\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \newtheorem{lemma}{Lemma} \begin{document} We consider unbounded representations of $*$-algebra generated by a pair of self-adjoint elements obeying the relation of the \begin{equation} \label{poly} [a,b] = i \,p(a), \end{equation} where $p(\cdot)$ is a polynomial. Since we are interested in unbounded representations, we need to discuss the operator sense of the releation more closely. To do it, we start with some formal calculations. \begin{lemma} Let $\phi(\cdot)$ be a polynomial. The following relations hold \begin{align} \phi(a)\, b & = b \,\phi(a) + i \phi'(a)\, p(a),\label{der} \\ a\,b^n & = \sum_{k=0}^n i^k C^k_n \,b^k \,\phi_{n-k}(a),\label{pow} \end{align} where the polynomials $\phi_i(\cdot)$ are defined as $\phi_0(x) =x$, $\phi_1(x) = p(x)$, $\phi_n(x) = \phi_{n-1}'(x) \,p(x)$, $n >1$. \end{lemma} \begin{proof} Relation \eqref{der} follows from \eqref{poly} directly, and implies, in its turn, \eqref{pow}. \end{proof} The relations obtained above enable one to get some formal relations wchich will be used below. Our aim is to obtain some formal relations for bounded functions of the generators, which have no sense in the $*$-algebra but will be assumed for the operators in any representations. Since the derived relations include only bounded operators, this eleminates the ambiguity with the sense of the relation in the unbounded case. First, we find, how the formal exponent of $b$ permutes with $a$. According to the Lemma above, we have \begin{align*} a \, e^{tb} & = a \sum_{n =0 }^\infty \frac{t^n b^n}{n!} = \sum_{n=0}^\infty \frac {t^n a b^n}{n! } = \sum _{n=0}^\infty \frac{t^n}{n!} \sum_{k=0}^n i^{n-k} C^k_n b^k \phi_{n-k}(a) \\ &= \sum_{n=0}^\infty \sum_{k=0}^n \frac{t^k }{k!} b^k \,\frac{t^{n-k}}{(n-k)!} i^{n-k} \phi_{n-k}(a) \\ &= \sum_{k=0}^\infty \frac {t^k b^k}{k!} \sum_{l=0}^\infty \frac{t^l i^l}{l!} \phi_l(a) = e^{tb} S_t(a), \end{align*} where we write \[ S_t(x) = \sum_{l=0}^\infty \frac{t^l i^l}{l!} \phi_l(x). \] Similarly to \cite{osbook}, we assume that in any representation the operator $A$ representing $a$ is self-adjoint, and replace it with its spectral projections. We have the relation for unbounded operators $A$, $B$ \begin{equation} \label{int} E_A(\Delta) \exp (it B} = \exp (itB) E_A (S_t^{-1}(\Delta)). \end{equation} We say that the pair of opertors $A$, $B$, satisfies \eqref{poly} if for any $t \in \mathbb{R}$, and any Borel $\Delta$ \begin{equation} \label{bound} E_A(\Delta) \exp (-itB) = \exp (-itB) \, E_A (S^{-1}_t (\Delta)), \end{equation} where $S_t(\cdot)$ is a continuous with respect to $t$ group of measurable transformations of $\mathbb{R} \cup \{\infty\}$ such that \[ S'_t(\lambda) |_{t=0} = f(B). \] The existence and uniqueness problems for $S_t(\cdot)$ are studied in \cite{dalet}. \end{document} %%% Local Variables: %%% mode: latex %%% TeX-master: t %%% End:

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