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There is possibly a huge literature on the subject but I am a newcomer on analytic representations and my need is rather specific. I simplify it below.

Let $A,B$ be two symbols (standing for annihilation and creation) and set $$ H_\mathbb{C}:=\frac{\mathbb{C}<A,B>}{<[A,B]-1>}=<A,B;[A,B]=1>_{\mathbb{C}-AAU} $$ where $\mathbb{C}<A,B>$ is the free (associative with unity i.e. AAU) algebra and $<[A,B]-1>$ the two sided ideal generated by $[A,B]-1$ (known as Heisenberg or Heisenberg-Weyl - one dimensional - algebra). My question is the following

Are there faithful and handy representations of $H_\mathbb{C}$ which allow the computation of one-parameter groups of its elements ?

The answers can be positive or even non-go theorems (see remarks below).

Remarks : (a) There are lots of variants of Fock spaces to represent faithfully $H_\mathbb{C}$, one of them is the Bargmann-Fock representation ($A=\frac{d}{dz},\ B=z$) which acts on : polynomials, series, entire functions (Hardy space $\mathcal{H}(\mathbb{C})$). The one on entire function allows for computation of the displacement operator as a one-parameter group through $$ e^{t\frac{d}{dz}}[f](z)=f(z+t) $$
the series converges for the topology of compact convergence on $\mathcal{H}(\mathbb{C})$.

(b) An easy non-go theorem is the following :

Theorem There exists no representation of $H_\mathbb{C}$ in a Banach algebra and hence in m-convex Fréchet algebras, the latter being projective limits of the former.

In particular, this proves that, although the space $H_\mathbb{C}=C^\omega(\mathbb{C},\mathbb{C})$ (the space of complex entire functions) can be endowed with the structure of a m-convex Fréchet algebra (the standard topology of compact convergence), the algebra of its continuous operators $\mathcal{L}(C^\omega(\mathbb{C},\mathbb{C}))$ admits no such structure.

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    $\begingroup$ Would the Stone-von Neumann uniqueness theorem or any other result guaranteeing the uiniqueness of the Fock representation -up to unitary equivalence- qualify for such a non-go theorem? $\endgroup$ – Konstantinos Kanakoglou Jul 9 '17 at 16:43
  • $\begingroup$ In order to be more specific, are you mainly interested in representations of the "integrated" forms of the Heisenberg-Weyl generators (i.e. their exponentials) or are you more interested in the representations of the Weyl algebra itself? $\endgroup$ – Konstantinos Kanakoglou Jul 9 '17 at 16:47
  • $\begingroup$ @KonstantinosKanakoglou [ qualify for such a non-go theorem?]---> Yes of course, because I am in search of knowing the possibility (or not) of representations of together Heisenberg-Weyl generators and their exponentials. $\endgroup$ – Duchamp Gérard H. E. Jul 9 '17 at 17:03
  • $\begingroup$ Duchamp, I have added one more tag. I think it is relevant. If however you disagree, feel free to retag. $\endgroup$ – Konstantinos Kanakoglou Jul 16 '17 at 16:04
  • $\begingroup$ @KonstantinosKanakoglou No this is OK. $\endgroup$ – Duchamp Gérard H. E. Jul 16 '17 at 16:32
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The Heisenberg-Weyl algebra or the Weyl algebra or the algebra of the Canonical Commutation Relations (CCR) is generated by the $p,q$ generators subject to the relation $$ [q, p] = i \hbar I \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$ This is isomporphic to the algebra of the OP under the "change of base": $$ A=\frac{1}{\sqrt{2}}(q-ip), \ \ \ \ B=\frac{1}{\sqrt{2}}(q+ip) $$

A first simple and well known result, in the sense of a non-go theorem, can be derived by direct inspection of the commutation relation:

The CCR admit no finite dimensional representations.

This follows easily by a trace argument on both sides of (1) (or of $[A,B]=1$) as long as Planck's constant $\hbar$ is different than zero. So, from the beginning we have to focus in infinite dimensional representations of the Weyl algebra and since the problem is motivated by physical considerations it is natural to consider its generators acting on Hilbert spaces.

An old result (see 1, 2) tells us that when dealing with representations of (1) in Hilbert spaces, the generators $q$ and $p$ cannot both be bounded: at least one of them has to be unbounded.
Given that a self-adjoint (and thus symmetric), unbounded operator in a Hilbert space cannot be defined on the whole of the space but only on a dense subspace of it (this is a direct consequence of the Hellinger–Toeplitz theorem), one can see that the above remark poses delicate questions on the domains of the above operators (and thus the representations of the Weyl algebra) and at the same time underlines the importance of the use of unbounded operators in the formulation of QM.

Initiating from the above remark, Stone and von Neumann studied the integrated forms of $p,q$ i.e. the unitary, one-parameter groups
$$ \begin{array}{cccccc} U(s) = exp(-i s p) & & & & & V(t) = exp(-i t q) \end{array} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) $$ The above expressions are well defined because $p, \ q$ are self-adjoint (see 2,3). $V(t)$ acts on $f(x)$ by multiplication with $exp(-itx)$, while $U(s)$ acts by horizontal translations (shifts) of $f(x)$ i.e. $$ \begin{array}{cccccc} (U(s)f)(x) = f(x-s) & & & & & (V(t)f)(x) = e^{-itx}f(x) \end{array} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) $$ for any square integrable complex function of a real variable, i.e. for any $f(x) \in L_{2}(- \infty, \infty)$. The operators $U(s)$, $\ \forall s \in \mathbb{R}$ and $V(t)$, $\ \forall t \in \mathbb{R}$, are unitary and thus bounded. Their domain is now the whole of the Hilbert space (and not only a dense subspace as was the case for the $p,q$ generators of the Weyl algebra).
Furthermore, $U(s)$, $V(t)$ satisfy $$ U(s)V(t) = e^{ist} V(t)U(s) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4) $$ for all $s \in \mathbb{R}$ and for all $t \in \mathbb{R}$. (4) is frequently called the integrated form of the Weyl relations or the integrated form of the CCR.

The Stone-von Neumann uniqueness theorem, determines conditions -the main one is (4)- which guarantee the uniqueness of the representation (3) (and not directly the uniqueness of the usual Fock representation of $p,q$). In other words, it tells us that, given the operators $U(s), \ V(t)$ on $L_{2}(- \infty, \infty)$, satisfying (4) (plus some technicalities on the domains) then their action is unitarily equivalent to the representation (2), (3).

There have been lots of studies (and references) on the Stone-von Neumann uniqueness theorem, its origins, its extensions, its descendants and on the exact relation between the Weyl generators and their integrated forms. If you are interested in such aspects, I could provide further references on such topics.

Remark 1: Formally, the transition between the CCR (1) and their integrated form (3), (4) can be understood as follows: $\bullet$ If we start from the CCR (1) together with (2) written in power series: $$ \begin{array}{cccc} U(s) = exp(-i s p) = \sum_{n=0}^{\infty}\frac{(-isp)^{n}}{n!} & & & V(t) = exp(-i t q) = \sum_{n=0}^{\infty}\frac{(-i t q)^{n}}{n!} \\ \\ \end{array} \ \ \ \ \ \ \ (5) $$ then one can deduce the integrated form of the CCR (3), (4) by applying the Baker–Campbell–Hausdorff formula. $\bullet$ For the converse, the CCR (1) follow formally on taking $\frac{\partial^2}{\partial t \partial s}$ of (4) at $s=t=0$.
However, this description is rather an indication than a strict proof of equivalence between the Weyl algebra and its integrated form: The reason is that $p, \ q$ are unbounded operators and hence the power series expressions cannot be valid on the whole of the space. Consequently, in general, the integrated form of the Weyl relations is not equivalent (at least not in some obvious way) with the Weyl relations themselves (i.e. the CCR).

Remark 2: For the case of the infinite degrees of freedom Weyl algebra, there is no analogue -at least to my knowledge- of the Stone-von Neumann theorem (its applicability is limited to the finite generators case). On the contrary, it has been shown (see 4) that, the CCR admit infinitely many, non-equivalent, irreducible representations on a Hilbert space, which are in bijection with the set of real numbers.
For interesting examples of irreducible CCR representations, which are inequivalent to the usual Fock space representation, one can look at books on the Foundations of Quantum Mechanics such as: 5, vol. II, p.251-252.

References:

  1. A. Wintner, "The unboundedness of quantum mechanical matrices", Phys. Rev., v.71, 1947, p. 738-739

  2. C. R. Putnam, "Commutation Properties of Hilbert Space Operators and Related Topics", Springer-Verlag, Berlin, (1967)

  3. M. Reed and B. Simon, "Methods of Modern Mathematical Physics", vol.I, Academic Press (1975).

  4. A. S. Wightman, S. S. Schweber, "Configuration space methods in relativistic quantum field theory", Phys. Rev., v.98, 3, (1955), p.812-837

  5. Galindo-Pascual, Quantum mechanics, v.I, II, Springer, 1990, Translated from the spanish by: J.D. Garcia, L. Alvarez-Gaume

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  • $\begingroup$ This is very interesting, it seems well-known and I have to look at this closer. What is maybe less standard is that I need one-parameter groups of all elements in the CCR (that I call $\mathcal{H}_\mathbb{C}$) i.e. for $h\in \mathcal{H}_\mathbb{C}$, how about $e^{th}$ ? $\endgroup$ – Duchamp Gérard H. E. Jul 9 '17 at 19:18
  • $\begingroup$ Well not only generators but all elements (if possible) $\endgroup$ – Duchamp Gérard H. E. Jul 9 '17 at 20:10
  • $\begingroup$ ahh ..so $h$ should be an arbitrary monomial of the CCR generators or some linear combination of such monomials, right? $\endgroup$ – Konstantinos Kanakoglou Jul 9 '17 at 20:11
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    $\begingroup$ Yes indeed. Thank you already for your detailed answer though. $\endgroup$ – Duchamp Gérard H. E. Jul 9 '17 at 20:19
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    $\begingroup$ As a first comment: I think that Baker-Campbell-Hausdorff formula could be of some help to your queries. I am not sure it will provide a definitive answer though. So, i will try to give some second thought on it and i will -hopefully- come back. $\endgroup$ – Konstantinos Kanakoglou Jul 9 '17 at 20:27

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