Suppose $H$ is a self-adjoint operator acting on a separable Hilbert space and $H$ has a discrete spectrum with eigenvalues converging to $+\infty$. I want to investigate under what condition the operator $$ \rho := e^{-\beta H} $$ is of trace-class?
What I have done so far is that if $\{\lambda_n\}$ are the eigenvalues and $\{f_j\}$ is an ONB then
$$ tr(e^{-\beta H}) = \sum_j \langle e^{-\beta H} f_j, f_j \rangle = \sum_j e^{-\beta \lambda_j} $$ Is there any restriction that can be imposed on those eigenvalues for that this last quantity is finite?
Thanks.
$\lambda_j \rightarrow +\infty$ is certainly not enough. For example, let $\lambda_j = \frac{ln(j)}{\beta}$, then $\sum_j e^{-\beta\lambda_j} = \sum_j \frac{1}{j} = +\infty$. But basically everything converging to $+\infty$ faster than $\frac{ln(j)}{\beta}$ works. More precisely, if $\lim \inf_{j \rightarrow \infty} \frac{\beta\lambda_j}{ln(j)} > 1$, then the infinite sum converges.