Problem
In Ceperley's 95 article on path integral Monte Carlo approach I have encountered $\hat{\rho}:L^{2}(R^{3N})\to L^{2}(R^{3N})$
$\hat{\rho} = e^{-\beta \hat{H}}$,
where $\hat{H}$ is a Hamiltonian operator and $\beta=\frac{1}{k_{B}T}$ is a constant. Then the image of some function $\psi$ is
$\hat{\rho}\psi(R_{1})=\sum\limits_{i=1}^{\infty} e^{-\beta E_{i}}\Phi_{i}(R_{1})<\Phi_{i},\psi> = \sum\limits_{i=1}^{\infty} e^{-\beta E_{i}}\Phi_{i}(R_{1})\int\limits_{\mathbb{R}^{3N}} \Phi_{i}^{*}(R_{2})\psi(R_{2})\,\mathrm{d}R_{2}$,
where $E_{i}$ and $\Phi_{i}$ are eigenvalues and eigenvectors of the Hamiltonian operator (just suppose the spectrum is discrete).
We can also suppose that either $\Phi_{i}$ and $\psi$ are not just from $L^{2}$ but from the Schwartz space. Moreover, the operator $\hat{\rho}$ is a trace class operator (which means that $\sum\limits_{i=1}^{\infty}e^{-\beta E_{i}}\Phi_{i}(R)\Phi_{i}^{*}(R)$ converges absolutely for any $R$).
In the article, they swapped the sum with the integral so they got
$\int\limits_{\mathbb{R}^{3N}} \sum\limits_{i=1}^{\infty} e^{-\beta E_{i}}\Phi_{i}(R_{1}) \Phi_{i}^{*}(R_{2})\psi(R_{2})\,\mathrm{d}R_{2}$.
I am struggling to prove that it can be done.
My attempt
I have been trying to prove that the partial sums uniformly converge but I have not managed to do that. I also have not found the dominating function for the Lebesgue's dominated convergence theorem.
I see that $\vert\int\limits_{\mathbb{R}^{3N}}\Phi_{i}^{*}(R_{2})\psi(R_{2})\,\mathrm{d}R_{2}\vert \leq \Vert\psi\Vert$ and $\lim\limits_{i\to\infty} E_{i} = 0$ but I can not prove that it will work even with the term $\Phi_{i}(R)$ inside.
Could you please provide me some advice?
