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2
= <~z,p(~z)>
= <~z,q(~z)>
It is n o t
hard
to s h o w
=
that
the
(Ap) 2 =
More
e-½lzl2-exp
specific,
_
denotes
let
z
T h e n we h a v e
,p2 (~z)>
= <~z
= <~z,q2(~Z)>.
that Ap-Aq
i.e.
where
deviation.
~z yield
,
uncertainty
= ~
is m i n i m i z e d
, in the
Heisenberg
relation.
44
Example I .19B,
one
can
3.15:
In t h e
compute
the
algebra
F ~ n = L 2 ( ~ n)
,
defined
in
coherent
vector
z
,
,
exp
z6C n
function ( e x p z)
-) =
e-½11~II2
•
e v~<"
I
z>
o
~(. )
: Rn
) C
°
example as
the
C h a p t e r 4:
The W i c k orderinq and the Weyl relations
Definition operator for
a
Bk
Consider
, k=1,2,...,n
bk6H
ordering)
4.1:
We
define
:BIB2...Bn:
,
of
the
of
il,i2,...,i n
is
the
Wick
a
BI,B2,...,B n
form either
ordering
BIB2...B n
:BIB2;..Bn: where
operators
= BiiBi2
permutation
x,yeK
a+(bk )
or
a(bk)
called
the
normal
•
"
•
Bi n
of
;
the
numbers
e(bi)
1,2,...,n
,
a p p e a r on the right
e+(bj)
We then extend the ordering
For
each
setting
chosen in such a way that all the operators side of the operators
(often
,
linearly.
we shall often write , x+y
shortly
for e+(x)
Example 4.2: By induction :(x+y*)n:
(the
binomial
formula
Lemma commutation I 2
4.3:
for
For
it is easy to show that n ~ (~)'e+(xk)'e(Y n-k)
k=O commuting operators!).
every
pair
x,y6H
and
n6~
we
have
the
rules
[~(y)n,a+(x) ] = n.
3
=
+ a(y)
= -- n.
[ (x+y*) n , ~+(x) ] = n-
Proof:
, e(y)]
All
= - n - < y , x > - ( 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=0 [in]
2)
:(x+y*)n:
k
= n!.
(-½
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)
(~
= ( n - 1 ) !-
k!- (n-l-2k)
!
: (x+y*)n-I -2k) :~+(x )
k=O n-1 2 (i
(n-1)
1-
+
k!- (n-1-2k)'
* n-I -2k (x)
: x+y
}
:
k=0 n-3 2 (~
+ (n-l)!-
(n-1-2k).
:(x+y
* n-2-2k )
k=0 n-1 2
(i
(n-1)!-
a+
n-1-2k (x)
: x+y*)
k=0 n-1 2
+ (n-1)! i=l
<~,x>)i_ 1 (i(-~),(n+1-2i
!(n+]-2i)
* n-2i )
47 n-1 2 = n!- ~ n~2k k!.(n-2k)! (½
~+(x)
:(x+y*)n-l-2k:
k=O n-1 2 + n! ~
i (½
i=I n-1 2 = n!- ~ n-2k (½
We return to the first identity (x+y*) n = ( x + y * ) n - ~ + ( x )
+ (x+y*)n-ly *
n-| 2 = n'. ~ n~2k (~
2 n~2k (1
48 n-1 2 = n !- ~
n - 2 k (½
~+(x)
:(x+y*)n-1-2k:
k=0 n-1 2 2-i (½
+ 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=0 n-1 2 + n!. ~
2.k (~
k=0 and s i n c e
[½n]
= ½(n-l)
for
[½n]~ (~
The the
reader.
: (x+y
O
L k=0
case The
n
odd
. "')n-2k)
k!" (n-2k) ! of
even
proof
n
of
,
the
which second
is v e r y m u c h identity
is
the left
same, to
the
is
left
to
reader
as
well.
Definition
4.5:
For
x,y6H
exp(x+y*)
we d e f i n e =
~
on
F0H
the o p e r a t o r s
(x+y*)n/n!
n=O
:exp(x+y
* ):
=
~
:(x+y
*)n : / n !
n=O
The defined.
following
theorem
ascertains
that
the
operators
are
well
4g Theorem 4.6: define
(cf.
exp(x+y*)
evaluated
[19,I.16])
and
on elements
For every
:exp(x+y*):
of
F0K
,
x,yeH
are
and on
the series which
absolutely F0K
we
summable
have
the
when
operator
identity exp(x+y*)
Proof: For
n>m
We
consider
= e ½
elements
:exp(x+y*):
of the
form
o
am
a6H
i
i
we have
:(x+y*)n:(am)
=
n ~ (~)~+(xn-k)y*k(am)
=
k=0 m
k=0 n)a+(xn-k ) m! (k (m-k)!
=
m ~ (~)~+(xn-k)y*k(am)
m-k
k=0 m = ~ (~)( n! n-k m-k n_k)!-
We estimate n •'-~"
:(x+y * )n:(am)
n=m <_ ~ n=m
n=m o~ n=m oo
m m m -< ~ ~ (k)" n=m k=0
~(n-k)!
"
xn-k.am-k
m ~ (~).
n=m k=0 m =
~ [ (~) j
k=0 for
n=m
~n-~j
!
k=I,2 .... ,m . Consider the absolutely
summable series
<
meS 0
50 e½
62p= I
~ (½
and
4.4 we
get e½
TM)
=
~ 6p.(½
n=0 co
co
(½
=
~
( x +n! y
)n(am)
=
exp(x+y
*
) (a m )
n=O
Proposition
4.7: For :exp(x+y
on
x,y6H *
): =
we have the identity +
(exp x
exp y
*
F0H .
Proof: For
f6FoH
• :exp(x+y ) : (f) =
~
consider the serles
:(x+y*)n:(f) n!
n=0
_ -
~~
~n e+(x) k .y * (n-k) k! (n-k) ! (f) n=0 k=0
e+(x)n. = * n: ~ an~.,n(f ) e+(exp x) exp y (f) n=O n=O
= ~
x,yeK
Proposition
4.8:
we have on
F0K
exp(x+y*)
(Simple
Campbell--Baker--Hausdorff
the identities = e½
x) exp y*
= e-½
formula)
For
51 Proof:
The first
4.7 and theorem
4.6.
identity
is an easy
Then according
consequence
to corollary
3.12 we get
+ (exp x) exp y * = e -
identity
of proposition
x)
follows.
The operators exp(x+y are defined
only on
F0K
to
FIH
so that theorem
on
FIH
as well.
To every
e6H
.
) , :exp(x+y
It is easy to extend
4.6,
proposition
we assign
By the Campbell--Baker-Hausdorff
) : FIH formula
W e = e-½1eI2.e+(exp FI~
the above
identities
4.7 and proposition
4.8 hold
the operator
W e = exp(e-e
on
):
, FI~ (proposition
4.8) we have
e).exp(-e*)
.
Lemma
4.9: For every
e,f6H
we have on
FIH
the relation
W e Wf = e -i'Im(<e'f>).we+ f
Proof:
Using the Campbell-Baker-Hausdorff
W e Wf = exp(e-e = e
) exp(f-f
-½1eI2-e+(exp
and by using
corollary
e) exp(-e*)
e -½1f12 • ~+(exp
e)-e-<e'f>.a+(exp
-
).We+f
f).exp-e*.exp-f*
exp-(e+f) ,
= e -½(<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
basis
} in
FH
.
{en}n6 ~
in
the
Hilbert
54
M = { exp Consider According
to
x
I x6K
f6M 1
theorem
is r e a l
and
Ixl
It
is
sufficient
to
f
can
be
in
1 .12B
written
f = ~ ar'e~
} .
show the
f
that
=
0
form
.
r
Let
x6~
denote
a real
element
with
Ixl
exp
(i.e.
x 6 M)
Using
w
the
fact
that
0 = <exp
exp
x,f>
x
is
multiplicative,
= ~ ar.<ex p x,e~>
we get
= ~ ar-<~,ex p x*(e~)>
r
r rI
= ~ ar'<~,(<x,e1>o
+ el)
rn> "..." ( < X , e n > ®
+ en)
r
rl = ~ ar.<~,<x,e1> 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 +
]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)
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) =
m<~n,a>a 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>
+
+
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~ <~,a>n am-2n
for m>2n for m<2n
0 The
case
for
n-1
n=1
is clear.
Take
n>1
and assume
that
the result
We have
(hl) n a m = hi[
(h~) n-1
= hi[ {
am
]
m! (m-2n+2)! o
<~,a>n-1
a m-2n+2
for m>2n+2 for m<2n+2
]
holds
58 m! (m-2n+2):
(h.) n am : {
m-2n+2
h*_[ a
]
for m>2n+2
for m<2n+2 !
0 m! = { (m-2n+2
for m>2n for m<2n
0 which implies * )n a TM ( h
m <[,a>n am-2n (m~2n) !
~(
for m_>2n
= -
0
for m<2n
By using this result we get [m/2] 6*(a TM) ~ I i~* n a TM m = ,-~n__) n.~ _ ~ (_~h*)n a__n! n=0 n=0 [m/2] (_½)n m!
Theorem 5.6B: On *
[0 K
we have
+
[6_,a
*
(x)]
=
-
_*
6_
for all
x
x6H
Proof: From theorem 5.3B we get [(~h~)m,~+(x)]
= m. (~h~) m-I ~*
for all
me~ 0
Hence [ ( - ~i~*.m n_~ ,~
Since
and
e+(x)
x*
are
* + (x)] = [6_,e
+ (x)]
closed
~
=
_m(_½h~)m-]
operators,
I *m-] -m (-~h_)
~*
we g e t
~* /m! = -6 ~ x*
m=]
We will
extend
the operator
5_
to the whole
theorem 5.6B as the intertwining * + * 6_ (x) = (~+(x) -- ~(~))6_ For
n6~
this is extended to an intertwining *
on F0H *
6_ a+(X n) = (~+(X) -- e(~))n 6_
,
FIK
Rewrite
59 We w r i t e
briefly n
eXPn(X ) = ~
xk/k!
,
k=0 which
gives
that +
6L eXPn(e+(x)) The
limit
on the r i g h t
side
= eXPn(= exists
*
(x) -- =(x))
on
F0K
,
6_
so that we m a y d e f i n e
on
F0K 6* e x p ( e + ( x ) ) By the s i m p l e •
= lim 6 L e X P n ( e + ( x ) ) n
Campbell--Baker--Hausdorff
+
formula
-- =(x))
we o b t a i n
6L
on F0K
*
6_ exp a (x) = e x p ( e + ( x )
-- e(x))
= exp(--½<x,x>) This w a y
= lim e X P n ( ~ + ( x ) n
6_
commutation
extends relation
6_
e+(exp
x) e ( e x p ~) 6*
to the w h o l e from t h e o r e m
FIK
It is e a s y to p r o v e
5.6B holds
on
FIK
that
the
in
[20,
.
C: The @ - p r o d u c t
The chapter
~-product,
4],
cf.
discussed
paragraph,
paragraph *
+
[6_,~ can be e a s i l y
rewritten
(x)6_
Definition
5.1C:
For
we have p r o v e d * ~* (X) ] = -6_
appears
the c o m m u t a t i o n
rule
in the form of i n t e r t w i n i n g
= 6*(~+(x)
f6FoH
f~ by i n d u c t i o n
this
[10] as well.
In the p r e c e d i n g
which
in
:
+
we d e f i n e
FIH
,
)
the o p e r a t o r
FIH
as follows.
I) The o p e r a t o r
~
is the identity.
2) For
x6H
we have
3) Let
:-:
denote
and
n68
,
x~ = ~+(x)
+ x
the W i c k ordering.
we d e f i n e xn~ = :(~+(x)
+ ~*)n:
For p o w e r s
xn ,
x6H
60
4) W e e x t e n d
It on the
is e a s y
choice The
for
x6H
of
the operation
to
check
that
representation
commutation
linearly
to
FOH
the
definition
of
f6FoH
from theorem
.
of
f9
does
not
depend
.
5.6B changes
to
.
Lemma
5.2C:
For
every
f,g6FoH
the
operators
f9
and
g9
commute.
Proof:
By using
the
fact
that
the operator
6*
is
invertible
we
get 6_(xg)(yg) which
f~
concludes
5.3C:
contains
FIH f9
For every
f6FoH
,
FIH
,
and i.e.
< f~a,b
Proof: and
n6~ 0
< f~a,b
= a+(x)a+(y)6
,
the proof
Lem~a
the operator
= a+(x)6*_(yg)
It
is
on
the domain the
dual
n : < ~
> = < a,T~b
sufficient
to
>
for all
consider
> = <
:(x+~*)n:a,b
(~)~+(xk)(x*)n-ka,b
>
k=O n
= < a, ~
(~)a+(xn-k)(x*)nb
>
k=O = < a,xngb
the dual
operator
is
operator
identical
of
with
we have
.
> = < xn~a,b
of
> = < a,[gb
>
.
>
a,b6FiH
elements
f=x n
for
x£H
61
Thus
the
by the same
operator
f@
is closable.
denote
the
closure
symbol.
Definition
5.4C:
~--multiplication
In
F0H
we
Remark
5.5C:
variables.
the
introduce
commutative
by d e f i n i n g f~g = (f~)(g~)~
both
We w i l l
The
Since
for
~--multiplication (f~)~
= f ,
f,gEFoH
is
we have
.
associative
and
linear
in
the i d e n t i t y
f@g = (f@ g •
Proposition Leibniz
rule
5.6C:
relative a (f#g)
Proof: sufficient [a * ,xn~]
The
annihilation
operators
to the V - m u l t i p l i c a t i o n ,
= (a f)~g + f~(a g)
We shall
prove
to c o n s i d e r
that
f = xn
for
[a*,f~] x6K
i.e.
with
obey for
f,geFoH
= (a'f)@ and
n6~ 0
on a6K
n ~ (~) a* +(xk)
(~*)n-k
k=0 n = ~ (~)(k
on
:(x + x * ) n l a *
k=0 :(x
=
=
+
)n:a*
n n ~ (k)k
k=0 n ~ (n)k
we have
FOH .
o
n-k + a + ( x k ) ( ~ * ) n - k a
the
.
= a * :(x + ~* )n : - :(x + ~ * ) n :a * =
F0H
*)
It
is
62 n-I *
n
[a ,x 4]
~ ~
=
n + -i (i+l) ( i + I ) < a , x > a (x i) (x*) n-I
i=0 =
Theorem equipped
n :(x + x*)n-l:
5.7C:
with
The
6_
is
into
a
homomorphism
F0K
with
of
the
F0K
original
i.e.
6 _*( f @ g )
(61x n)
operator
@--multiplication
multiplication,
Proof:
= (a*xn),
= (61f) ( 6 _*g )
We shall
denotes
prove
that
multiplication
for all
6_(xn~) by
f,geFoH
.
= (6*xn)6_ _
the
element.
on
The
F0K
,
proof
where
goes
by
induction. n=0
n=l
:
Since
:
e~=~+(e)
By v i r t u e
,
we e a s i l y
get
6_(e~)
= 6*e+(e)_
= 6_ = (6_e)6_
of t h e o r e m 6 _*( x ~ )
Now the i n d u c t i o n
follows
xn~ which
.
5.6B we get =
+ (x)61 =
(61x ) 61
: An easy c a l c u l a t i o n
= (x~)(xn-1,)
shows
- (n-1)<~,x>(xn-2~)
that ,
implies
61(xn,)
= 61((x,)(xn-1@))
- (n-1)<~,x>61(xn-2,)
= 51(x~)6~(xn-1~)
- (n-1)<x,x>(61xn-2)61
= ~+(x)61(xn-l@)
- 61(~*xn-I~)
61
: e+(x)(6*xn-l)6 * - 61(x*xn-l~)
= [a+(X)(61x
6*
n-I ) - 61(x*xn-l@)]61
= ( ~ + ( X ) ( 6 1 X n-I ) + [6 * ,~ + (X) ] x n - I
:
(61a+(x).n-1)61
Though
the
~-multiplication
= (61xn)61
operators is
* )5_
defined
f@ only
, on
f6FoH F0K
,
act
We w i l l
on
FIK
extend
it
, to
the the
63
whole
F IH
.
Notice
that the operator 6 _* =
~
I * n (-~h_) /n!
n=0 is i n v e r t i b l e
on
F IH
with
the i n v e r s e
A, 6_ =
~
1
*
(+~h)n/n!
n=0
Definition setting
5.8C -
for e v e r y p a i r
We
extend
5.9C:
@--multiplication
over
E
f1.E6_*gl I
T h e C--product is still
commutative
on
FIH
relation 6_(f~g) holds
for all
Proposition
the C-product
5.10C:
For
For
zeH
between
x,y6~
(exp x ) C ( e x p y)
Proof:
= (6 f)(6_g)
f,g6F]H
L e t us e x a m i n e
coherent
vectors.
we h a v e
= e<X'Y>exp(x
we s h a l l c a l c u l a t e h~exp
z =
+ y)
6_(exp
z)
Since
z
we get (-½h~)nexp
z = (-½
z ,
thus g i v i n g
-½<~,z> 6*(exp
z)=
~ n=0
Hence we get
FIK
f,geFiK
fog :
Remark
the
(-½
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<~,x> -½<~,y> = e
e
exp
(x+y)
-½(<~,x> + <~,y>) = e
exp(x+y) -I < x , x > + < y , y > )
+½<x+y,x+y>
= e which
, 6_(exp(x+y))
.e
implies
+½(<~,y> (exp x)~ e x p y)
+ <~,x>)
= 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
FIH
•
by the f o r m u l a
= <6_f,6_g>
.
In a d d i t i o n x6H
.
can be s h o w n to h o l d on the w h o l e
):
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
<'>o
coincides
the
creation
*
= <6_f,e
+
with
the
= <e(x)6~f,6_g> = <6_f,6~(a+(X)
= <6_f,6*(x@)g>_ *
= <6"_f , 6 _*e o+( X ) g >
=
for all
f,g6FoK
So(X )
to
annihilation
*
(x)6~g>
of
The d u a l o p e r a t o r
We have
operator
+ e(x))g>
,
operator
65
and
the
commutation relation + [ a o ( X ) , e o ( Y ) ] = [a(x),a+(y)+a(y)] = <x,y>~
Hence
(FOK,<,>o)
algebra
with
called
vacuum
e
the @-picture Likewise
Bose
equipped
algebra
we
of shall
VIK,<,>
= <x,Y>oe with
and F0 H call
= [e(x),a+(y)]
base
the
for
all
x,y6K
@--multiplication
space
H
This
. is
a
algebra
new will
Bose be
. FIK,<,> o
the
@--picture
of
the
extended
Chapter
6:
In by
Segal
method
by
The
this
chapter
[16]
and
which
we
wave
will
Bargmann
this
Definition
complex
representation
describe
[I].
Concerning
representation
6.1:
For
the
is
f6FK
we
representation, the
priority,
introduced
define
introduced cf.
originates
the
value
[4]. from
f
of
The [20].
in
z6H
setting f[z]
Since of
the
f6FH
coherent is
Lemma
vectors
= <exp
are
total
all
6.2:
For
every
We
start
by
pair
showing
e(f)exp x6H
< ~(f)exp
Since
FH
,
the
representation
the
f,g6FiH
we
have
that
= f[z].g[z]
for
f6FiH
z = f[z]-exp
we z
have
.
w
and z,x
that
z = f exp
n6~ 0 n
set
we
get,
> = < exp
by
z,fx n
using
that
exp
z f)-
= f[z]<exp
z,x n
> = < f[z]-exp
xn
and
n6~ 0 }
x6H
z
is
> =
total
is
z = f[z]-exp
z
in
,
and = <exp
z,fg>
=
=
exp
z,g>
= f[z]-<exp
z,g>
an
isomorphism,
z * f)(expz *X n ) >
> = < o,(exp
= < ~,(exp
f exp
(f-g)[z]
f[. ]
zeH
Proof:
For
in
injective.
(f-g)[z] for
z,f>
= f[z]-g[z]
z,f> >
. FK
,
we
get
67
Notice
which
that
means
polynomial
that of
use
in
K
be
of the
f = xn
be
an
for x 6 H
= <exp
z,xn>
this
case
f[-]
a
zeK
,
then
=
k-dimensional
orthonormal
n6~
a
grade
homogeneous
n
.
basis
complex
in
~
.
Hilbert
space
Identifying
~
and
with
let
Ck
by
transformation K 9 z
where
and
f[z]
the v a r i a b l e
Let k {ei}i= I
if
zi
=
integration considered
<ei,z>
over
K
as the
, (Zl,Z2,...,Zk) ,
i=1,2,...k,
as the
we
corresponding
2k-dimensional
6 Ck
can
define
Lebesgue
Euclidean
the
Lebesgue
integration
space.
We
have
the
over
~k
following
lemma.
Lepta Let
k
6.3:
denote
Consider
the
a
dimension
finite of
= -k.[
K
dimensional .
For
all
of
subspace
f,geFoK
f[z].g[z].exp(_izl2)dz
we
H
.
have
,
k where
9 z =
and
(Zl,Z2, .... Zk)
zj = xj + i.yj
Izl 2 = dz
Izll
2 +
£ e
of
~2k
we
orthonormal
that
basis
in
Proof:
of
b y the
get
+
j=1,2,...k
...
k
indicates
integration
orthonormal
transformation the
IZk[2
+
= dXldYldX2dY2...dXkdY
change
transformation,
for
Iz212
the L e b e s g u e
Since
6 ~k
integration
over
basis
theorem does
not
~2k
induces
for the depend
a
unitary
Lebesgue
integral
on
the
choice
total
set
K
It is s u f f i c i e n t
to p r o v e
the
{ en- I n 6 ~ k
} ,
lemma
on the
of
68
where
{el,e2,..,ek}
is a n
orthonormal
n nl e- = e I It h a s
been
proved
n!
Using
and
= nl ! n 2 ! . . n k !
lemma
6.2 we
en[z]
in
and
K
n2 nk e 2 ...e k
that < e~,e~
where
basis
> = n!-6
6
is
~,~
the
'
Cronecker
delta
symbol.
compute
= < exp
z,e~
n2
nl
> =
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~.
Since
n!/2 to
the
e~[z]
first
identity
we
e~[z]-exp(-Izl2)dz the
n # m
hvis
n = m
= "]-[ ~ - r . ! - 6 i=l
concludes
hvis
get
~k which
.
I
C returning
m>n
=
z
ri,
~
Si
k
.r!.6 -
_r,s
'
proof.
! In defined what
appendix
on
we
have
Hilbert--Schmidt
follows
can
Consider dimensional
2
be
found
K
in t h e
can
extend
considering
the
k
to
the
orthogonal
H
.
Since ~ ~ H
k
: H
H
h
is of
where using
projection p~
of
functional
: ~ whole
H
gauss!an
measure The
~
details
, of
appendix.
linear
of
k =
the
enlargements
a continuous
subspace
introduced
, K
the
defined the
a6~
on
a
finite
form
,
same
expression,
i.e.
6g and s e t t i n g
for
x6K k(x)
By
the
same
functional
on
projection
PK
procedure ~
= X(pXx)
we
We c h o o s e
extends
k
extend ~
to
a
continuous
in s u c h a w a y t h a t
to an o r t h o g o n a l
projection
pK
linear
the o r t h o g o n a l ,
and d e f i n e :
Since
f[-]
the s p a c e
K ,
for e v e r y
f6FOK
set
of
reformulate
.
all
FOE
finite
lemma
Corollary
~
~ C
of
.
linear
functionals
in this w a y the f u n c t i o n
We d e n o t e
that
:
a polynomial
we extend
Observe the
is
the e x t e n s i o n
by
is a u n i o n of all dimensional
f[. ] f r o m
K
on
to
7[ ° ]
FOX
subspaces
defined
,
where
of
H
K
runs over
Hence
we
can
6.4
and
of
FH
6.3.
6.4:
For all
=
f,g6FoH
we h a v e
f[z].g[z]
~(dz)
K
Theorem
6.5:
u ,r~× KCH~ U extends
(cf.
[20]
I dim(~)
formula < ~
is c a l l e d
Proof: the fact t h a t
We under
9 f
, ~[-]
6 L2(~)
to an i s o m e t r y FK 9 f
which
IB3) T h e i s o m e t r y
this
~ 7[-]
6 L2(~)
,
the c o m p l e x w a v e r e p r e s e n t a t i o n .
The
theorem
V0H
cannot
is d e n s e
expect
isometry
is
is
an in
easy FH
consequence
of
corollary
.
the
isometry
not
invariant
to be u n i t a r y under
as the
complex
image
conjugation.
The
70 relation
between
multiplication For expand
standard
of c o m p l e x
an
feFH
the
arbitrary
multiplication
polynomials orthonormal
in the basis
comes
and
from
{en}n6 ~
in
the
lemma H
usual
6.2.
,
we
can
1.12B,
~ ar .e ~
f =
F0H
directly
basis
from t h e o r e m
in
.
r
By the P a r s e v a l
identity
if12 =
we get
~ ar. er
[2 = ~
ar.er[2
= ~ r!. larl2
r
r
r
thus FH = { f = ~ ar-e-r
~ r!-,ar,2
r If we e x p a n d
geFH
< ~ }
r
in the same basis, g =
as
"e L
,
S
we get
For f[z]
= <exp
f6FH
we have
z,f>
=
= ~ ar'br'[ ! r
by s e t t i n g
[ = (r],r2,...,rn)
z> =
z> 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
<
~ (k+2) '~ l a l k+2 •
lhl k
(k+2)
!
"
Ihl
k=O i.e. ~0
I exp(h'a)-e h
-a
I <
~ ~ "lhl k=0 (~'-k-'i-2) !
,
co
which
- since
the
lal k+2
series k=0
as
h
tends to
Notice
is
summable
-- converges
to
V(k+2) !
0
that
also for complex
the above t
For
argument f6FiH
C 3 t
works
not only
for real
t
but
the function
) f_(t-a + b) e C
is a c t u a l l y holomorphic. The
above
definition
together
with
the
proved
lemma
gives
the
identity aaf_(- ) = (a f)_(-)
We introduce
the following notation.
Given an a r b i t r a r y
subspace
- ,
K
we denote by
•
X
of
K
the real part of
invariant K
under the conjugation
relative
to the conjugation•
74
In p a r t i c u l a r
Lemma Hilbert c F0H
H
denotes
7.4:
space
the real p a r t of
denote
Let
H ,
invariant
a finite
under
dimensional
subspace Then
conjugation
of
for
the
f,g6FoK
we have
1 = (2~)-2!k •# f _ ( x ) - g _ ( x ) - e x p ( - ~ I xl2)dx
,
K
where
k
is
indicates
the
measure
on
the
dimension
integration
of
the
with
real
part
respect
to
the
K
of
~
and
normalized
dx
Lebesgue
K.
Proof:
For
f,geF0~
we d e f i n e
(f,g)
Choosing <ej,x>
= ( 2 ~ ) - ½ k . l~ f _ ( x ) . g _ ( x ) . e x p ( _ ½ i x i 2 ) d x • d K k I orthonormal basis {ei}i= in K and writing
an
,
for
we get (f,g}
We will prove
x 3.
show
= (2=)-½k[ that
f_(x).g_(x).exp(-½1xl2)-dXl...dXk
= (f,g)
for
all
f,geFoK
It
is
easy
to
that (o,~)
=
I
k Furthermore,
for
x =
~ xie i 6 K
we h a v e
i=I ei_(x )
=
< e x p x,6 _ e I>
= < e x p x , e .l>
= <x,ei>
= xi6~
,
and t h e n we get ( e i , e j} = ( 2 ~ ) - ½ k [~ k e.l- (x) -ej_(x). e x p ( - ½ 1 x l 2 ) ' d x
+x lldx,
:
J~k so
that
sufficient
i
= (f,g)
respect
b
for
f,g6~
To
dxk:
conclude
the
ij, proof
it
is
to s h o w t h a t (b-f,g}
where
3
is the d u a l to the o r i g i n a l
= (Zf,b g)
for all
to the o p e r a t o r inner
product
f,g6FoK
,
of m u l t i p l i c a t i o n <,>
By u s i n g
by
beK
theorem
with
5.6B we
75 obtain *
+
(bf)_(x) = <exp x,6*(bf)> = <exp x,6 ~ (b)f> : <exp x, (e+(b)~_* - D*e(b))f> = <exp x,~+(b)6*f> W
- <exp x,6_*a(b)f> --W
= b[x]" (6_f)[x] - (b f)_(x) = <x,b>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)
&k
-- (2~)-½k!k(b*f)_(x)g_(x).exp(-½1xl2).dx = (2~)-~k~ f_(x)
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[
-- (2~)-½k[ =
(b*f)_(x)-g_(x).exp(-½1xl2)dx
2v)-½k[
- (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
H
x = Re(z)
z = x+i-y choose
onto
H
K
Then
in
such
extends
to
6 K
and
a way
where
an
and
z6K
y = Im(z) = <x,->
the
orthogonal
6 K
,
we have
+ i
orthogonal
projection
projection
PK
PK
of
onto
of
H K
we d e f i n e
cf.
~ C
. By s e t t i n g
We
: K
appendix The
= <x,PK->
+ i
,
2.
extensions
will
be m a r k e d
by
Then we
reformulate
lemma
7.4.
Corollary H
,
F0H
invariant we h a v e
7.5:
Let
under
the
X
denote
a
conjugation
finite -
dimensional
For
every
subspace
pair
f,g6FoK
of C
that
=
J~~_qx)g_(x)V H ( d X ) H
Theorem
7.6:
{
(cf.
[20]
dim(~)<
Theorem
~
F0K
U
_
XCH
K invariant
can be e x t e n d e d
to a u n i t a r y
we
shall
Proof: prove within
that
call
The
the
the
9 f
7_(-)
)
6 L2(~H )
transformation , ~_(. ) 6 L 2 ( ~ H ) ,
real
extension
extension
isometry
under
FH 9 f which
4C1 ) The }
is
wave
to
representation.
an
isometry
surjective.
is
This
trivial. we
do
by
It
remains
looking
to
at
~ Since
functions
cylinder
depending
sets on
a
form finite
a
base
number
for of
the
Borel
variables
sets
will
be
in
~
,
dense
in
78 L2(~H) in
Hence
n
it is sufficient
variables,
ne~
,
to observe
is dense
in
that the set of polynomials
L2(~n,~)
,
where
~
is the
measure "Y (dx) But
this
is
= (2~)-~n.exp(_½( x 2+ 1 • .+x2))dx I . .dXn a
well
known
fact
and
the
theorem
follows. I
We gaussian
have
measure,
corresponds
to
multiplication ~--multiplication original
identified
the
and
in
the
space
of
of
polynomials
in
FOX
multiplication
space
this
FH
as
an
L2--space
representation
real
variable
in
the
the
polynomials.
L2--space
in
F0H
into
the p o i n t w i s e
in the case of the complex wave representation.
space The
corresponds
so that the r e p r e s e n t a t i o n
over
the F0H
pointwise to
the
does not turn the multiplication
as
C h a p t e r 8:
B o s e a l q e b r a s of o p e r a t o r s
The results of this and the next c h a p t e r h a v e t h e i r o r i g i n book of L o u i s e l l preliminaries notes,
they
[11]. Some of those results was p r i m a r i l y i n t e n d e d for
of
[21].
are
the author of
form.
they
here
fit
so well
instead,
to this
following
the
set
of
lecture
suggestions
of
[21]. is a d e s c r i p t i o n
compositions
normal
Since
presented
Our goal are
in a
of
The
the
of a special
creation
operators
are
and
type of operators,
annihilation
linearly
generated
operators
which in
by o p e r a t o r s
the
of the
form +
W
(a)e(f) Let
I
= af
for
a,feFoH
d e n o t e the i d e n t i t y operator.
.
We i n t r o d u c e
the space of
operators n
f
k=1 These FIH
and w i t h We
product P
operators
and
F0H
make :pQ:
Q
defined
on
the w h o l e
FIH
with
out
of
0(FoK )
P,Q e O ( F o H )
:-:
denotes
an
algebra
where
PQ
by
introducing
W
the Wick o r d e r i n g from c h a p t e r 4.
*
)(bg
*
): = (ab)(fg)
for
a,b,f,g£FoK
.
M o r e o v e r we d e f i n e the H i l b e r t space K+H
•
=
{
+
(x) + ~(y)
= x+y
*
x,yeK
)
with inner p r o d u c t (x+y
in
,u+v ~ = <x,u> +
It is clear that
K+K
c O(FoK )
the
Wick
is the usual c o m p o s i t i o n of
The m u l t i p l i c a t i o n b e t w e e n the g e n e r a t o r s becomes :(af
values
as an invariant subspace.
of
and
are
for
x,y,u,v6K
.
80 I
Definition chapter algebra
8.1:
6, d e f i n e d O(FoH ) ~P'Q~N
Let
denote
on Hilbert--Schmidt
we d e f i n e
=
~(dz)
the
gaussian
enlargements
~
measure
of
H
from
In the
the i n n e r p r o d u c t
~exp(-z)-P-exp(z),exp(-z)-Q-exp(z)>-~(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 =
= ( a f * ,(x * b ) g * ) N + (af * ,b{y*g~*.. ) N follows
,bg
8.2
= <xa,b>
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>
+
g>
= ~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)
<(y f) a , b >
=
not
)
mix
elements
Particularly
the
By u s i n g
the
Leibniz
=
(bf)
- fy*b>
f)b>
W
=
to
-
operators
in
the
(y'f)
is
not
operator rule
W
(bf)>
and
we g e t
*
for
a,beFo~
*
b> = < a y , b f > - < f a , y b> + * - = < ( f * a + ( y ) - a + ( y ) f * ) a , b
=
> •
Hence *
*
*
(y f)
Lem~a
8.6:
For
= [f
P 6 O(FoH )
+
,e
(y)]
and
x6H
.
Proof:
We prove
Pa+(x)
= (e(x)
a(x)P
= (a(x)
the
identities af
af*a+(x)
= a(f*a+(x)) = a(x*f)*
a(x)af
•
*
= x*af
=
= (a(x)
Theorem
+ a
intertwinings
(x))P
+ a+(x))P
.
on t h e g e n e r a t o r s
+ a + ( x ) f *)
+ a + ( x ) a f * = (a(x*) )f * + ax
the
6 0(FOH )
= a- ( ( x ' f ) *
(x*a
we have
+
.
f.
+ a+(x))af *
= (x . a)f
w
+ a(xf)
•
+ a+(x))-af *
8.7:
On
the
algebra
O(FOH )
we
have
the
commutation
relation *
+
[a(x+y
Proof: [a(x+y
We
),a+(u+v = a(x+y = a(x+y
compute
*
(u+v
*
)] = ( x + y * , u + v
on g e n e r a t o r s
}N I .
af*,bg*
6 O(FoH )
)]af )
(u+v
)af
.[) ( u - a ) f . + W
= e(x+y
),e
a+(u+v*)-a(x+y*)af a(v-f)
W
)[(ua)f
.] - a +( u + v ).[ (x
W
] + a(x+y +
a
*
(u+v)[(x
* .
a)f
,
+ a(y
W
)[a(vf) *
] -
*
a)f
+
] - a
*
(u+v)[a(y
*
*
f)
]
.,.] f
83 *
*
*
= (X (ua))f
+ ua(y
*
*
f)
*
+ (x a)(vf)
(u-x a)f
*
- (x a ) ( v - f )
( u a ) ( y f)
*
+ a(y
(vf))
-
- a ( v y f)
= (x * (ua))f * + a ( y * (vf)) * - (u.x * a)f * - a ( v y * f ) * =
x
(ua)
- u-x*a
=[(x u)a]f
+ a
*
= <x,u>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 <-,a>
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
(,}' * *
=
representation
+ <e,uy>
to
(ag[z]).(bf)[z]
=
= (x+y
show
respect
wave
,bg }'
the b a s e
=
+
W
this
by
are
dual
using
the
we have
*
)af
M(x+y*)
*
,bg )'
- <(y a)g,bf>
-
f)>
g),bf>
+ (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>
*) :
n=l
=
):
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]
= <§:exp(z+z
):,§P)
6 ~ )
We
instead.
91 The last i d e n t i t y
(:exp(z+z
):,af
we v e r i f y
on the g e n e r a t o r s
}N = ( ( e x p
z)exp
= a[z]-f[z]
The
real
wave
z ,af
af
e F0(H+H
)N = <exp z , a > < 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
= <~,exp
z
exp-z
z)> = < e , e x p
z ((exp-z)P(exp
P exp
z>
vP[z]
= < e , W _ z P W zO>
z))>
=
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
the o p e r a t o r
creations
x)
y
and
the
f o r m u l a we get
*
show that
)
linear
§-i
left
and
x) e x p y
combinations
transforms
*
= exp y
of c o h e r e n t
operators
annihilations
to
in n o r m a l the
i.e. = g* +(f)
for all
+
e
f,g e FI (H+H*)
(exp x) vectors, form,
right,
to
we
i.e. the
93 Though
it
constitutes require
is
not
written
an element
subsequent
of
in
the
normal
FI(K+K* )
use of the canonical
form,
the
operator
To reach the normal commutation
relation.
g*a+(f) form would
94
Appendix
Halmos'
lem~a:
Let
the A
denote
a
exist J
,
an
linear
contraction
orthogonal
such
I:
: }{1
projection
We
explicitly
P
a
U
with
S =
(I-A'A) ½ !
and
T
(I-AA)
.
map
U
and
Then an
there
isometry
= xI •
A
defined
by
~ H 1
: H2
~ H2
H2
0
> K2 e
-S : H 1
~
KI
2
projection
P
is
identifying
the
spaces
K2
and
,
It
is
an
specifications. that
we
spaces.
maps.
~ HI
: H2 @
K1
:
P ( x 2 • xl)
This
unitary
the
: KI e K2
P
know
Hilbert
= P-U-J
: HI J(Xl)
0
two
,
define J
H2 •
~ H2
between
that
Proof:
The
lem~a
map
A
=
Halmos"
S
show
easy But
and
as
exercise to
T
prove
to that
intertwine
follows.
show the
with = TA
Consider
the =
= x2
that
the
maps
map
U
is
have
unitary,
A
AS
A.S 2 = A-(I-A*A~
) H2
. identity
(I-AA*I.A
= T2.A
.
the we
required need
to
95 By a s i m p l e
induction
in
n6~
this e x t e n d s
to
A . S 2n = A. s 2 ( n - I ) . S 2 = T 2 ( n - ] ) . A . S 2 = T 2 ( n - I ) . T 2 . A We h a v e
.
shown that A. p o I ( S 2) = p o l ( T 2 ) . A
where
= T2n.A
pol
denotes
a polynomial.
We n o w c h o o s e {Pn}n6 ~
converges
calculus
,
polynomials to
~
pn(- )
uniformly
on
on
[0,1] [0,1]
in s u c h a w a y that By
the
we get P n ( S 2) ~ Pn (T2)
n
S
in
~(H1)
~ T
in
~(~2)
Hence AS = TA For
x I6H I
.
we get
P-U-J(Xl)
= P-U(x1@
0) = P ( A x I @
(-SXl))
= A(Xl)
functional
g6
A p p e n d i x 2: G a u s s i a n m e a s u r e s
The
results
construction
of
in
the
this
appendix
gaussian
u n i q u e n e s s of the m e a s u r e
come
measure
and
on
from
[18].
infinite
some p r o p e r t i e s
We
provide
dimensional
the
spaces,
that are b o t h useful
and
have i l l u s t r a t i v e purposes. The n u m b e r i n g
is i n d e p e n d e n t
of the n u m b e r i n g
in o t h e r parts of
these notes.
A: The q a u s s i a n m e a s u r e on
R=
We d e f i n e the space
~
as the set of all real sequences,
N~ = { ~=(al }a2 '" "''an '"'"') We d e f i n e a t o p o l o g y in
~
a'e~ l
i.e.
f°r all ieN
by the m e t r i c IXn-Ynl
d(x,y)
= I~-~I
=
~ 2 -n
where
x,~
~
1+IXn-Ynl n=1 The well known real Hilbert space is a subset of
~
,
n= ] Writing first
n
~-n
positions,
for the subset of and i d e n t i f y i n g
by adding zeros a f t e r the first ~ In
~m
n
n
are
c o n s i s t i n g of zeros in the
the space positions,
~n
as a subset of
we get
we d e f i n e the c y l i n d e r sets
natural
numbers
~-n
and
B
runs
over
Borel
These c y l i n d e r sets form a base for the Borel sets of We d e f i n e by setting,
for
the g a u s s i a n n6~
~(B 8 ~ - n ) =
content
and a Borel set
~
sets
on the c y l i n d e r
B C ~n
dxldx2...dx n
in
~n
~,I'I sets
,
(2~)-n/2 f e x p ( _ ~ ( x ~ + . . + x ~ ) ) d X l . . d x n B
where
m
= ~n @ ~ - n
B 8 ~ where
~
denotes the L e b e s g u e i n t e g r a t i o n over
~n
in
97
We
have
that ~(~)
and
for
every We
fixed
shall
is a c o n t e n t
on
n6~
~
use
the
the
cylinder
= I ,
becomes
a measure
Kolmoqorov
extension
sets
in
~(~) and
if
~
uniquely
is
to We
a
measure
a Borel
denote
measure
the
For
a > 0
a(B
~ ~-n)=
on
measure
we
by
,
~n
theorem such
saying
that
then
~
if
that
= I ,
~n on
~
on
for
every
~,I"
I
the
same
ne~
symbol
,
~
extends
.
define (2~)-n/2
~ exp(_~(x~+..+x~)/a)dxl
..dx n
B
It
is
known
pairwise
that
for
singular.
different
For
In p a r t i c u l a r
a's
simplicity
we
shall
measures
will
need
Theorem
]A:
numbers.
We
define
be denoted
Consider the
by
the
the measures
and
a
corresponding
we develop
a = ] These
the
~
theory for
the
measures for
~ =
are
1
numbers
a = ½ and
~2
sequence
.
(an)n6 ~
of
positive
real
set co
E = { x6~ ~
~ anX2
< ~ }
.
n=l
If t h e
series
an
is
summable
then
~(E)=I
,
otherwise
~(E)=0
•
n=l
Proof: indicator
We
function
construct of
E
,
a I
E fk
net
of
Given
" :
functions k>0 ~ ~
by
anXnl n=1
we
converging define
the
to
the
function
g8
It is easy to prove that I 0 fA(--x) = and that
a positive real number
fA(x)
The
, IE(_X)
pointwise for
X
if
xCE
if
x6E
I
~0 .
functions N
2
fk,N(X) = exp(-X ~ anXn)
for N6N
n=l
are obviously integrable over sequence.
By the
Lebesgue
~ ~ fk(x)~(dx) in
case
the
limit
Notice number
of
Am
theorem
and for fixed X of
monotone
form a decreasing
convergence
we get
that
= ~ N ~ N---~ lim fX,N(X) ~(dx) = N---~ lim ~ ~fk'N(X)~ (dx) exists.
that
in
variables
the
case
x 1 ,x2,...,x
where n
f
the
depends
integration
only
on
reduces
to
a
finite
N
~(dx) = (2~)-N/2exp(-~
x2)dxldx2...dXN
.
n=]
Using the substitution
t = ~l+2Xan-X
we compute
N
fk,N(X)7(dx)
= (2v)-N/2 ~
N = (2~) -N/2 ~
N
exp(-k ~ anX2)exp(-½ ~ x2)dXldX2...dXN NN n=l n=]
[ ~ exp((-kan+½)X2n)dXn ]
n=l
[R
~t 2 ,
dt
N
= "1-[ ( 1 + 2 • a n ) - 1 / 2 n=l
since
~ exp(-½t2)dt IR
m
If
an
= ~ ,
then
n=|
IT (1+2ha) n
= ~ ,
thus
n=l
~
fx(x)'Y(dx)
= 0
I:O
Otherwise we have that an n=l
and by the inequality for non--negative
x
= ~
•
99 0 < in(1+x)
< x
we get t h a t co
I _< ~
<
(1+2ka n)
n=l
This
implies
that
0 ~
fx(x)~(dx)
if
= {
n=1 (l+2kan)
if
L n=1
n=1 By u s i n g
the L e b e s g u e
~(E)
= ~
an
= ~
t h e o r e m of d o m i n a t e d
I E ( X ) ~ ( d x ) = lim
~
<
an
oo
convergence
w e get
fh(x)~(dx)
co
If
a n
< ~
we have that
,
n=1
which
concludes
The
lim ~ (l+2Aan) k--+0 n= I the proof.
difficulty
in d e v e l o p i n g
the
= I
theory
of g a u s s i a n
measures
in
0~
is where
due
to
the
~(6 2) = I ,
fact
that
we have,
contrary as s h o w n
to
the
finite
in the above,
dimensional
case
that
7(6 2 ) = 0 . To
shorten
the
notation
we
define
the
on
measures
~n
,
n
n6~
,
by ~n(dX)
If we i d e n t i f y by as
~-n
the
the m e a s u r e
the p r o d u c t
in = (2~) -2 . e x p ( - ½ ( x ~ + . . . + x 2 ) ) d x I ...dx n
the s p a c e s gaussian ~
measure
on
R ~ = ~n ~ ~ - n
measure ~
,
~n ® ~ m - n
on
~-n
and ,
the m e a s u r e ' n
~-n
~n × ~ ¢ - n
constructed can
be
,
and d e n o t e
in the same w a y identified
with
100
Definition integrable
with
2A:
Let
respect
to
f
denote
a
7
For
n6~
(Enf)(x)
= T
function
defined
~
on
and
we d e f i n e
f(x+s)'~-n(dS)
'
cO-- n
where
x61R n
By
and
the
integrable
s611t m - n
Fubini
function
,
and
IRn
theorem over
is
Enf
(~n,~n)
treated
is well
as
a
subset
defined,
and
of
[R~
E f n
is
an
with
!"f (Enf)(x)'~n(dx) ( - s ) ' ~ ( d =s co ~- ) n
Lena
3A: For
feLl (w)
and
n6~
we have that
llEnflll ~ llfllI
Proof: that
fl0
Assume
that
Then we have
f
is a n o n - n e g a t i v e
that
E f > 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 <
lIEn( f-g)1[ I +
= llEn(f-g)lll
Corollary integrable every
with
finite
is c o n s t a n t
7A:
a.e.
+ llf-glll
Let
respect
number in
IIf-Englll
f to
denote ~
of v a r i a b l e s ~
-< 2"llf-glll
< e
a function If
f
(depending
is
defined
on
constant
only
on
the
~
with
which
is
respect
to
tail),
then
f
102
Proof: on
the
first
constant all
Considered n
with
m,n6~
as a t a m e
variables.
respect
to
function
Hence
every
E f n
finite
~
,
En f
constant
a.e.
on
is
number
of
depends Since
variables,
we
only f
is
have
for
that
Enf(Xl,X2,...,Xn)
= [
f(x+s)'~_n(d~)
v ~ _ n
= [
--
--
f(x+s)'~_m(d~)
v ~ _ m
--
R
= Emf(Xl,X2,...,Xm)
--
Thus f = l i m Enf
in
LI(~)
n
is o b v i o u s l y
constant
Remark: corollary The
Since
7A holds
a.e.
the arctan
of a m e a s u r a b l e
for every measurable
following
corollary
is
function
is i n t e g r a b l e ,
f
often
called
the
Borel
set
Kolmogorov
zero--one
law.
Corollary indicator finite
which
B: T h e
8A:
function
Let
of
B
B ,
number
of v a r i a b l e s ,
Proof:
From
linear
IB then
corollary
it is o b v i o u s
that
IB:
, B
is
constant
is e i t h e r
function
a.e.
functionals
k
a
is
7A the
IB = 0
measurable
Definition
denote
or
on
called
a
in
with
~ respect
of m e a s u r e
IB 1B = I
is
If
zero
constant
k
is d e f i n e d
2)
is
linear
on a l i n e a r on
E
set
functional
on
a.e.
~
linear
E C ~
and measurable
o r one.
by
measurable
fulfills
I)
every
a.e.,
co
if
to
the
with
of
full measure
respect
to
1
103
In what
follows
we
shall
provide
linear measurable functionals defined on
Proposition a,x6~
2B:
(The
a representation
for
these
n6~
and
~
Kolmogorov
inequality)
For
we define n Sn(X ) =
akx k • k=1
For arbitrary
6>0
we have that n
n
m<_n k=l
k=l
Proof: We start by noticing that I ~ t.exp(-½t2)dt = 0 and I ~ For
arbitrary,
men , we have n n a2 = ~ a 2 ~/~ ~ i
k=1
k=1 n = ~ ~ k=1
t2exp(-½t2)dt = I
pairwise
disjoint,
measurable
sets
QmC~ ~
,
x .expl_ x /dx k
~ a2 x2"exp(-½x2)dXk
=
~ ~ k=1
a2 x2 ~(dx)
n =~
( ~ a k x k )2 (d_x) = ~
(Sn(X))2 (d_x)
k=1
and since the r e l a t i o n
S2n = s2m + 2Sm(Sn-Sm)
estimate
+ (Sn-Sm)2
implie s the
s2 -> 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
such that
x_6Qm })
= e2~( ' x6~ ~
there exist
m
such that
f(x)=m ~) J
= e2~( ~ X6~ ~ which concludes
Theorem
sup ISm(X) l > 6 }) , m!n the proof.
3B:
(The
Kolmogorov
large
number
theorem)
Consider
a
c0
sequence
of positive
numbers
(an)n£ ~
If
~ a2 < ~ n=1
,
then
the
105 series k(x) =
converges
a.e.
in
~ anX n n=1 with respect to
~
Proof: We shall prove that the measure
of
00
{ x6~
~
anXn
diverges
}
n=l
is zero. n we define
For
m6~
Sn(X ) =
~ akXk k=1
~(x) = lim sup Sn(X ) n s(x) = lim inf Sn(_X ) n we have
s(x) - s(x) = lim sup Sn(X ) - lim inf Sn(X ) n n < p>m sup Sp(X) - q2m inf Sq(X) = p>m sup Sp(X) + q_>m sup -s q (x) =
sup {Sp(X)p,q>_m
Sq(X)}
_< p, supq_>m [ Sp(X)
- Sq(X)
< p>mSUp ISp(X) - Sm(--x)l + q>mSUp ISm(X ) - Sq(X) l = 2 sup p_>m
For arbitrary
Isp(_X) e>O
Sm(X)[
we define
M = { x6~ ~ Hence for all
~(x)
- s(x){
> 26 }
m6~ M
c
M m := { x6~ m
sup I~(~) - ~(~)I > ~ } p>m
and thus ~(M) < ~(Mm) Using the Kolmogorov
inequality,
for all we compute
me~
[
106
(M)
~ lim
sup
v(Mm)
sup
~({
sup m
~ a k2 /e2 k= m
have
seen
m F
lim
xe~ ~
m lim
We
that
sup n>m
ISn(X)
= 0
.
series
- Sm(X)
of t h e
form
co
X(x)
0o
=
anX n
with
an
n=l converges functional. We
a.e.
in
The
~
converse
introduce
some
,
the
number
I
Theorem co
4B:
Then
])
we
~
thus
takes
Consider
(0,
defines
a
linear
measurable
as w e l l .
notation.
_'en > = 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
<
n=l
Proof: For
x = (x],x2,..,Xn,..)6~ ~
we define
~(n) = (0'0''''0'Xn+l'Xn+2 ''')6~ Thus we have that n =
For
~ Xkek + ~(n) k=~ we get that
x6~(k) n
n
h(x) =
~ Xkk(ek) + k(X(n)) = ~ Xkk(ek) + An(X) , k=1 k=1 and has the same = h(X(n)) is measurable for all n
where
An(X)
domain
of definition
are i n t e g r a b l e
over
For arbitrary
as
k
Hence
exp(i-k(x))
and
exp(i-kn (x) )
Nm u > 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
<_ e x p ( - ½ u 2 ~
exp(i.u-A(x))~(dx)
Ik(ek)l 2)
for all
n
.
k=l co
Assume
IX(e n) 12 = ~
that
This y i e l d s
n=l
~ exp(i-u-X(x))~(dx)
= 0
IR
for all
u > 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]
i=I .... m
orthogonal
for d i f f e r e n t
indices
i .
'
115 C: L i n e a r t r a n s f o r m a t i o n s
we maps in
shall ~m
extend
in
~m
unitary
transformations
62
of
onto
6
2
to
that p r e s e r v e the g a u s s i a n measure.
Definition
IC:
A weak
measurable
linear
transformation
in
is a map A
: ~D
EA9
Z = (Y]'Y2 .... ym,.. ) = A x 6 ~
,
where I)
The d o m a i n of d e f i n i t i o n
2)
The m a p
3)
Every coordinate
A
is a linear set of full measure.
is linear.
functional
in
function
I) A c c o r d i n g
Ym
: EA
~ ~
is a linear m e a s u r a b l e
~m
There are some comments
~
EA
in c o n n e c t i o n with this definition.
to the earlier
paragraph
concerning
functionals
in
we have that
Ym(X) The
transformation
{amn}m,n6~
=
~ amnX n n=1
A
is
w h e r e the rows
with
then
given
{amn}n6~
2) By the m a x i m a l d o m a i n
{amn}n6N662 by
the
are elements
(infinite) in
matrix
62
for a linear f u n c t i o n a l
h
in
R
we
Ym
and
u n d e r s t a n d the set
{ ~6~
I
~ k(en)Xn
converges }
n=1 If we set
let
Em
denote
E = D Em m=1
,
the m a x i m a l
EA ~ E ,
the whole
for the
functional
we get that EA c E
If
domain
and
v(E)
= I
we e x t e n d in the obvious m a n n e r the t r a n s f o r m a t i o n
E ,
thus g e t t i n g
A
m a x i m a l l y defined.
A
to
116
3) A
is
It
follows
uniquely
determined
from
the
determined
b y its
values
earlier
by
its
on the
results
values
L2 c
that 62
on
EA
and
It
is
that
indeed
set {en}n6 ~
,
where en =
(0,0,..,0,I,0...)
T coordinate 4) the
rows
linear
The of
converse
holds
a matrix
A
=
as w e l l ;
in
Let
a linear
denote
we
extend
~
by
in
{amn}m, n 6 ~
A
bounded
to
the
rows
are
co
elements
~. i< A e n , e m
n=1
a weak
measurable of
linear
definition
transformation
for
the
matrix
=
of
,
by the Parseval
identity
=
Let
transformation.
We
by
{Umn
is i n v a r i a n t
Proof:
under
U
extend =
the
U
It is o b v i o u s
to
t2
~
a weak
n6~
B = { x6~ ~ that
:
=
B that
and
and
U(B)
U
* 2 IIA emll < ~
for are
denote
measurable the
linear gaussian
a map
unitary in
measure
.
ak,bk6~
ak<Xk
62
Then
transformation
Consider
show
2 e m >I
defined.
2C:
matrix
I< e n , A n=l
is w e l l
Theorem
to
*
>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<
(U*~)k ~ bk
for k=1 .... n j~ ,
denote the extension of the adjoint of
U
in the matrix
of
U
are the columns
U .
in the matrix
of
Since the rows U*
,
that c0
U(B) = { _x6~
1 ak<
~ UikX i <_ b k i=]
for k=1 .... n } .
We define the functionals c0
kk(X ) =
~ UikX i i=I and observe that since the rows in U ~(U(B))
= ~< x 6 ~
I ak< kk(X)_< b k bI
k=1 ,..,n , are pairwise orthogonal,
for k=1 .... n }
bn
(2~) -½n ~ .. ~ exp(-½(x1+ . 2 .+x 2))dxl . .dXn
=
aI
an
D: The qaussian measure on
C=
Consider the linear space e
=
_z = (z] ,z2,...,Zn,..
z.6Cl for all
)
i6~ }
and the Hilbert space
1
=
and introduce
in
~n , •
cn where
n6~ ,
n=1 the gaussian measure
=
~2n
dXldYl...dxndY n
we get
118
=
(Zl,Z2,...,Zn)
Izl 2 = and
Izl I2 +
...
6 Cn
+
and
IZn 12
dXldYl...dxndY n
indicates
all
extended
E:
former
results
to t h i s
Hilbert--Schmidt
Let
the
gaussian
on
C
measure
denote
a
enlargement
containing
inclusion
Lebesgue
integration
measure
are
easily
enlarqements
H,<,>
Hilbert-Schmidt space
concerning
new gaussian
the
2n
in
Then
zk = x k + i-y k 6
H
mapping
K,~,>
as
of
K
real
a
of
linear
into
~
or
complex
H,<,> dense
is
subset
Hilbert itself and
is a Hilbert--Schmidt
~ llenll 2
space.
a new such
A
Hilbert that
operator,
the
i.e.
<
n = ]
for
every
denote
the
orthonormal norms
basis
{en}ne ~
corresponding
to
in
the
~
,
inner
where
products
II'II
and
~II'II
<,>
and
<,>
respectively. We H
which
shall is
orthogonality note
that
is a t o t a l In
need a
known
fact
complete
if
{en}n6 ~
set
in
what
that
for this
is
there
orthonormal
is p r e s e r v e d
Hilbert-Schmidt well
the
an
exists system
basis
orthonormal
when
an o r t h o n o r m a l in
~
considered
basis
in
H
IE:
,
in then
in
i.e.
the
H,<,>
(we
{en}n6 ~
~). follows
we
enlargement
shall of
keep
on
H,<,>
denoting
We
shall
There
exists
a constant
~lJXlJ < K" JlxJl
for all
K
fulfilling x6}{
.
by
need
results.
Len~ma
,
basis
~,~,> the
a
following
119
Lena linear
in
2E: the
constant
K
Let
k
first
: HxH
term
, C
and
denote
linear
in
the
a map second.
there
If
is
conjugate
there
exists
a
fulfilling Ik(x,y) l < K. IIxll-IlYll for all
then
which
exists
a linear
bounded A
x,y6H
,
operator -~
: H
K
fulfilling h(x,y)
We h a v e
the
following
Theorem which
3E:
fundamental
There
is a c o m p l e t e
Proof:
=
exists
orthogonal
IE t h e r e
exists
there
exists
an
orthonormal
system
It is e a s i l y
seen
operator
that
= <x,y>
in
~
basis
{bn}n6 ~
in
H
.
: K×H
J
if
J
= <Jx,y>
that
K
, fulfilling
< K. llxIl-llyll
an o p e r a t o r
<x,y>
and
.
result.
a constant
Ih(x,y)l by w h i c h
x,y£K
We define k(x,y)
By l e m m a
for all
for all : H
for all
is a s t r i c t l y
{en}ne ~
~ H
denotes
x,y6H with
x,yeH
positive an
. (hence
orthonormal
self-adjoint) basis
in
H
,
then
tr(J)
=
~ <en,Jen>
=
n=1
=
~ <Jen,en>
=
n=1
~ ~llenrl 2 < ~
~ ~en,en > n=1
,
n=l i.e.
J
is
a
trace
Now c h o o s e eigenvectors
for
class an J
operator.
orthonorraal with
basis
{bn}ne N
eigenvalues {Xn}ne N ,
Xn
>0
.
in
~
consisting
of
120
We then have 0~
I J
The series
{An}n6 ~
is convergent,
i.e.
since
IN n < m n=1
is of trace class. 2
The orthogonality
follows easily for
n,m6~
Corollary H,<,>
and
4E:
There
a convergent
{bn}n6 ~
exists
series
is total in
an
of
orthonormal
strictly
~{,<,> .
basis
positive
{bn}n6 ~
numbers
in
{An}n6 ~
such that ~ = {
Proof:
We choose
~ Xnb n n=1
~ kn" IXn [2 < ~ } n=]
{bn}n6 ~
and
{An}n6 ~
as constructed
in the
proof of theorem 3E and the corollary follows.
F: The qaussian measure on Hilbert-Schmidt
Using eigenvalues
the orthonormal {kn}ne ~
basis
enlarqements
{bn}ne ~
in
H
from the above paragraph we get
~bn,X> = <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
<'>2
exists
a Hilbert-Schmidt
123 Proof:
in
H,<,>
Use
such
theorem
that
3E
to
{en}ne ~
choose
are
an
orthonormal
pairwise
orthogonal
basis with
{en}n6 ~ respect
to
Then
<'>l
<en,-> I = Expanding
aeK
we
[lenll~-<en,->
•
get
a =
~ an-e n
with
=
an
<en,a>
and
lan 12
<
n=] and
the
functional
A
can be e x p r e s s e d
h =
as
an -<e n , ->
= 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
<- ]~ 2 n" n=1
i-Ia
k"
<ek' x> 2
k=m. mn+1-1
]~ k=m~
ak'e k
=
w e get
~ ~ 2n" n=l
2
-][x[[ 2 = ][xl[ 2
<
mn+ I -I ~ ak'ek k=mn ~
mn+1-1
]~ 2 n" ]~ n=1
2 , x >
k=mn
[ak 12 < °°
natural
124
<',->.
is well defined on
H .
We define the inner product
<'>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 n" ~ lak 12 < ~ , n=1 k=mo it is easy to see that
2
<" ''>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 )~ <
( ~ p:l
the
number
Itxll
-< C n"
Cn
does
for every not
depend
x6H on
the
.
choice
of
the
basis
{ep}p6~ For
x,yeH
we define
the inner
product
o0
~<x,y>
=
~ an.<X,y> n n:1
and the n o r m co
~[Ixll 2 = where
{an}ne ~
is a s e q u e n c e
2 ~ a n "II×[In , n=]
of s u f f i c i e n t l y
fast d e c r e a s i n g
numbers.
Choose an
=
2-n.
Cn 2
We c o m p u t e CO
~ii xj[2 =
~ an" ]jx[]2 < n=1
~
~ an. C2. 11xlj2 < n=1
oO
~ 2-n. [jxlj2 <_ Hxj12 < ~ n=1
,
127
and c o n c l u d e
<,>
that
Let
~,~<,>
be an o r t h o n o r m a l co
co
~ I}ep I]2 =
:
basis
.
Take
obvious
the
in
completion
H,<,>
of
H,~<,>
and
let
{ep}pe ~
T h e n we h a v e
e I]n2 ~ a n" IIp n=1
co
co
~ an
~
n=1 We
denote
co
~ p=1
p=]
is w e l l d e f i n e d .
have
p:] to
n6~
co
~ an "C2 =
~ 2 -n < ~
n=]
prove We
co
IIepl[n 2 = that
must
n=]
~
is
show
a Hilbert--Schmidt
that
~
is
finer
enlargement
than
~n
of
It
is
t h a t the i n c l u s i o n m a p p i n g I : H,
<,>
~ Hn,<,> n
is c o n t i n u o u s . It
remains
one--to--one. A s s u m e - norm
a n d that
to
prove
that
that
the
{Xp}p6 ~ c ){
IiXplIo
~ 0
extension
of
I
over
is a C a u c h y - s e q u e n c e
Then
for e v e r y
~
in the
is -II'II
n6~
P llXp - Xqlln We k n o w t h a t hence
for e v e r y
I{Xpll 0
> 0
n
> 0 P,q the e n l a r g e m e n t Kn
is f i n e r
than
]{0
and
implies
P
IIxpHn
~0 P
Since e0
ILXpLl = : [ an" IL~pLln2 )~ n:l
we get by l e m m a
2F t h a t
IIXp II which Since
proves the
injection
that
the i n c l u s i o n
inclusion of
~
into
mapping Kn
of
~0
P mapping Hn
of
~
into
m u s t be o n e - t o - o n e
We a r e r e a d y to p r o v e
the a n n o u n c e d
into H0
K0 is
is o n e - t o - o n e . one-to-one,
as w e l l .
prlncipal
result.
the
128
Theorem
5F:
space
there
Hilbert finer
than
To
every
always
pair exists
any enlargements
Proof:
Let of
{en}ne ~
H,<,>
in
Normalize
a Hilbert
{en}n6 ~
the pair.
and
H2,<,>2
for
every
Hi,<,> I
n
define
enlargement
a pair
Choose
of
an
orthogonal
of
that
a is
Hilbert--Schmidt
orthonormal
system
in
basis
HI,<,> I
by s e t t i n g
bn = and
be
K,<,>
is a c o m p l e t e
in HI,<,>]
enlargements
a Hilbert--Schmidt
space
that
Hilbert-Schmidt
from
HI,<,> ]
enlargements
of
the
IIenlI[1-en continuous
linear
functional
k
n
in
by h n :
Note
that
kn
is c o n t i n u o u s hn =
By
lemma
H3,<,> 3
of
functionals For
3F
and
that
{hn}n6 ~
are
is
the
we d e f i n e
corresponding
4F
and in
we
find
finer
than
continuous
<x,y> 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 -
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