Bounds on Matrix Elements of Unbounded Operators Let $H$ be a separable Hilbert space und $T$ a possibly unbounded densely defined linear operator. (One could probably assume that it's ess. self-adjoint but I would like to avoid this assumption.)
Let $\{e_n\}$ be an orthonormal basis of $H$. The matrix elements of $T$ w.r.t. this basis are
\begin{equation}
 T_{mn}:=\langle e_m, Te_n\rangle. 
\end{equation}
(I guess that the basis must lie in the domain of definition of the operator. For the following discussion it should be possible to assume, that the Hilbert space is $L^2(\mathbb{R})$ and the operator is defined at least on the Schwarz functions)
My question is for which complex numbers $y,z$ the series
\begin{equation}
 f(y,z):=\sum_{m,n} T_{mn}\frac{y^mz^n}{\sqrt{m!n!}}
\end{equation}
is convergent. I would like to find conditions for $T$ such that the function $f$ is defined everywhere in $\mathbb{C}^2$ and holomorphic. In the paper
(R.J. Glauber, Coherent and incoherent states of radiation field, Phys. Rev. 131 (1963) 2766-2788)
the author claims without reference that for arbitrary operators (edit: exact quote is "The operators that occur in quantum mechanics", sorry about imprecise paraphrasing)
\begin{equation} 
|T_{mn}|\leq M n^j m^k
\end{equation} 
for fixed numbers $M, j ,k$ and that this implies convergence of the series everywhere. Cauchy-Schwarz doesn't seem to help to prove this. How can I show this result? I'm very happy about any references on the mathematical properties of matrix elements and thank you very much in advance for your answers. 
 A: Under assumptions the only constraint on these matrix elements is that for any vector $x$ in the domain of $T$, its image $Tx$ should have finite norm. If we assume that all basis vectors are in the domain of $T$ (which is basis dependent statement and doesn't follow from the assumptions) then we have
$$ T e_n = \sum_m e_mT_{mn}, $$
but the norm of this vector is just sum of coefficient in orthonormal basis squared. Therefore we have that
$$ \sum_m |T_{mn}|^2 < \infty .$$
Since this series is convergent, we must have that for every $n$ sequence $T_{mn}$ converges to $0$ as $m \to \infty$ (this condition is necessary, but not yet sufficient because decay of coefficients needs to be sufficiently rapid).
Now in the opposite direction: assume that set of matrix elements $T_{mn}$ satisfying condition derived above is given. Choose domain of $T$ to be set of all finite linear combinations of basis vectors and define $T$ by the formula above. This domain is dense so assumptions you mentioned are satisfied. Therefore we conclude that condition I derived is the best you can get. Of course this is much weaker than the inequality you mention, and it was already shown in comments that it is not true in general. This is in accord with standard principle in functional analysis: If you don't make some strong assumptions about the operator and allow it to be unbounded then it can as pathological as you imagine.
If we assume that operator $T$ is symmetric (just in the sense that matrix elements satisfy $T_{mn}=T_{nm}^*$, not in the compicated functional-analyst sense) then we also get the same estimate for sequence $T_{mn}$ as $n \to \infty$ with $m$ fixed.
If more is known about the domain of $T$ then in principle more estimates can be constructed, basically due to the fact that image of any vector needs to have a finite norm. 
I am quite sure that if assumptions such as essential self-adjointness (or even much weaker: normality, closability) are made then much more rigid estimates can be made about these matrix elements. Sadly I am not able to give you any details off the top of my head. In general keep in mind that difference between general unbounded operator and a self-adjoint operator is huge. 
