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I want to check two properties of trace class operators which I believe to be true an even have some skectches of proofs. In what follows, $A: D(A) \to \mathscr{H}$ is a densely defined self-adjoint operator on a Hilbert space $\mathscr{H}$.

Suppose $e^{-tA}$ is trace class, $t > 0$. Then, it is compact, by the spectral theorem, also self-adjoint. Thus, there is a complete orthonormal set $\{\psi_{n}\}_{n\in \mathbb{N}}$ composed of eigenvectors of $e^{-tA}$ with eigenvalues, say, $e^{-tA}\psi_{n} = \lambda_{n}\psi_{n}$. Because $e^{-tA}$ is trace class and self-adjoint, its eigenvalues are all positive and satisfy $\lambda_{n} \to 0$ as $n \to \infty$.

Question 1: Does it follow that $\{\psi_{n}\}_{n\in \mathbb{N}}$ is also a set of eigenvectors of $A$, with eigenvalues $\mu_{n} = -\frac{1}{t}\ln\lambda_{n}$?

I believe the answer is yes, and my reasoning is the following. By the spectral theorem, every eigenvector of $A$ is also an eigenvector of $e^{-tA}$ and if $\mu_{n}$ is an eigenvalue of $A$ its associated eigenvalue of $e^{-tA}$ will be $\lambda_{n} = e^{-t\mu_{n}}$. Conversely, if $e^{-tA}$ has an eigenvalue $\lambda_{n} \ge 0$ with associated eigenvector $\psi_{n}$, then $\mu_{n} =-\frac{1}{t}\ln\lambda_{n}$ must be an eigenvalue of $A$ with eigenvector $\psi_{n}$, since by the spectral theorem $e^{-tA}\psi_{n} = e^{-t\mu_{n}}$.

Is my proof correct?

For the second question, suppose $A$ has a countable set of eigenvalues $\{\mu_{n}\}_{n\in \mathbb{N}}$ and a trace class operator $B$ has a set of eigenvalues $\{e^{-t\mu_{n}}\}_{n\in \mathbb{N}}$. Question 2: Does it follow that $B = e^{-tA}$?

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I don't think your proof works. For starters, there is a priori no reason for $A$ to have eigenvalues/eigenvectors at all. Also, it is true that $e^{-tA}$ is positive, but that's not just because it is selfadjoint, but rather because (as you say) its eigenvalues are of the form $e^{-t\lambda_n}$ with $\lambda_n$ real; alternatively, $e^{-tA}$ is positive because it has a selfadjoint square root.

But the most important problem is that because $A$ is densely defined, you have no guarantee whatsoever that $A\psi$ exists if $e^{-tA}\psi=\lambda\psi$.

As for your second question, the answer is no. Having the same set of eigenvalues, even if counting multiplicities and the operators are selfadjoint, does not give you equality. The best you can say is that if two selfadjoint operators have the same set of eigenvalues, counting multiplicities, and they both have an orthonormal basis of eigenvectors, then they are unitarily equivalent.

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