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=
=
It is n o t
hard
to s h o w
=
that
the
(Ap) 2 =
More
e-½lzl2-exp
specific,
_
2
denotes
let
z
T h e n we h a v e
,p2 (~z)>
= - ( x + y *)n-1
identities
are
easy
to v e r i f y
by using
the operator
46
identity [An,B]
that holds
A n-k. [A,B]'A k-1
k=1 operators
for a r b i t r a r y
Proposition
n ~
=
4.4:
For e v e r y
A
and
x,y£H
B
and
we h a v e
n6~
[in] *n
1)
(x+y)
= n!-
~
(i) k *)n-2k) : k!- (n-2k) l : (x+y
k=0 [in]
2)
:(x+y*)n:
k
= n!.
(-½) (x+y)n-2k k ! - ( n - 2 k ) .t k=O
Proof: follow
We will
at once,
prove
where
I)
by induction.
n>1
n=0
and
n=1
From
the
by d e f i n i t i o n
A 0 = I = the i d e n t i t y Take
The cases
operator.
T h e n we h a v e * n
(x+y) Assuming induction
n
hypothesis
= (x+y*) to
be
n-I
odd,
and lemma
+
(x) + (x+y*)
we
get
[
n-1
y
*
n-I 2
]
4.3 we get
n-1 2
(x+y
*)n-la+
(x)
(~) k
= ( n - 1 ) !-
k!- (n-l-2k)
!
: (x+y*)n-I -2k) :~+(x )
k=O n-1 2 (i) k =
(n-1)
1-
+
k!- (n-1-2k)'
* n-I -2k (x)
: x+y
}
:
k=0 n-3 2 (~) k kl-(n-1-2k)!
+ (n-l)!-
(n-1-2k).-
:(x+y
* n-2-2k )
k=0 n-1 2
(i) k kZ-(n-1-2k)!
(n-1)!-
a+
n-1-2k (x)
: x+y*)
k=0 n-1 2
+ (n-1)! i=l
)i_ 1 (i(-~),(n+1-2i
!(n+]-2i):(x+y
* n-2i )
47 n-1 2 = n!- ~ n~2k k!.(n-2k)! (½) k
~+(x)
:(x+y*)n-l-2k:
k=O n-1 2 + n! ~
i (½) i-I *)n-2i i!. (n-2i) ! : (x+y
i=I n-1 2 = n!- ~ n-2k (½) k + * n-1-2k n k!- (n-2k) ! (x) : ( x + y ) k=O n-1 2 + n! ~ 2-i (½) i * n i!. (n-2i) I : (x+y n-2i: i=O
We return to the first identity (x+y*) n = ( x + y * ) n - ~ + ( x )
+ (x+y*)n-ly *
n-| 2 = n'. ~ n~2k (~) k n-l-2k • k!-(n-2k)! ~+(x) :(x+y*) : k=O n-1 2 + n' ~ 2-i (½) i * n-2i )n-ly* • n i!.(n-2i)! :(x+y : + (x+y* i=0 and by using the induction h ~ o t h e s i s on the last t e ~ , we get n-1 : n!-
2 n~2k (1)k l ~ * n-l-2k . ~+,x, :(x+y ) k!-(n-2k)! L k=O n-1 2 2-i (½) i :(x+y*) n-2i + n!n i!. (n-2i)! i=O n-1 2 (½) k : (x+y*)n-1-2k) : + (n-1)'- ~ kl. (n-1-2k) I k=O
48 n-1 2 = n !- ~
n - 2 k (½)k ki.in_2k)! n
~+(x)
:(x+y*)n-1-2k:
k=0 n-1 2 2-i (½) l n i!-(n-2i)!
+ n!- ~ i=0 n-1 2
:(x+y*)n-2i
I< )k * n-1-2k * n - 2 k (z y , x > :(x+y ) : y n k:. (n-2k) !
+ n!k:0 Summing
the f i r s t
and
last t e r m s g i v e s
n-1 2
= n!- ~
n - 2 k (I~ k . [ (n+m-2k)! !]½ (n-k)! [(n-k)!-(m-k) -(nV--(-(-(-(-(-(-(~-k)!-V(m-k)!k=O Ixln-k. lal m-k m k .V(n+m_2k),.ixln-k. lalm-k (n-k)!
n=m k=0 m =
~ [ (~) j)n nl = ~ Ep "{~e+(exp
x) exp y*
= e-½.a+(exp
-
).We+f
f).exp-e*.exp-f*
exp-(e+f) ,
= e -½(<e ' f> - ).e-½1e+fl 2 -~+(exp(e+f)) (<e,f>
f) exp(-f*)
3.~2 we get
= e -½ (lel2+IfI2+2<e'f>)-a+(exp(e+f))
-~
we get
)
= e-½(lel2+Ifl2).e+(exp
= e
formula
exp-(e+f) *
= e-i'Im(<e'f>).We+f
52
Lemlaa
4.10:
For
every
e6K
we
have
on
FIH
, W
= W
e
W e W_e
Proof:
For
<Wef,g> The
other
is
unitary we
get
is
now
transformation
of
clear FIH
Weyl
Theorem
e)
of
the
F]H
FH
exp(-e*)f,g> of
that
onto
I
have
a consequence
transformation the
we
= < e - ~Il e l 2 - e + ( e x p
relation
It
f,g6FiH
=
-e
.
onto
lemma
=
4.9.
operator We
extend
FK
.
We We
As
a
is by
a
unitary
continuity
corollary
to
to
lemma
a
4.9
relations.
4.11 :
For
every
e,f6H
we
have
on
FH
the
Weyl
relations We-W f =
Example element
holds
e6H
on
the
4.12: .
Then
whole
e-i'Im(<e,f>).We+f
Consider it
FH
is
.
a
unitary
easy
to
show
FU-W e
"FU* =
operator that
W(Ue)
the
U identity
of
and
an
Chapter
5:
In the
Some
this
chapter
operator
6_
we
special
are
This
operators
primarily
operator
representation
introduced
in c h a p t e r s
A:
in Hilbert
spaces
Conjuqation
Definition
5.1A:
conjugate--linear
A
concerned
links
the
6 and
7.
conjugation
in
with
complex
a Hilbert
investigating and
real
space
wave
~
is
a
mapping : K
,H
fulfilling I)
is a n i n v o l u t i o n
2)
<x,y>
=
Just
after
for a l l
proving
procedure
can
be
we
the
conjugation
extend
abandon
in t h i s
extension holds
is
them
applied
case
.
proposition to
the
x,yeH
conjugate to
standard
multiplicative
on
2.1
we
mentioned
linear
isometries
a conjugation
on
notation
as t o o
the
F-
subalgebras
the
F0H
that
the
as w e l l . whole
same Hence
FH
.
We
complicated.
The
and
,
and
in
the
FIK
invariant.
An element
feFK
will
be c a l l e d
real
if
f = f
Theorem Hilbert
5.2A:
space
H
Consider Then
the
{ exp x is t o t a l
in
FH
Proof: space
K
,
,
i.e.
Choose
and
the
the
r>0
and
a
conjugation
set
I x6H
a real
consider
fixed
span
is r e a l of t h e
orthonormal set
and set
Ixl "..." ( < X , e n > ®
+ en)
r
rl = ~ ar. r
. -...
>r <x,e n
r1 n.~>
= ~ ar.<X,el > r
rn -....<X,en>
Letting x = s]e I + s2e 2 +...+ where
n6~
and
Sl,S2,...,Sn6~
are
2
sI + s we
get
by
setting
Sk=
0
for
such
+...+s~
Sne n that
< r
2
,
k>n r]
0 =
,
ar-S ]
rn "...'s n
r Since
(Sl,S2,...,Sn)
run
over
the
ball
a
= 0
for
all
with
radius
r
we
conclude
that
by which
B:
The
f = 0
functors
h*
and
,
6*
denote
a
Hilbert
--
conjugation
r
.
(H,)
Let
r
space 00
and
an
orthonormal
basis
{en}n= I
with
a
fixed
chosen
55 Definition
5.1B: We define the m a p p i n g h_ : FI~
,
V~
by the series co *
h* = ~ a(en)e(en)
=
n=l
The
lower
index
--w
~ en en n=l
-
points
out
that
the
definition
depends
on
the choice of the conjugation. We shall need the following properties
Theorem depend
5.2B:
The
operator
on the choice of the basis
I)
F0H
is an invariant
2)
h~
(a
3)
h
(amexp
TM)
well-defined
a m-2
is well H , h_
for
a,z6H
and
does
not
depend
h_
and
does
not
and ,
i.e.
aeg
and
h_(FOH ) c F0K
.
men 0 + a
]exp
z
m6~ 0 .
I) is evident and it follows and
defined
= [ m ( m - 1 ) < a , a > a m - 2 + 2 m < a , z > a m-1
with
Proof:
in
space under
= m(m-1) z)
h
of the o p e r a t o r
on
the
from 2) that the operaZor chosen
orthonormal
basis.
prove 2) notice that ~n*(a
TM) =
ma m-1
which implies W
--W
en en
(am )
= m ( m - 1 ) < e n , a > < e n , a > a m-2
and h~(a m) = (~ < ~ , e n > < e n , a > ) m ( m - 1 ) a m - 2
= m ( m _ 1 ) < ~ , a > a m-2
n=l
3)
is similar,
Theorem
5.3B:
thought a little more technical.
For all
x6H
[(~-h~)m,~+(x)]
and
m6~
the c o m m u t a t i o n
.lh*.m-1 = m(~ _)
~*
relation
is To
56 holds
on the w h o l e
Proof: f=ak.exp
z
Take
with
*-* enen(xakexp
FIN
.
m=1
a,z6K
is
It
and
k6~ 0 .
-
sufficient
consider
elements
We compute,
-
z) = < X , e n > < e n , a > . k - a k lexp z + < X , e n > < e n , Z > - a + <en,a><en,x>-k.ak-]exp -
k
exp
z
z *-*
+ <e n , z > < e n , x > - a k e x p which
to
k
z + e+(X)enen(a
exp z)
implies
*-* + enen~ ( x ) [ a k e x p
+ *-* k z] - ~ (X)enen[a exp z] =
= kak-lexp
z. [<X,en><en, a> + < a , e n > < e n , X > ]
+ akexp
z-[<X,en><en,Z>
+ <en,X>]
+
Since
a (x)
½(h:e+(x) which
is a c l o s e d
- e+(x)h:)-akexp
operator
z = k<x,a>ak-lexp
over
z + ak<x,z>exp
n6~
z ,
gives [½h:,a+(x)](akexp
It
we get by s u m m a t i o n
is
easy
to
check
that
for
z) = x * ( a k e x p
arbitrary
z)
operators
A,B
and
operators
x*
m6~
we
have m = ~ A m-k
jAm,B]
[A,B]
A k-1
k=1 Using commute
this
identity
we get on
and
the
fact
that
the
and
h*
define
the
FIN k
[(~lh*_).k ,a+(x)]
Definition
=
5.4B:
~ (½h:) k-j x* j=l
(cf.
[20]
_J' J - ] (2i h *
paragraph
transformation w
5_ : FOH-----o
FOH
by the s e r i e s
5_ = (-~h_)n/n! n=O
=
k(~lh*_) k - 1 ~ *
2A)
We
57 The
operator
6_
can
be g e n e r a l i z e d
[20]) by taking in place of a conjugation L : K which
is
conjugate
real-self--adjoint
linear.
has a dual o p e r a t o r
a+(6L ) ,
-
a continuous
6L
(cf.
mapping
; H
Then
Hilbert--Schmidt
to an o p e r a t o r
in
the
strict
case
when
contraction,
L
is
the operator
which is d e n s e l y defined.
a 8L
In physics
the
elements 6 L = a+(6L).~ when n o r m a l i z e d
provide
called u l t r a c o h e r e n t
Proposition and
m6~
the so-called
squeezed
states;
in [20] they are
vectors.
5.5B: The o p e r a t o r
is well d e f i n e d and for
6
a6K
we have [m/2]
I
8_(a m) = .
(_½)n
m! (m-2n)!
<ara> n m-2n n! a
n=0 and e s p e c i a l l y
~(~) 6_(x)
Proof:
=
= x
for all
xeK
.
It is enough to prove that [m/2] 6~(a m) =
~ n=0
(_½)n
We start by proving that for all m! (m-2n)!
(h~) n a m = {
m! (m-2n)!
<ara> n am-2n n!
m,n6~ n am-2n
for m>2n for m1
and assume
that
the result
We have
(hl) n a m = hi[
(h~) n-1
= hi[ {
am
]
m! (m-2n+2)! o
n-1
a m-2n+2
for m>2n+2 for mexp(x
we s h a l l c a l c u l a t e h~exp
z = exp
+ y)
6_(exp
z)
Since
z
we get (-½h~)nexp
z = (-½)nexp
z ,
thus g i v i n g
-½ 6*(exp
z)=
~ n=0
Hence we get
FIK
f,geFiK
fog :
Remark
the
(-½)nexp
z/n!
= e
exp z .
a n d the
64
6 ~ ( e x p x ~ e x p y) = (6[exp x ) ( 6 1 e x p
y)
-½ -½ = e
e
exp
(x+y)
-½( + ) = e
exp(x+y) -I < x , x > + < y , y > )
+½<x+y,x+y>
= e which
, 6_(exp(x+y))
.e
implies
+½( (exp x)~ e x p y)
+ )
= e
exp(x+y)
= e<X,Y>exp(x+y)
.
Slnce e<X'Y>exp(x+y)
= (exp x) e < X ' Y > (
e x p y)
= (exp x) e x p x * ( e x p y) =
the o p e r a t o r
:exp(x+x
): e x p y
,
identity (exp x)~ = : e x p ( x + x
holds
on the c o h e r e n t This
identity
Let us d e f i n e
vectors
in
FH
f,geFoK
a new inner product
with
a~(x) e(x)
respect
with .
o for all
):
we
denote
+ a o ( X ) = x~
by
to the B - m u l t i p l i c a t i o n .
respect
to
=
nk
•
....
,
and
•
e~[z]
e~[z]
exp(-Izl2)dz
=
~k nI =
nk ..
m -
mk - e x p ( - I z l 2 )dz
I ..
k k ~ = ]~-
i=l Without
_Zn i "Z m i - e x p ( - { z i { 2 ) d x i d Y i
C
loss
;
of g e n e r a l i t y
we
assume
~n-zm'exp(-Izl2)-dxdy
that
= 2~.
= = z> = ~ ar" <e--,exp r
r rI =
~r.<el,z >
r
r2 .<e2,z >
r "''''<en'Z>
n
r rl =
ar'Zl
r2
rn
"z2
" " "'Zn
'
r where
z i = <ei,z>
L2--space function
L2(C) on
,
the
the
on
Notice
i6~
function
elements
representation
functions
for
Identifying f[- ]
can be
K
as
considered
the
weiqhted
as an e n t i r e
~ .
Therefore wave
6C
are
of
the
often
image
of
FH
under
the
conjugate
continuous
functions
called
the
complex
holomorphic
H that
we
extend
some
on
H
to
71
continuous
functions
on
the measure-theoretical on a measure
zero
Example pointed
out
traditional Considering Fourier
set
6.6:
enlargements point
of v i e w
This
is
a
Fourier
transformation
transform
of
f
f
6
in
that
amounts space
to Cn
in the
the
complex
FC n z6C n
composition
wave of
set
L 2 ( R n)
= f[i~]
awkward
functions
find
space then
defined
2.11.
coincides the
from
2).
example
and
It
was
with
the
L2(R n )
evaluating
the
that
,
representation with
is
appendix
of
in
we easily
f[- ]
It
F(--i) ~
=
•
(cf.
operator
(~f)[~] i.e.
K
continuation
that
function
the
of
since we extend
to a f u l l m e a s u r e
earlier
a
~
Fourier
a 90 d e g r e e s
transformations rotation
of
the
Chapter
7:
The real wave representation
This by S e g a l
chapter
f
the
representation
[15]. We a p p l y the t e c h n i q u e
Definition of
concerns
in
xeK
7.1:
For
every
developed
feFiK
we
7.2:
the
V-value
f_(x)
For
f,g6FiH
of lemma
= <exp x,6_f>
we h a v e t h a t
= f_(x)-g_(x)
From chapter
hence by virtue
*
= (6_f)[x]
(f~g)_(x)
f*g_(x)
[20].
define
*
Proof:
in
originally
setting f_(x)
Lemma
introduced
for all
xeH
.
5 we h a v e t h a t
6.2 w e get
= (61(f,g))[x ] =((61f)(61g))[x ] = (61f)[x](61g)[x ] : f_(x)-g_(x)
For f_(')
arbitrary
in d i r e c t i o n
aeH a
and
7.3:
= ~-~ d f_(t-a
For f i x e d
feFIH
d d-~ f _ ( t - a
Proof: sufficient
Since
to p r o v e
the
define
the
derivative
of
and
+ - )it= 0
a,b6K
*
+ b)
(a f ) _ ( t . a
operators
a
we h a v e + b)
and
6
commute,
that
d <exp(t.a dt This
we
setting
Oaf_(')
Lemma
f6FiH
can be r e d u c e d exp(h-a)-~ h
+ b),f>
= <exp(t-a
to p r o v i n g a
+ b),a
f>
that in
FH
for
h--~0
it
is
73 since the operator
is closed.
a+(exp x)
We compute co
exp(h-a)-e h
--
a
=
exp(h-a)
h- -~ a- h
= hl [ 2 (h'a)nn! - ~ - h-a ] n=0
co
_ I 2
co
(h'a)k+2 (k+2) .'
h-a) n 1 ~ ( n' -h
k=0 n=2 a k+2 h k = ~ (k+2) !.h . k=O We have for Ihl -< 1 co
co
I
~ I
exp(h-a)-e h
ak+2 hk (k+2)! h I
I
- a
k=O co
f_(x) - (b f)_(x) Since b= Re(b) + i-Im(b) , where
Re(b) - b +2 ~
we can assume that
and b6K
Im(b) = b2.i-~ ' is real and that
without loss of generality llbll=1
We have {bf,g) = (2~)-½k~ (bf)_(x).g_(x).exp(_~Ixl2).dx ~k = (2~)-½k[ [<x,b>f_(x) - (b*f)_(x)]g_(x)-exp(-½1xI2)-dx = (2~)-½k[ f_(x)g_(x).exp(-½1xl2).dx
&k
-- (2~)-½k!k(b*f)_(x)g_(x).exp(-½1xl2).dx = (2~)-~k~ f_(x)g_(x)-exp(-½1xl2)-dx ~k Choosing the first basis vector
eI = b
- (b*f,g)
.
we get
(f,b*g) = (2~)-½kF f_(x)(b*g)_(x).exp(-½1xl2).dx = (2~)-½k!kf_(x). [d~ g_(t'b + x)lt=o-eXp(-½ Ix,2)'dx = (2~)-½k~ ~k
f---~-~'O~ g--(x)'exp(-½1xl2)dx1"''dXk I
Since f_(x-----7.~ig_(x).exp(_½,x,2)dXl
=.
~ ~a-Xl 0 [f~-(x)"exp(-~ Ix I2 )]g-(x)dXl
and and -
a ~-~ig_(x)'exp(
=_
-½
Ixl2)dXl
=
~ ~-~1[f_(x)'exp( a _½ Ixl 2)]g_(x)dx I
76
--T
[a
f_(x).exp(_½1xi2),
xlf_(x).exp(_½1xl2)]g_(x)dxl
= ~ xlf-(x)'g-(x)'exp(-½1xl2)dXl
returning (f,b , g) =
T ~--~1f_(x), g_(x) •exp(-½ Ix 12)dx]
to the integration
I ~ 2~) -~k
over
~k
,
we get
a _½ ix I2)d x f_(x).~-~ig_(x).exp( ~k
=
2v)-½k~
xlf_(x) • g_(x) • exp(-½ Ix 12)dx ~k
_ (2v)-½k~ =
a f_(x), g_(x), exp(_½ 1xl 2)dx ~k ~--~I
2v)-½k[~k <e I 'x>f-(x) "g-(x) "exp(-½ Ix 12)dx
- (2v)-½k!k(eTf)_(x).g_(x)-exp(-½1xl2)dx =
2=)-½k[
f_(x) • g_(x) • exp(-½ 1xl 2)dx
-- (2~)-½k[ =
(b*f)_(x)-g_(x).exp(-½1xl2)dx
2v)-½k[ f_(x)-g_(x).exp(-½1xl2)dx
- (b*f,g)
= (bf,g)
We
introduce
Hilbert-Schmidt
the
enlargements
is done in appendix
functionals
continuous If
~H
=
~H
of the real Hilbert
defined on
as
in chapter
on finite
'
sitting
space
H ,
on
as it
,
functionals
6 we
dimensional
extend
continuous
subspaces
of
linear H
,
to
H .
is a finite dimensional
the conjugation to extend
way
functionals K
H
measure
2.
In a similar real
gaussian
and if
K
of the form
subspace
denotes
of
H ,
its real part
invariant ,
under
then we have
77
-~(dz)
.
K
The theorem
subscript
N
in
~'}N
refers
to the
normal
ordering
(see
8.8). Take
P = af*
and
Q = bg*
Since
f exp
z = f[z]exp
z ,
we
get <exp(-z).P.exp(z),exp(-z).Q-exp(z)>
Lemma
8.2: F o r
a,b,f,g6FoK (af
Proof:
It
representation
is
,bg }N = < a , b > < f , g >
an
easy
consequence
of
the
complex
wave
we o b s e r v e
that
~'}N
and
~,~
are e q u a l
on the
P e
e(FoX)
H+K
Lemma within
that
a n d the above.
In p a r t i c u l a r , subspace
we h a v e
= f[z]g[z]
8.3:
FOK,
Let
P
denote
the
conjugation
in
e
(S
Proof:
of
) P
6 O(FOH )
O(FoH),(,} N ,
i.e.
*
,T~ N = ~T
A complex
conjugate--linear
operator
The operation O(F0~ ) 9 P
is a c o m p l e x
dual
for all
conjugation
involution <x,y>
This we check
,S} N
S,T 6 O ( F o H )
,
as d e f i n e d
in c h a p t e r
fulfilling =
for all
on the g e n e r a t o r s
x,y
af*,bg*
. 60(FoH
)
5, is a
81
/(af
) , b g )N = ~ifa , b g ~'N = < f , b > < a , g > =
Definition the
creation
8.4:
For
= (af
,gb ~ N = ( ( b g
every
pair
x,y6K
) ,af ~ N
"
we define
operator +
*
a (x+y)
: O(FoK )
' O(rOK)
by e+(x+y*)P and
the
annihilation
=
:(x+y*)P:
for P e O ( £ 0 K )
operator
~(x+y ) : O(FoK ) as the dual of
Notice (x+y) e+(x+y*) c(x)
is
and
,
and
+
a
e(x+y*)
8.5:
(x+y)
the
case
generators
af
y
=
0
multiplication ,
the
operators
by
operators ~+(x)
and
an e x t e n s i o n .
£ O(FoH )
we h a v e
that
(x-a)f
elements
af
60(FoH
*
+ a(y-f)
I = 0 (x * a)f * + a ( y * f) *
We o n l y lemma
verify
2).
For
the
= ~(xa)f*
+ a(yf)
)
we
+
, b g )N =
it
of W i c k
known
is m e r e l y
*
=
(a+(x+y*)af*,bg*~N
Then
of
to the w e l l
the notation
*
(x+y)af
by using
in
reduce
On the
~ ( x + y * )af * =
get
that
operator
I = x+y
*
Proof:
is the
*
~(x+y*)
2)
obvious
*
+
e
c+(x+y*)
therefore
Lemma
I)
a+(x+y *)
that
It
~ 0(FoH)
that
~(x+y
)bg
b>
+
= ~af * ,(x * b ) g * + b ( y * g) * ) N
= (x b ) g
+ b ( y g)
"
82
Care expression the
should for
be
taken
a(x+y
same as
f*a+(y)
=
W
= af
+ af
It is o b v i o u s
operator
in
the
I .
*
= (x+y
*
that every operator
operators
*
,u+v )N-af
~+(x+y*)
f r o m the a l g e b r a ,
H e n c e we h a v e the f o l l o w i n g
Theorem inner product
- (vy f)
+ a[(y*v)f]*
*
polynomial
(y*(vf))
8.8:
O(FOH )
equipped
~')N
is a B o s e a l g e b r a
x,y6K
the
annihilation
We
creation
operators
denote
O(FoK)'~'~N
operators ~(x+y
by
the
identity
with
the W i c k m u l t i p l i c a t i o n
with
the v a c u u m +
;
and
are
is a
theorem.
*
~+H
,
D(VOH )
~
I
and
a n d the b a s e
*
(x+y)
with
the
dual
)
O ( F o K ) '{'}N
the
completion
of
the
space
-
Consider
for
x+y
6 H+H
the c o h e r e n t
:exp(x+y in the H i l b e r t
space
By v i r t u e
):
~(FOK )
of p r o p o s i t i o n
:exp(x+y
vectors
*
~
): =
4.7 the i d e n t i t y
:(x+y
* n * ) :/n! = e + ( e x p x ) e x p y
n=0 holds
on e v e r y e l e m e n t :exp(x+y We
i.e.
the
denote
by
from
FIK
,
and h e n c e we get
): = e + ( e x p x ) e x p y O(FIH )
the
extended
s p a n of e l e m e n t s
:P-exp(x+y ) :
in
O(FOH )
Bose
algebra
of
O(ro~)
,
84 for
P 6 0(FoH ) , x+y
6 H+K
Then we have
O(FI~ ) = span } ~+(a-exp x)(b-exp y)* = span
Our goal called
a+(f)g *
f,g6FiH } .
is to construct
a Bose algebra
and an inner product
§ ,
new algebra is the algebra Let
{an}n6 ~
a,b6FoH , x,y6~ }
~,>
denote an orthonormal
basis in
Let
the
O(FIH )
which
are
--
symbol
§
is an orthonormal real
with
denote
the
respect
to
the
in
/5_
inverse
of
conjugation
relative
to
the
algebra
By using the real basis written above
* ~( En)2 + ~(Fn)2 = 2.~(en)~(en)
we get
co
§ = exp(-½ ~ [~(En)2+~(Fn)2]) n=l the
consisting
complex
00
and
H+H*
W
an)
basis
operator
and the conjugation
and the fact that
We define
2 -~- (e n + en) .
elements
H
*
F n = 2 -~ (-i)" (e n I nEN}
of this
O(EOH),:-:,{,) N
En
{En,F n
a multiplication,
such that the ,-picture
=
Here
with
operator
= exp(- ~ ~(en)~(en) ) , n=1
§-1
§-i = exp(+ ~ ~(en)e(en) ) n=l
Then we define the new multiplication p§Q = §[:(§-Ip)(§-~Q):] according
to theorem
inner product
Since product H+H
(,}
(,}
§ for
5.7C and definition of
P,Q e O(F]K)
in
O(FI}{ )
setting
p,Q e O(FI~ ) 5.8C.
Moreover
we define
the
setting
(p,Q} = ( §-Ip,§-IQ }N " §-i is the identity operator on K+K* , , restricted to H+H and the original inner
the
inner
product
on
are identical. For
product
(,)
the algebra
O(FOH )
as an algebra
we use the notation O(FIK),§,~ ,}
with
multiplication
V0(H+~* )
and write
§
and
FI (~+H*)
inner for
85
We F0(~+H
shall
),(,)
write
, the
)
for
the
base
8.9: . H+H
F0(H+H ,
the
),§,(,}
creation
is a Bose
their
dual
annihilation
the
space
Proof:
We m u s t
algebra
with
the v a c u u m
- ~(y+x*)
operators
M(x+y
4(x+y
of
operators
M + ( x + y *) = e+(x+y*) and
completion
.
Theorem I
F(H+H
) = a(x+y
)
s h o w the following:
) I = 0 *
[4(x+y
+
*
),4
4 + ( x + y *)
*
(u+v)]
= (x+y
generates
*
,u+v ) I
from the v a c u u m
I
the w h o l e
rO(~{+){ )
O(~o~{ )
=
and f i n a l l y S+(x+y*)
and
From
4(x+y
lemma
)
are dual
8.5 we have M(x+y
and
from
the
*
[4(x+y
conlmutation
+
operators
on
F0(~{+K
) I = ~(x+y
relation
in
) I = 0 ,
theorem
8.7
we h a v e
.
(U+V)]
= [ ~ ( x + y * ) , ~ + ( x + y *) - ~ ( y + x = [~(x+y*),a+(x+y*)] = (x+y
Since
•
that
*
),4
),(,)
4+(x)a
,u+v
= xa
)]
= (x+y~,u+V*)N.I
)I
and
N+(y
)f
=
(yf)
for all
x,y6H +
a,feFoH
,
it
the i d e n t i t y
is
easily
seen
operator
I
that
the
generate
creation
the w h o l e +
It operators
remains on
to
F0(K+K
w
prove ),(,~
6*e+(X) which
in
the
algebra
that
F1 ( H + H )
algebra
the
fact
that
the
= O(FIK )
and
and
4(x+y
commutation
- e(x))6
§
F0(H+K
for corresponds
4(x+y
and
)
*
§ M + ( x + y *) = (4+(x+y *) - 4 ( y + x * ) ) § By u s i n g
N (x+y)
*
~ (x+y)
We n e e d
= (~+(X)
operators
and
*
)
from
x6H
)
are
dual
theorem
5.6B
, to
. commute
and
applying
86
the o p e r a t o r
a+(x+y *
we get + . (x+y)§
= §~+(x+y*)
,
and h e n c e § - I M + ( x + y * ) = ~ + ( x + y * ) § -I For
P,Q 6 F0(H+H
)
and
x+y
£ H+K
-
(M+(x+Y*)P,Q)
= (§
I M + ( x + y * ) P , § - X Q ) N = (a
= ~§ =
Definition of a f i n i t e
-1
P,e(x+y
*
)
5-1
Q)N
~
~§
*
(x+y)§-IP,§-~Q) -
8.]0:
We
denote
~(H+H
by
N
IP'M(x+Y*)§-~Q)N
{§-~P,§-~M(x+y*)Q) N = (P,M(x+y*)Q)
number
)
.
the
set
of
polynomials
of v a r i a b l e s
T h e n we d e f i n e
we h a v e
+
and
where
a,b6H
.
the t r a n s f o r m a t i o n : F0(K+K
, 9(K+K
) = 0(FOK )
)
by (~P)[z] where
z,(exp-z)P(exp
for all
z)>
z6H
,
p 6 F0(H+H* )
It t u r n s function,
cf. Using
out
If H
extensions
~P[. J
the g e n e r a t o r s
we to
that
correspond
to t h e W i g n e r
distribution
[21,25].
(~P)[z]
from
= <exp
= <exp
extend
the
P = af
z,(exp-z)af functions
Hilbert-Schmidt
by
we h a v e t h a t
~(H+H*) ,
(exp z)> = a [ z ] - f [ z ]
~P[-]
6 ~(K+~*)
enlargements
we g e t
H
and
in denote
the the
usual space
that 1
! where
~
is the g a u s s i a n
Theorem
8.11:
measure,
F o r all
defined
on
P,Q 6 F0(H+H* )
l i p , Q~ =
7rP[ z ] .TrQ[ z ] . ~ ( d z ) K
~
.
we h a v e t h e i d e n t i t y
way of
87
where then
for
simplicity
of
automatically
understood
to a H i l b e r t - S c h m i d t
Proof:
notation
we
have
that
enlargement
the
H
omitted
the
functions
usual
are
It is
extended
from
H
.
We define =
(P,Q)'
~P[z].~Q[zl-~(dz) K
Thus
for
af
,bg
e F0(K+H
v(af
)[z]-v(bg*)[z]
)
w e get
that
= a[z]f[z]-b[z]g[z] = a[z]g[z]-b[z]f[z]
and
then
by v i r t u e
of
the
complex
(af Using ~x+y
the
elements
* , U + V *~'
=
from
+ <xv,®>
<X,U>
= <x,u>
It operators
remains with
generators
af
*
*
(M+(x+y*)af
+
to
,bg
*
+
,bg )'
= (e
H+H
that
6 F0(H+K
*
)
= <xag,bf>
= (af
,u+v }
M+(x+y*)
and
We
prove
*
For
x+y 6K+H *
- ~(y+x
+
b)f>
,(x b)g
get
+
*
(x+y)af
we
,
we get
(,}' * *
=
to
(ag[z]).(bf)[z]
=
= (x+y
show
respect
wave
,bg }'
the b a s e
=
+
+ (af
,b(y g)
= (af * , M ( x + y * )bg*}
While
proving
Corollary
theorem
8.12:
On
8.11
the
the
following
generators
af
that (af
,bg } = < a g , b f >
corollary
,bg
was
e F0(H+~
shown:
)
we
have
88 Lemma
8.13: The operation F0(H+H
is a complex
conjugation (S
,T>
) 9 P in
,P
£ F0(H+H
(F0(H+K),(,))
= (T*,S)
for all
,
)
i.e.
S,T 6 F0(~+H
)
Proof: As usual we check the identity on the generators af
,bg
6 F0(H+H
)
We have ((af
)
,bg
)
= l[fa
,bg
= ~af
,(gb
Notice
> = ) )
=
= ((bg)
that the same result
,af
>
concerning
the algebra
0(FOK),(,) N
has been proved in lemma 8.3. w
From the d e f i n i t i o n
Theorem complex
same,
8.14:
conjugation
In
of
F0(H+H
the
Bose
the
)
we get the following
algebra
~--product
and
F0(H+H the
),§,(,)
Wick
product
results.
with
the
are
the
thus P~Q = :PQ:
Theorem
for all
8.15: The Bose algebra
the Bose algebra
F0(H+H*),(,)
p,Q 6 F0(H+~
8(FoH),(,> N
)
is the ~--picture of
Chapter
9:
In
Wave
this
F(H+K
)
,
lemma
8.] 3.
representations
chapter
equipped
In c h a p t e r
we
with
shall the
:exp(x+y for the c o h e r e n t
vectors
Let us d e n o t e
in
*
study
complex
8 we d e r i v e d
of
F (H+K)
some
wave
representations
conjugation
,
discussed
of in
the e x p r e s s i o n
+
): =
(exp x ) e x p
y*
[(FOH )
by co
exp§(x+y*)
=
~ (x+y*)§n/n! n=0
the c o h e r e n t
vectors
of
(x+y*)§n
F(~+H*)
For
P 6 F0(~+~* ) = O(F0~ )
= §(:(§-~(x+y*))n:)
we have
= §(:(x+y*)n:)
and thus c0
(exp§(x+y),p)
=
~ ((x+y*)§n/n!,p~ n=O
=
~ ~§ ( : ( x + y * ) n : / n ! )
, p~
n=O =
~{: (x+y
*)n:
/n!,§- ~P}N
n=0 = {:exp(x+y
):'§-~P)N
= (§:exp(x+y):,P}
Hence exp§(x+y In c h a p t e r
) = §:exp(x+y
8 we c o m p u t e d
the r e d u c e d
§ = exp(-
): expression
for the o p e r a t o r
e(en)e(en) ) ,
n=l where
{en}ne ~
is an o r t h o n o r m a l
basis
in
K .
T h e n we get
90 co
co
e(en)e(en):exp(x+y
): =
n=l
e(en)~{en,X+y
)N:exp(x+y
):
n=l co
=
~ (en, x+Y*)N" (en, x + Y * ) N :exp (x+Y*) : n=]
=
~ <en, x><en,Y> :exp (x+Y*) : n=l
=
~ <en,X>:exp(x+Y
*) :
n=l
= :exp(x+y
):
and hence *
exp§(x+y P 6 F(H+H
For
)
) = e - < y , x > . .exp(x+y
the value
P[x+y and
hence
the
complex
of
P
in
representation
):
x+y*
] = (expg(x+y),P~
wave
*
6 H+K
is
, with
p,Q
6
F(H+H
gives
)
the i d e n t i t y (P,Q} where
~
examine of
transformation
~
.
1
of
)
the H+H*
the
P 6
omit
the
FO(•+?{
functions
equipped
as the d o m a i n
with
,
are
correspondence
extended
of integration.
between
by
real ,
wave
and the
the
real
part
of
with
H+H
respect
to
the
, i.e.
)
notation
) = §P[z+z
the
the c o n j u g a t i o n
we
compute
P.(z+z
)
I zeK }
the from
qJ-value chapter
of 7 and
P
z+z
in
use
P(z+z
We have P(z+z
to
.
~ = { z+z* For
1
] - Q [ x + y * ] ' ~ ( d x ) ' ~ K ( d~y )
that
closer
F(H+H
denote
conjugation
P[x+y
enlargement
representation
We
I
indicates
Hilbert--Schmidt
We will
=
] = ~.expg(z+z
= (:exp(z+z
):'P)N
),§P)
= ~P[z]
= < e x p
= ~(af
) ,
z,f>
)[z]
representation
of
F(H+H
)
yields
for
P,QeF(K+K*) (P,Q) where
the
functions
=
P(z+z
are e x t e n d e d
)-Q(z+z
)-~(z+z
) ,
to a H i l b e r t - S c h m i d t
enlargement
of
over which we integrate. Let with
H~
denote
the s p a c e
H
considered
as a real H i l b e r t
space
inner product ~
T h e n the u n i t a r y
=
Re()
=
½(
+
)
mapping ]
*
HA 9 z transforms
the
measure
Hilbert--Schmidt (P,Q)
~
enlargements
=
P(z+z
of
F(~+H
,
)
9.1:
We denote
The Weyl
the
H ,
).~N(z+z
fundamental
measure
i.e.
for
) =
get
for
the
operator
we h a v e
~P[z].~Q[z].~(dz)
~
extends
to a u n i t a r y
mapping
b y the s a m e symbol.
We ,
e6K
,
on
FIH
is of the f o r m ,
(exp e) e x p ( - e
Campbell-Baker-Hausdorff )
)
on
(~)
the e x t e n s i o n
p 6 FI(K+H
sitting
I
W e = e-½1]e][ by
,
P,Q 6 F 0 ( H + H
2. + Thus
~
theorem.
The t r a n s f o r m a t i o n L2
onto
of
).Q(z+z
and we get the f o l l o w i n g
Theorem
into
and
z6H
formula
and
) some
calculations
we
92 ~P[z]
= <exp
z,(exp-z)P(exp
= = < e , e x p
z ((exp-z)P(exp
P exp
z>
vP[z]
= < e , W _ z P W zO>
z))>
= Thus
by which
~P
operator
W
in
--Z
On t h e *
since
is the s o - c a l l e d
vacuum
expectation
v a l u e of the
Z
generators
af
*
~(af and
P W
zeH
,
6 FO(?~+H
)
we have
*
"k
) [z] = ~ ( f a the
)[z]
elements
af
~P For
P,Q
7rP[z]-TrQ[z]
= f[z]-a[z] are
= ~P
6 I" 1 ( K + K )
= P(z+z
and
)-Q(z+z
:PQ:(z+z
in
for all z6H
F(?{+H
)
P 6 F(K+K
)
l-§Q[z+z
] = §(:PQ:)[z+z
) = ~:PQ:[z]
= ~(af ,
)[z]
,
we get
we have
) = §p[z+z
= (§P)§(§Q)[z+z =
total
= a[z]-f[z]
]
]
,
and h e n c e ~(:PQ:) We
finish
introduced
the
x+y
Using
chapter
in c h a p t e r exp§(x+y
for
= (~P)'(~Q)
for all
returning
8. C o n s i d e r
*
P,Q e F I ( K + K to
the
operators
§
and
§-~
the i d e n t i t y
) = §:exp(x+y
*
): = e
-
:exp(x+y
*
):
6 K+K
the
fact
that
:exp(x+y*):
Campbell-Baker--Hausdorff
=
a+(exp
§-le+(exp
x) e x p y
Approximating
f,g£FiK
anti-normal
to
the
operator, §-1(fg*)
exp
*
= e 0 n
llEnflll = I ( E n f ) ( ~ ) ' ~ n ( d ~ ) n An a r b i t r a r y
f
can be w r i t t e n
f+
and
f-
= J f(s)-~(ds) ~
uniquely
are n o n - n e g a t i v e
assume
= llfllI
as ,
integrable
f+-f-
i.e.
and
f = f+ _ fwhere
function,
functions
such
that
= 0
Ifl = f+ + f-
.
Then we have llEnfUl = IIEn(f + - f-)lll ~ llEn(f+)lll + llEn(f-)lll = llf+lll + llf-llI
= IIf÷÷ f-lll = llifllll = ilfltl
Definition there
exists
an
4A: A n6~
function
f
defined
such that f = fOPn
,
where
~n Pn
: ~
on
~
is c a l l e d
tame
if
101 denotes
the
natural
Notice the
first
projection.
that
n
if
f6L1 (~)
variables,
is
a tame
function,
depending
only
on
sets
in
then E f = f n
Lemma
5A:
Proof: ,
Since
every
of
of
tame
the
Borel
combinations functions
The
cylinder
function
6A:
sets
are d e n s e
sets can
indicator
cylinder
Theorem
functions
form
be
(Jessen)
L I (~)
a base
for
the
approximated
functions are
in
of
by
cylinder
Borel finite
sets.
linear
Indicator
tame.
For
f6L I (~)
Enf
~ f
we have
in
that
L 1 (~)
n~
Proof:
Consider
6>0
Find
a tame
I] f - g l l l As
g
first
is tame, N
there
variables.
exists For
an
n_>N
function
g6L1(7)
such
that
< 6/2 N6~
such
we h a v e
that
g
depends
only
on the
that
Eng = g
and HEnf
- fl] l
S2m + 2Sm(Sn-Sm) we can continue n _> ~ ~ [s2 + 2Sm(Sn-Sm)]~(dx) m= I Qm as the inverse images by the We construct the sets Qm ' m6~ I
function
104
f(x) = the smallest
k6~
= inf { k6~
such that
ISk(X)[
]Sk(X) l > £
> 6 }
Hence Qm : f-] (m Notice coordinates
that
the
= { x6~
function
I f(x) = m } .
IQm
only
function
only
on
the
Since the functions
sm
first
m
Xl,X2,..,x m .
We return to the integration. depend
depends
on
the
(Sn-Sm)
first depends
m
variables
Xl,X2,...,x m
only on the variables
and
,
and
IQ~ the
Xm+1,Xm+2,...,x n
,
we get by the Fubini theorem
f
Sm(Sn-Sm)~(dx)
= ~ (IQm'Sm)(Sn-Sm)~(d~) N~
Qm
= ~ IQm'Sm~(d~) Sn-S m
being linear in the variables We estimate
then
f(x) = m ,
= 0 ,
Xm+],Xm+2,..,x n
the second expression using the fact that if and hence s~(dx)
n
~ (Sn-Sm)7(d~)
Sm(X) 2 ~
> 6 . 62~(dx)
xeQ m ,
Then we get = 62~(Q m)
~Qm Qm we return to the first estimate n n n
~ a~ ! ~ k=1 m=1
s~(dx)
!
m
~ 62~(Qm)= m=1
62~(m~i
Qm)
= 62~( ~ x_6~~
there exist
m = x n <X
on
and
is t r u e
en = where
n=!
Let
..
us d e f i n e
,0,I,0,
the
n-th
for
-x =
a
linear
..
) ,
position
,
and
( x 1 ' x 2 ' ..,x n '" .)6~
measurable
functional
k
defined
have
I k ( e n ) I 2 < co
n=1
i 2)
k ( x_ )
co k ( e n ) < X_, e n >
=
a.e.
in
N
n=1 Moreover,
the
Before the
representation
proving
translation
the
is u n i q u e .
2
theorem
we
need
some
preparations.
We
define
operator T
Z
for y6R ~
by setting (T f ) ( ~ ) where
f
denotes
a
function
defined
: f(~-Z) on
N
, Moreover,
we
introduce
a
107
subspace ~0
of
=
~
,
x6~
there
Theorem following
5B:
exists
an
Consider
a
statements
are
~(X
Proof: where
~
is
It
is
the
such
measurable
that
xn
set
B
=
0
for
C
}
n>N
~
Then
the
equivalent,
(I) (2)
N6N
~(B)
> 0
.
+ B)
> 0
for
all
to
see
that
easy
gaussian
measure
Z£~ 0
the
~n
theorem
in
the
holds
finite
in
the
case
dimensional
space
~n Let are
us
cylinder
first sets.
prove
the
Consider
restricted
the
form
of
theorem
5B,
where
B
set co-- n
C = B ~ ~ for
an
ne~
arbitrary consider m result
and y6~ 0
cases.
< n
:
> n
Borel
there
two
Then
follows m
a
by :
the
We
B •
an
regard
y
finite
~m-n
is
prove
the
=
a
B C
exists
rewrite C
where
we
set
B
Assume
m6~
as
such
being
dimensional the
set
~-n
~
~n
Borel
C
in
that
in
~(C)
>
y6~ m
the
space
~n
0
.
We
have
,
and
For to
the
case. as
= B ~ ~m-n
set
that
8) E ~ - m
Rm
Then
we
, refer
to
the
case
m
~n To integrable an
m6~
function such
that
full f
version
: ~
y6~ m
~
.
If
of
the
theorem,
Consider n > m
we
have
TMEn(f ) = EnTz(f ) , where
(Enf)(zl,z
2 .... z n)
= ~
f(~+~)~-n ~-n
and
z = (zl,z2,..,Zn) By
the
Fubini
theorem
the
function
(d~)
[6~ 0
take ,
a
then
non-negative there
exists
108 E
is i n t e g r a b l e
over
~n
with
n
f
: ~n
respect
(Enf)(Z)~n(dZ)
, C
to = ~
~n Notice
that
~n
and
f(x)~(dx) ~
f 2 0
implies
that
Enf 2 0
Assume that
f(£)~(dx)
> 0
R
Hence (Enf)(z)~(dz)
> 0
for all
n6~
.
~n U s i n g the t h e o r e m
for the
finite dimensional
case and the r e l a t i o n
TzEn(f ) = EnTz(f ) , we get t h a t
~mn
(EnTxf)(z)~(dz)
for all
> 0
n6~
.
Hence 0 < ~n(EnTxf)(£)~(d~) Setting
f = IB ,
Corollary
the t h e o r e m
6B:
= ~mTZf(~)~(d~)
follows.
For a r b i t r a r y
linear
measurable
X
we
measure.
We
functional
have that
~0 c ~(x) where
~(k)
denotes
Proof: will prove
~(k)
the d o m a i n
of
contains
linear
a
,
k
subset
E
of
full
that ~0 c E c ~(k)
Assume define
that
there
exists
an
~0 £ ~0
\ E .
the sets E t = tx0 + E = {tx0+ x
] ~6E}
For positive
t
we
109 Since
E
is a linear
different
indices.
set,
the
sets
Et
measures.
is
a
family
of
This contradicts
Proposition N
. T h e n we h a v e
are
pairwise
disjoint
for
Using theorem 5B, we get ~(Et)
Hence
Et
> 0 .
pairwise
disjoint
the fact that
7B: Denote by
h
~(~)
sets
with
positive
=
a linear measurable
functional
on
that
tk(en)l 2
0
we get n
exp(i-u-h(~))~(d~)
= ~ exp[i.u
R
k=l
~
= k~=nl ~ e x p [ i . u . h ( e k ) t -
n
it2 ]
dt.
~
exp(i.U.kn(X,)v(dx )
;
= -~ exp(-~u2[X(ek)l k=] Elementary
~ h(ek)xk]exp(i-U-kn(~))~(d~)
2)
computation
exp(i-U'Xn(~))~(d~) ascertains
that
110
# exp[i.u.h(ek)t
exp(-½u21A(ek)l 2
- ½t 2] dt
2~T Using
this we get that n
0
By the d o m i n a t e d
lim ~ exp(i.lk(x))~(dx) iR~ n--~
= ~
convergence
lim iit ~
= ~
theorem
we get
exp(i.~A(x))~(dx)
n------~
1.~(dx)
= 1 ,
0o
IR
which
is a c o n t r a d i c t i o n .
Proposition
8B: C o n s i d e r k(en)
a linear m e a s u r a b l e
= 0
for all
n6~
functional
h
•
If
,
then h = 0
Proof:
k(en)
= 0
for all
a.e.
n6~
implies
that
u
Let
Since
E
denote
X
the d o m a i n
is linear,
is s y m m e t r i c ,
of d e f i n i t i o n
and d e f i n e
I
h(x)
~ 0 }
E- = { x6E
1
h(x)
~ 0 } .
we have
that
E + = -E-
, and
since
the m e a s u r e
we get
it is o b v i o u s
=
x
w(E-)
.
that
~(E +) + ~(E-) Since
k ,
E + = { x6E
w(E +)
Then
for
+ E+ = E+
for
> ~(E)
every
= I
x6~ 0
and
likewise
for
E-
,
111
both
I
and
I
E+ of
variables.
either
0
are
constant
From
or
the
1 ,
Kolmogorov
and
likewise ~ ( E +)
and
with
respect
to
every
finite
number
E-
zero-one with
law
E-
= ~(E-)
we
get
that
w(E + )
is
Then
= I ,
hence ~({ x e E
We
are
Proof remains
to
now
k(x)
ready
(Theorem
prove
2),
= 0 })=
to p r o v e
4B):
~(E+
theorem
Since
I)
D E-)
= I
4B.
amounts
to
proposition
7B,
it
i.e. o~
k(x)
=
~ h(en)<X,en >
a.e.
n=l
Since
by p r o p o s i t i o n
7B
IX(en)I
2
< ~
,
n=]
theorem
3B a s c e r t a i n s
that
k(en)<X,en > n=l
converges We m u s t
a.e.
prove
in
N~
that
and
A =
h
A(ek)
defines o
=
For
a linear keel
we
measurable
functional
A
.
have
~ A ( e n ) < e k , e n > = A(ek) n=l
By p r o p o s i t i o n
8B we
conclude 00
h(x)
= A(x)
h ( e n ) < X_, e n >
=
a.e.
n=l
For measure
of
arbitrary sets
of
functionals
the
form
kl,..,k n
,
we wish
to
calculate
the
112
{ x6~ ~ where
ak,b k
I ak< k k ( X ) < -
are r e a l n u m b e r s . [A>c]
Lemma
in
~
Consider
9B:
with
values
to a f u n c t i o n
in
bk
= { x6~ ~
If
} ,
often use the shorter
I ~(x)>c
a sequence •
, k=1,.-,n
We shall
}
{f n}n= I
{fn}n~ I
notation
of m e a s u r a b l e
converges
almost
functions everywhere
f , i.e. fn
t h e n to e v e r y
n
c6~
~ f
there
pointwise exists
for a.e.
a sequence
ck
_x6~ ~
,
{Ck}k6 ~
fulfilling
~ c k
and ~[f>c]
Proof : zero--measure
Assume
= lim lim ~ [ f n > C k ] k n
that
[ f=c ] co
set on w h i c h
{fn}n= I
=
0
does
and not
denote
converge
by
to
f
M .
the T h e n we
get [fn>C] pointwise
on
the
By u s i n g
set
~
\
(M U
the d o m i n a t e d
n
[f>c]
[f=c])
,
convergence
~[fn>C]
i.e.
almost
everywhere
on
t h e o r e m w e get
~ ~[f>c] n
The
case when
W e find a s e q u e n c e
the
set
{Ck}ke ~
ck
~ c
and
[f=c]
has p o s i t i v e
of real n u m b e r s ~[f=ck]
= 0
measure
now
follows.
fulfilling
for e v e r y
ke~
.
k This
is
possible,
uncountable
family
contradicting
the
of
since
there
disjoint
fact t h a t the m e a s u r e
would
sets ~
with
otherwise positive
is finite.
exist
an
measure,
113
By sets
applying
[f>ck]
the
above
and using ~[f>c]
The measure
established
the dominated
= lim v[f>ck] k--~
to
convergence
the
zero--measure
theorem,
we get,
= lim lim ~[fn_>Ck] k--~ n--~
of the sets ~[f1>cl , f2>c2
where
result
fl ' f2''''fm
are
,.., fm>Cm ] ,
measurable
functions~
can
be
calculated
in a
similar way. We shall ~
, denoted
apply
by
the
k .
lemma with
a linear
It has been proved
earlier
cO
X(_x)
by
Then by lemma
AN
the
=
anX n
with
in
that
an
n=1
sum of the terms
with
indices
from
I
to
N
.
9B we get v[k>c]
where
functional
cO
n=1 We denote
measurable
the sequence
{Ck}k6 ~
We now calculate n ~[kn->Cm ] = ~{ x6~m
1
= lim lim V[hN>_Ck] k---~ N---~ converges
to
c6~
,
.
~[hN_>Ck]
~ akx k -> c m } k=1
= (2~) -½n ~
lM'exp(-½(x2+'''+x2))dx1"'dXn
'
~n where n M = { xE~n By
choosing
spanned
an
orthogonal
[
~ akXk _> Cm } . k=l transformation in
sending
the
line
by ( al,a 2 .... a n )/lla_nll2 , n into
IIaoII2 = vector
k=1 in ~n
,
the
line
spanned
by
the
we get by using the transformation
first
natural
theorem
basis
114 co
=
exp(-½t2)dt
= (2~) -~
exp(-½t2)dt
.
Cm/I] -an II 2
{ t_>Cm/I[--an ]I2 } 0o
We
choose
k
with
Ilxl12
2
=
1
,
.
i.e
~
a n2
1
=
By
first
letting
n=l
n ,
and
m
afterwards
,
go to
infinity
the
above
expression
reduces
to ~[k>c]
= (2=) -~ ~ exp( -½t 2 )dt
,
C
with
k
being
a
linear
measurable
functional
in
Nm
with
m
IX(en)l 2 =
1
n=]
The e x p r e s s i o n
can e a s i l y
~[kl>Cl .... km>Cm]
be e x t e n d e d
= (2~)-½m i
... i e x p ( - ½ ( x ~ + . . + x ~ ) ) d x l
cI where
kl,..,k m
denote
linear
to -.dx m ,
cm
measurable
functionals
in
Nm
all
,
with 0o
Iki(en) I2 =] and
the
i
One
.
vectors hereby
A property
simple
n=l i
{a n obtains
set
for
= Xi(en)}n6~ an
of the m e a s u r e
orthogonal
orthogonal
theoretical give
argument
kl,..,h m
= (2~) -~m ~
are m e a s u r a b l e Iki(en)
different in
together
tRN
with
with
the
i {a n = k i ( e n ) } n £ ~
N_~m .
additive
bm . .. ~ e x p ( - ½ ( x 1 2+ . . + x ~ ) ) d X l . . d X m am
functionals
12 =1
for
in
~
i=1,..,m
with ,
n=1 and
indices
us the e x p r e s s i o n
aI where
for
transformation
bI ~[a]12
n=1
the extension
We have
measurable
os
amn2 =
the
a weak
as
by mn
the
is
taken
~ 62
domain
a Since
A
are
transformation
: 62
uniquely
extending
given
then
I
L2
of
~
A Then
n
if e l e m e n t s
{amn}m,n6~
transformation
A
number
,
k=1 , . . , n
.
k=1 .... n }
.
sets
of e q u a l
We define
measure.
117
bI bn ~(B) = (2~)-½n ~ .. ~ exp(-~(x1+..+Xn))dx I 2 2 I ..dx n aI
an
and that U(B) = (t ~ 6 ~ where
ak
= <Jbn,X> = hn
for
Let us define co
t2 = { X 6 C ~
~ ,Xn,2 < ~ } n=l
~2 = { X 6 ~ ~
~ kn,Xnl2 < m } n=l
x6K
(cf.
[23])
and the positive
121 By i d e n t i f y i n g ~
t 2
~A~2 via
the
orthonormal
measure
~
on the
,
a
basis =
½,1
corresponding
{bn}n6 ~
in
on the
Borel
,
Borel
sets
in
~(~) a
Since
unitary
weakly
measurable
is i n v a r i a n t under
the
spaces. the
under
the
showing
we h a v e basis to
show
selected
Hilbert--Schmidt
that
HI
enlargements
then
the
get
62
and
the
unitary
between
does
measure
not
two
~
can
different
Hilbert
depend with
does
This
~
is i n v a r i a n t
for i d e n t i f y i n g
gaussian
orthogonal measure
7~
maps ~
to
the
the m e a s u r e
denote a
extend
~2
enlargement.
exist
gaussian u the m e a s u r e
as
then
in
used the
H2
there
and
~
the m e a s u r e
K
that
and
of
that
in
can d e f i n e
of
onto
transformations,
like
if
62
extensions
we
2) = I
these
orthonormal would
of
,
transformations
In p a r t i c u l a r ,
chosen
= ~(Z
, sets
linear
corresponding
We on
transformations
~2
H
not be
on
62 depend
done
by
Hilbert-Schmidt
Hilbert--Schmidt
enlargement
fulfilling c H.
i=I,2
.
1
If in H2
C
,
f
denotes
and
fl
and
respectively,
almost
then
Hilbert-Schmidt is f i n e r
extends
f2
are fl
function
continuous
and
f2
are
defined
on
H
with
extensions
of
f
to
equal
on
~
,
values KI
hence
and equal
everywhere.
Definition
KI
a continuous
|F:
Let
Hi, I
enlargements
than
K2
to a c o n t i n u o u s
of
if the
and
a Hilbert
and space
identity I : H -
, H
one-to-one
map
I : H], I
~ H2, 2 .
H2, 2 H,
We
denote say
that
122
It is o b v i o u s a constant
C > 0
that
such
if
HI
is f i n e r
K2 ,
than
then
there
exist
that
JIxJl2 ~ c-JJxlJl for
all
norm
of
xeH I
.
The
in
smallest
C
fulfilling
the
above
is
the
operator
~(Kl,K2)
Lelmaa 2F:
Consider
seminorms
{]]-][n}n6 ~
and define
ll'li*
in
well
known
setting co
n=] If a s e q u e n c e
fulfills
{Xp}pe ~
IIXp - XqII.
, 0 p,q
and
for every
nE~
JlXpitn
. 0 , P
then
llXpl].
, 0 P
Proof: method
of
The
proof
verification
amounts
that
the
to
an
adjustment
countable
direct
sum
of of
the
Hilbert
spaces
is c o m p l e t e .
Lemma K,
.
3F:
Let
HI, I
denote
is a c o n t i n u o u s enlargement H2
2)
k
Hilbert--Schmidt
enlargement
of
If k =
])
a
linear
H2, 2 is
finer
functional, of
than
is c o n t i n u o u s
H,
: H
, C
then
there
such that
HI in t h e m e t r i c
of
l
<en,-> I = Expanding
aeK
we
[lenll~-<en,->
•
get
a =
~ an-e n
with
=
an
<en,a>
and
lan 12
= n=]
numbers
Choose
a
with
m]
strictly
increasing
= I
that
such
n=1 For
x,yeH
we
then
<xcy>.
sequence
{mn}n6 ~
of
k=m~
define
m~+~-1
m0+~-1
k=mn
k=m~
= n=1
and
mn+~-] n=1 Since
by the
Cauchy--Schwarz
k=m0
inequality
÷
II x II .2
=
i 2n. mi n=l
k=mn
[ak 12 < °°
natural
124
2
in
for
<x,y> 2 = <x,y>. + <x,y>] and the corresponding
For
norm,
H
setting x,y6H
I["I12 '
2 IIx II~ = IIx I1.2 ÷ fixrlI consider the expression
xeH
o0
k(X) =
~ an.<en,X> n=1 mn+~-I
co
ak'<ek'x> = n=l k=mo
mn+
~ (v~)-n" (v~)n ~ ak'<ek 'x> n=1 k=m~
Using the Cauchy--Schwarz
inequality, we get mN+1-1 h(x)l -< ~ (v~)-n" (v~) n ~ ak'<ek ,x> n=1 k=m~ -< [ ~ 2-n]~" n=]
= .e.
k
llxJl.
[ ~ 2n n:1
~ ak'<ek ,x> k=mn
in the metric of
Since ~0
~0
[Iep[l 2. = p=1
]
_< IIxH2 ,
is continuous
CO
mn+
I --I
2
~ ~ 2 n" ~ ak" <ek, ep > p=] n =1 k=m~ ~ mn.~-] 2
:
p=1 n=1 ~
n=l p=l
I --I
k=m~ mn+~-1 k=m~
n=1 p=1 k=mn ~o mn+1-1 2
125 2
IIep II2 Then of
the H,
completion
H2
is a H i l b e r t - S c h m i d t
enlargement
.
It r e m a i n s that
< ~
p=1 H, 2 of
to p r o v e
that
H2
is finer
than
It is o b v i o u s
HI
the i n c l u s i o n m a p p i n g I : K, 2
~ H, I
is c o n t i n u o u s . To p r o v e is s u f f i c i e n t
that
the
continuous
to s h o w that
for each
[]Xp
1)
-
of
sequence
..$Xp~pe~ C H
xqll2
,
I
is o n e - t o - o n e ,
extension
0
P,g and
2)
IlxpU]
~ o
w
~ o
.
P we
have
11xp II 2 P
From
2) it follows
that
for e v e r y
k6~
<ek,Xp> l
~ 0 P
and in p a r t i c u l a r
that
<ek,Xp> for e v e r y
ke~
= l]ekll?2-<ek,Xp>1
By l i n e a r i t y
we get that
m~+~-I 2n" For
n6~
and
x6~
~ k=m~
we d e f i n e
2
ak" <ek'Xp> [
~ 0 P
the s e m i n o r m s
[Ix I[n2 = 2n"
[I" []n '
mo÷~-1 ~ ak" <ek'x> k=m n
Then by lemma
> 0 P for e v e r y
2F w e get that
ILXpll.
~ 0 P
2
n£~
such
that
it
126
and t h e r e b y
II Xp II 2
0
,
P
Lemma
4F:
Hilbert-Schmidt a
fixed
denotes
{Kn,n}ne~
enlargements
Hilbert-Schmidt
Hilbert--Schmidt n6~
If
of a H i l b e r t
enlargements
enlargement
~,~
finer
sequence
K,
space
K0, 0
a
all
then
,
than
of
finer
there
Kn,n
than
exist for
a
every
.
Proof: the i n c l u s i o n
Take
an
orthonormal
mappings
from
basis
~, II" II to
{en}n6 ~ ]{, If- IIn
in
K,
Since
are H i l b e r t - S c h m i d t ,
we
have 00
Cn : It follows
that
~o
that
llxlln Note
2
Ilepll n )~
with
K,
IIenlI]-<en,->
lemma
K,
x,yeH
in
the
in
inner
a
hence H
.
Hilbert--Schmidt H2, 2
and
enlargement
such
that
the
H3, 3 product
= <x,y> I + <x,y> 3
norm
IJxjj : IJxlJ + IJxjj Then
we d e f i n e
K
to be the
~llenll 2 = n=1 which
makes The
the
inner
inclusion
completion
~
l'en'l~ +
n=] product
well
of
~
H,~
.
llenll~ < "
We h a v e
,
n=] defined.
mappings I : H,
H],]
I : K,
~ K3, 3
and
are
obviously That
continuous. their
extensions
are
one-to-one
follows
from
the
fact
that
129
the
functionals
the and
If- U3 hence
-
norm
dense
When a complex
{hn}ne ~ and
~n
Hilbert
continuous these
gaussian
space 7~
in b o t h
functionals
(Hi,i)'
considering
and
H,
will
then
where
we
a real
z =
denote
• e x p ( - ( I Z l 12 +
(Zl,Z2,...,Zn)
6 Cn
in
let
and
(H,)'
H,
denote
space.
the m e a s u r e
...
+
with
dz = dxldYldX2dY2...dxndY
In a s i m i l a r
n
in
[Znl2))dz
~2 c
given
way
~H
denotes
,
zk = x k + i.y k 6
indicates
the
the
in
Lebesgue
~2n
measure
in
~2
c
given
on
by I (2v)-~n-exp(-½(x~
where and
dense
always
Hilbert
integration
~n
are
If-IfI - n o r m
(H3,3)~
measures
and
the
by -n
and
that
in b o t h
The m e a s u r e on
are
x =
(Xl,X2,...,Xn)
dx = dxldx2..0dx n
+
...
2 + Xn))d ~
,
6 ~n indicates
the L e b e s g u e
integration
in
~n
I
Observe produced
that
the m e a s u r e
by considering ~
}{
~C
as a real
= ½(
on a c o m p l e x Hilbert
+ )
=
Re
Hilbert
space
with
.
space inner
}{
is
product
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Subject
i n d e x
real wave representation 2,3,53, annihilation operator ],3,6,8,10, 72,77,90,91 21,25,61,64,79,8],83,85,92 anti-commutator I second quantization 30 anti-normal 92 Stone 29,30 base space 1,2,4,]0,23,65,83,85, tame 100,101,102 total 16,53,66,67,92,118,120 87 Bose albebra 1,2,3,4,6,]0,11,20, ultracoherent 57 vacuum 1,4,10,20,23,65,83,85,92 22,23,65,79,83,84,85,88 - extende 36,65,83 value of 66,90 weak measurable linear trans-- Fock space 1,2,3 formation 115,116,121 Campbell--Baker--Hausdorff 50,5], 59,91,92 Weyl 1,45,52,91 coherent 2,33,34,43,44,63,64,66, Wick 22,45,59,79,81,83,88 83,89,92 Wiener 16,20 commutation 9,10,11,12,45,55,59, Wigner 3,86 60,65,82,85,93 complex wave representation 2,53, 66,69,70,71,78,80,87,90 conjugation 1,2,3,53,54,55,57,69, 73,74,76,77,80,84,88,89,90 creation operator 1,3,6,8,10,2], 25,64,79,8],83,85,92 cylinder set 96,97,]01,107 derivation 1,6,10,]],30 Fourier transformation 25,31,32,7] free commutative algebra 1,4 - product 1,4,18 gaussian content 96 measure 3,68,76,78,80,86,96,99, 107,115,116,117,118,120,121,]22 Halmos 27,94 Heisenberg 33,43 Hermite 2,31,32 Hilbert--Schmidt enlargement 68,71,76, 80,86,87,90,91,118,120,121,122, 125,126,127,128 Kolmogorov extension theorem 97 -- inequality 103,105 -- large number theorem 104 - zero one law 102,111 Leibniz rule 6,30,31,36,37,39,61,82 linear measurable functional 102,103, 106,108,109,110,111,113,114,115, I]7,122,123,128,129 Nelson 27 one-parameter group 29 4-- picture 65,84,88 product 59,61,62,63,64,65,78,88 value 72,90 -
---