• Ei tuloksia

5 Regularity of States and the Construction of Abstract Annihilation and Creation Operators

5.1 Regular States

The generators of the Weyl algebra are all unitary operators, and, if we fixf ∈ H, then we can consider a semi-group given by the collection of operators{W(tf)}t∈R. In the case where we con-structed the Weyl operators via the operators Φ(·), it is obvious that this collection of operators is, in fact, strongly continuous. However, the strong continuity of the above semi-group is not guaranteed. We note that if this semi-group is strongly continuous, then, by Stone’s theorem, there will exist a self-adjoint operator ˜Φ(f) such thatW(tf) =eitΦ(f)˜ . This self-adjoint operator is, of course, the infinitesimal generator of the time evolution, and, as in the case of the operators Φ(·), we can recuperate the annihilation and creation operators by considering linear combinations of ˜Φ(·).

Motivated by this discussion, we introduce the concept of a regular representation. Let (π,H) be a representation of theC-algebra generated by a collection of elements{W(f) :f ∈H} satis-fying the Weyl relations. We say thatπis a regular representation if the mappingst7→π(W(tf))

are strongly continuous for all f ∈H. We say that a stateω is regular if the associated cyclic representation (πω,Hω,Ωω) is regular.

As stated, we can consider the infinitesimal generators of the induced time-evolutions, and thus recuperate the Φ(·)-type operators. Using linear combinations of these operators, we can then define the creation and annihilation operators. As is usually the case with (possibly) unbounded operators, we must pay attention to the domains of these infinitesimal generators. The follow-ing proposition deals with the technical issues of the domains of the operators, and proves some relevant and expected properties of these operators.

Proposition 5.1. Let Abe the CCR algebra over the Hilbert space H with the symplectic form σ(f, g) =−12Imhf, giandωa regular state onAwith the associated cyclic representation(πω,Hω,Ωω).

Let f ∈ H and define Φω(f) to be the infinitesimal generator of the strongly continuous unitary semi-group{W(tf)}t∈R.

It follows that for any finite dimensional subspaceM ofHthe operators{Φω(f),Φω(if), f ∈M} have a common dense set of analytic vectors. Define the domains

D(aω(f)) =D(Φω(f))∩D(Φω(if)) =D(aω(f)). (5.1) It follows that the operatorsaω(f)andaω(f), defined by

aω(f) =Φω(f) +iΦω(if)

2 , aω(f) =Φω(f)−iΦω(if)

2 , (5.2)

are densely defined, closed, (aω(f))=aω(f), and, for all ψ∈D(aω(f)), we have

||Φω(f)ψ||2+||Φω(if)ψ||2= 2||aω(f)ψ||2+||f||2||ψ||2 . (5.3) Proof. The idea and motivation of this proof comes from the theory of semi-groups. We are going to construct the operators using a Bochner integral with a Gaussian weight. The main tool of this proof is integration by parts and some estimates for exponentials and Hermite polynomials. In fact, most of this proof is going to be spent proving technical details and estimates.

Let M be a finite dimensional subspace of H with dimension m. Let {fj}mj=1 be an orthonor-mal basis ofM andf = (f1, ..., fm). Define an operatorRn onHωby

Rn =n π

mZ

Rm×Rm

dsdte−n(|s|2+|t|2)πω(W(hs+it,fi)) . (5.4) In the above operator we are using the inner product brackets are meant to be understood as

hs+it,fi:=

m

X

j=1

(sj+itj)fj∈ H. (5.5)

Of course, in this case we have the inner product of a vector in Rm+iRm and a vector inHm, but we should interpret the vector inHmas a vector ofCm. This operator is well-defined since it is basis independent. To see this, let{gj}mj=1 be another orthonormal basis ofM and letgbe the corresponding row vector. The change of basis betweengandf can be represented by an unitary matrixM. Explicitly, we haveg=Mf. By unitarity,Mis a surjective inner product preserving

operator with determinant 1. Using these observations, we have Z

Rm×Rm

dsdte−n(|s|2+|t|2)πω(W(hs+it,fi)) (5.6)

= Z

MRm×MRm

dsdte−n(|Ms|2+|Mt|2)πω(W(hMs+iMt,gi))

=

det MT

2Z

Rm×Rm

dsdte−n(|s|2+|t|2)πω(W(hs+it,gi))

= Z

Rm×Rm

dsdte−n(|s|2+|t|2)πω(W(hs+it,gi)) .

Letψ∈ Hω. Defineψn =Rnψ. We will show that||ψn−ψ|| →0. First, note that n

π mZ

Rm×Rm

dsdte−n(|s|2+|t|2)= 1. (5.7) This implies that

ψn−ψ=n π

mZ

Rm×Rm

dsdte−n(|s|2+|t|2)ω(W(hs+it,fi))−1]ψ . (5.8) Let ε >0. By strong continuity of the regular representation, there exists δ > 0 such that for (s,t)∈B(0, δ), we have

||[πω(W(hs+it,fi))−1]ψ|| ≤ε . (5.9) In the previous remark the existence of such aδ >0 follows from noticing that

πω(W(hs+it,fi)) =

m

Y

j=1

πω(W(sjfj))

m

Y

j=1

πω(W(itjfj)) (5.10) and using strong continuity separately on each operator, then choosing the smallest radius.

We have

||ψn−ψ|| ≤n π

mZ

Rm×Rm

dsdte−n(|s|2+|t|2)||[πω(W(hs+it,fi))−1]ψ||

≤ε+ 2||ψ||n π

mZ

(Rm×Rm)\B(0,δ)

dsdte−n(|s|2+|t|2)

=ε+ 2||ψ||n π

m C(2m)

Z δ

dr r2m−1e−nr2

=ε+ 2||ψ||

1 π

m C(2m)

Z

dr r2m−1e−r2 .

→ε, n→ ∞.

In the above calculation, the first two inequalities follow from splitting the integration domain and utilizing strong continuity, the final lines consist of doing a change of variables of the form

√nr7→rand taking the limit. Sinceε >0 was arbitrary, we have||ψn−ψ|| →0 as desired.

Next, we will show thatψn are, in fact, analytic vectors of Φω(·). First, we note that

W(hs+it,fi) =W

m

X

j=1

(sj+itj)fj

=

m

Y

j=1

W((sj+itj)fj). (5.11)

This follows since the vectorsfj are orthogonal. Next, leth6= 0, we compute

W((sj+h+itj)fj) =eihtj2 W(hfj)W((sj+itj)fj). (5.12) Fork∈ {1, ..., m}, we definehk∈Rmby (hk)k=h, and the components are 0 otherwise. By the above computation, we have

W(hs+hk+it,fi) =eihtk2 W(hfk)W(hs+it,fi) (5.13)

=e

ihhk,ti

2 W(hhk,fi)W(hs+it,fi).

In the calculation of this equation, we again used the orthogonality of fj to commute W(hfk) through to the beginning of the product. Now, we compute

Z

Rm×Rm

dsdte−n(|s|2+|t|2)πω(W(hs+it,fi)) (5.14)

= Z

Rm×Rm

dsdte−n(|s+hk|2+|t|2)πω(W(hs+hk+it,fi))

= Z

Rm×Rm

dsdte−n(|s+hk|2+|t|2)e

ihhk,ti

2 πω(W(hhk,fi))πω(W(hs+it,fi))

. (5.15)

This computation shows that Z

Rm×Rm

dsdt e−n(|s+hk|2+|t|2)eihhk2,tiπω(W(hhk,fi))−e−n(|s|2+|t|2)1

|hk| πω(W(hs+it,fi)) = 0. (5.16) In order to save space, we define

w(s,t) =e−n(|s|2+|t|2) . (5.17) We have

e−n(|s+hk|2+|t|2)eihhk2,tiπω(W(hhk,fi))−e−n(|s|2+|t|2)1

|hk| (5.18)

=eihh2k ,tiw(s+hk,t)πω(W(hhk,fi))−w(s,t)1

|hk| .

Telescoping, we have

eihhk ,t2 iw(s+hk,t)πω(W(hhk,fi))−w(s,t)1

|hk| =e

ihhk ,ti

2 w(s+hk,t)−w(s,t)

|hk| πω(W(hhk,fi)) (5.19) +e

ihhk ,ti

2 w(s,t)πω(W(hhk,fi))−1

|hk| . Applying these computations, we have

πω(W(hhk,fi))−1

|hk|

Z

Rm×Rm

dsdte

ihhk ,ti

2 w(s,t)πω(W(hs+it,fi)) (5.20)

=−πω(W(hhk,fi)) Z

Rm×Rm

dsdte

ihhk ,ti

2 w(s+hk,t)−w(s,t)

|hk| πω(W(hs+it,fi)) .

Applying this equation, we have Now, writing out the vectorhk, we can identify the left side

πω(W(hhk,fi))−1

|hk| ψn= πω(W(hfk))−1

h ψn . (5.22)

The right-hand side would converge to the strong derivative of the mapping t 7→ πω(W(tfk)).

Sinceω was a regular state, if the above norm limit exists ash→0, we haveψn ∈D(Φω(fk)).

First, notice that we have the pointwise limit lim

|hk|→0eihh2k ,tiw(s+hk,t)−w(s,t)

|hk| =−2nskw(s,t). (5.23)

The limiting function is clearly integrable since

| −2nskw(s,t)| ≤2n|s||w(s,t)|, (5.24) and the exponential of |s|2 will always converge faster than any polynomial of |s|. Second, for

|hk| ≤1, we have the bound is a∗-homomorphism, we have

By the dominated convergence theorem for Bochner integrals, we have

lim In this notation, we have

πω(W(hhk,fi))

We are thus left with 4 terms to evaluate. By strong continuity, we know that By a trivial application of dominated convergence, we have

lim The first and third terms thus cancel each other out.

For the second term, we have

The pointwise limit of the integrand as |hk| →0 is 0, and, by dominated convergence using the prior bounds, we have

lim

We have already shown what the fourth term converges to. Compiling these results, we have

|hlimk|→0πω(W(hhk,fi))

Which shows that lim By earlier remarks, we have actually shown that

Φω(fkn =−in there exists scalars{λk}mk=1 such that

g=

Leth6= 0, and fork∈ {1, ..., m}definehk as before. Using orthogonality, and the fact thatπωis a homomorphism, we have

πω(W(hg))−1 Now, observe that

and, using orthogonality, we have

Combining the two inequalities above, and telescoping appropriately, we have

h→0lim

”method” that was used to compute Φ(fkn was basically a glorified version of integration by parts. Let k, l ∈ {1, ..., m}. Iterating the same proof, it is not difficult to see that Φω(fk) and Φω(fl) commute. Because of this commutation, we have

ω(g))pψn = X To make the next step clearer, we note that

Φω(fkn=in In general, we have

m

We have and, subsequently, we have

P analytic vectors for everyg.

First, note that

kl

Using the above to separate the integrand appropriately, we have Z

In the above, to get the exponent p, we used the fact that we are summing over indices which satisfyk1+k2+...+km=p. We see that an appropriate bound can be found by bounding

Recall that the derivative is related to thek-th Hermite polynomialHk by

k

∂xk

e−x2

= (−1)ke−x2Hk(x). (5.56)

We thus have

It has been shown by Jack Indritz in [7] that

|Hk(√

ns)| ≤(2kk!)12ens

2

2 . (5.58)

We thus have

Applying this bound, we have Z

Applying this bound, we have

Gathering these results, we have n Applying this bound to eq. (5.52), we have

P

Applying Cauchy-Schwartz, we have X In the above, we used two identities. The first one is simply

X

and the second one is

X

k1+...+km=p

1 =

p+m−1 m−1

. (5.67)

The second identity is the answer to the combinatorial question : Givenmbars, how many ways are there to distribute pstars in between the m bars? It is thus a matter of applying the stars and bars method which gives us our identity.

We thus have 2m2||ψ||

P

X

p=0

t maxl∈{1,...,m}l|

n12212p p!

X

k1+k2...+km=p

p k1, k2, ..., km

m Y

l=1

kl!

!12

(5.68)

≤ 2m2||ψ||

((m−1)!)12

P

X

p=0

t

max

l∈{1,...,m}l|

(mn)12212 p

((p+m−1)!)12

p! .

Finally, we have

((p+m)!)12

(p+1)!

((p+m−1)!)12 p!

= (p+m)12

p+ 1 →0, p→ ∞. (5.69)

This implies that the series

P

X

p=0

t

max

l∈{1,...,m}l|

(mn)12212 p

((p+m−1)!)12

p! (5.70)

converges asP → ∞. By earlier observations, this implies that

X

p=0

tp

p!||(Φω(g))pψn||<∞. (5.71) To summarize what we have shown. For anyψ∈ Hω, there exists a collection of vectors{ψn}n∈N

such that each ψn is an analytic vector of Φω(g), for all g ∈ M, and ψn → ψ. To put it more succinctly, for everyg∈M, the operators Φω(g) all have and share a dense set of analytic vectors.

By definitions of the domains of aω(g) and aω(g), we immediately see that these are densely defined operators. Next, letφ∈D(aω(g)) andη∈D(aω(g)). A simple calculation shows that

hη, aω(g)φi=haω(g)η, φi . (5.72) The adjoint of aω(g) is the unique operator which satisfies the above equation. Sinceaω(g) also satisfies this equation, we see that (aω(g)) is an extension ofaω(g). Sinceaω(g) is a densely de-fined operator, it follows that (aω(g)) is also a densely defined operator, and the operatoraω(g) thus has a densely defined adjoint which implies thataω(g) is closable. Applying the same logic toaω(g), we see that it too is closable.

Letφ∈D(aω(g)). A simple calculation shows that

||Φω(g)φ||2+||Φω(ig)φ||2=||aω(g)φ||2+||aω(g)φ||2. (5.73) Going further, using the unitarity of the given operators, we have

ω(g)φ,Φω(ig)φi= lim

h→0

πω(W(hg))−1

h φ,πω(W(hig))−1

h φ

(5.74)

= lim

h→0

πω(W(−hig))−1 h

πω(W(hg))−1

h φ, φ

.

Applying the commutation relations, and telescoping appropriately, we have πω(W(−hig))−1

h

πω(W(hg))−1 h =e−ih

2||g||2

2 −1

h2 W(−hig)W(hg) (5.75)

ω(W(hg))−1 h

πω(W(−hig))−1

h .

Going back to the original equation, we have πω(W(−hig))−1

h

πω(W(hg))−1

h φ, φ

=e−ih2||g||2−1

h2 hW(−hig)φ, W(−hg)φi (5.76) +

πω(W(−ihg))−1

h φ,πω(W(−hg))−1

h φ

.

In the above, we used the commutation relations to compute W(−hig)W(hg) =e−iImhhig,hgi2 W(h(1−i)g) =e−ih

2||g||2 2 e−ih

2||g||2

2 W(hg)W(−hig) (5.77)

=e−ih2||g||2W(hg)W(−hig). Lettingh→0, we have

ω(g)φ,Φω(ig)φi=−i||g||2||φ||2+hΦω(ig)φ,Φω(g)φi . (5.78) Equipped with this equality, we have

||aω(g)φ||2− ||aω(g)φ||2=

Φω(g) +iΦω(ig)

√2 φ

2

Φω(g)−iΦω(ig)

√2 φ

2

(5.79)

=ihΦω(ig)φ,Φω(g)φi − hΦω(g)φ,Φω(ig)φi 2

−ihΦω(g)φ,Φω(ig)φi − hΦω(ig)φ,Φω(g)φi 2

=−||g||2||φ||2 . Combining the above equation and eq. (5.73), we have

||Φω(g)φ||2+||Φω(ig)φ||2= 2||aω(g)φ||2+||g||2||φ||2 . (5.80) We are now in a position to show that the operatorsaω(g) are closed. Letφ∈D(aω(g)) such that there exists a sequence {φn}n∈N such that φn →φ and aω(g)φn →η for some η ∈ Hω. By the previous equation, we have

||Φω(g)(φn−φm)||2+||Φω(ig)(φn−φm)||2= 2||aω(g)(φn−φm)||2+||g||2||φn−φm||2 (5.81) The operators Φω(·) are self-adjoint and thus they are closed. Furthermore, we haveφn−φm ∈ D(Φω(g))∩D(Φω(ig)), andφn−φm→0, n, m→ ∞. We have

||Φω(g)(φn−φm)||2+||Φω(ig)(φn−φm)||2→0, n, m→ ∞. (5.82) This implies that the sequence with terms Φω(g)φn is a Cauchy sequence and hence converges.

Because Φω(·) are closed operators andφ∈D(Φω(g)), we must have

Φω(g)φn→Φω(g)φ, n→ ∞, (5.83)

and the same for Φω(ig). Reusing the previous equation, we have

||Φω(g)(φn−φ)||2+||Φω(ig)(φn−φ)||2= 2||aω(g)(φn−φ)||2+||g||2||φn−φ||2 . (5.84)

Lettingn→ ∞, we have

||aω(g)(φn−φ)|| →0, n→ ∞, (5.85)

which implies that η = aω(g)φ. The operator aω(g) is thus closed. The same argument shows thataω(g) is closed.

We have shown that (aω(g)) is an extension of aω(g). Unfortunately, the proof that aω(g) is also an extension of (aω(g)) and thus aω(g) = (aω(g)) would lead us too far astray from our main topic. The proof of this relies on the Stone-von Neumann theorem regarding the uniqueness of some finite dimensional algebras. Furthermore, one must introduce the concept of normal states and normal representation which will not play a part in the rest of this thesis. For the interested reader, it is encouraged to examine the full proof of the Stone-von Neumann theorem along with the corollary of our desired result in [1, p. 34, Corollary 5.2.15].