gives the canonical partition function Q(N V T ).

D. Classical Virial Theorem (canonical ensemble derivation) Again, let xi and xj be specic components of the phase space vector x = (p1  ::: p3N  q1  ::: q3N ). Consider the canonical average given by

@H i hxi @x j

@H i = 1 C Z dxx @H e H hxi @x i @x Q N j  j1 @  Z 1 = Q CN dxxi ;  @x e ;

But

(x)

  @xi xi @x@ e H (x) = @x@ xi e H (x) ; e H (x) @x j j j  H (x)  @ = @x xi e ;
ij e H (x) j ;

;

;

;

Thus,

H (x)

;

j

;

@H i = ; 1 C Z dx @ x e H  + 1
C Z dxe hxi @x i Q N Z @x Q ij N j j Z  @  1 C + kT
ij = ; Q N dx dxj @x xi e H j Z  = dx xi e H  + kT
ij ;

0

0

;

(x)

(x)

;

;

1

xj =

;1

Several cases exist for the surface term xi exp(;H (x)): 7

(x)

H (x)

1. xi = pi a momentum variable. Then, since H  p2i , exp(;H ) evaluated at pi = 1 clearly vanishes. 2. xi = qi and U ! 1 as qi ! 1, thus representing a bound system. Then, exp(;H ) also vanishes at qi = 1. 3. xi = qi and U ! 0 as qi ! 1, representing an unbound system. Then the exponential tends to 1 both at qi = 1, hence the surface term vanishes. 4. xi = qi and the system is periodic, as in a solid. Then, the system will be represented by some supercell to which periodic boundary conditions can be applied, and the coordinates will take on the same value at the boundaries. Thus, H and exp(;H ) will take on the same value at the boundaries and the surface term will vanish. 5. xi = qi , and the particles experience elastic collisions with the walls of the container. Then there is an innite potential at the walls so that U ;! 1 at the boundary and exp(;H ) ;! 0 at the bondary. Thus, we have the result

@H i = kT
hxi @x ij j

The above cases cover many but not all situations, in particular, the case of a system conned within a volume V with reecting boundaries. Then, surface contributions actually give rise to an observable pressure (to be discussed in more detail in the next lecture).

8

G25.2651: Statistical Mechanics Notes for Lecture 5

I. TEMPERATURE AND PRESSURE ESTIMATORS From the classical virial theorem

@H i = kT
hx @x i

ij

j

we arrived at the equipartition theorem:

*X

p2 =1 2m

+

N

i

i

i

= 32 NkT

where p1  ::: p are the N Cartesian momenta of the N particles in a system. This says that the microscopic function of the N momenta that corresponds to the temperature, a macroscopic observable of the system, is given by N

K (p1  ::: p ) = N

X p2 N

i

=1 2m

i

i

The ensemble average of K can be related directly to the temperature 2 hK (p  ::: p )i = 2 hK (p  ::: p )i T = 3Nk 1 1 3 3nR K (p1  ::: p ) is known as an estimator (a term taken over from the Monte Carlo literature) for the temperature. An estimator is some function of the phase space coordinates, i.e., a function of microscopic states, whose ensemble average gives rise to a physical observable. An estimator for the pressure can be derived as well, starting from the basic thermodynamic relation:  @ ln Q(N V T )   @A  = kT P =; N

N

@V

with

Q(N V T ) = C

N

Z

@V

N T

Z

dx e; (x) = C
H

N

d p N

N T

Z V

d r e; (p r) N


H

The volume dependence of the partition function is contained in the limits of integration, since the range of integration for the coordinates is determined by the size of the physical container. For example, if the system is conned within a cubic box of volume V = L3 , with L the length of a side, then the range of each q integration will be from 0 to L. If a change of variables is made to s = q =L, then the range of each s integration will be from 0 to 1. The coordinates s are known as scaled coordinates. For containers of a more general shape, a more general transformation is i

i

i

s = V ;1 3 r In order to preserve the phase space volume element, however, we need to ensure that the transformation is a canonical one. Thus, the corresponding momentum transformation is  = V 1 3p With this coordinate/momentum transformation, the phase space volume element transforms as d p d r = d d s =

i

=

i

N

N

i

N

1

i

N

Thus, the volume element remains the same as required. With this transformation, the Hamiltonian becomes

H=

X p2 X V ;2 3  2 13 13 + U ( r  ::: r ) = 1 2m 2m + U (V s1 ::: V s ) N

N

i

=1

N

=1

i

i

=

i

=

=

N

i

i

and the canonical partition function becomes

Z

Q(N V T ) = C

d  N

N

( "X ;2 3 2 #) V  1 3 1 3 d s exp ; 2m + U (V s1  ::: V s )

Z

N

=

N

=1

i

=

=

N

i

i

Thus, the pressure can now be calculated by explicit di erentiation with respect to the volume, V : P = kT 1 @Q

Q @V

# " Z Z X  2  ;2 3 X kT 2 @U ; 5 3 = QC d  d s 3 V ; 3V s
@ (V 1 3 s ) e; 2 m " X 2 =1 X # =1 Z Z  p ;  @U e; (p r) = kT C d p d r r
Q 3V =1 m 3V =1 @ r *X  2 +  p 1 = 3V +r
F =1 m = h; @H @V i N

N

N

N

N

=

i

=

i

N

=

i

N

N


H

i

i

i

N

N

i

i


H

i

i

i

i

N

i

i

i

i

i

Thus, the pressure estimator is

X  p2 + r
F ( r ) (p1  ::: p  r1  ::: r ) = (x) = 31V =1 2m N

N

i

N

i

i

i

i

and the pressure is given by

P = h(x)i For periodic systems, such as solids and currently usedPmodels of liquids, an absolute Cartesian coordinate q is ill-dened. Thus, the virial part of the pressure estimator q F must be rewritten in a form appropriate for periodic systems. This can be done by recognizing that the force F is obtained as a sum of contributions F , which is the force on particle i due to particle j . Then, the classical virial becomes i

i

i

i

i

X N

=1

r
F = i

X N

i

i

i

=1

2

r


X

i

j

6=

F

ij

ij

3

i

X X X X = 12 4 r
F + r
F 5 6= 6= i

2

i

ij

j

j

ji

j

i

i

3

j

X X X X = 12 4 r
F ; r
F 5 6= 6= X X 1 1 =2 (r ; r )
F = 2 r
F 6= 6= i

ij

i

j

i

i j i

j

j

i

j

j

ij

i

j

ij

ij

i j i

ij

j

where r is now a relative coordinate. r must be computed consistent with periodic boundary conditions, i.e., the relative coordinate is dened with respect to the closest periodic image of particle j with respect to particle i. This gives rise to surface contributions, which lead to a nonzero pressure, as expected. ij

ij

2

II. ENERGY FLUCTUATIONS IN THE CANONICAL ENSEMBLE In the canonical ensemble, the total energy is not conserved. (H (x) 6= const). What are the uctuations in the energy? The energy uctuations are given by the root mean square deviation of the Hamiltonian from its average hH i:

q

p

E = h(H ; hH i)2 i = hH 2 i ; hH i2

@ ln Q(N V T ) hH i = ; @ Z hH 2 i = Q1 C dxH 2 (x)e; 1 Z @2


H

N

= QC

(x)

dx @ 2 e; (x)
H

N

@2 Q = Q1 @ 2

  @ 2 ln Q + 1 @Q 2 = @ 2 Q2 @ 



@ 2 ln Q + 1 @Q 2 = @ 2 Q @  2 @ ln Q + @ ln Q 2 = @ 2 @ Therefore

2

@ ln Q hH 2 i ; hH i2 = @ 2 But

@ 2 ln Q = kT 2C @ 2

V

Thus,

p

E = kT 2C

V

Therefore, the relative energy uctuation E=E is given by

p

E = kT 2C E E

V

Now consider what happens when the system is taken to be very large. In fact, we will dene a formal limit called the thermodynamic limit, in which N ;! 1 and V ;! 1 such that N=V remains constant. Since C and E are both extensive variables, C  N and E  N , E  p1 ;! 0 as N ! 1 V

V

E

N

But E=E would be exactly 0 in the microcanonical ensemble. Thus, in the thermodynamic limit, the canonical and microcanonical ensembles are equivalent, since the energy uctuations become vanishingly small.

3

III. ISOTHERMAL-ISOBARIC ENSEMBLE A. Basic Thermodynamics The Helmholtz free energy A(N V T ) is a natural function of N , V and T . The isothermal-isobaric ensemble is generated by transforming the volume V in favor of the pressure P so that the natural variables are N , P and T (which are conditions under which many experiments are performed { \standard temperature and pressure," for example). Performing a Legendre transformation of the Helmholtz free energy

@A A~(N P T ) = A(N V (P ) T ) ; V (P ) @V

But

@A @V = ;P

Thus,

A~(N P T ) = A(N V (P ) T ) + PV  G(N P T ) where G(N P T ) is the Gibbs free energy. The di erential of G is

 @G 

dG = @P

 @G 

N T

dP + @T

 @G 

N P

dT + @N

dN P T

But from G = A + PV , we have

dG = dA + PdV + V dP but dA = ;SdT ; PdV + dN , thus

dG = ;SdT + V dP + dN Equating the two expressions for dG, we see that

 @G 

V = @P  @G  S = ; @T  @G   = @N

N T

N P

P T

B. The partition function and relation to thermodynamics In principle, we should derive the isothermal-isobaric partition function by coupling our system to an innite thermal reservoir as was done for the canonical ensemble and also subject the system to the action of a movable piston under the inuence of an external pressure P . In this case, both the temperature of the system and its pressure will be controlled, and the energy and volume will uctuate accordingly. However, we saw that the transformation from E to T between the microcanonical and canonical ensembles turned into a Laplace transform relation between the partition functions. The same result holds for the transformation from V to T . The relevant \energy" quantity to transform is the work done by the system against the external pressure P in changing its volume from V = 0 to V , which will be PV . Thus, the isothermal-isobaric partition function can be expressed in terms of the canonical partition function by the Laplace transform: 4

Z1 (N P T ) = V1 dV e;


P V

0 0

Q(N V T )

where V0 is a constant that has units of volume. Thus, Z1 Z (N P T ) = V N1!h3 dV dxe; ( (x)+ ) 0 0
H

PV

N

The Gibbs free energy is related to the partition function by G(N P T ) = ; 1 ln (N P T )



This can be shown in a manner similar to that used to prove the A = ;(1= ) ln Q. The di erential equation to start with is

G = A + PV = A + P @G @P

Other thermodynamic relations follow: Volume:



N P T ) V = ;kT @ ln (@P

 N T

Enthalpy:

@ ln (N P T ) H = hH (x) + PV i = ; @ Heat capacity at constant pressure

 @ H 

C = @T P

Entropy:

2

N P

@ ln (N P T ) = k 2 @

 

S = ; @G @T

N P

 = k ln (N P T ) + H T

The uctuations in the enthalpy H are given, in analogy with the canonical ensemble, by p H = kT 2C P

so that

p

H = kT 2C H H p  H  1= N which vanish in the thermodynamic limit. so that, since C and H are both extensive, H= P

5

P

C. Pressure and work virial theorems As noted earlier, the quantity ;@H=@V is a measure of the instantaneous value of the internal pressure Pint . Let us look at the ensemble average of this quantity Z1 Z @H hPint i = ; 1 C dV e; dx @V e; (x) 0 Z1 Z 1 @ e; (x) ; dV e dxkT @V = C Z 10 1 @ Q(N V T ) = dV e; kT @V 0 Doing the volume integration by parts gives @    1 Z 1 ; hPint i = 1 e; kTQ(N V T ) 1 ; dV kT e Q(N V T ) 0  0 @V Z1 dV e; Q(N V T ) = P = P 1 0 Thus,
P V

N


H


P V

N


H


P V


P V


P V


P V

hPint i = P This result is known as the pressure virial theorem. It illustrates that the average of the quantity ;@H=@V gives the xed pressure P that denes the ensemble. Another important result comes from considering the ensemble average hPint V i

hPint V i = 1

Z1 0

dV e;


P V

@ Q(N V T ) kTV @V

Once again, integrating by parts with respect to the volume yields e; kTV Q(N V T )1 ; 1 Z 1 dV kT  @ V e; 1 hPint V i =  0  0 @V  Z Z 1 1 = 1 ;kT dV e; Q(V ) + P dV e; V Q(V ) 0 0 = ;kT + P hV i
P V


P V


P V



Q(N V T )


P V

or

hPint V i + kT = P hV i This result is known as the work virial theorem. It expresses the fact that equipartitioning of energy also applies to the volume degrees of freedom, since the volume is now a uctuating quantity.

6



       
 
  


  


      
    
          
   
  
    

   
  
       
   
  
  


 

    
  
     
        
   

 
  

 
  
      

         
      

    

          





            


 
 
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> N1 and V2 >> V1 so that system 2 is a particle reservoir. The total particle number and volume are V = V1 + V2 N = N1 + N2 The total Hamiltonian H (x
N ) is H (x
N ) = H1 (x1
N1 ) + H2 (x2
N2 ) If the systems could not exchange particles, then the canonical partition function for the whole system would be Z 1 Q(N
V
T ) = N !h3N dxe;(H1 (x1 N1 )+H2 (x2 N2)) = N1N!N! 2 ! Q1(N1
V1
T )Q2(N2
V2
T ) where

Z Q1 (N1
V1
T ) = N !h1 3N1 dxe;H1 (x1 N1 ) 1 Z Q2 (N2
V2
T ) = N !h1 3N2 dxe;H2 (x2 N2 ) 2

However, N1 and N2 are not xed, therefore, in order to sum over all microstates, we need to sum over all values that N1 can take on subject to the constraint N = N1 + N2 . Thus, we can write the canonical partition function for the

whole system as

Q(N
V
T ) =

N X N1 =0

f (N1
N ) N1N!N! 2 ! Q1 (N1
V1
T )Q2(N2
V2
T )

where f (N1
N2 ) is a function that weights each value of N1 for a given N . Thus, f (0
N ) is the number of congurations with 0 particles in V1 and N particles in V2 . f (1
N ) is the number of congurations with 1 particles in V1 and N ; 1 particles in V2 . etc. Determining the values of f (N1
N ) amounts to a problem of counting the number of ways we can put N identical objects into 2 baskets. Thus, f (0
N ) = 1 f (1
N ) = N = N !=1!(N ; 1)! f (2
N ) = N (N ; 1)=2 = N !=2!(N ; 2)! etc. or generally, f (N1
N ) = N !(NN;! N )! = N N!N! ! 1 1 1 2 which is clearly a classical degeneracy factor. If we were doing a purely classical treatment of the grand canonical ensemble, then this factor would appear in the sum for Q(N
V
T ), however, we always include the ad hoc quantum 3

correction 1=N ! in the expression for the canonical partition function, and we see that these quantum factors will exactly cancel the classical degeneracy factor, leading to the following expression:

Q(N
V
T ) =

N X N1 =0

Q1(N1
V1
T )Q2(N2
V2
T )

which expresses the fact that, in reality, the various congurations are not distinguishable from each other, and so each one should count with equal weighting. Now, the distribution function (x) is given by 1 e;H (x N ) (x
N ) = N !hQ3N(N
V
T )

which is chosen so that

Z

dx(x
N ) = 1

However, recognizing that N2
N , we can obtain the distribution for 1 (x1
N1) immediately, by integrating over the phase space of system 2: Z 1 (x1
N1 ) = Q(N
1V
T ) N !h1 3N1 e;H1 (x1 N1) N !h1 3N2 dx2 e;H2(x2 N2) 1 2 where the 1=N1!h3N1 prefactor has been introduced so that N Z X

N1 =0

dx1 (x1
N1 ) = 1

and amounts to the usual ad hoc quantum correction factor that must be multiplied by the distribution function for each ensemble to account for the identical nature of the particles. Thus, we see that the distribution function becomes 1 (x1
N1) = QQ2 ((NN
2
V
V2T
T) ) N !h1 3N1 e;H1 (x1 N1 ) 1 Recall that the Hemlholtz free energy is given by A = ; 1 ln Q



Thus,

or

Q(N
V
T ) = e;A(N V T ) Q2(N2
V2
T ) = e;A(N2 V2 T ) = e;A(N ;N1 V ;V1 T ) Q2 (N2
V2
T ) = e;(A(N ;N1 V ;V1 T );A(N V T )) Q(N
V
T )

But since N >> N1 and V >> V1 , we may expand:

@A N ; @A V +    A(N ; N1
V ; V1
T ) = A(N
V
T ) ; @N 1 @V 1 = A(N
V
T ) ; N1 + PV1 +   

Therefore the distribution function becomes

1 (x1
N1 ) = N !h1 3N1 e
N1 e;PV1 e;H1 (x1 N1 ) 1 1 e
N1 e;H1 (x1 N1 ) = N !h1 3N1 ePV 1 1

4

Dropping the \1" subscript, we have



1 1 
N ;H (x N ) (x
N ) = ePV N !h3N e e We require that (x
N ) be normalized: 1 Z X

1

"1 X

ePV N =0

N =0



dx(x
N ) = 1 #

1 e
N Z dxe;H (x N ) = 1 N !h3N

Now, we dene the grand canonical partition function

Z (
V
T ) =

1 X

N =0

1 e
N Z dxe;H (x N ) N !h3N

Then, the normalization condition clearly requires that Z (
V
T ) = ePV ln Z (
V
T ) = PV kT

Therefore PV is the free energy of the grand canonical ensemble, and the entropy S (
V
T ) is given by

@ @ ( PV ) S (
V
T ) = = k ln Z (
V
T ) ; k ln Z (
V
T )

@T

@


V

We now introduce the fugacity dened to be

= e



V

Then, the grand canonical partition function can be written as Z 1 1 X X 1 N Z (
V
T ) = N !h3N dxe;H (x N ) = N Q(N
V
T ) N =0 N =0 which allows us to view the grand canonical partition function as a function of the thermodynamic variables , V , and T . Other thermodynamic quantities follow straightforwardly: Energy: 1 N Z X E = hH (x
N )i = Z1 dxH (x
N )e;H (x N ) 3N N ! h N =0
@ ln Z (
V
T ) =;

@

Average particle number:

V


@ ln Z (
V
T ) hN i = kT @ VT

This can also be expressed using the fugacity by noting that

@ = @ @ =  @ @ @ @ @

Thus,

hN i = @ @ ln Z (
V
T ) 5

C. Ideal gas Recall the canonical partition function expression for the ideal gas:

"
3=2 #N 1 V 2

m Q(N
V
T ) = N ! h3 

Dene the thermal wavelength ( ) as

2
1=2 ( ) = 2h

m

which has a quantum mechanical meaning as the width of the free particle distribution function. Here it serves as a useful parameter, since the canonical partition can be expressed as


N Q(N
V
T ) = N1 ! V3 The grand canonical partition function follows directly from Q(N
V
T ): 1 1 V
N X Z (
V
T ) = N ! 3 = eV =3 N =0

Thus, the free energy is

PV = ln Z = V kT 3

In order to obtain the equation of state, we rst compute the average particle number hN i

hN i = @ @ ln Z = V3 Thus, eliminating in favor of hN i in the equation of state gives

PV = hN ikT as expected. Similarly, the average energy is given by
E = ; @ ln Z

3V @ = 3 hN ikT = @ V 4 @ 2 where the fugacity has been eliminated in favor of the average particle number. Finally, the entropy (
V
T ) S (
V
T ) = k ln Z (
V
T ) ; k @ ln Z@








3 = 25 hN ik + hN ik ln VhNi
V which is the Sackur-Tetrode equation derived in the context of the canonical and microcanonical ensembles.

D. Particle number
uctuations In the grand canonical ensemble, the particle number N is not constant. It is, therefore, instructive to calculate the uctuation in this quantity. As usual, this is dened to be

p

N = hN 2 i ; hN i2 Note that 6

"1 #2 1 X X N @ 1 1 @ 2 N N Q(N
V
T ) ; Z 2 N Q(N
V
T ) @ @ ln Z (
V
T ) = Z N =0 N =0 = hN 2 i ; h N i 2 Thus,

@ @ ln Z (
V
T ) = (kT )2 @ 2 ln Z (
V
T ) = kTV @ 2 P (N )2 = @ @ @ 2 @ 2

In order to calculate this derivative, it is useful to introduce the Helmholtz free energy per particle dened as follows: a(v
T ) = 1 A(N
V
T )

N

where v = V=N = 1= is the volume per particle. The chemical potential is dened by

@A = a(v
T ) + N @a @v = @N @v @N = a(v
T ) ; v @a @v

Similarly, the pressure is given by

@A = ;N @a @v = ; @a P = ; @V @v @V @v

Also,

@ = ;v @ 2 a @v @v2

Therefore, and

  @P = @P @v = @ 2 a v @ 2 a ;1 = 1 @ @v @ @v2 @v2 v   @ 2 P = @ @P @v = 1 v @ 2 a ;1 = ; 1 @ 2 @v @ @ v2 @v2 v3 @P=@v

But recall the denition of the isothermal compressibility: 1 T = ; V1 @V @P = ; v@p=@v Thus, @2P = 1 

@

and

N = and the relative uctuation is given by

v2 T

2

r

hN ikTT v

r

N = 1 hN ikTT  p 1 ! 0 ashN i ! 1 N hN i v hN i Therefore, in the thermodynamic limit, the particle number uctuations vanish, and the grand canonical ensemble is equivalent to the canonical ensemble. 7

G25.2651: Statistical Mechanics Notes for Lecture 8

I. STRUCTURE AND DISTRIBUTION FUNCTIONS IN CLASSICAL LIQUIDS AND GASES So far, we have developed the classical theory of ensembles and applied it to the ideal gas, for which there was no potential of interaction between the particles: U = 0. We were able to derive all the thermodynamics for this sytem using the various ensembles available to us, and, in particular, we could compute the equation of state. Now, we wish to consider the general case that the potential is not zero. Of course, all of the interesting physics and chemistry of real systems results from the specic interactions between the particles. Real systems can exhibit spatial structure, undergo phase transitions, undergo chemical changes, exhibit interesting dynamics, basically, a wide variety of rich behavior. Consider the two snapshots below. On the left is shown a conguration of an ideal gas, and on the right is shown a conguration of liquid argon. Can you see any inherent structure in the snapshot of liquid argon? While it may not be readily apparent, there is considerable structure in the liquid argon system that is clearly not present in the ideal gas. One way of quantifying spatial structure is through the use of the radial distribution function g(r), which will be discussed in great detail later. For now, it is su cient to know that g(r) is a measure of the probability that a particle will be located a distance r from a another particle in the system. The gure below shows the function g(r) for the ideal gas and for the liquid argon systems. It can be seen that the radial distribution function for the ideal gas is completely featureless signifying that it is equally likely to nd a particle at any distance r from a given particle. (Since the probability is uniform, g(r) 1=r2 for small r. This is the particularly normalization condition on g(r) that gives rise to uniform probability. Hence its rapidly rising behavior for small r.) For the liquid argon system, g(r) exhibits several peaks, indicating that at certain radial values, it is more likely to nd particles than at others. This is a result of the attractive nature of the interaction at such distances. The plot of g(r) also shows that there is essentially zero probability of nding particles at distances less than about 2.5 A from each other. This is due to the presence of very strong repulsive forces at short distances. Sometimes, the structure can be readily seen in snapshots of congurations. Consider the following snapshot of a system of water molecules: The red spheres are oxygen atoms, the grey spheres are hydrogen atoms, the green lines are hydrogen bonds, and the reddish-grey lines are covalent bonds. A good deal of structure can be seen in the form of a complex network of hydrogen bonds. This high degree of structure is characteristic of water and gives rise to the ease which with water can form stable, organized structures around other molecules. In water, one can ask several questions related to structure. For example, what is the probability that an oxygen atom will be found at a distance r away from another oxygen atoms? What is the probability that a hydrogen atom will be located at a distance r from an oxygen atom, etc. The plot below shows the radial distribution functions corresponding to these two scenarios. The peak in the O-O radial distribution function occurs at roughly 2.8 A which is the well known average hydrogen bond length in water. Of course, one could also ask about structure from the point of view of a hydrogen atom and obtain two other representations of structure in water. One can also ask questions pertinent to chemical reactions. The following three snapshots depict a proton transfer reaction in water upon solvation of an excess proton: While the reaction dynamics is expected to be very complicated, involving the breaking of hydrogen bonds and other uctuations away from the rst solvation shell, it might be possible to characterize certain features of the reaction by a single, special coordinate that somehow describes the motion of the proton through the hydrogen bond. Such a coordinate is called a reaction coordinate and can, in general, be a funtion of all of some subset of the cartesian positions of the other atoms in the system:

qreac = q(r1
r2
:::
rN ) There are many things one can do with such a coordinate, such as study its dynamics or obtain its distribution and hence its free energy prole. In the case of the latter, the free energy prole describes the \eective" potential at a given temperature that the proton sees as it moves through the hydrogen bond. Computing this free energy prole, one might obtain a plot that looks as follows: 1

In this case, the free energy prole is obtained with and without quantum eects (which cannot be neglected for protons). For the classical prole, the energy dierence between the top of the barrier and the wells is the eective activation energy. Quantum mechanically, this activation energy is reduced due to zero-point motion (a concept we will discuss in the context of quantum statistical mechanics).

A. General distribution functions and correlation functions We begin by considering a general N -particle system with Hamiltonian 3N X

p2i + U (r
:::
r ) 1 N 2m i=1

H=

For simplicity, we consider the case that all the particles are of the same type. Having established the equivalence of the ensembles in the thermodynamic limit, we are free to choose the ensemble that is the most convenient on in which to work. Thus, we choose to work in the canonical ensemble, for which the partition function is Z

Q(N
V
T ) = N !h1 3N

d3N p d3N re;

P3N p2i i=1 2m e; U (r1
:::
rN )

The 3N integrations over momentum variables can be done straighforwardly, giving Z 1 Q(N
V
T ) = N !3N dr1


drN e; U (r1
:::
rN ) = NZ!N3N where  = function

p

h2 =2m is the thermal wavelength and the quantity ZN is known as the congurational partition ZN =

Z

dr1


drN e;

U (r1
:::
rN )

The quantity

e;

U (r1
:::
rN )

ZN

dr1


drN  P (N ) (r1
:::
rN )dr1


drN

represents the probability that particle 1 will be found in a volume element dr1 at the point r1 , particle 2 will be found in a volume element dr2 at the point r2 ,..., particle N will be found in a volume element drN at the point rN . To obtain the probability associated with some number n < N of the particles, irrespective of the locations of the remaining n + 1
:::
N particles, we simply integrate this expression over the particles with indices n + 1
:::
N :
Z 1 (n) ; U (r1
:::
rN ) P (r
:::
r )dr


dr = dr


dr e dr


dr 1

n

1

n

n+1

ZN

N

1

n

The probability that any particle will be found in the volume element dr1 at the point r1 and any particle will be found in the volume element dr2 at the point r2 ,...,any particle will be found in the volume element drn at the point rn is dened to be (n) (r1
:::
rn )dr1


drn = (N N;!n)! P (n)(r1
:::
rn )dr1


drn which comes about since the rst particle can be chosen in N ways, the second chosen in N ; 1 ways, etc. Consider the specal case of n = 1. Then, by the above formula, Z (1) (r1 ) = Z1 (N N;! 1)! dr2


drN e; U (r1
:::
rN ) N Z N =Z dr2


drN e; U (r1
:::
rN ) N

2

Thus, if we integrate over all r1 , we nd that 1

V

Z

dr1 (1) (r1 ) = N V =

Thus, (1) actually counts the number of particles likely to be found, on average, in the volume element dr1 at the point r1 . Thus, integrating over the available volume, one nds, not surprisingly, all the particles in the system.

3

G25.2651: Statistical Mechanics Notes for Lecture 9

I. DISTRIBUTION FUNCTIONS IN CLASSICAL LIQUIDS AND GASES (CONT'D) A. General correlation functions A general correlation function can be de ned in terms of the probability distribution function (n) (r1
:::
rn ) according to g(n) (r
:::
r ) = 1 (n) (r
:::
r ) 1

n

n

n

1

Z nN ! V = Z N n (N ; n)! drn+1


drN e; U (r1
:::
rN ) N

Another useful way to write the correlation function is Z nN ! V (n) g (r1
:::
rn ) = Z N n (N ; n)! dr01


dr0N e; U (r1
:::
rN ) (r1 ; r01 )


(rn ; r0n ) N *Y + n nN ! V 0 = Z N n (N ; n)! (ri ; ri ) N i=1 r1
:::
rN i.e., the general n-particle correlation function can be expressed as an ensemble average of the product of -functions, with the integration being taken over the variables r01
:::
r0N . 0

0

0

0

B. The pair correlation function Of particular importance is the case n = 2, or the correlation function g(2) (r1
r2 ) known as the pair function. The explicit expression for g (2) (r1
r2 ) is 2 g(2) (r1
r2 ) = N 2V(NN;! 2)! h(r1 ; r01 )(r2 ; r02 )i Z 2 V ( N ; 1) = dr3


drN e; U (r1
:::
rN )

correlation

NZN = N (N 2; 1) h(r1 ; r01 )(r2 ; r02 )ir1
:::
rN 0

0

In general, for homogeneous systems in equilibrium, there are no special points in space, so that g(2) should depend only on the relative position of the particles or the di erence r1 ; r2 . In this case, it proves useful to introduce the change of variables r = r1 ; r 2
R = 12 (r1 + r2 ) r1 = R + 12 r
r2 = R ; 21 r

Then, we obtain a new function g~(2) , a function of r and R: Z 2 g~(2)(r
R) = V (N ; 1) dr3


drN e; U (R+ 21 r
R; 12 r
r3
:::
rN )

NZN      N ( = N ; 1)  R + 1 r ; r0  R ; 1 r ; r0 2

2

1

1

2

2

r1
:::
rN 0

0

In general, we are only interested in the dependence on r. Thus, we integrate this expression over R and obtain a new correlation function g~(r) de ned by Z 1 g~(r) = dRg~(2)(r
R)

V N ; 1) Z dRdr


dr e; U (R+ 12 r
R; 21 r
r3
:::
rN ) = V (NZ 3 N N Z = (N Z; 1) dRdr3


drN e; U (R+ 21 r
R; 12 r
r3
:::
rN ) N

For an isotropic system such as a liquid or gas, where there is no preferred direction in space, only the maginitude or r, jrj  r is of relevance. Thus, we seek a choice of coordinates that involves r explicitly. The spherical-polar coordinates of the vector r is the most natural choice. If r = (x
y
z ) then the spherical polar coordinates are

x = r sin  cos  y = r sin  sin  x = r cos  dr = r2 sin drdd where  and  are the polar and azimuthal angles, respectively. Also, note that

r = rn where

n = (sin  cos 
sin  sin 
cos ) Thus, the function g(r) that depends only on the distance r between two particles is de ned to be Z g(r) = 41 sinddg~(r) Z = (4N ;Z1) sindddRdr3


drN e; U (R+ 21 rn
R; 12 rn
r3
:::
rN ) N  0)   ( r ; r ( N ; 1) = 4 rr0 r


R
r3
:::
rN 0

0

0

0

0

0

Integrating g(r) over the radial dependence, one nds that 4

Z1 0

drr2 g(r) = N ; 1  N

The function g(r) is important for many reasons. It tells us about the structure of complex, isotropic systems, as we will see below, it determines the thermodynamic quantities at the level of the pair potential approximation, and it can be measured in neutron and X-ray di raction experiments. In such experiments, one observes the scattering of neutrons or X-rays from a particular sample. If a detector is placed at an angle  from the wave-vector direction of an incident beam of particles, then the intensity I () that one observes is proportional to the structure factor

+ *X  eikrm    *Xm + 1 ik rm ;rn 2

I ()  N1 =N

 S (k)

m
n

e

(

)

where k is the vector di erence in the wave vector between the incident and scattered neutrons or X-rays (since neutrons and X-rays are quantum mechanical particles, they must be represented by plane waves of the form exp(ik
r)). 2

By computing the ensemble average (see problem 4 of problem set #5), one nds that S (k) = S (k) and S (k) is given by Z1 4  drr sin(kr)g(r) S (k) = 1 +

k

0

Thus, if one can measure S (k), g(r) can be determined by Fourier transformation.

C. Thermodynamic quantities in terms of g(r) In the canonical ensemble, the average energy is given by

@ ln Q(N
V
 ) E = ; @ ln Q(N
V
 ) = ln ZN ; 3N ln ( ) ; ln N ! Therefore,

@ ; 1 @ZN E = 3 N @ ZN @ Since

 h  = 2

= 2m @ 1 @ = 2 Thus,

1 2

Z 1 3 dr1


drN U (r1
:::
rN )e U (r1
:::
rN ) E = 2 NkT + Z N = 32 NkT + hU i

In order to compute the average energy, therefore, one needs to be able to compute the average of the potential hU i. In general, this is a nontrivial task, however, let us work out the average for the case of a pairwise-additive potential of the form X u(jri ; rj j)  Upair (r1
:::
rN ) U (r1
:::
rN ) = 12 i
j
i6=j i.e., U is a sum of terms that depend only the distance between two particles at a time. This form turns out to be an excellent approximation in many cases. U therefore contains N (N ; 1) total terms, and hU i becomes XZ 1 hU i = 2Z dr1


drN u(jri ; rj j)e; Upair (r1
:::
rN ) N i
j
i6=j Z = N (2NZ ; 1) dr1


drN u(jr1 ; r2 j)e; Upair (r1
:::
rN ) N The second line follows from the fact that all terms in the rst line are the exact same integral, just with the labels changed. Thus,

3

 Z Z hU i = 12 dr dr u(jr ; r j) N (NZ ; 1) dr


drN e; N Z 1 = 2 dr dr u(jr ; r j) (r
r ) Z N dr dr u(jr ; r j)g (r
r ) = 2V 1

2

1

2

1

2

1

2

3

(2)

2

2

1

2

1

2

1

(2)

Upair (r1
:::
rN )



2

1

2

Once again, we change variables to r = r1 ; r2 and R = (r1 + r2 )=2. Thus, we nd that

Z 2 N hU i = 2V 2 drdRu(r)~g(2) (r
R) Z Z 2 = 2NV 2 dru(r) dRg~(2)(r
R) 2 Z =N 2V dru(r)~g (r) Z1 2 =N dr4r2 u(r)g(r) 2V 0

Therefore, the average energy becomes

Z1 3 N E = 2 NkT + 2 4 drr2 u(r)g(r) 0 Thus, we have an expression for E in terms of a simple integral over the pair potential form and the radial distribution function. It also makes explicit the deviation from \ideal gas" behavior, where E = 3NkT=2. By a similar procedure, we can develop an equation for the pressure P in terms of g(r). Recall that the pressure is given by P = 1 @ ln Q  @V N = Z1 @Z N @

The volume dependence can be made explicit by changing variables of integration in ZN to

si = V ;1=3 ri

Using these variables, ZN becomes

Z

ZN = V N ds1


dsN e; U (V 1=3 s1
:::
V 1=3 sN ) Carrying out the volume derivative gives

N @ZN = N Z ; V N Z ds


ds 1 X ri
@U e; U (V 1=3 s1
:::
V 1=3 sN ) 1 N 3V @V V N @ r i i=1 Z X N 1 = V ZN +  dr1


drN 3V ri
Fi e; U (r1
:::
rN ) i=1

Thus,

*

N 1 @ZN = N +  X ZN @V V 3V i=1 ri
Fi

+

Let us consider, once again, a pair potential. We showed in an earlier lecture that 4

N X i=1

ri
Fi =

N X X i=1 j =1
j 6=i

ri
Fij

where Fij is the force on particle i due to particle j . By interchaning the i and j summations in the above expression, we obtain

2

N X

3

X X ri
Fi = 12 4 ri
Fij + rj
Fji 5 i=1 i
j
i6=j i
j
i6=j

However, by Newton's third law, the force on particle i due to particle j is equal and opposite to the force on particle j due to particle i:

Fij = ;Fji Thus,

2

N X

3

X X X X ri
Fi = 12 4 ri
Fij ; rj
Fij 5 = 21 (ri ; rj )
Fij  12 rij
Fij i=1 i
j
i6=j i
j
i6=j i
j
i6=j i
j
i6=j

The ensemble average of this quantity is

*

+

*

+

Z N X X  X   r
F = r
F = d r rij
Fij e; Upair (r1
:::
rN ) i i ij ij 1


drN 3V i=1 6V i
j
i6=j 6V ZN i
j
i6=j

As before, all integrals are exactly the same, so that

*

+

N  X N (N ; 1) Z dr


r r
F e; Upair (r1
:::
rN ) r
F = 1 N 12 12 3V i=1 i i 6V ZN   Z Z = 6V dr1 dr2 r12
F12 N (NZ ; 1) dr3


drN e; Upair (r1
:::
rN ) N Z  = 6V dr1 dr2 r12
F12 (2) (r1
r2 ) Z 2 N = 6V 3 dr1 dr2 r12
F12 g(2)(r1
r2 )

Then, for a pair potential, we have pair = ;u0 (jr ; r j) (r1 ; r2 ) = ;u0 (r ) r12 F12 = ; @U 1 2 12 @r jr ; r j r 12

1

2

12

where u0 (r) = du=dr, and r12 = jr12 j. Substituting this into the ensemble average gives

*

+

Z N 2  X N 0 (2) 3V i=1 ri
Fi = ; 6V 3 dr1 dr2 u (r12 )r12 g (r1
r2 )

As in the case of the average energy, we change variables at this point to r = r1 ; r2 and R = (r1 + r2 )=2. This gives

*

+

N  X N 2 Z drdRu0 (r)rg~(2) (r
R) r
F = ; 3V i=1 i i 6V 3 Z 2 0 = ; N 6V 2 dru (r)rg~(r) Z1 2 3 0 = ; N 6V 2 0 dr4r u (r)g(r)

5

Therefore, the pressure becomes

P = ; 2 Z 1 dr4r3 u0 (r)g(r) kT 6kT 0

which again gives a simple expression for the pressure in terms only of the derivative of the pair potential form and the radial distribution function. It also shows explicitly how the equation of state di ers from the that of the ideal gas P=kT = . From the de nition of g(r) it can be seen that it depends on the density and temperature T : g(r) = g(r
T ). Note, however, that the equation of state, derived above, has the general form

P kT = + B

2

which looks like the rst few terms in an expansion about ideal gas behavior. This suggests that it may be possible to develop a general expansion in all powers of the density about ideal gas behavior. Consider representing g(r
T ) as such a power series:

g (r 
T ) =

1 X j =0

j gj (r T )

Substituting this into the equation of state derived above, we obtain

1 P = +X Bj+2 (T ) j+2 kT j =0 This is known as the virial equation of state, and the coecients Bj+2 (T ) are given by

1 Bj+2 (T ) = ; 6kT

Z1 0

dr 4r3 u0(r)gj (r T )

are known as the virial coe
cients. The coecient B2 (T ) is of particular interest, as it gives the leading order deviation from ideal gas behavior. It is known as the second virial coecient. In the low density limit, g(r
T )  g0 (r T ) and B2 (T ) is directly related to the radial distribution function.

6

G25.2651: Statistical Mechanics Notes for Lecture 10

I. DISTRIBUTION FUNCTIONS AND PERTURBATION THEORY A. General formulation Recall the expression for the con gurational partition function:

Z ZN = dr1

drN e; U (r1
:::
rN )

Suppose that the potential U can be written as a sum of two contributions U (r1 ::: rN ) = U0 (r1 ::: rN ) + U1 (r1 ::: rN ) where U1 is, in some sense, small compared to U0 . An extra bonus can be had if the partition function for U0 can be evaluated analytically. Let

ZN (0) =

Z

drN e; U0(r1
:::
rN )

dr1

Then, we may express ZN as

(0) Z Z N ZN = Z (0) dr1 drN e; U0(r1
:::
rN ) e; U1 (r1
:::
rN ) N = ZN (0) e; U1 (r1
:::
rN ) 0 0 means average with respect to U0 only. If U1 is small, then the average can be expanded in powers of U1 : 2 3 e; U1 = 1  U +  U 2  U 3 + h

where

h

i

i

h

i0

;

= The free energy is given by

h

1 i0

1 (  )k X U1k k ! k=0 ;

h

2!

h

1 i0

;

3!

h

1 i0

i0







(0) 1 ; U1 A(N V T ) = 1 ln NZ!N3N = 1 ln NZN!3N  ln e Separating A into two contributions, we have A(N V T ) = A(0) (N V T ) + A(1) (N V T ) where A(0) is independent of U1 and is given by (0)
1 Z N (0) A (N V T ) =  ln N !3N and A(1) (N V T ) = 1 ln e; U1 ;

;

;

;

;



h

i0

1 (  )k X = 1 ln U1k k ! k=0 ;

h

1

;

h

i0

h

i0

We wish to develop an expansion for A(1) of the general form

A(1) =

1 (  )k;1 X !k k=1 k ! ;

where !kPare a set of expansion coe cients that are determined by the condition that such an expansion be consistent k k with ln 1 k=0 (  ) U1 0 =k!. Using the fact that h

;

h

i

ln(1 + x) = we have that ;

1 X

k=1

k

( 1)k;1 xk ;

1 !  1 ! X (  )k k (  )k U k 1 1 ln X  k=0 k! 1 0 =  ln 1 + k=0 k! U1 0 ;

h

i

;

;

h



i

1 1 (  )l X X = 1 ( 1)k;1 k1 U1l l ! k=1 l=1 ;

;

;

h

!k i0

Equating this expansion to the proposed expansion for A(1) , we obtain 1 X k=1

1 ! 1 X (  )l l k X U = (  )k !kk! k l=1 l! 1 0 k=1

(;1)k;1 1

;

h

i

;

This must be solved for each of the undetermined parameters !k , which can be done by equating like powers of  on both sides of the equation. Thus, from the  1 term, we nd, from the right side: Right Side :

;

!1 1!

and from the left side, the l = 1 and k = 1 term contributes: Left Side :

;

 U1 0 h

from which it can be easily seen that

1!

i

!1 = U1 0 h

i

Likewise, from the  2 term,

2 !

Right Side :

2! 2 and from the left side, we see that the l = 1 k = 2 and l = 2 k = 1 terms contribute: Left Side : Thus,

2 ; U 2 2

1 i0

h

;h

U1 20 i



!2 = U12 0 U1 20 h

i h

i

For  3 , the right sides gives: Right Side :

3 !

3! 3 the left side contributes the l = 1 k = 3, k = 2 l = 2 and l = 3 k = 1 terms: 2

;

3 Left Side : 6 U13 + ( 1)2 31 (  U1 0 )3 21 h

;

Thus,

i

;

;

h

i

 U1 0 + 12  2 U12

;

;

h

i

h


2 i

!3 = U13 0 + 2 U1 30 3 U1 0 U12 0 h

i

h

i

;

h

i h

i

Now, the free energy, up to the third order term is given by

2 A = A(0) + !1 2 !2 + 6 !3 (0)
;  2;  = 1 ln NZN!3N + U1 0 2 U12 0 U1 20 + 6 U13 0 3 U1 0 U12 0 + 2 U1 30 + In order to evaluate U1 0 , suppose that U1 is given by a pair potential X U1 (r1 ::: rN ) = 12 u1 ( ri rj ) ;

h

;

h

i

;

h

i

;h

i

h

i

;

h

i h

i

h

i

i

j

i6=j

Then, h

U1 0 = Z 1(0) i

Z

N

dr1

drN 12

X i6=j

u1 ( ri j

;

;

rj j)e;

Z Z = N (N (0)1) dr1 dr2 u1 ( r1 r2 ) dr3 2ZN 2 Z N = 2V 2 dr1 dr2 u1( r1 r2 )g0(2) (r1 r2 ) 2V Z 1  = 4r2 u (r)g (r)dr ;

j

j

;

;

j

j

U0 (r1
:::
rN )

drN e; U0(r1
:::
rN )

j

1 0 2 0 The free energy is therefore given by Z1 (0)
A(N V T ) = 1 ln NZN!3N + 21 2 V 4r2 u1 (r)g0 (r)dr 0 ;

;

 ; U2 2

h

1 i0

;h

U1 20 i



B. Derivation of the Van der Waals equation As a speci c example of the application of perturbation theory, we consider the Van der Waals equation of state. Let U0 be given by a pair potential: X U0 (r1 ::: rN ) = 12 u0 ( ri rj ) i6=j j

with



;

j

r> r

u0 (r) = 0

1



This potential is known as the hard sphere potential. In the low-density limit, the radial distribution function can be shown to be given correctly by g0 (r) exp( u0 (r)) or  r> g0 (r) = 10 r 

;



= (r ) u1 (r) is taken to be some arbitrary attractive potential, whose speci c form is not particularly important. Then, the full potential u(r) = u0 (r) + u1 (r) might look like: ;

3

u(r)

r

σ

FIG. 1.

Now, the rst term in A(1) is

A

(1)



Z1 1 2 = 2 V 4r2 u1 (r)g0 (r) 0 Z1 = 12 2 V 4r2 u1 (r) (r ) 0 ;

= 22 V where

a = 2 ;

Z1 

Z1 

r2 u1 (r)dr

aN

 ;

drr2 u1 (r) > 0

is a number that depends on and the speci c form of u1 (r). Since the potential u0 (r) is a hard sphere potential, ZN (0) can be determined analytically. If were 0, then u0 would describe an ideal gas and

ZN (0) = V N However, because two particles may not approach each other closer than a distance between their centers, there is some excluded volume: 4

d d/2

d/2

FIG. 2.

If we consider two hard spheres at closest contact and draw the smallest imaginary sphere that contains both particles, then we nd this latter sphere has a radius :

d

FIG. 3.

Hence the excluded volume for these two particles is 4  3 3 and hence the excluded volume per particle is just half of this: 2  3 b 3 Therefore Nb is the total excluded volume, and we nd that, in the low density limit, the partition function is given approximately by 

ZN (0) = (V Thus, the free energy is

A(N V T )



 ;

;

1 ln (V



5

Nb)N

Nb)N  aN 2 N !3N V ;

;

If we now use this free energy to compute the pressure from

P=

;

@A
@V N
T

we nd that

P = N aN 2 kT V Nb kTV 2 2 = 1  b a kT ;

;

;

;

This is the well know Van der Waals equation of state. In the very low density limit, we may assume that b



h

which essentially states that the greater certainty with which a measurement of X or P can be made, the greater will be the uncertainty in the other.

F. The Heisenberg picture In all of the above, notice that we have formulated the postulates of quantum mechanics such that the state vector (t) evolves in time but the operators corresponding to observables are taken to be stationary. This formulation of quantum mechanics is known as the Schr
odinger picture. However, there is another, completely equivalent, picture in which the state vector remains stationary and the operators evolve in time. This picture is known as the Heisenberg picture. This particular picture will prove particularly useful to us when we consider quantum time correlation functions. The Heisenberg picture species an evolution equation for any operator A, known as the Heisenberg equation. It states that the time evolution of A is given by dA = 1 A H ] dt ih While this evolution equation must be regarded as a postulate, it has a very immediate connection to classical mechanics. Recall that any function of the phase space variables A(x p) evolves according to j

i

dA = A H dt f

g

where ::: ::: is the Poisson bracket. The suggestion is that in the classical limit (h small), the commutator goes over to the Poisson bracket. The Heisenberg equation can be solved in principle giving f

g

A(t) = eiHt=h Ae;iHt=h = U y (t)AU (t) where A is the corresponding operator in the Schrodinger picture. Thus, the expectation value of A at any time t is computed from

A(t) =  A(t) 

h

i

h

j

j

i

where  is the stationary state vector. Let's look at the Heisenberg equations for the operators X and P . If H is given by j

i

2 H = 2Pm + U (X )

6

then Heisenberg's equations for X and P are

dX = 1 X H ] = P dt ih m dP = 1 P H ] = @U dt ih @X Thus, Heisenberg's equations for the operators X and P are just Hamilton's equations cast in operator form. Despite ;

their innocent appearance, the solution of such equations, even for a one-particle system, is highly nontrivial and has been the subject of a considerable amount of research in physics and mathematics. Note that any operator that satises A(t) H ] = 0 will not evolve in time. Such operators are known as constants of the motion. The Heisenberg picture shows explicitly that such operators do not evolve in time. However, there is an analog with the Schrodinger picture: Operators that commute with the Hamiltonian will have associated probabilities for obtaining di erent eigenvalues that do not evolve in time. For example, consider the Hamiltonian, itself, which it trivially a constant of the motion. According to the evolution equation of the state vector in the Schrodinger picture, (t) =

j

i

X i

e;iE t=h Ei Ei (0) i

j

ih

j

i

the amplitude for obtaining an energy eigenvalue Ej at time t upon measuring H will be

Ej (t) =

h

j

i

=

X X i i

e;iE t=h Ej Ei Ei (0) i

h

j

ih

j

i

e;iE t=h ij Ei (0) i

h

= e;iE t=h Ej (0) j

h

j

j

i

i

Thus, the squared modulus of both sides yields the probability for obtaining Ej , which is jh

Ej (t) 2 = Ej (0) j

ij

jh

j

ij

2

Thus, the probabilities do not evolve in time. Since any operator that commutes with H can be diagonalized simultaneously with H and will have the same set of eigenvectors, the above arguments will hold for any such operator.

7

G25.2651: Statistical Mechanics Notes for Lecture 13

I. PRINCIPLES OF QUANTUM STATISTICAL MECHANICS The problem of quantum statistical mechanics is the quantum mechanical treatment of an N -particle system. Suppose the corresponding N -particle classical system has Cartesian coordinates

q1 ::: q3N and momenta

p1 ::: p3N and Hamiltonian

H=

3N X

p2i + U (q ::: q ) 1 3N i=1 2mi

Then, as we have seen, the quantum mechanical problem consists of determining the state vector (t) from the Schr odinger equation j

i

@ (t) H (t) = i h @t Denoting the corresponding operators, Q1 ::: Q3N and P1 ::: P3N , we note that these operators satisfy the commuj

i

j

i

tation relations:

Qi Qj ] = Pi Pj ] = 0 Qi Pj ] = i hIij and the many-particle coordinate eigenstate q1 :::q3N is a tensor product of the individual eigenstate q1 ::: q3N : q1 :::q3N = q1 q3N j

i

j

j

i

j

i

j

i

j

i

i

The Schr odinger equation can be cast as a partial dierential equation by multiplying both sides by q1 :::q3N : h

@ q :::q (t) q1 :::q3N H (t) = i h @t 1 3N # 3N X

h2 @ 2 + U (q ::: q ) (q ::: q t) = i h @ (q ::: q t) 1 3N 1 3N 3N 2 @t 1 i=1 2mi @qi j

h

" ;

j

j

i

h

j

i

where the many-particle wave function is (q1 :::: q3N t) = q1 :::q3N (t) . Similarly, the expectation value of an operator A = A(Q1 ::: Q3N P1 ::: P3N ) is given by h

Z

A = dq1

h

i



j

i



dq3N  (q1 ::: q3N )A q1 ::: q3N hi @q@ ::: hi @q@ (q1 ::: q3N ) 1 3N

1

A. The density matrix and density operator In general, the many-body wave function (q1 ::: q3N t) is far too large to calculate for a macroscopic system. If we wish to represent it on a grid with just 10 points along each coordinate direction, then for N = 1023, we would need 23 1010 total points, which is clearly enormous. We wish, therefore, to use the concept of ensembles in order to express expectation values of observables A without requiring direct computation of the wavefunction. Let us, therefore, introduce an ensemble of systems, with a total of Z members, and each having a state vector ( ) ,  = 1 ::: Z . Furthermore, introduce an orthonormal set of vectors k ( k j = ij ) and expand the state vector for each member of the ensemble in this orthonormal set: h

j

j

i

h

j

i

i

i

(

j

)

i

=

X

k

Ck( ) k j

i

The expectation value of an observable, averaged over the ensemble of systems is given by the average of the expectation value of the observable computed with respect to each member of the ensemble:

A = Z1

h

Z X

( ) A (

h

i

=1

j

)

j

Substituting in the expansion for ( ) , we obtain X A = Z1 Ck( ) Cl( ) k A l j

i

i

h

i

=

k
l  Z X 1X

Z

k
l

=1

Let us dene a matrix

lk = and a similar matrix

h

j

j

!

Cl Ck ( )

Z X =1

i

( )

Cl( ) Ck(

h

k A l j

j

i

)

Z X ~lk = Z1 Cl( ) Ck( =1

)

Thus, lk is a sum over the ensemble members of a product of expansion coecients, while ~lk is an average over the ensemble of this product. Also, let Akl = k A l . Then, the expectation value can be written as follows: X X A = 1  A = 1 (A) = 1 Tr(A) = Tr(~A) h

j

j

i

kk Z Z k where  and A represent the matrices with elements lk and Akl in the basis of vectors k . The matrix lk is h

i

Z k
l lk kl

fj

ig

known as the density matrix. There is an abstract operator corresponding to this matrix that is basis-independent. It can be seen that the operator

=

Z X

(

j

=1

)

(

ih

)

j

and similarly Z X ~ = Z1 ( =1 j

2

)

ih

(

)

j

have matrix elements lk when evaluated in the basis set of vectors k . fj

l  k =

h

j j

Z X

i

l (

h

=1

)

j

Z X

( ) k =

ih

j

ig

i

=1

Cl( ) Ck(

)

= lk

Note that  is a hermitian operator

y =  so that its eigenvectors form a complete orthonormal set of vectors that span the Hilbert space. If wk and wk represent the eigenvalues and eigenvectors of the operator ~, respectively, then several important properties they must satisfy can be deduced. Firstly, let A be the identity operator I . Then, since I = 1, it follows that X 1 = 1 Tr() = Tr(~) = w j

i

j 

Pk .

h i

Z

k

k

Thus, the eigenvalues of ~ must sum to 1. Next, let A be a projector onto an eigenstate of ~, A = wk wk Then Pk = Tr(~ wk wk ) But, since ~ can be expressed as j

h

i

j

X

~ =

ih

j

wk wk wk j

k

ih

ih

j

and the trace, being basis set independent, can be therefore be evaluated in the basis of eigenvectors of ~, the expectation value becomes

Pk =

h

i

=

X

wj

h

j X

i
j = wk

X

j

wi wi wi wk wk wj j

i

ih

j

ih

j

i

wi ij ik kj

However,

Pk = Z1

h

Z X

i

( ) wk wk (

h

=1

j

ih

Z X ( ) wk = Z1 jh

j

=1

j

ij

2



)

i

0

P

Thus, wk 0. Combining these two results, we see that, since k wk = 1 and wk 0, 0 wk 1, so that wk satisfy the properties of probabilities. With this in mind, we can develop a physical meaning for the density matrix. Let us now consider the expectation value of a projector ai ai ai onto one of the eigenstates of the operator A. The expectation value of this operator is given by 

j

ih

i





j  P

Z X 1 ( ai = Z

hP



h

=1

)

jP

ai j

( )

i

Z X 1 =Z ( ) ai ai ( h

j

ih

=1

j

)

i

Z X 1 =Z ai ( jh

j

) 2 ij

=1

But ai ( ) 2 Pa(i ) is just probability that a measurement of the operator A in the th member of the ensemble will yield the result ai . Thus, jh

j

ij



P 1X ( ) = ai Z Pai

hP

i

=1

3

or the expectation value of ai is just the ensemble averaged probability of obtaining the value ai in each member of the ensemble. However, note that the expectation value of ai can also be written as P

P

X

Pai ) = Tr( ai i = Tr(~

hP

= =

X

k
l X k

k

j

wk kl wk ai ai wl h

wk ai wk jh

j

X

wk wk wk ai ai ) =

ih

ih

j

ih

j

wl wk wk wk ai ai wl

h

k
l

j

j

ih

j

ih

j

i

i

2

ij

Equating the two expressions gives Z X 1X ( ) P = wk ai wk a i Z h

i

jh

k

=1

j

2

ij

The interpretation of this equation is that the ensemble averaged probability of obtaining the value ai if A is measured is equal to the probability of obtaining the value ai in a measurement of A if the state of the system under consideration were the state wk , weighted by the average probability wk that the system in the ensemble is in that state. Therefore, the density operator  (or ~) plays the same role in quantum systems that the phase space distribution function f (;) plays in classical systems. j

i

B. Time evolution of the density operator The time evolution of the operator  can be predicted directly from the Schr odinger equation. Since (t) is given by

(t) =

Z X j

=1

( ) (t) ( ) (t) ih

j

the time derivative is given by Z @ = X @t =1



@ () @t  (t) j

 i

h

j

Z h X = i1 h H ( ) (t) ( ) (t) j

i



@ ( ) (t)  (t) +  (t) @t ( )

h

( )

j

i

( ) (t)

j ;j

=1

h

 h

 j

( ) (t) H

h

i

j

= i1h (H H ) = i1 h H ] ;

@ = 1 H ] @t i h

where the second line follows from the fact that the Schr odinger equation for the bra state vector ( ) (t) is h

;

@ ( ) (t) = ( ) (t) H i h @t h

j

h

j

j

Note that the equation of motion for (t) diers from the usual Heisenberg equation by a minus sign! Since (t) is constructed from state vectors, it is not an observable like other hermitian operators, so there is no reason to expect that its time evolution will be the same. The general solution to its equation of motion is (t) = e;iHt=h (0)eiHt=h = U (t)(0)U y (t) The equation of motion for (t) can be cast into a quantum Liouville equation by introducing an operator 4

iL = i1 h ::: H ]

In term of iL, it can be seen that (t) satises

@ = iL @t (t) = e;iLt (0) ;

What kind of operator is iL? It acts on an operator and returns another operator. Thus, it is not an operator in the ordinary sense, but is known as a superoperator or tetradic operator (see S. Mukamel, Principles of Nonlinear Optical Spectroscopy, Oxford University Press, New York (1995)). Dening the evolution equation for  this way, we have a perfect analogy between the density matrix and the state vector. The two equations of motion are

@ (t) = i H (t) @t

h @ (t) = iL(t) @t j

i

;

j

i

;

We also have an analogy with the evolution of the classical phase space distribution f (; t), which satises

@f = iLf @t ;

with iL = ::: H being the classical Liouville operator. Again, we see that the limit of a commutator is the classical Poisson bracket. f

g

C. The quantum equilibrium ensembles At equilibrium, the density operator does not evolve in time thus, @=@t = 0. Thus, from the equation of motion, if this holds, then H ] = 0, and (t) is a constant of the motion. This means that it can be simultaneously diagonalized with the Hamiltonian and can be expressed as a pure function of the Hamiltonian

 = f (H ) Therefore, the eigenstates of , the vectors, we called wk are the eigenvectors Ei of the Hamiltonian, and we can write H and  as j

H= =

i

X

j

Ei Ei Ei j

i X

ih

j

f (Ei ) Ei Ei j

i

i

ih

j

The choice of the function f determines the ensemble. 1. The microcanonical ensemble

Although we will have practically no occasion to use the quantum microcanonical ensemble (we relied on it more heavily in classical statistical mechanics), for completeness, we dene it here. The function f , for this ensemble, is

f (Ei )E = (Ei (E + E )) (Ei E ) where (x) is the Heaviside step function. This says that f (Ei )E is 1 if E < Ei < (E + E ) and 0 otherwise. The partition function for the ensemble is Tr(), since the trace of  is the number of members in the ensemble: ;

(N V E ) = Tr() =

X

i

;

;

(Ei (E + E )) (Ei E )] ;

5

;

;

The thermodynamics that are derived from this partition function are exactly the same as they are in the classical case:

S (N V E ) = k ln (N V E )  1 @ ln  T = k @E ; ;

N
V

etc. 2. The canonical ensemble

In analogy to the classical canonical ensemble, the quantum canonical ensemble is dened by

 = e;H f (Ei ) = e;Ei Thus, the quantum canonical partition function is given by

Q(N V T ) = Tr(e;H ) =

X

i

e;Ei

and the thermodynamics derived from it are the same as in the classical case: A(N V T ) = 1 ln Q(N V T )

@ ln Q(N V T ) E (N V T ) = @

@ ln Q(N V T ) P (N V T ) = 1 @V ;

;

etc. Note that the expectation value of an observable A is A = 1 Tr(Ae;H ) h

i

Q

Evaluating the trace in the basis of eigenvectors of H (and of ), we obtain X X A = Q1 Ei Ae;H Ei = Q1 e;Ei Ei A Ei h

i

i

h

j

j

i

h

i

j

j

i

The quantum canonical ensemble will be particularly useful to us in many things to come. 3. Isothermal-isobaric and grand canonical ensembles

Also useful are the isothermal-isobaric and grand canonical ensembles, which are dened just as they are for the classical cases: (N P T ) = Z

( V T ) =

Z

1

0 1 X

N =0

dV e;PV Q(N V T ) = eN Q(N

V T) =

6

1 X

N =0

Z

1 0

dV Tr(e;(H +PV ) )

Tr(e;(H ;N ) )

D. A simple example { the quantum harmonic oscillator As a simple example of the trace procedure, let us consider the quantum harmonic oscillator. The Hamiltonian is given by 2 H = 2Pm + 12 m!2 X 2 and the eigenvalues of H are   1 En = n + 2 h! n = 0 1 2 ::: Thus, the canonical partition function is

Q( ) =

1 X

n=0

e;(n+1=2)h! = e;h !=2

1 X ; ;h ! n e n=0

This is a geometric series, which can be summed analytically, giving ;h  !=2 Q( ) = 1e e;h ! = eh !=2 1e;h !=2 = 21 csch( h!=2) The thermodynamics derived from it as as follows: 1. Free energy: The free energy is ;  A = 1 ln Q( ) = h 2! + 1 ln 1 e;h ! ;

;

;

;

2. Average energy: The average energy E = H is h

i





@ ln Q( ) = h! + h!e;h! = 1 + n h! E = @

2 1 e;h ! 2 ;

3. Entropy The entropy is given by

;

h

i

;h ! ;  S = k ln Q( ) + ET = k ln 1 e;h ! + hT! 1 e e;h ! ;

;

;

Now consider the classical expressions. Recall that the partition function is given by Z ; Q( ) = h1 dpdxe

Thus, the classical free energy is



p2 1 2m + 2

m!2 x2



 1=2   2 1=2 = 2 = 1 = h1 2 m

m!2

!h h!

Acl = 1 ln( h!)

In the classical limit, we may take h to be small. Thus, the quantum expression for A becomes, approximately, in this limit: h ! + 1 ln( h!) AQ 2

and we see that

h! AQ Acl 2 The residual h !=2 (which truly vanishes when h 0) is known as the quantum zero point energy. It is a pure quantum eect and is present because the lowest energy quantum mechanically is not E = 0 but the ground state energy E = h !=2. ;!

;

;!

!

7

G25.2651: Statistical Mechanics Notes for Lecture 14

I. DERIVATION OF THE DISCRETIZED PATH INTEGRAL We begin our discussion of the Feynman path integral with the canonical ensemble. The expressions for the partition function and expectation value of an observable A are, respectively Q(N V T ) = Tr(e; H ) hAi = Q1 Tr(Ae; H ) It is clear that we need to be able to evaluate traces of the type appearing in these expressions. We have already derived expressions for these in the basis of eigenvectors of H . However, since the trace is basis independent, let us explore carrying out these traces in the coordinate basis. We will begin with the partition function and treat expectation values later. Consider the ensemble of a one-particle system. The partition function evaluated as a trace in the coordinate basis is

Q(
) =

Z

dxhxje; H jxi

We see that the trace involves the diagonal density matrix element hxje; H jxi. Let us solve the more general problem of any density matrix element hxje; H jx0 i. If the Hamiltonian takes the form

2 H = 2Pm + U (X ) K + U then we cannot evaluate the operator exp(;
H ) explicitly because the operators for kinetic (T ) and potential energies (U ) do not commute with each other, being, respectively, functions of momentum and position, i.e., K U ] 6= 0 In this instance, we will make use of the Trotter theorem, which states that given two operators A and B , such that A B ] 6= 0, then for any number , h
B=2P
A=P
B=2P iP e
(A+B) = Plim e e e !1

Thus, for the Boltzmann operator,

h ; U=2P ; K=P ; U=2P iP e e; (K +U ) = Plim e e !1

and the partition function becomes

Z h ; U=2P ; K=P ; U=2P iP Q(
) = Plim dx h x j e e e jxi !1

De ne the operator in brackets to be : Then,

= e; U=2P e; K=P e; U=2P

Z Q(
) = Plim dxhxj P jxi !1 1

In between each of the P factors of , the coordinate space identity operator

I=

Z

dxjxihxj

is inserted. Since there are P factors, there will be P ; 1 such insertions. the integration variables will be labeled x2 ::: xP . Thus, the expression for the matrix element becomes

Z

hxj jx0 i = dx2    dxP hxj jx2 ihx2 j jx3 ihx3 j    jxP ihxP j jx0 i =

Z

dx2    dxP

P Y i=1

hxi j jxi+1 ijx1 =xxP +1 =x

0

The next step clearly involves evaluating the matrix elementx

hxi j jxi+1 i = hxi je;

U (X )=2P e; P 2 =2mP e; U (X )=2P jxi+1 i

Note that in the above expression, the operators involving the potential U (X ) act on their eigenvectors and can thus be replaced by the corresponding eigenvalues:

hxi j jxi+1 i = e;

(U (xi )+U (xi+1 )=2

hxi je;

P 2 =2mP jxi+1 i

In order to evaluate the remaining matrix element, we introduce the momentum space identity operator

I=

Z

dpjpihpj

Letting K = P 2 =2m, the matrix remaining matrix element becomes

hxi je; K=P jxi+1 i = =

Z Z

dphxi jpihpje; P 2 =2mP jxi+1 i dphxi jpihpjxi+1 ie; p2 =2mP

Using the fact that

hxjpi = p 1 eipx=h 2h

it follows that

hxi je; K=P jxi+1 i = 1

Z

ip(xi ;xi+1 )=h ; p2 =2mP

e 2h dpe The remaining integral over p can be performed by completing the square, leading to the result



mP 1=2 exp ; mP (x ; x )2 2
h2 2
h2 i+1 i Collecting the pieces together, and introducing the P ! 1 limit, we have for the density matrix " P
#
P=2 Z X mP mP
 2 hxje; H jx0 i = Plim dx2    dxP exp ; 2 (xi+1 ; xi ) + 2P (U (xi ) + U (xi+1 )) x =xx =x !1 2
h2 i=1 2
h 1 P +1

hxi je;

K=P jxi+1 i =

The partition function is obtained by setting x = x0 , which is equivalent to setting x1 = xP +1 and integrating over x, or equivalently x1 . Thus, the expression for Q(
) becomes

" P
#
P=2 Z X 1 2 mP 1 2 Q(
) = Plim m!P (xi+1 ; xi ) + P U (xi )  dx1    dxP exp ;
!1 2
h2 2 i=1 xP +1 =x1 2

0

where we have introduced a \frequency"

p

!P =
Ph When expressed in this way, the partition function, for a nite value of P , is isomorphic to a classical con guration integral for a P -particle system, that is a cyclic chain of particles, with harmonic nearest neighbor interactions and interacting with an external potential U (x)=P . That is, the partition function becomes

Q(
) 

Z

where

Ue (x1 ::: xP ) =

dx1    dxP e; Ueff (x1 :::xP ) P 1 X i=1

1 2 2 2 m!P (xi+1 ; xi ) + P U (xi )



Thus, for nite (if large) P the partition function in the discretized path integral representation can be treated as any ordinary classical con guration integral. Consider the integrand of Q(
) in the limit that all P points on the cyclic chain are at the same location P x. Then the harmonic nearest neighbor coupling (which is due to the quantum kinetic energy) vanishes and (1=P ) Pi=1 U (xi ) ! U (x), and the integrand becomes

e; U (x) which is just the true classical canonical position space distribution function. Therefore, the greater the spatial spread in the cyclic chain, the more \quantum" the system is, since this indicates a greater contribution from the quantum kinetic energy. The spatially localized it is, the more the system behaves like a classical system. It remains formally to take the limit that P ! 1. There we will see an elegant formulation for the density matrix and partition function emerges.

3

G25.2651: Statistical Mechanics Notes for Lecture 15

I. THE FUNCTIONAL INTEGRAL REPRESENTATION OF THE PATH INTEGRAL A. The continuous limit In taking the limit P so that P

! 1, it will prove useful to de
ne a parameter " =
Ph

! 1 implies " ! 0. In terms of ", the partition function becomes m
Q(
) = !1 lim !0 2" h

P=

P

2

"

Z

"

 

X m x +1 ; x dx1  dx exp ; h" 2 " =1 P

i

i

2

P

i

!# + U (x )  P +1 = 1 i

x

x

We can think of the points x1  ::: x as speci
c points of a continuous functions x( ), where x = x( = (k ; 1)") such that x(0) = x( = P") = x( =
h ): P

k

x

x4 x2

x P+1

x1

0

2 ε 4ε

Pε FIG. 1.

1

τ

Note that lim !0

 x +1 ; x  k

"

"

and that the limit

k

 x(k") ; x((k ; 1)")  dx = lim = !0

" 

" X m x +1 ; x lim !1 !0 h =1 2 " P

P

"

"

"

i

i

2

d

#

+ U (x ) i

i

is just a Riemann sum representation of the continuous integral

"  2

1 j Z  d m dx h 0 2 d h

Finally, the measure P

lim !0 !1 "

#

+ U (x( ))

 m  2 dx1  dx 2"h2 P=

P

represents an integral overa all values that the function x( ) can take on between  = 0 and  =
h such that x(0) = x(
h). We write this symbolically as Dx(). Therefore, the P ! 1 limit of the partition function can be written as

Q(
) = =

Z I

Z

"

Z  m
1 dx D x() exp ; h d 2 x_ 2 + U (x( )) (0)= 0 " Z 
# 1 m 2 Dx() exp ; h 0 d 2 x_ + U (x( )) (  )=x

x h

x

h

#

x

h

The above expression in known as a functional integral. It says that we must integrate over all functions (i.e., all values that an arbitrary function x( ) may take on) between the values  = 0 and  =
h. It must really be viewed as the limit of the discretized integral introduced in the last lecture. The integral is also referred to as a path integral because it implies an integration over all paths that a particle might take between  = 0 and  =
h such that x(0) = x(
h , where the paths are paramterized by the variable  (which is not time!). The second line in the above expression, which is equivalent to the
rst, indicates that the integration is taken over all paths that begin and end at the same point, plus a
nal integration over that point. The above expression makes it clear how to represent a general density matrix element hxj exp(;
H )jx0 i:

hxje; jx0 i = H

Z

(  )=x0

x h

(0)=x

x

"

Z  m
1 Dx() exp ; h 0 d 2 x_ 2 + U (x( )) h

#

which indicates that we must integrate over all functions x( ) that begin at x at  = 0 and end at x0 at  =
h:

2

x

x’

x

βh

0

τ

FIG. 2.

Similarly, diagonal elements of the density matrix, used to compute the partition function, are calculated by integrating over all periodic paths that satisfy x(0) = x(
h) = x:

3

x

x’

x τ

βh

0 FIG. 3.

Note that if we let
= it=h, then the density matrix becomes

(x x0  it=h) = hxje;



iH t=h

jx0 i = U (x x0  t)

which are the coordinate space matrix elements of the quantum time evolution operator. If we make a change of variables  = is in the path integral expression for the density matrix, we
nd that the quantum propagator can also be expressed as a path integral:

U (x x0  t) = hxje;



iHt=h

jx0 i =

Z

( )=x0

x t

Dx() exp

i Z

t


ds m2 x_ (s) ; U (x(s))



h 0 Such a variable transformation is known as a Wick rotation. This nomenclature comes about by viewing time as a complex quantity. The propagator involves real time, while the density matrix involves a transformation t = ;i
h to the imaginary time axis. It is because of this that the density matrix is sometimes referred to as an imaginary time path integral. (0)=x

x

B. Dominant paths in the propagator and density matrix Let us
rst consider the real time quantum propagator. The quantity appearing in the exponential is an integral of 1 mx_ 2 ; U (x)  L(x x_ ) 2 which is known as the Lagrangian in classical mechanics. We can ask, which paths will contribute most to the integral

Z

t

0

i Z h ds m2 x_ 2 (s) ; U (x(s)) = dsL(x(s) x_ (s)) = S x] t

0

4

known as the action integral. Since we are integrating over a complex exponential exp(iS=h), which is oscillatory, those paths away from which small deviations cause no change in S (at least to
rst order) will give rise to the dominant contribution. Other paths that cause exp(iS=h) to oscillate rapidly as we change from one path to another will give rise to phase decoherence and will ultimately cancel when integrated over. Thus, we consider two paths x(s) and a nearby one constructed from it x(s) + x(s) and demand that the change in S between these paths be 0 S x + x] ; S x] = 0 Note that, since x(0) = x and x(t) = x0 , x(0) = x(t) = 0, since all paths must begin at x and end at x0 . The change in S is

S = S x + x] ; S x] =

Z

0

t

dsL(x + x x_ + x_ ) ;

Z

t

0

dsL(x x_ )

Expanding the
rst term to
rst order in x, we obtain

@L x ; Z L(x x_ ) = Z ds @L x_ + @L x

ds L(x x_ ) + @L  x _ + @ x_ @x @ x_ @x 0 0 0 The term proportional to x_ can be handled by an integration by parts:  Z d @L Z @L Z @L d  ds @ x_ x_ = ds @ x_ dt x = @L x @ x_ 0 ; 0 ds dt @ x_ x 0 0 because x vanishes at 0 and t, the surface term is 0, leaving us with Z d @L @L

S = ds ; dt @ x_ + @x x = 0 0 S =

Z

t

t

t

t

t

t

t

t

Since the variation itself is arbitrary, the only way the integral can vanish, in general, is if the term in brackets vanishes:

d @L @L dt @ x_ ; @x = 0

This is known as the Euler-Lagrange equation in classical mechanics. For the case that L = mx= _ 2 ; U (x), they give

d (mx_ ) + @U = 0 dt @x mx = ; @U @x

which is just Newton's equation of motion, subject to the conditions that x(0) = x, x(t) = x0 . Thus, the classical path and those near it contribute the most to the path integral. The classical path condition was derived by requiring that S = 0 to
rst order. This is known as an action stationarity principle. However, it turns out that there is also a principle of least action, which states that the classical path minimizes the action as well. This is an important consideration when deriving the dominant paths for the density matrix, which takes the form

(x x0 
) =

Z

(  =x0

x h

(0)=x

x

The action appearing in this expression is

S x] = E

Z



h

0

"

Z  m
1 Dx() exp ; h 0 d 2 x_ ( ) + U (x( )) h

h i Z d m2 x_ 2 + U (x( )) =



h

0

#

dH (x x_ )

which is known as the Euclidean action and is just the integral over a path of the total energy or Euclidean Lagrangian H (x x_ ). Here, we see that a minimum action principle is needed, since the smallest values of S will contribute most to the integral. Again, we require that to
rst order S x + x] ; S x] = 0. Applying the same logic as before, we obtain the condition E

E

E

5

d @H ; @H = 0 d @ x_ @x @ U (x ) mx = @x

which is just Newton's equation of motion on the inverted potential surface ;U (x), subject to the conditions x(0) = x, x(
h ) = x0 . For the partition function Q(
), the same equation of motion must be solved, but subject to the conditions that x(0) = x(
h ), i.e., periodic paths.

II. DOING THE PATH INTEGRAL: THE FREE PARTICLE The density matrix for the free particle 2 H = 2Pm

will be calculated by doing the discrete path integral explicitly and taking the limit P The density matrix expression is

(x x0 
) =

 mP  2 Z lim dx2  dx !1 2
h2

"

P=

P

P

! 1 at the end.

X exp ; mP2 (x +1 ; x )2 2
h =1 P

i

i

i

#   1= x

P +1 =x0

x x

Let us make a change of variables to

u1 = x1 u = x ; x~ k

k

k

x~ = (k ; 1)xk +1 + x1 k

k

The inverse of this transformation can be worked out explicitly, giving

x1 = u1 x = k

+1 X k;1

P

l

The Jacobian of the transformation is simply 01 B0 J = det B B@ 0 0



=1

P ;k+1 l ; 1 u + P u1

;1=2 1 0 0



l

0 ;2=3 1 0



 1  CC  CA = 1    

0 0 ;3=4 1

Let us see what the eect of this transformation is for the case P = 3. For P = 3, one must evaluate (x1 ; x2 )2 + (x2 ; x3 )2 + (x3 ; x4 )2 = (x ; x2 )2 + (x2 ; x3 )2 + (x3 ; x0 )2

According to the inverse formula,

x1 = u 1 x2 = u2 + 21 u3 + 13 x0 + 32 x

Thus, the sum of squares becomes

x3 = u3 + 23 x0 + 13 x 6

(x ; x2 )2 + (x2 ; x3 )2 + (x3 ; x0 )2 = 2u22 + 32 u23 + 31 (x ; x0 )2 = 2 ;2 1 u22 + 3 ;3 1 u23 + 31 (x ; x0 )2 From this simple exmple, the general formula can be deduced:

X P

(x +1 ; x )2 = i

=1

i

i

X k 2 1 02 k ; 1 u + P (x ; x ) P

k

k

=2

Thus, substituting this transformation into the integral gives

(x x0 
) =

" X #

m P m (x ; x0 )2 2 exp ; u exp ; 2
h2 2
h2

 m 1 2 Y  m P 1 2 Z du2  du 2
h2 2
h2 P

=

P

=

k

k

P

k

k

=2

where

k

=2

m = k ;k 1 m k

and the overall prefactor has been written as

 mP 

2

P=

2
h

2

 m 1 2 Y  m P 1 2 = =

P

=

k

2
h

2

k

=2

2
h

2

Now each of the integrals over the u variables can be integrated over independently, yielding the
nal result

(x x0 
) =

 m 1 2 m

0 2 exp ; 2 (x ; x ) 2
h2 2
h =

In order to make connection with classical statistical mechanics, we note that the prefactor is just 1= , where

 2
h2 1 2 
h2 1 2

= = =

=

m

2m is the kinetic prefactor that showed up also in the classical free particle case. In terms of , the free particle density matrix can be written as (x x0 
) = 1 e; ( ; )2 2



 x

0

x

=

Thus, we see that represents the spatial width of a free particle at
nite temperature, and is called the \thermal de Broglie wavelength."

7

G25.2651: Statistical Mechanics Notes for Lecture 16

I. THE HARMONIC OSCILLATOR { EXPANSION ABOUT THE CLASSICAL PATH It will be shown how to compute the density matrix for the harmonic oscillator: 2 H = 2Pm + 12 m!2 X 2 using the functional integral representation. The density matrix is given by


(x x   ) = 0

Z x( h )=x

0

x(0)=x

"

# Z h
1 1 1 2 2 2 Dx( ) exp ; h d 2 mx_ + 2 m! x 0

As we saw in the last lecture, paths in the vicinity of the classical path on the inverted potential give rise to the dominant contribution to the functional integral. Thus, it proves useful to expand the path x( ) about the classical path. We introduce a change of path variables from x( ) to y( ), where

x( ) = xcl ( ) + y( ) where xcl ( ) satis es

mx cl = m!2xcl subject to the conditions

xcl ( h) = x

xcl(0) = x

0

so that y(0) = y( h) = 0. Substituting this change of variables into the action integral yields

S=

Z h 0

Z h



d 12 mx_ 2 + 21 m!2 x2







= d 12 m(x_ cl + y_ )2 + 21 m!2 (xcl + y)2 0  Z h   Z h  = d 12 mx_ 2cl + 12 m!2 x2cl + d 21 my_ 2 + 21 m!2 y2 0 0 +

Z h 0



d mx_ cl y_ + m!2 xcl y



An integration by parts makes the cross terms vanish: Z h 0





d mx_ cly_ + m!2xcl y = mx_ cl yj0 h +

Z h 0





d ;mx cl + m!2xcl y = 0

where the surface term vanishes becuase y(0) = y( h) = 0 and the second term vanishes because xcl satis es the classical equation of motion. The rst term in the expression for S is the classical action, which we have seen is given by Z h





m! h(x2 + x 2 )cosh( h!) ; 2xx i d 21 mx_ 2cl + 21 m!2 x2cl = 2sinh(  h!) 0 Therefore, the density matrix for the harmonic oscillator becomes 0

1

0





m! (x2 + x 2 )cosh( h!) ; 2xx
(x x   ) = I y] exp ; 2sinh(  h!) 0

0

where I y] is the path integral

I y] =

Z y( h )=0

y(0)=0

0

# Z h
2  m m! 1 2 2 Dy( ) exp ; h 2 y_ + 2 y 0 "

Note that I y] does not depend on the points x and x and therefore can only contribute an overall (temperature dependent) constant to the density matrix. This will aect the thermodynamics but not any averages of physical observables. Nevertheless, it is important to see how such a path integral is done. In order to compute I y] we note that it is a functional integral over functions y( ) that vanish at  = 0 and  =  h. Thus, they are a special class of periodic functions and can be expanded in a Fourier sine series: 0

X 1

y ( ) =

n=1

cn sin(!n  )

where

!n = n  h Thus, we wish to change from an integral over the functions y( ) to an integral over the Fourier expansion coecients cn . The two integrations should be equivalent, as the coecients uniquely determine the functions y( ). Note that y_ ( ) =

X 1

n=1

!n cn cos(!n  )

Thus, terms in the action are:

Z h 0

d m1 y_ 2 = m2

XX 1

1

n=1 n =1

cn cn !n !n 0

Z h

0

0

d cos(!n  ) cos(!n  ) 0

0

Since the cosines are orthogonal between  = 0 and  =  h, the integral becomes Z h 0

d m1 y_ 2 = m2

X 1

n=1

Z h

c2n !n2

0

Similarly,

d cos2 (!n  ) = m2

X 1

n=1

c2n !n2

Z h 1 0

Z h 0





X d 21 + 12 cos(2!n ) = m4 h c2n !n2 n=1 1

m h !2 X c2 2 2 m! y = n 2 4 1

n=1

The measure becomes

Dy(t) !

Y 1

n=1

p

dcn 4=m!n2

which, is not an equivalent measure (since it is not derived from a determination of the Jacobian), but is chosen to q give the correct free-particle (! = 0) limit, which can ultimately be corrected by attaching an overall factor of m=2 h2 . With this change of variables, I y] becomes

I y] =

YZ 1

1

n=1

;1

 Y 1=2  2 ! dc m n n 2 2 2 p exp ; 4 (! + !n )cn = 2 2 4=m!n2 n=1 ! + !n 1

The in nite product can be written as 2

Y 1

n=1

"






2 n2 = 2 h2 = Y 1 +  2 h2 !2 2 n2 !2 + 2 n2 = 2 h2 n=1 1

#

;

1

the product in the square brackets is just the in nite product formula for sinh( h!)=( h!), so that I y] is just s

 h ! I y] = sinh(  h!)

q

Finally, attaching the free-particle factor m=2 h2 , the harmonic oscillator density matrix becomes: 



r m! m! (x2 + x 2 )cosh( h!) ; 2xx
(x x   ) = 2 hsinh( exp ;  h!) 2sinh( h!) Notice that in the free-particle limit (! ! 0), sinh( h!)   h! and cosh( h!)  1, so that   r m m 2
(x x   ) ! exp ; 2 (x ; x ) 2 h2 2 h 0

0

0

0

0

which is the expected free-particle density matrix.

II. THE STATIONARY PHASE APPROXIMATION Consider the simple integral:

I = lim

Z

1

dx e f (x) ;

!1

;1

Assume f (x) has a global minimum at x = x0 , such that f (x0 ) = 0. If this minimum is well separated from other minima of f (x) and the value of f (x) at the global minimum is signi cantly lower than it is at other minima, then the dominant contributions to the above integral, as  ! 1 will come from the integration region around x0 . Thus, we may expand f (x) about this point: f (x) = f (x0 ) + f (x0 )(x ; x0 ) + 21 f (x0 )(x ; x0 )2 +    Since f (x0 ) = 0, this becomes: f (x)  f (x0 ) + 12 f (x0 )(x ; x0 )2 Inserting the expansion into the expression for I gives 0

0

00

0

00

I = lim e f (x0 )

Z

1

;

!1

dx e 2 f (x0 )(x x0)2 ;

;1



00

;



1=2 = lim f 2(x ) e f (x0 ) 0 Corrections can be obtained by further expansion of higher order terms. For example, consider the expansion of f (x) up to fourth order: 1 f (iv) (x )(x ; x )4 f (x)  f (x0 ) + 21 f (x0 )(x ; x0 )2 + 16 f (x0 )(x ; x0 )3 + 24 0 0 Substituting this into the integrand and further expanding the exponential would give, as the lowest order nonvanishing correction: ;

00

!1

00

I = lim e

f (x0 )

000

Z

1

;

!1

;1

dx e

;

2

 f (x0 )(x x0 )2 1 ; 00

;

3

 f (iv) (x )(x ; x )4 0 0 24



This approximation is known as the stationary phase or saddle point approximation. The former may seem a little out-of-place, since there is no phase in the problem, but that is because we formulated it in such a way as to anticipate its application to the path integral. But this is only if  is taken to be a real instead of an imaginary quantity. The application to the path integral follows via a similar argument. Consider the path integral expression for the density matrix:


(x x   ) = 0

Z x( h )=x

0

x(0)=x

We showed that the classical path satisfying

SE x]=h

Dx]e

;





mx cl = @U @x x=xcl x(0) = x x( h) = x is a stationary point of the Euclidean action SE x], i.e., SExcl ] = 0. Thus, we can develop a stationary phase or 0

saddle point approximation for the density matrix by introducing an expansion about the classical path according to

x( ) = xcl ( ) + y( ) = xcl ( ) +

X

cn n ( )

n

where the correction y( ), satisfying y(0) = y( h) = 0 has been expanded in a complete set of orthonormal functions f n ( )g, which are orthonormal on the interval 0  h] andsatisfy n (0) = n ( h) = 0 as well as the orthogonality condition: Z h 0

d n ( ) m ( ) = mn

Setting all the expansion coecients to 0 recovers the classical path. Thus, we may expand the action S x] (the \E" subscript will henceforth be dropped from this discussion) with respect to the expansion coecients:



@S

1 X @ 2 S

S x] = S xcl ] + @c c + j 2 jk @cj ck c =0 cj ck +    j c =0 j X

f g

Since

f g





Z h

S x] = d 12 mx_ 2 + U (x( )) 0 the expansion can be worked out straightforwardly by substitution and subsequent dierentiation: S x] = @S = @cj

Z h 0

Z h 0

2

X d 4 12 m x_ cl + cn _ n n "

d m(x_ cl +

X

!2

+ U (xcl +

X

n

0

n

n

0

= mx_ cl j j0 h + =0

Z h 0

" @ 2 S = Z h m _ _ + U j k @cj @ck 0

d ;mx cl + U (xcl )] j 0

xcl +

00

4

cn n )5 !

X cn _ n ) _ j + U xcl + cn n j

Z h h i @S

_ = d m x _

+ U ( x )

cl j cl j @cj c =0 0 f g

3

X

n

!

cn n ] j k

#

#



Z h h i @ 2 S

_ j _ k + U (xcl ) j k = d m



@cj @ck c =0 0 00

f g

= =

Z h 0

Z h 0

h

d ;m j

k + U (xcl ( )) j k 00



i 

2 d j ( ) ;m dd 2 + U (xcl ( )) k ( ) 00

where the fourth and eighth lines are obtained from an integration by parts. Let us write the integral in the last line in the suggestive form:

@ 2 S

@ 2 + U (x ( ))j i =  = h

j ; m j cl k jk

@cj @ck c =0 @ 2 00

f g

which emphasizes the fact that we have matrix elements of the operator ;md2=d 2 + U (xcl ( )) with respect to the basis functions. Thus, the expansion for S can be written as X S x] = S xcl ] + 12 cj jk ck +    jk 00

and the density matrix becomes


(x x   ) = N 0

Z Y

j

pdcj e 2 h

Scl (xx 

;

0

)

e

;

1 2

P

 j
k cj jk ck =h

where Scl(x x   ) = S xcl ]. N is an overall normalization constant. The integral over the coecients becomes a p generalized Gaussian integral, which brings down a factor of 1= det:
(x x   ) = N e Scl (xx  ) p 1 det 1 S (xx  ) q cl = Ne  ; det ;m dd22 + U (xcl ( )) 0

0

0

;

0

;

00

where the last line is the abstract representation of the determinant. The determinant is called the Van Vleck-Paulideterminant. If we choose the basis functions n ( ) to be eigenfunctions of the operator appearing in the above expression, so that they satisfy Morette





;m dd 2 + U (xcl ( )) n ( ) = n n ( ) 2

00

Then, jk = j jk = j (x x   ) jk and the determinant can be expressed as a product of the eigenvalues. Thus, Y
(x x   ) = N e Scl (xx  ) p 1 j (x x   ) j 0

0

0

;

0

The product must exclude any 0-eigenvalues. Incidentally, by performing a Wick rotation back to real time according to  = ;it= h, the saddle point or stationary phase approximation to the real-time propagator can be derived. The derivation is somewhat tedious and will not be given in detail here, but the result is 1 i=2 U (x x  t) = e hi Scl (xx t) q ; e 2 d det ;m dt2 ; U (xcl (t)) 0

0

;

00

5

where xcl (t) satis es





mx cl = ; @U @x x=xcl xcl (0) = x xcl (t) = x

0

and  is an integer that increases by 1 each time the determinant vanishes along the classical path.  is called the Maslov index. It is important to note that because the classical paths satisfy an endpoint problem, rather than an initial value problem, there can be more than one solution. In this case, one must sum the result over classical paths: X 1 U (x x  t) = e hi Scl (xx t) i=2 q ;  2 d det ;m dt2 ; U (xcl (t)) classical paths 0

0

;

00

with a similar sum for the density matrix.

6

G25.2651: Statistical Mechanics Notes for Lecture 17

I. EXPECTATION VALUES OF OBSERVABLES Recall the basic formula for the expectation value of an observable A: A = Q(1 ) Tr(Ae; H ) Two important cases pertaining to the evaluation of the trace in the coordinate basis for expectation values will be considered below: h

i

A. Case 1: Functions only of position If A = A(X ), i.e., a function of the operator X only, then the trace can be easily evaluated in the coordinate basis: Z 1 A = dx x A(X )e; H x h

Q

i

h

j

j

Since A(X ) acts to the left on one of its eigenstates, we have Z A = 1 dxA(x) x e; h

Q

i

h

j

Hj

i

x

i

which only involves a diagonal element of the density matrix. This can, therefore, be written as a path integral:

" P

# X 1 2 mP P=2 Z dx dx A(x ) exp 1 2 1 P 1 2 m!P (xi+1 xi ) + P U (xi ) 2
h2 i=1 However, since all points x1  :: xP are equivalent, due to the fact that they are all integrated over, we can make P equivalent cyclic renaming of the coordinates x1 x2 , x2 x3 , etc. and generate P equivalent integrals. In each, the function A(x1 ) or A(x2 ), etc. will appear. If we sum these P equivalent integrals and divide by P , we get an h

A = Q1 Plim !1 i

;

!

expression:

A = Q1 Plim !1

h

i

mP 2
h2


P=2 Z

dx1

;

!

P X 1 dxP P A(xi ) exp

" ;

i=1


P X 1 m!2 (x 1 2 2 P i+1 xi ) + P U (xi )

#

;

i=1

This allows us to de ne an estimator for the observable A. Recall that an estimator is a function of the P variables x1  ::: xP whose average over the ensemble yields the expectation value of A: P X 1 aP (x1  ::: xP ) = P A(xi ) i=1 Then

A = Plim a !1 p x1 :::xP

h

i

h

i

where the average on the right is taken over many con gurations of the P variables x1  ::: xP (we will discuss, in the nex lecture, a way to generate these con gurations). The limit P can be taken in the same way that we did in the previous lecture, yielding a functional integral expression for the expectation value: ! 1

h

I A = Q1 i

"

D

#

Z h x( ) 1h dA(x( )) exp 0 1


#

"

;

1 Z h d 1 mx_ 2 + U (x( )) h 0 2

B. Case 1: Functions only of momentum Suppose that A = A(P ), i.e., a function of the momentum operator. Then, the trace can still be evaluated in the coordinate basis: Z A = 1 dx x A(P )e; H x h

i

Q

h

j

j

i

R

However, A(P ) acting to the left does not act on an eigenvector. Let us insert a coordinate space identity I = dx x x between A and exp( H ): Z 1 dxdx0 x A(P ) x0 x e; H x A = j

ih

j

;

h

Q

i

h

j

j

ih

j

j

i

Now, we see that the expectation value can be obtained by evaluating all the coordinate space matrix elements of the operator and all the coordinate space matrix elements of the density matrix. A particularly useful form for the expectation value can be obtained if a momentum space identity is inserted: Z A = 1 dxdx0 dp x A(P ) p p x0 x0 e; H x h

i

Q

h

j

j ih j

ih

j

j

i

Now, we see that A(P ) acts on an eigenstate (at the price of introducing another integral). Thus, we have Z Z A = 1 dpA(p) dxdx0 x p p x0 x0 e; H x

Q Using the fact that x p = (1=2
h) exp(ipx=h), we nd that Z Z 1 A = 2
hQ dpA(p) dxdx0 eip(x;x )=h x0 e; h

h

i

h

j ih j

ih

j

j

i

j i

x

Hj

0

h

i

h

j

i

In the above expression, we introduce the change of variables 0 r = x +2 x s = x x0 Then Z Z A = 2
1h Q dpA(p) dr dseips=h r 2s e; H r + 2s De ne a distribution function Z W (r p) = 2
1h dseips=h r 2s e; H r + 2s Then, the expectation value can be written as Z 1 A = drdpA(p) (r p) ;

h

i

h

h

h

i

Q

;

;

j

j

j

j

i

i

W

which looks just like a classical phase space average using the \phase space" distribution function W (r p). The distribution function W (r p) is known as the Wigner density matrix and it has many interesting features. For one thing, its classical limit is

W (r p) = exp



;

p2 + U (r) 2m




which is the true classical phase space distribution function. There are various examples, in which the exact Wigner distribution function is the classical phase space distribution function, in particularly for quadratic Hamiltonians. Despite its compelling appearance, the evaluation of expectation values of functions of momentum are considerably more dicult than functions of position, due to the fact that the entire density matrix is required. However, there are a few quantities of interest, that are functions of momentum, that can be evaluated without resorting to the entire density matrix. These are thermodynamic quantities which will be discussed in the next section. 2

II. THERMODYNAMICS FROM PATH INTEGRALS Although general functions of momentum are dicult (though not intractable) to evaluate by path integration, certain functions of momentum (and position) can be evaluated straightforwardly. These are thermodynamic quantities such as the energy and pressure, given respectively by

E = @@ ln Q(  V ) @ ln Q(  V ) P = 1 @V ;

We shall derive estimators for these two quantities directly from the path integral expression for the partition function. However, let us work with the partition function for an ensemble of 1-particle systems in three dimensions, which is given by

mP Q(  V ) = Plim !1 2
h2


3P=2 Z

dr1

drP exp

"

;

P X 1 m!2 (r 2 P i+1

;

i=1


#

1 ri ) + U (ri ) P 2

Using the above thermodynamic relation, the energy becomes E = 1 @Q ;

Q@

" P
X 1 2 mP 3P=2 Z dr dr exp 1 P 2 2 m!P (ri+1 2
h i=1 # P P X X 1 m!2 (r 1 2 2 P i+1 ri ) + P U (ri )

= Q1 Plim !1

"



3P 2

;

;

;


#

1 ri ) + U (ri ) P 2

;

i=1

i=1

= Plim " (r  ::: rP ) !1 P 1 h

i

where

"P (r1  ::: rP ) = 32P

;

P X 1 i=1

2 2 m!P (ri+1

;

ri )2

P X + P1 U (ri ) i=1

is the thermodynamic estimator for the total energy. Similarly, an estimator for the internal pressure can be derived using P = kT@ ln Q=@V . As we have done in the past for classical systems, the volume dependence can be made explicity by introducing the change of variables: rk

= V 1=3 sk

In terms of the scaled variables sk , the partition function expression reads:

mP Q(  V ) = Plim !1 2
h2


3P=2

VP

Z

ds1

dsP exp

"

;

P X 1 2 2=3 2 m!P V (si+1 i=1


#

;

1 si )2 + U (V 1=3 si ) P

Evaluating the derivative with respect to volume gives the internal pressure: P = 1 @Q

Q @V

= Q1 Plim !1

"



P V

;


3P=2

"

Z

P X 1 m!2 V 2=3 (s V ds1 dsP exp i+1 P 2 i=1 # P P X @U 1 m!2 V ;1=3 X 1 1 2 ; 2=3 (si+1 si ) P 3 P @ (V 1=3 s ) 3 V si

mP 2
h2

P

;

;

i=1

;

i=1

3

i

;


#

1 si ) + U (V 1=3 si ) P 2

= Q1 Plim !1

"



P V

;

"


3P=2 Z

P X 1 m!2 (r dr1 drP exp 2 P i+1 i=1 # P P X 1 m!2 X 1 @U 2 3V P i=1 (ri+1 ri ) 3V P i=1 @ ri si

mP 2
h2

;

;

;


#

1 ri ) + U (ri ) P 2

;

= Plim p (r  ::: rP ) !1 P 1 h

i

where

pP (r1  ::: rP ) = PV

;

P  1 X 2 3V i=1 m!P (ri+1

;

1 @U ri ) + ri P @r 2



i

is the thermodynamic estimator for the pressure. Clearly, both the energy and pressure will be functions of the particle momenta, however, because they are related to the partition function by thermodynamic dierentiation, estimators can be derived for them that do not require the o-diagonal elements of the density matrix.

III. PATH INTEGRAL MOLECULAR DYNAMICS (OPTIONAL READING) Consider once again the path integral expression for the one-dimensional canonical partition function (for a nite but large value of P ):

Q( ) = mP 2 2
h


P=2 Z

dx1

dxP exp

"

;


# P X 1 1 m!2 (x 2 2 P i+1 xi ) + P U (xi )

(1)

;

i=1

(the condition xP +1 = x1 is understood). Recall that, according to the classical isomorphism, the path integral expression for the canonical partition function is isomorphic to the classical con guration integral for a certain P particle system. We can carry this analogy one step further by introducing into the above expression a set of P momentum integrations:

Z

Q( ) = dp1

dpP dx1

dxP exp

"

;


P X p2i + 1 m!2 (x 1 2 2m0 2 P i+1 xi ) + P U (xi ) i=1 ;

#

(2)

Note that these momentum integrations are completely uncoupled from the position integrations, and if we were to carry out these momentum integrations, we would reproduce Eq. (1) apart from trivial constants. Written in the form Eq. (2), however, the path integral looks exactly like a phase space integral for a P -particle system. We know from our work in classical statistical mechanics that dynamical equations of motion can be constructed that will generate this partition function. In principle, one would start with the classical Hamiltonian

 P  2 X 1 1 p i 2 2 H= 2m0 + 2 m!P (xi+1 xi ) + P U (xi ) i=1 ;

derive the corresponding classical equations of motion and then couple in thermostats. Such an approach has certainly been attempted with only limited success. The diculty with this straightforward approach is that the more \quantum" a system is, the large the paramester P must be chosen in order to converge the path integral. However, if P is large, the above Hamiltonian describes a system with extremely sti nearest-neighbor harmonic bonds interacting with a very weak potential U=P . It is, therefore, almost impossible for the system to deviate far harmonic oscillator solutions and explore the entire available phase space. The use of thermostats can help this problem, however, it is also exacerbated by the fact that all the harmonic interactions are coupled, leading to a wide variety of time scales associated with the motion of each variable in the Hamiltonian. In order to separate out all these time scales, one must somehow diagonalize this harmonic interaction. One way to do this is to use normal mode variables, and this is 4

a perfectly valid approach. However, we will explore another, simpler approach here. It involves the use of a variable transformation of the formed used in previous lectures to do the path integral for the free-particle density matrix. Consider a change of variables: u1 = x1 uk = xk x~k k = 2 ::: P where x~ = (k 1)xk+1 + x1 ;

;

i

k

The inverse of this transformation can be worked out in closed form: x1 = u1 P X 1) u xk = u1 + ((kl 1) l l=k ;

;

and can also be expressed as a recursive inverse: x1 = u1 xk = uk + k k 1 xk+1 + k1 x1 ;

The term k = P here can be used to start the recursion. We have already seen that this transformation diagonalized the harmonic interaction. Thus, substituting the transformation into the path integral gives:

Z

Q( ) = dp1

dpP du1

duP exp

"

;


# P X p2i + 1 m !2 u2 + 1 U (x (u  ::: u )) P 2m0i 2 i P i P i 1 i=1

The parameters mi are given by

m1 = 0 mi = i i 1 m ;

Note also that the momentum integrations have been changed slightly to involve a set of parameters m0i . Introducing these parameters, again, only changes the partition function by trivial constant factors. How these should be chosen will become clear later in the discussion. The notation xi (u1  ::: uP ) indicates that each variable xi is a generally a function of all the new variables u1  ::: uP . A dynamics scheme can now be derived using as an eective Hamiltonian:

 P  2 X 1 1 p i 2 2 H= 2m0i + 2 mi !P ui + P U (xi (u1  ::: uP )) i=1

which, when coupled to thermostats, yields a set of equations of motion m0 u = m !2 u 1 @U _ u_ i i

;

i P i

Q i = mi u_ 2i 1

;

P @ui

;

i i

;

(3)

These equations have a conserved energy (which is not a Hamiltonian):

H0 =

 P  X 1 m0 u_ 2 + 1 m !2 u2 + 1 U (x (u  ::: u )) + 1 Q _ 2 + 1

i i 1 P i 2 i i 2 P i P 2 i i=1

Notice that each variable is given its own thermostat. This is done to produce maximum ergodicity in the trajectories. In fact, in practice, the chain thermostats you have used in the computer labs are employed. Notice also that the 5

time scale of each variable is now clear. It is just determined by the parameters mi . Since the object of using such dynamical equations is not to produce real dynamics but to sample the phase space, we would really like each variable to move on the same time scale, so that there are no slow beads trailing behind the fast ones. This eect can be produced by choosing each parameter m0i to be proportional to mi : m0i = cmi . Finally, the forces on the u variables can be determined easily from the chain rule and the recursive inverse given above. The result is f

g

P 1 @U = 1 X @U P @u1 P i=1 @xi   1 @U = 1 (k 2) @U + @U P @ui P (k 1) @uk;1 @xk ; ;

where the rst (i = 1) of these expressions starts the recursion in the second equation. Later on, when we discuss applications of path integrals, we will see why a formulation such as this for evaluating path integrals is advantageous.

IV. PATH INTEGRALS FOR N -PARTICLE SYSTEMS If particle spin statistics must be treated in a given problem, the formulation of the path integral is more complicated, and we will not treat this subject here. The extension of path integrals to N -particle systems in which spin statistics can safely be ignored, however, is straightforward, and we will give the expressions below. The partition function for an N -particle system in the canonical ensemble without spin statistics can be formulated essentially by analogy to the one-particle case. The partition function that one obtains is

"N Y mI P
3P=2 Z (1) Q(N V T ) = Plim drI !1 I =1 2
h2

drI

(P )

#

exp

"

;

N P X X 1 2 (i+1) 2 mI !P (rI i=1

;

I =1

) + P1 U (r(1i)  ::: r(Ni) )

(i) 2

rI

!#

Thus, it can be seen that the N -particle potential must be evaluated for each imaginary time discretization, however, there is no coupling between separate imaginary time slices due arising from the potential. Thus, interactions occur only between particles in the same time slice. From a computational point of view, this is advantageous, as it allows for easily parallelization over imaginary time slices. The corresponding energy and pressure estimators for the N -particle path integral are given by

P ( pP (

f

r(1)  ::: r(P ) ) = (1)

r

f

 ::: r

(P )

3NP 2

) = NP V

;

P X N X 1 i=1 I =1



(i) (i+1) 2 rI ; rI +

 1 X X m !2 r(i) 3V i=1 I =1 I P I P

;

2 2 mI !P

N

6

;



(i+1) 2

rI

P 1X (i) (i) P U (r1  ::: rN ) i=1

+ P1 r(Ii)

r

(i)

rI

U





       
 
   


   


 
         
           

             
    
     


    

 
   
 
 

      
   

 
   
     
                  
  
       
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     (

   Tc

P

T = Tc T < Tc

Inflection pt.

.

Pc

Linear regime

ρG

ρ

ρ

ρ

L

c

FIG. 2.

The dashed curve corresponds to the gas-liquid coexistence curve. Below the critical isotherm, the gas-liquid coexistence curve describes how large a discontinuous change in the density occurs during rst-order gas-liquid phase transition. At the in ection point, which corresponds to the critical point, the discontinuity goes to 0. As noted above, the divergences in thermodynamic derivative quantities occur in the same way for systems belonging to the same universality class. These divergences behave as power laws, and hence can be described by the exponents in the power laws. These exponents are known as the critical exponents. Thus, the critical exponents will be the same for all systems belonging to the same universality class. The critical exponents are dened as follows for the gas-liquid critical point: 1. The heat capacity at constant volume dened by

CV = @E @T




V

2 = ;T @@TA2




V

diverges with temperature as the critical temperature is approached according to

CV
jT ; Tc j;

2. The isothermal compressibility dened by

T = ; V1 @V @P


T

@ = 1 @T


T

diverges with temperature as the critical temperature is approached according to

T
jT ; Tc j;
2

3. On the critical isotherm, the shape of the curve near the in ection point at ( c
Pc ) is given by P ; Pc
j ; c j sign( ; c ) >0 4. The shape of the coexistence curve in the -T plane near the critical point for T < Tc is given by  L ; G
(Tc ; T ) These are the primary critical exponents.

A. Review of the Van der Waals theory Recall that the Van der Waals equation of state was derived earlier by perturbation theory. The unperturbed Hamiltonian describes a system of hard spheres and is given by

H0 =

N 2 X X pi + u0 (jri ; rj j) i=1 2m i<j

and a perturbation of the form

X

H1 = where u0 (r) is the hard-sphere potential given by

i<j

u1 (jri ; rj j)



r> r

0 u 0 (r ) = 1

and u1 (r) is an arbitrary attractive potential. In the low density limit, we had

g(r)  e;u0 (r) = (r ; ) and the free energy was determined to be

  )N ; aN 2 A(N
V
T )  ; 1 ln (VN;! Nb 3N V

with

b = 23 3 Z1 a = ;2 dr r2 u1 (r) 

The equation of state takes the form The critical point is dened by the conditions:

P = 1 ;kT b ; a

@P @V = 0

2

@2P = 0 @V 2

which leads to the following values for the critical pressure, temperature and density: 8a 1 Pc = 27ab2 Tc = 27 c= b 3b The critical exponents predicted by the theory are as follows: 3

1. The internal energy is given by

2 E = 32 NkT + aN V

from which it can be seen that the heat capacity is


@E CV = @T = 32 Nk
jT ; Tcj0 V

) = 0.

2. The isothermal compressibility can be expressed as

1 T = ; V (@P=@V )

and


1 8 a @P  @V V =Vc = 4Nb2 27b ; kT
(T ; Tc )

so that

T
(T ; Tc);1

)=1 3. By Taylor expanding the equation of state about the critical pressure and density, it is easy to show that 1 (P ; P )
const + ( ; )3 +    c

kT

)=3

c

4. The exponent can be computed using the Maxwell construction (see problem set 11), which attempts to x the fact that the Van der Waals equation has an unphysical region where @P=@V > 0. The Maxwell construction is illustrated below:

P

Van der Waals eqn

a2

Tie line

a1

VG

VL FIG. 3.

4

V

When the Maxwell construction is carried out, it can be shown that L

;


(Tc ; T )1=2

G

) = 1=2. The following table compares the Van der Waals exponents to the experimental critical exponents:

α

β

γ

δ

VdW

0

1/2

1

3

Exp.

0.1

0.34

1.35

4.2

FIG. 4.

Thus, one sees that the Van der Waals theory is only qualitatively correct, but not quantitatively. It is an example of a mean eld theory.

II. MAGNETIC SYSTEMS AND THE ISING MODEL Imagine a cubic lattice in which particles carrying spin S are placed on the lattice sites with jSj = h =2 as shown below:

FIG. 5.

5

Such a model describes ferromagnetic materials, which can be magnetized by applying an external magnetic eld . In the absence of a eld, the unperturbed Hamiltonian takes the form X H0 = ; 12 i  Jij  j ij

h

where J is a tensor and  is a spin vector such that jj = 1. Quantum mechanically,  would be the vector of Pauli matrices. In general, the spins can point in any spatial direction, a fact that makes the problem dicult to solve. A simplication introduced by Ising was to allow the spins to point in only one of two possible directions, up or down, e.g., along the z -axis only. In addition, the summation is restricted to nearest neighbor interactions only. In this model, the Hamiltonian becomes X H0 = ; 12 Jij i j hiji where hi
j iindicates restriction of the sum to nearest neighbor pairs only. The variables i now can take on the values 1 only. The couplings Jij are the spin-spin \J " couplings. In the presence of a magnetic eld, the full Hamiltonian becomes

H = ; 12

X hiji

Jij i j ; h

N X i=1

i

which describes a uniaxial ferromagnetic system in a magnetic eld. The parameters T and h are experimental control parameters. Dene the magnetization per spin as N X m = N1 i i=1

Then the phase diagram looks like:

h

Spin up phase

T Spin down phase

Tc

FIG. 6.

6

where the colored lines indicate a nonzero magnetization at h = 0 below a critical temperature T . The persistence of a nonzero magnetization in ferromagnetic systems at h = 0 below T = Tc indicates a transition from a disordered to an ordered phase. In the latter, the spins are aligned in the direction of the applied eld before it is switched o. If h ;! 0+ , then the spins will point in one direction and if h ;! 0; , it will be in the opposite direction. A plot of the isotherms of m vs. h yields:

m T < TC T = TC T>TC h

FIG. 7.

Notice an in ection point along the isotherm T = Tc, at h = 0, where @m=@h ;! 1. The thermodynamics of the magnetic system can be dened in analogy with the liquid-gas system. The analogy is shown in the table below: Gas-Liquid

Magnetic

P

h

V

;M = ;Nm

T = ; V1 @V @P

 = @m @h

A = A(N
V
T )

A = A(N
M
T )

G = A + PV (P )

G = A ; hM (h)

@A P = ; @V

@A h = @M

V = @G @P
2 CV = ;T @@TA2 V

M = ; @G @h 2 CM = ;T @@TA2

7


M

2 CP = ;T @@TG2




2 Ch = ;T @@TG2

P

where  is the magnetic susceptibility. The magnetic exponents are then given by

Ch
jT ; Tcj; h = 0 limit ;

jT ; Tcj h = 0 limit h
jmj sign(m) at T = Tc m
(Tc ; T ) T < Tc

8


h

G25.2651: Statistical Mechanics Notes for Lecture 26

I. MEAN FIELD THEORY CALCULATION OF MAGNETIC EXPONENTS The calculation of critical exponents is nontrivial, even for simple models such as the Ising model. Here, we will introduce an approximate technique known as mean eld theory. The approximation that is made in the mean eld theory (MFT) is that uctuations can be neglected. Clearly, this is a severe approximation, the consequences of which we will see in the nal results. Consider the Hamiltonian for the Ising model: X X H = 21 Jij i j + h i i hi ji The partition function is given by (N
h
T ) =

X

XX


1
2




e



h P 1 2

hi j i



+h i

j

P i




i

i

N

Notice that we have written the partition function as an isothermal-isomagnetic partition function in analogy with the isothermal-isobaric ensemble. (Most books use the notation Q for the partition function and A for the free energy, which is misleading). This sum is nontrivial to carry out. In the MFT approximation, one introduces the magnetization N X m = N1 i i=1 h

i

explicitly into the partition function by using the identity i j = ( i m + m)( j m + m) = m2 + m( i m) + m( j m) + ( i m)( j m) The last term is quadratic in the spins and is of the form ( i )( j ), the average of which measures the spin uctuations. Thus, this term is neglected in the MFT. If this term is dropped, then the spin-spin interaction term in the Hamiltonian becomes: 1 XJ 1 XJ m2 + m( + ) 2m2
i j ij i j 2 hi j i 2 hi ji ij X
= 12 Jij m2 + m( i + j ) hi j i ;

;

;

;

;h

;

i

;h



;

i

;

;

P

We will restrict ourselves to isotropic magnetic systems, for which j Jij is independent of i (all sites are equivalent). P ~ , where z is the number of nearest neighbors of each spin. This number will depend on the number Dene j Jij Jz of spatial dimensions. Since this dependence on spatial dimension is a trivial one, we can absorb the z factor into the ~ . Then, coupling constant and redene J = Jz 1 X J = 1 NJ 2 i 2 



where N is the total number of spins. Finally, 1 X J m( + ) = Jm X i 2 hi ji ij i j i 1

and the Hamiltonian now takes the form X 1 XJ + h ij i j i 2 hi ji i

;! ;

1 NJm2 + (Jm + h) X i 2 i

and the partition function becomes (N
h
T ) = e;NJm =2 2

X



= e;NJm

2


1 X

X




e(Jm+h)

N

e(Jm+h)


P

!N


=
1 2 =2 ; NJm =e 2cosh (Jm + h)]N

The free energy per spin g(h
T ) = G(N
h
T )=N is then given by g(h
T ) = 1 ln (N
h
T ) ;

N

= 12 Jm2 The magnetization per spin can be computed from

;

1 ln 2cosh (Jm + h)]







@g m = @h h = tanh (Jm + h) 0, one nds a transcendental equation for m m = tanh(mJ ) ;

Allowing h

;!

which can be solved graphically as shown below:

2

i




i

tanh(βJm)

f(m)=m

− m0 m m0

FIG. 1.

Note that for J > 1, there are three solutions. One is at m = 0 and the other two are at nite values of m, which we will call m0. For J < 1, there is only one solution at m = 0. Thus, for J > 1, MFT predicts a nonzero magnetization at h = 0. The three solutions coalesce onto a single solution at J = 1. The condition J = 1 thus denes a critical temperature below which (J > 1) there is a nite magnetization at h = 0. The condition J = 1 denes the critical temperature, which leads to 

kTc = J To see the physical meaning of the various cases, consider expanding the free energy about m = 0 at zero-eld. The expansion gives

g(0
m) = const + J (1 J )m2 + cm4 where c is a (possibly temperature dependent) constant with c > 0. For J > 1, the sign of the quadratic term is negative and the free energy as a function of m looks like: ;

3

g(0,m)

m m0

−m 0 FIG. 2.

Thus, there are two stable minima at m0 , corresponding to the two possible states of magnetization. Since a large portion of the spins will be aligned below the critical temperature, the magnetic phase is called an ordered phase. For J > 1, the sign of the quadratic term is positive and the free energy plot looks like: 

4

g(0,m)

m FIG. 3.

i.e., a single minimum function at m = 0, indicating no net magnetization above the critical temperature at h = 0. The exponent  can be obtained directly from this expression for the free energy. For T < Tc, the value of the magnetization is given by 

@g  @m m=m0 = 0 which gives

2J (T T )m + 4cm3 = 0 c 0 0 T 1 = m0 (Tc T ) 2 ;



;

Thus,  = 1=2. From the equation for the magnetization at nonzero eld, the exponent  is obtained as follows: m = tanh (Jm + h)  (Jm + h) = tanh;1 m 

3



Jm h kT m + m3 + = mk(T T ) + kT m3 ;



c

3 where the second line is obtained by expanding the inverse hyperbolic tangent about m = 0. At the critical temperature, this becomes ;

h



m3

so that  = 3. For the exponent , we need to compute the heat capacity at zero-eld, which is either Ch or Cm . In either case, we have, for T > Tc, where m = 0, 5

G = NkT ln 2 ;

so





2 Ch = T @@TG2 = 0 from which is it clear that  = 0. For T < Tc, Ch approaches a dierent value as T on T Tc is the same, so that  = 0 is still obtained. ;

j

;

!

Tc , however, the dependence

j

Finally, the susceptibility, which is given by

1 = @m @h = @h=@m but, near m = 0, 3 h = mk(T Tc ) + kT 3m @h 2 @m = k(T Tc) + kTm As the critical temperature is approached, m 0 and T Tc ;1 which implies = 1. ;

;

!

 j

;

j

The MFT exponents for the Ising model are, therefore

=0

 = 1=2

=1

=3

which are exactly the same exponents that the Van der Waals theory predict for the uid system. The fact that two (or more) dissimilar systems have the same set of critical exponents (at least at the MFT level) is a consequence of a more general phenomenon known as universality, which was alluded to in the previous lecture. Systems belonging to the same universality class will exhibit the same behavior about their critical points, as manifested by their having the same set of critical exponents. A universality class is characterized by two parameters: 1. The spatial dimension d. 2. The dimension, n, of the order parameter. An order parameter is dened as follows: Suppose the Hamiltonian H0 of a system is invariant under all the transformations of a group . If two phases can be distinguished by the appearance of a thermodynamic average  , which is not invariant under , then  is an order parameter for the system. G

h

h

i

G

i

The Ising system, for which H0 is given by

X H0 = 21 Jij i j ;

hi ji

is invariant under the group Z2 , which is the group that contains only two elements, an identity element and a spin reection transformation: Z2 = 1
1. Thus, under Z2 , the spins transform as ;

i

From the form of H0 is can be seen that H0 However, the magnetization

!

!

i

i

! ;

i

H0 under both transformations of Z2 , so that it is invariant under Z2 . X = 1N i m = N1 i h

6

i

is not invariant under a spin reection for T < Tc, when the system is magnetized. In a completely ordered state, with all spins aligned, under a spin reection m m. Thus, m is an order parameter for the Ising model, and, since it is a scalar quantity, its dimension is 1. Thus, the Ising model denes a universality class known as the Ising universality class, characterized by d = 3, n = 1 in three dimensions. Note that the uid system, which has the same MFT critical exponents as the Ising system, belongs to the same universality class. The order parameter for this system, by the analogy table dened in the last lecture, is the volume dierence between the gas an liquid phases, VL VG , or equivalently, the density dierence, L G . Although the solid phase is the truly ordered phase, while the gas phase is disordered, the liquid phase is somewhere in between, i.e., it is a partially ordered phase. The Hamiltonian of a uid is invariant under rotations of the coordinate system. Ordered and partially ordered phases break this symmetry. Note also that a true magnetic system, in which the spins can point in any spatial direction, need an order parameter that is the vector generalization of the magnetization: ! ;

;

;

N X 1 m= N i h

i=1

i

Since the dimension of the vector magnetization is 3, the true magnetic system belongs to the d = 3, n = 3 universality class.

II. EXACT SOLUTIONS OF THE ISING MODEL IN 1 AND 2 DIMENSIONS Exact solutions of the Ising model are possible in 1 and 2 dimensions and can be used to calculate the exact critical exponents for the two corresponding universality classes. In one dimension, the Ising Hamiltonian becomes:

H=

N X

;

i=1

Ji i+1 i i+1 h ;

N X i=1

i

which corresponds to N spins on a line. We will impose periodic boundary conditions on the spins so that N +1 = 1 . Thus, the topology of the spin space is that of a circle. Finally, let all sites be equivalent, so that Ji i+1 J . Then, 

N X

H= J ;

The partition function is then

i=1

(N
h
T ) =

X

X

i i+1 h ;




i=1

i


P P 1 e J

+1 + 2 h =1 (
+
+1 ) i


1

N X

i

N

i

i

i

i

N

In order to carry out the spin sum, let us dene a matrix P with matrix elements: P 0 = e J

+h(
+
)=2] 1 P 1 = e(J +h) 1 P 1 = e(J ;h) 1 P 1 = 1 P 1 = e;J Thus, the matrix P becomes is a 2 2 matrix given by 0

h

j

h j

j

0

i

j i

h; j

j;

i

h j

j;

i

h; j

j i







1 jPj 2 ih 2 jPj 3 i

h

 (J +h) ;J P = e e;J ee(J ;h)

so that the partition function becomes X (N
h
T ) = =

X


1




X


1

h

N

h

1 jPN j 1 i

;

= Tr PN

7

N ;1 jPj N ih N jPj 1 i

A simple way to carry out the trace is diagonalize the matrix, P. From det(P I) = 0 ;

the eigenvalues can be seen to be 

= eJ cosh(h) 

q

sinh2 (h) + e;4J

 




where + corresponds to the choice of + in the eigenvalue expression, etc. The trace of the PN is then ;

Tr PN = N+ + N;

We will be interested in the thermodynamic limit. Note that + > ; for any h, so that as N over N; . Thus, in this limit, the partition function has the single term: (N
h
T )

;!

! 1

, N+ dominates

N+

Thus, the free energy per spin becomes

g(h
T ) = kT ln +   q = J kT ln cosh(h) + sinh2 (h) + e;4J ;

;

;

and the magnetization becomes 



@g m = @h = @ (@h) ln + h)cosh(h) sinh(h) + sinh( 2 (h)+e 4 sinh q = cosh(h) + sinh2 (h) + e;4J ;

p

;
J

which, as h 0, since cosh(h) 1 and sinh(h) 0, itself vanishes. Thus, there is no magnetization at any nite temperature in one dimension, hence no nontrivial critical point. While the one-dimensional Ising model is a relatively simple problem to solve, the two-dimensional Ising model is highly nontrivial. It was only the pure mathematical genius of Lars Onsager that was able to nd an analytical solution to the two-dimensional Ising model. This, then, gives an exact set of critical exponents for the d = 2, n = 1 universality class. To date, the three-dimensional Ising model remains unsolved. Here, the Onsager solution will be outlined only and the results stated. Consider a two-dimension spin-lattice as shown below: !

!

!

8

n+1 . . . .

2

1

. . . . . . . . . . . . . . . . . . . . . . . . 1

2

......

. n+1

FIG. 4.

The Hamiltonian can be written as X X H = J ( i j i+1 j + i j+1 i j ) h ij ;

;

ij

ij

where the spins are now indexed by two indices corresponding to a point on the 2-dimensional lattice. Introduce a shorthand notation for H :

H=

n X j =1

E (j
j+1 ) + E (j )]

where

n X

E (j
k )

 ;

E (j )

 ;

i=1 n X

J

ij ik

i=1

ij i+1 j ; h

and j is dened to be a set of spins in a particular column:

j

 f

1j
:::
nj g

Then, dene a transfer matrix P, with matrix elements: 9

X

ij

j

j P k = e; E(  )+E( )]

h

j

j

j

i

j

k

which is a 2n 2n matrix. The partition function will be given by 

 = Tr (Pn ) and, like, in the one-dimensional case, the largest eigenvalue of P is sought. This is the nontrivial problem that is worked out in 20 pages in Huang's book. In the thermodynamic limit, the nal result at zero eld is:   Z  q kT 1 2 2 g(T ) = kT ln 2cosh(2J )] 2 d ln 2 1 + 1 K sin  0 where 2 K = cosh(2J )coth(2 J ) The energy per spin is ;

;

;

Z  sin2  "(T ) = 2J tanh(2J ) + 2K dK d d 0 (1 + ) ;

where

q

 = 1 K 2 sin2  ;

The magnetization, then, becomes n

m = 1 sinh(2J )];4 ;

o1=8

for T < Tc and 0 for T > Tc , indicating the presence of an order-disorder phase transition at zero eld. The condition for determining the critical temperature at which this phase transition occurs turns out to be 2tanh2 (2J ) = 1 kTc 2:269185J 

Near T = Tc, the heat capacity per spin is given by        C (t) = 2 2J 2 ln 1 T + ln kTc k  kTc Tc 2J Thus, the heat capacity can be seen to diverge logarithmically as T Tc. ;

;

The critical exponents computed from the Onsager solution are

!

 = 0 (log divergence)  = 18

= 74  = 15 which are a set of exact exponents for the d = 2, n = 1 universality class.

10

1 + 4

 ;



G25.2651: Statistical Mechanics Notes for Lecture 27

I. THE EXPONENTS AND
Consider a spin-spin correlation function at zero
eld of the form 1 X X e;
H i j = i j h

Q

i

N

1

If i and j occupy lattice sites at positions ri and rj , respectively, then at large spatial separation, with r = the correlation function depends only r and decays exponentially according to

G(r)

i j i ; h i ih j i

 h



ri rj ,

j

;

j

e;r= rd;2+

for T < Tc. The quantity  is called the correlation length. Since, as a critical point is approached from above, long range order sets in, we expect  to diverge as T Tc+ . The divergence is characterized by an exponent  such that !



T Tc ;

 j

;

j

At T = Tc, the exponential dependence of G(r) becomes 1, and G(r) decays in a manner expected for a system with long range order, i.e., as some small inverse power of r. The exponent  appearing in the expression for G(r) characterizes this decay at T = Tc. The exponents,  and  cannot be determined from MFT, as MFT neglects all correlations. In order to calculate these exponents, a theory is needed that restores uctuations at some level. One such theory is the so called LandauGinzberg theory. Although we will not discuss this theory in great detail, it is worth giving a brief introduction to it.

II. INTRODUCTION TO LANDAU-GINZBERG THEORY The Landau-Ginzberg (LG) theory is a phenomenological theory meant to be used only near the critical point. Thus, it is formulated as a macroscopic theory. The basic idea of LG theory is to introduce a spin density
eld variable S (x) de
ned by

S (x) =

N X

i

i=1

Then the total magnetization is given by

M=

Z

(x

;

xi )

dd xS (x)

Since the free energy at zero
eld is A = A(N M T ), there should be a corresponding free energy density A(T S (x)) =

a(x) such that

A(N M T ) =

Z

dd xa(x)

It is assumed that a(x) can be represented as a power series according to

a(x) = a0 + a1 (T )S 2(x) + a2 (T )S 4 (x) + 1

such that a1 and a2 are both positive for T > Tc and a1 vanishes at T = Tc. These conditions are analogous to those exhibited by the total free energy in MFT above and near the critical point (see previous lecture). By symmetry, all odd terms vanish and are, therefore, not explicitly included. In the presence of a magnetic
eld h(x), we have a Gibbs free energy density given by

g(x) = a(x) h(x)S (x) = a0 + a1 (T )S 2(x) + a2 (T )S 4(x) h(x)S (x) In addition, a term a3 ( S )2 with a3 > 0 is added to the free energy in order to damp out local uctuations. If there are signi
cant local uctuations, then ( S )2 becomes large, so these
eld con
gurations contribute negligibly to the ;

;

r

partition function, which is de
ned by

Z= =

Z

D

Z D

S ] exp S ] exp

Z

r

 d xg(x)

;

Z

d




d ;

 d x a0 + a1 (T )S (x) + a2 (T )S (x) + a3 ( S )

;

2

4

r

2

;



h(x)S (x)




Thus, the LG theory is a
eld theory. The form of this
eld theory is well known in quantum
eld theory, and is known as a d-dimensional scalar Klein-Gordon theory. In terms of Z , a correlation function at zero
eld can be de
ned by 

2 ln Z   2 h(x)h(x0 ) h=0 Thus, by studying the behavior of the correlation function, the exponents  and  can be determined. For the choice a2 = 0, the theory can be solved exactly analytically. This is known as the Gaussian model, for 1 0 hS (x)S (x )i =

which

Z=

Z D

S ] exp

Z

d ;

 d x a0 + a1 (T )S (x) + a3 ( S )

;

2

r

2

;

h(x)S (x)






which leads to values of  and  of 0 and 1/2, respectively. These values are known as the \classical" exponents. They are independent of the number of spatial dimensions d. Dependence on d comes in at higher orders in S . Comparing these to the exact exponents from the Onsager solution, which gives  = 1=4 and  = 1, it can be seen that the classical exponents are only qualitatively correct. Going to higher orders in the theory leads to improved results, although the theory cannot be solved exactly analytically. It can either be solved numerically using path integral Monte Carlo or Molecular Dynamics or analytically perturbatively using Feynman diagrams.

III. RENORMALIZATION GROUP AND THE SCALING HYPOTHESIS A. General formulation The renormalization group (RG) has little to do with \group theory" as it is meant mathematically. Also, there is no uniqueness to the renormalization group, so the use of \the" in this context is misleading. Rather, the RG is an idea that exploits the physics of systems near their critical point which leads to a procedure for
nding the critical point. It also oers an explanation of universality, perhaps the closest thing there is to a proof of this concept. Finally, through the scaling hypothesis, it generates relations, called scaling relations satis
ed by the critical exponents. It does not allow actual determination of speci
c exponents. However, given a numerical calculation or some other method of determining a small subset of exponents, the scaling relations can be used to determine the remaining exponents. In order to see how the RG works, we will consider a speci
c example. Consider a square spin lattice:

2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

FIG. 1.

which has been separated into 3 3 blocks as shown. Consider de
ning a new spin lattice from the old by application of a coarse-graining procedure in which each 3 3 block is replaced by a single spin. The new spin is an up spin if the majority of spins in the block is up and down is the majority point down. The new lattice is shown below: 



3

.

.

.

. FIG. 2.

Such a transformation is called a block spin transformation. Near a critical point, the system will exhibit long-range ordering, hence the coarse-graining procedure should yield a new spin lattice that is statistically equivalent to the old spin lattice. If this is so, then the spin lattice is said to possess scale invariance. What will be the Hamiltonian for the new spin lattice? To answer this, consider the partition function at h = 0 for the old spin lattice using the Ising Hamiltonian as the starting point:

Q=

X X 1

N

e;
H0 (

1 :::

N)



Tr e;
H0 (

1 :::

N)

The block spin transformation can be expressed by de
ning, for each block, a transformation function: 

0 P9

i=1 i > 0 0 otherwise When inserted into the expression for the partition function, T acts to project out those con
gurations that are consistent with the block spin transformation, leaving a function of only the new spin variables 10  ::: N0 , in terms of which a new partition function can be de
ned. To see how this works, let the new Hamiltonian be de
ned through:

T ( 0  1  ::: 9 ) = 1

f

e;
H0 (f g) = Tr 0

"

Y

0

blocks

0g

#

T ( 0  1  ::: 9 ) e;
H0 (f g)

That this is a consistent de
nition follows from the fact that X

T ( 0  1  ::: 9 ) = 1

0

Thus, tracing both sides of the projected partition function expression over 0 yields: Tr e;
H0 (f g) = Tr e;
H0 (f g) which states that the partition function is preserved by the block spin transformation, hence the physical properties are also preserved. Transformations of this type should be chosen so as to preserve the functional form of the Hamiltonian, for if this is done, then the transformation can be iterated in exactly the same way for each new lattice produced by the previous iteration. The importance of being able to iterate the procedure is that, in a truly ordered state formed at a critical point, the iteration transformation will produce exactly the same lattice as the previous iteration, thus signifying the existence of a critical point. If the functional form of the Hamiltonian is preserved, then only its parameters are aected by the transformation, so we can think of the transformation as acting on these parameters. If the original Hamiltonian contains parameters K1  K2  ::: K (e.g., the J coupling in Ising model), then the transformation yields a Hamiltonian with a new set of parameters K0 = (K10  K20  :::), such that the new parameters are functions of the old parameters 0

0

0



4

K0 = R(K)

The vector function R characterizes the transformation. These equations are called the renormalization group equations or renormalization group transformations. By iterating the RG equations, it is possible to determine if a system has an ordered phase or not and for what values of the parameters such a phase will occur.

B. Example: The one-dimensional Ising model For the one-dimensional Ising model: N X

H0 = J ;

i=1

i i+1

De
ne: = H K = J

H0

so that H0

= K

N X

;

i=1

i i+1

and the partition function becomes

Q = Tr e;H0 We will consider a simple block spin transformation as illustrated below:

σ1

. σ1

σ2

. σ2

3

2

1

σ3

. σ3

σ1

σ2

.

. σ5

σ4

σ3

. σ6

σ1

. σ7

σ2

. σ8

FIG. 3.

The
gure shows the one-dimension spin lattice numbered in two dierent ways { one a straight numbering and one using blocks of three spins, with spins in each block numbered 1-3. The block spin transformation to be employed here is that the spin of a block will be determined by the value of the spin in the center of the block. Thus, for block 1, it is the value of spin 2, for block 2, it is the value of spin 5, etc. This rather undemocratic choice should be reasonable at low temperature, where local ordering is expected, and spins close to the center spin would be expected to be aligned with it, anyway. The transformation function, T for this case is

T ( 0 1 2 3) = 

The new lattice will look like: 5

0

2

.

.

. σ 3’

σ2 ’

σ’ 1

FIG. 4.

with 10 = 2 , 20 = 5 , etc. The new Hamiltonian is computed from

XXX e;cHz (f g) = 0

X;

0

1

=

2

3

XXXX 1

3

4

N



eK

0 1 2



0

1 1

 K e

0 2 5

eK

0

1 3

eK

1 2

3 4

eK

eK

2 3

eK

3 4

eK

4 5

0

4 2

6

The idea is then to
nd a K 0 such that when the sum over 3 and 4 are performed, the new interaction between 10 and 20 is of the form exp(K 0 10 20 ), which preserves the functional form of the old Hamiltonian. The sum over 3 and 4 is XX K e 3

Note that

3 4

0

1 3

eK

3 4

eK

0

4 2

4

= 1. Then, since 

e = cosh + sinh = cosh 1 + tanh] e; = cosh sinh = cosh 1 tanh] ;

we can express exp(K

3 4

;

) as

eK

3 4

= coshK 1 +

3 4

tanhK ]

Letting x = tanhK , the product of the three exponentials becomes:

eK

0

1 3

eK

3 4

eK

0

4 2

When summed over

= cosh3 K (1 + 10 3 x)(1 + 3 4 x)(1 + 4 20 x) = cosh3 K (1 + 10 3 x + 3 4 x + 4 20 x + 10 32 4 x2 + 10 3

2 0 2 3 4 2

x + 10

2 2 0 3 3 4 2

x)

and 4 , most terms in the above expression will cancel, yielding the following expression:

XX 3

0 x2 +

3 4 2

eK

0

1 3

eK

3 4

eK

0

4 2

 = 2cosh3 K 1 + 10 20 x3



coshK 0 1 + 10 20 x0 ]

4

where the last expression puts the interaction into the original form with a new coupling constant K 0. One of the possible choices for the new coupling constant is tanhK 0 = tanh3 K K 0 = tanh;1 tanh3 K This, then, is the RG equation for this particular block spin transformation. With this identi
cation of K 0, the new Hamiltonian can be shown to be N0

0 0 g) = N 0 g(K ) ; K 0 X 0 0 H0 (f i i+1 i=1

where the spin-independent function g(K ) is given by 6

3
K 2 ln 2 g(K ) = 13 ln cosh coshK 0 3 ;

;

Thus, apart from the additional term, the new Hamiltonian is exactly the same functional form as the original Hamiltonian but with a dierent set of spin variables and a dierent coupling constant. The transformation can now be applied to the new Hamiltonian, yielding the same relation between the new and old coupling constants. This is equivalent to iterating the RG equation. Since the coupling constant K depends on temperature through K = J=kT , the purpose of the iteration would be to
nd out if, for some value of K , there is an ordered phase. In an ordered phase, the transformed lattice would be exactly the same as the old lattice, and hence the same coupling constant and Hamiltonian would result. Such points are called
xed points of the RG equations and are generally given by the condition

K = R(K)

The
xed points correspond to critical points. For the one-dimensional Ising model, the
xed point condition is 

K = tanh;1 tanh3 K



or, in terms of x = tanhK ,

x = x3 Since K is restricted to K 0, the only solutions to this equation are x = 0 and x = 1, which are the
xed points of the RG equation. To see what these solutions mean, consider the RG equation away from the
xed point: 

x0 = x3 Since K = J=kT , at high T , K 0 and x = tanhK RG equation as an iteration or recursion of the form !

!

0+ . At low temperature, K

! 1

and x

!

1;. Viewing the

xn+1 = x3n if we start at x0 = 1, each successive iteration will yield 1. However, for any value of x less than 1, the iteration eventually goes to 0 in some
nite (though perhaps large) number of iterations of the RG equation. This can be illustrated pictorially as shown below:

Stable

Unstable

.

.

x=1

x=0

K=

K=0 FIG. 5.

The iteration of the RG equation produces an RG ow through coupling constant space. The
xed point at x = 1 is called an unstable
xed point because any perturbation away from it, if iterated through the RG equation, ows away from this point to the other
xed point, which is called a stable
xed point. As the stable
xed point is approached, the coupling constant gets smaller and smaller until, at the
xed point, it is 0. The absence of a
xed point for any
nite, nonzero value of temperature tells us that there can be no ordered phase in one dimension, hence no critical point in one dimension. If there were a critical point at a temperature Tc , then, at that point, long range order would set it, and there would be a
xed point of the RG equation at Kc = J=kTc. Note, however, that at T = 0, K = , there is perfect ordering in one dimension. Although this is physically meaningless, it suggests that ordered phases and critical points will be associated with the unstable
xed points of the RG equations. 1

7

Another way to see what the T = 0 unstable
xed point means is to study the correlation length. The correlation is a quantity that has units of length. However, if we choose to measure it in units of the lattice spacing, then the correlation length will be a number that can only depend on the coupling constant K or x = tanhK :

 =  (x) Under an RG transformation, the lattice spacing increases by a factor of 3 as a result of coarse graining. Thus, the correlation length, in units of the lattice spacing, must decrease by a factor of 3 in order for the same physical distance to be maintained:  (x0 ) = 13  (x) In general, for block containing b spins, the correlation length transforms as  (x0 ) = 1  (x)

b

A function satisfying this equation is

 (x )



Since, for arbitrary b, the RG equation is

1 ln x

x0 = xb we have

 (x0 ) =  (xb )

1 ln xb = b ln1 x = 1b  (x) 

so that

1 asT 0 ln tanhK so that at T = 0 the correlation length becomes in
nite, indicating an ordered phase. Note, also, that at very low T , where K is large, motion toward the stable
xed point is initially extremely slow. To see this, rewrite the RG equation as

 (K )



;! 1

;!

tanhK 0 = tanh3 K = tanhK tanh2 K
cosh(2K ) 1 = tanhK cosh(2 K) + 1 ;

Notice that the term in brackets is extremely close to 1 if K is large. To leading order, we can say

K0



K

Since the interactions between blocks are predominantly mediated by interactions between boundary spins, which, for one dimension, involves a single spin pair between blocks, we expect that a block spin transformation in 1 dimension yields a coupling constant of the same order as the original coupling constant when T is low enough that there is alignment between the blocks and the new lattice is similar to the original lattice. This is reected in the above statement. 8

G25.2651: Statistical Mechanics Notes for Lecture 28

I. FIXED POINTS OF THE RG EQUATIONS IN GREATER THAN ONE DIMENSION In the last lecture, we noted that interactions between block of spins in a spin lattice are mediated by boundary spins. In one dimension, where there is only a single pair between blocks, the block spin transformation yields a coupling constant that is approximately equal to the old coupling constant at very low temperature, i.e., K K . Let us now explore the implications of this fact in higher dimensions. Consider a two-dimensional spin lattice as we did in the previous lecture. Now interactions between blocks can be mediated by more than a single pair of spin interactions. For the case of 3
3 blocks, there will be 3 boundary spin pairs mediating the interaction between two blocks: 0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FIG. 1.

1

. . . .

Since 3 boundary spin pair interactions mediate the block-block interaction, the result of a block spin transformation should yield, at low T , a coupling constant K roughly three times as large as the original coupling constant, K : 0

K

3K

0

In a three-dimensional lattice, using 3
3
3 blocks, there would be 32 = 9 spin pairs. Generally, in d dimensions using blocks containing bd spins, the RG equation at low T should behave as

K

0

T !0 K!1

bd 1K ;

The number b is called the length scaling factor. The above RG equation implies that for d > 1, K > K for low T . Thus, iteration of the RG equation at low temperature should now ow toward K = 1 and the xed point at T = 0 is now a stable xed point. However, we know that at high temperature, the system must be in a paramagnetic state, so the xed point at T = 1 must remain a stable xed point. These two facts suggest that, for d > 1, between T = 0 and T = 1, there must be another xed point, which will be an unstable xed point. To the extent that an RG ow in more than one dimension can be considered a one-dimensional ow, the ow diagram would look like: 0

Unstable

Stable

Stable

.

.

x=1 K=

.

K=K c

x=0

T=T c

K=0

FIG. 2.

Any perturbation to the left of the unstable xed point iterates to T = 0, K = 1 and any perturbation to the right iterates to T = 1 and K = 0. The unstable xed point corresponds to a nite, nonzero value of K = Kc and a temperature Tc , and corresponds to a critical point. To see that this is so, consider the correlation length evolution of the correlation length under the RG ow. Recall that for a length scaling factor b, the correlation length transform as  (K ) = 1  (K ) 0

b  (K ) = b (K ) Suppose we start at a point K near Kc and require n(K ) iterations of the RG equations to reach a value K0 between K = 0 and K = Kc under the RG ow: 0

Stable

.

x=1 K=

Unstable

. K=K c T=T c FIG. 3.

2

. K0

Stable

. x=0 K=0

If 0 is the correlation length at K = K0, which can expect to be a nite number of order 1, then, by the above transformation rule for the correlation length, we have

 (K ) = 0 bn(K) Now, recall that, as the starting point, K is chosen closer and closer to Kc, the number of iterations needed to reach K0 gets larger and larger. (Recall that near an unstable xed point, the initial change in the coupling constant is small as the iteration proceeds). Of course, if K = Kc initially, than an in nite number of iterations is needed. This tells us that as K approaches Kc, the correlation length becomes in nite, which is what is expected for an ordered phase. Thus, the new unstable xed point must correspond to a critical point. In fact, we can calculate the exponent  knowing the behavior of the RG equation near the unstable xed point. Since this is a xed point, Kc satis es, quite generally,

Kc = R(Kc) Near the xed point, we can expand the RG equation, giving

K  R(Kc) + (K ; Kc)R (Kc ) +    0

0

De ne an exponent y by

y = ln Rln(bKc ) 0

so that

K  Kc + by (K ; Kc ) 0

Near the critical point,  diverges according to

 jT ; Tcj

;

1 1 K ; Kc

K ; Kc K

;

;

K ; Kc Kc

;

Thus,

jK ; Kcj



;

but

 (K ) = b (K ) 0

which implies

jK ; Kc j

bjK ; Kc j = bjby (K ; Kc)j

;

0

;

;

which is only possible if

 = y1 This result illustrates a more general one, namely, that critical exponents are related to derivatives of the RG transformation.

II. GENERAL LINEARIZED RG THEORY The above discussion illustrates the power of the linearized RG equations. We now generalize this approach to a general Hamiltonian H0 with parameters K1 K2 :::  K. The RG equation

K = R(K) 0

3

can be linearized about an unstable xed point at bK according to 

Ka ; Ka  0



where

X

b

Tab (Kb ; Kb ) 

a Tab = @R @Kb K=K

The matrix T need not be a symmetric matrix. Given this, we de ne a left eigenvalue equation for T according to X

a

ia Tab = i ib

where the eigenvalues fg can be assumed to be real (although it cannot be proved). Finally, de ne a scaling variable, ui by

ui =

X

a

ia (Ka ; Ka ) 

They are called scaling variables because they transform multiplicatively near a xed point under the linearized RG ow:

ui = 0

X

a

ia (Ka ; Ka ) 0

=

XX

=

X

a b



ia Tab (Kb ; Kb ) 

b i i b

  (Kb ; Kb ) 

= i ui Since ui scales with i , it will increase if i > 1 and will decrease if i < 1. Rede ning the eigenvalues i according to

i = by

i

we see that

ui = by ui 0

i

By convention, the quantities fyig are called the RG eigenvalues. These will soon be shown to determine the scaling relations among the critical exponents. For the RG eigenvalues, three cases can be identi ed: 1. yi > 0. The scaling variable ui is called a relevant variable. Repeated RG transformations will drive it away from its xed point value, ui = 0. 2. yi < 0. The scaling variable ui is called an irrelevant variable. Repeated RG transformations will drive it toward 0. 3. yi = 0. The scaling variable ui is called a marginal variable. We cannot tell from the linearized RG equations if ui will iterate toward or away from the xed point. Typically, scaling variables are either relevant or irrelevant. Marginality is rare and will not be considered here. The number of relevant scaling variables corresponds to the number of experimentally `tunable' parameters or `knobs' (such as T and h in the magnetic system, or P and T in a uid system in the case of the former, the relevant variables are called the thermal and magnetic scaling variables, respectively). 4

III. UNDERSTANDING UNIVERSALITY FROM THE LINEARIZED RG THEORY In the linearized RG theory, at a xed point, all scaling variables are 0, whether relevant, irrelevant or marginal. Consider the case where there are no marginal scaling variables. Recall, moreover, that irrelevant scaling variables will iterate to 0 under repeated RG transformations, starting from a point near the unstable xed point, while the relevant variables will be driven away from 0. These facts provide us with a formal procedure for locating the xed point: i. Start with the space spanned by the full set of eigenvectors of T. ii. Project out the relevant subspace by setting all the relevant scaling variables to 0 by hand. iii. The remaining subspace spanned by the irrelevant eigenvectors of T de nes a hypersurface in the full coupling constant space. This is called the critical hypersurface. iv. Any point on the critical hypersurface belongs to the irrelevant subspace and will iterate to 0 under successive RG transformations. This will de ne a trajectory on the hypersurface that leads to the xed point as illustrated below:

K

3

Fixed point

. K1 FIG. 4.

5

K2

This xed point is called the critical xed point. Note that it is stable with respect to irrelevant scaling variables and unstable with respect to relevant scaling variables. What is the importance of the critical xed point? Consider a simple model in which there is one relevant and one irrelevant scaling variable, u1 and u2 , with corresponding couplings K1 and K2 . In an Ising-type model, K1 might represent the reduced nearest neighbor coupling and K2 might represent a next nearest neighbor coupling. Relevant variables also include experimentally tunable parameters such as temperature and magnetic eld. The reason u1 is relevant and u2 is irrelevant is that there must be at least nearest neighbor coupling for the existence of a critical point and ordered phase at h = 0 but this can happen whether or not there is a next nearest neighbor coupling. Thus, the condition u1 (K1 K2) = 0 de nes the critical surface, in this case, a one-dimensional curve in the K1 -K2 plane as illustrated below:

K

2

K1 FIG. 5.

Here, the blue curve represents the critical \surface" (curve), and the point where the arrows meet is the critical

xed point. The full coupling constant space represents the space of all physical systems containing nearest neighbor and next nearest neighbor couplings. If we wish to consider the subset of systems with no next nearest neighbor coupling, i.e., K2 = 0, the point at which the line K2 = 0 intersects the critical surface (curve) de nes the critical value, K1c and corresponding critical temperature and will be an unstable xed point of an RG transformation with K2 = 0. Similarly, if we consider a model for which K2 6= 0, but having a xed nite value, then the point at which this line intersects the critical surface (curve) will give the critical value of K1 for that model. For any of these models, K1c lies on the critical surface and will, under the full RG transformation iterate toward the critical xed point. This, then, de nes a universality class: All models characterized by the same critical xed point belong to the same universality class and will share the same critical properties. This need not include only magnetic systems. Indeed, a uid system, near its critical point can be characterized by a so called lattice gas model. This model is similar to the Ising model, except that the spin variables are replaced by site occupance variables ni , which can take on values 0 or 1, depending on whether a given lattice site is occupied by a particle or not. The grand canonical partition function is 6

Z=

X

n1 =0
1



X

nN =0
1



P i

e

ni 2

P

hi j i

Jij ni nj

and hence belongs to the same universality class as the Ising model. The critical surface, then, contains all physical models that share the same universality properties and have the same critical xed point. In order to see how the relevant scaling variables lead to the scaling relations among the critical exponents, we next need to introduce the scaling hypothesis.

IV. THE SCALING HYPOTHESIS Recall that the RG transformation preserves the partition function: Tr0 e

0 (  0 g
K 0 )

;H0 f

= Tr e

(  g
K)

;H0 f

For the one-dimensional Ising model, we found that the block spin transformation lead to a transformed Hamiltonian of the form

H0 (f g K ) = N g(K ) ; K 0

0

0

0

0

0

N X i=1

i i+1 0

0

Thus, H0 (f g K ) contains a term that is of the same functional form as H0 (f g K ) plus an analytic function of K . De ning the reduced free energy per spin from the spin trace as f (K ), the equality of the partition functions allows us to write generally: 0

0

e

;

Nf (fK g)

= Tr e

(  g
fK g)

;H0 f

=e

;

N 0 g(fK g)

Tr e

(  0 g
fK 0 g)

;H0 f

=e

N 0 g(fK g)

;

e

;

N 0 f (fK 0 g)

which implies that

f (fK g) = g(fK g) + NN f (fK g) 0

0

If b is the size of the spin block, then

N = b dN 0

;

from which

f (fK g) = g(fK g) + b df (fK g) ;

0

Now, g(fK g) is an analytic function and therefore plays no role in determining critical exponents, since it does not lead to divergences. Only the so called singular part of the free energy is important for this. Thus, the singular part of the free energy fs (fK g) can be seen to satisfy a scaling relation:

fs (fK g) = b d fs (fK g) ;

0

This is the basic scaling relation for the free energy. From this simple equation, the scaling relations for the critical exponents can be derived. To see how this works, consider the one-dimensional Ising model again with h 6= 0. The free energy depends on the scaling variables through the dependence on the couplings K . For h = 0, we saw that there was a single relevant scaling variable corresponding to the nearest neighbor coupling K = J=kT . This variable is temperature dependent and is called a thermal scaling variable ut . For h 6= 0 there must also be a magnetic scaling variable, uh. These will transform under the linearized RG as

ut = by ut uh = by uh 0

t

0

h

where yt and yh are the relevant RG eigenvalues. Therefore, the scaling relation for the free energy becomes

fs (ut uh) = b dfs (ut uh ) = b dfs (by ut by uh) ;

0

0

7

;

t

h

Now, after n iterations of the RG equations, the free energy becomes

fs (ut uh) = b

nd

;

fs (bny ut bny uh) t

h

Recall that relevant scaling variables are driven away from the critical xed point. Thus, let us choose an n small enough that the linear approximation is still valid. In order to determine n, we only need to consider one of the scaling variables, so let it be ut . Thus, let ut0 be an arbitrary value of ut obtained after n iterations of the RG equation, such that ut0 is still close enough to the xed point that the linearized RG theory is valid. Then,

ut0 = bny ut t

or

n = y1 logb uut0 = logb uut0 t t t

1=yt

and

fs (ut uh ) = uut t0

d=yt

fs ( ut0 uh jut =ut0j

y =yt

; h

)

Now, let

t = T ;T Tc We know that fs must depend on the physical variables t and h. In the linearized theory, the scaling variables ut and uh will be related linearly to the physical variables: ut = t ut0 t0 uh = hh 0

Here, t0 and h0 are nonuniversal proportionality constants, and we see that ut ! 0 when t ! 0 and the same for uh. Then, we have 0 fs ut0 jt=th=h j0 y =y



fs (t h) = jt=t0j




d=yt

h

t

The left side of this equation does not depend on the nonuniversal constant ut0, hence the right side cannot. This means that the function on the right side depends on a single argument. Thus, we rewrite the free energy equation as 0 fs (t h) = jt=t0jd=y  jt=th=h y =y 0j




t

h

t

The function  is called a scaling function. Note that the dependence on the system-particular variables t0 and h0 is trivial, as this only comes in as a scale factor in t and h. Such a scaling relation is a universal scaling relation, in that it will be the same for all systems in the same universality class. From the above scaling relation come all thermodynamic quantities. Note that the scaling relation depends on only two exponents yt and yh. This suggests that there can only be two independent critical exponents among the six,    , and . There must, therefore, be four relations relating some of the critical exponents to others. These are the scaling relations. To derive these, we use the scaling relation for the free energy to derive the thermodynamic functions: 1. Heat Capacity:

Ch

@2f @t2

but 8

jtjd=y

2

t;

h=0

jtj

Ch

;



Thus,

= 2 ; yd

t

2. Magnetization:

m = @f @h

jt=t0 jd=y jt=t0 jy =y

jtj(d

t

h

h=0

y )=yt

; h

t

but

m jtj Thus,

= d ;y yh t

3. Magnetic susceptibility:

@2f @h2

 = @m @h

jtj(d

2yh )=yt

;

but

 jtj



;

Thus,

 = 2yhy; d t

4. Magnetization vs. magnetic
eld: (d m = @f @h = jt=t0 j

y )=yt

; h




0

h=h0 jt=t0 jy =y h



t

but m should remain nite as t ! 0, which means that  (x) must behave as xd=y 0

m jt=t0 j(d

y )=yt

; h

(h=h0 )

(h=h0)(d

jt=t0j

But

m h1=  = d ;yhy h 9

y )=yt

; h

yh (d;yh )=(yt yh )

(d;yh )=yh

Thus,

1

h;

as x ! 1, since then

From these, it is straightforward to show that the four exponents , ,  , and  are related by

+ 2 +  = 2 + (1 + ) = 2 These are examples of scaling relations among the critical exponents. The other two scaling relations involve  and  and are derived from the scaling relation satis ed by the correlation function:

G(r) = b

;

2(d;yh )

G(r=b by t) t

G(r) = jt=t0 j

2(d;yh )=yt

This leads to the relations






jt=t0 j

 = y1 t  = d + 2 ; 2yh

and the scaling relations:

= 2 ; d  =  (2 ; )

10

r



1=yt

;