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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.

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    $\begingroup$ The claim is false. Let $\{|n\rangle\}_{n=0,1,2, \ldots}$ be a Hilbert basis and define, in strong sense, the unbounded operator $T := \sum_{n \in \mathbb N}e^n |n\rangle \langle n|$ with its natural domain. The matrix elements of $T$ with respect to the said basis (which is trivially included in $D(T)$) are $T_{nm} = \delta_{n,m} e^n$. There are no constant $j,k, M$ satisfying the inequality you wrote for every $n,m$. $\endgroup$ – Valter Moretti Aug 16 '16 at 19:08
  • $\begingroup$ I'm confused about what the actual question is. Do you want a proof that if $|T_{mn}|\leq M n^j m^k$ then $f(y,z)$ is defined and holomorphic everywhere? Or conditions on $T$ such that $f$ exists and is holomorphic? It's OK to have several related questions, but please be precise about exactly what claims and criteria you are interested in. $\endgroup$ – Emilio Pisanty Aug 17 '16 at 14:08
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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.

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  • $\begingroup$ Thank you very much for your answer. I'm not sure if I understand the statement you derrived correctly, in particular what you mean by "the condition derived above" and "under assumptions". The way it seems to me, you established that for arbitrary densely defined operators their matrix elements (w.r.t. a basis in the domain of definition) are square summable in the first index. I'm probably missing something, but I'm not sure how to use this to find conditions on the operator, such that my series in two variables in analytic on $\mathbb{C}^2$. $\endgroup$ – Adomas Baliuka Aug 16 '16 at 19:56
  • $\begingroup$ Your question was about bounds on matrix elements of a densely defined operator. I derived such an estimate and showed that this is the strongest you can get assuming only dense domain. It is much weaker than what was claimed in the paper. In general function you wrote down will not be analytic or even well defined. I believe that the key to understanding what aurhors meant is in phrase "The operators that occur in quantum mechanics". It shows that they had in mind some much more narrow class of operators which is well behaved. I am not sure which exactly they had in mind. $\endgroup$ – Blazej Aug 16 '16 at 20:40

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