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Let $A$ be an unbounded self-adjoint operator on $L^2(\mathbb{R})$ and we are assuming the $L^2$ functions to be complex-valued.

We further assume (e.g. compactness of resolvent) that there exists an orthonormal eigenbasis $\{ \phi_n \}$ of $L^2(\mathbb{R})$ with real eigenvalues $\lambda_n$: \begin{equation} \int_{\mathbb{R}} \overline{\phi_n}(x) \phi_m(x) dx=\delta_{m,n} \text{ , } A\phi_n=\lambda_n \phi_n \end{equation}

Moreover, the completeness relation is found in physics liteature as follows: \begin{equation} \sum_{n}\phi_n(x) \overline{\phi_n}(y)=\delta(x-y) \end{equation} where $\delta(x-y)$ is the Dirac-delta distribution. My questions are as follows:

  1. What is the most rigorous way I can interpret the completeness relation above? Perhaps in the sense of tempered distributions?

  2. Since $A$ is self-adjoint, functional calculus allows us to define $e^{-A}$ as a bounded operator on $L^2(\mathbb{R})$. Then, is it possible to carry out a computation like the below? \begin{equation} e^{-{A_x}} \delta(x-y)=\text{some smooth function in }x,y = \sum_{n} e^{-{A_x}} \phi_n(x) \overline{\phi_n}(y)=\sum_n e^{-\lambda_n} \phi_n(x)\overline{\phi_n}(y) \end{equation}

where I used the notation $A_x$ to mean that the operator $A$ acts on the $x$ argument only.

  1. If the computation in item 2 is possible, in what sense should I understand it?

These seem quite subtle problems between math and physics.. Could anyone please clarify for me?

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    $\begingroup$ You might want to be assuming also that $A$ has compact resolvent, to assure existence of an orthonormal basis of eigenvectors. $\endgroup$
    – paul garrett
    Commented Jun 24, 2022 at 18:39
  • $\begingroup$ Ok I will asume that as well. $\endgroup$
    – Isaac
    Commented Jun 24, 2022 at 18:59
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    $\begingroup$ No, self-adjointness alone is not enough, either for bounded or unbounded (self-adjoint) operators. For example, multiplication by $1/(1+x^2)$ is bounded self-adjoint, but has no eigenvectors. Likewise, multiplication by $x^2$ is unbounded, and has self-adjoint extensions... but has no eigenvectors. But I guess this is just part of the hypotheses you want. Or have "generalized eigenvectors", doing distributional stuff, but then the sum is not a literal sum, etc. $\endgroup$
    – paul garrett
    Commented Jun 24, 2022 at 19:04
  • $\begingroup$ OK, I edited my question. $\endgroup$
    – Isaac
    Commented Jun 24, 2022 at 19:25
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    $\begingroup$ The fact that you start with an operator $A$ is irrelevant to the question. You simply have a basis $\{\phi_n(x)\}_{n\in\mathbb N}$ of $L^2(\mathbb R)$, and you're asking whether $\sum_{n}\phi_n(x) \overline{\phi_n}(y)=\delta(x-y)$ in the sense of distributions on $\mathbb R^2$ (I think that the answer should be yes, but I'm not sure). The "completeness" that you're referring to is the statement that $\{\phi_n(x)\}_{n\in\mathbb N}$ forms a genuine orthonormal basis of $L^2(\mathbb R)$, and not just a set of orthonormal linearly independent vectors. $\endgroup$
    – André Henriques
    Commented Jun 24, 2022 at 23:46

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