# Proof $\exp(-\beta H)$ trace-class operator

Let $$H=\frac{p^2}{2}+\frac{x^2}{2}\, : D(H) \to L^2(\mathbb{R})$$ be the Hamiltonian of the harmonic oscillator with $$m=\hbar=\omega=1$$. Prove that $$\exp(-\beta H)$$ is a trace-class operator if $$\beta>0$$.

We know that $$A$$ is a trace-class operator if $$\sqrt{|A|}$$ is a Hilbert-Schmidt operator or equivalently if $$A$$ is compact and

$$\sum_{\lambda~\in \text{ sing}(A)} \lambda m_\lambda < \infty\; ,$$

where $$m_\lambda$$ is the multiplicity of $$\lambda$$. We know that $$\lambda\in \text{sing}(\exp(-\beta H))$$ is of the form

$$\exp\left(-\beta \left( n+\frac{1}{2}\right)\right)$$

with $$m_\lambda=1$$ and $$n\in \mathbb{N}$$. So we have

$$\sum_{\text{sing}(\exp(-\beta H)}\lambda m_\lambda=\sum_{n=0}^{\infty} \exp\left(-\beta \left( n+\frac{1}{2}\right)\right) \leq \sum_{n=0}^{\infty}\frac{1}{\beta^2\left(n+\frac{1}{2}\right)^2} <\infty.$$

Then it only remains to prove that $$\exp(-\beta H)$$ is compact. I have tried to prove that

$$\sum_{k=0}^n \frac{(-\beta H)^k}{k!}$$

is compact $$\forall n$$. In this way, using the fact that the space of compact operator is a Banach space, we can conclude. I cannot figure out how to prove this.

Observe that, from the spectral decomposition of $$e^{-\beta H}$$ we have that $$e^{-\beta H} \psi - \sum_{n=0}^N e^{-\beta (n+1/2)}|n\rangle \langle n|\psi\rangle = \sum_{n=0}^{+\infty} e^{-\beta (n+1/2)}|n\rangle \langle n|\psi\rangle-\sum_{n=0}^N e^{-\beta (n+1/2)}|n\rangle \langle n|\psi\rangle= \sum_{n=N+1}^\infty e^{-\beta (n+1/2)}|n\rangle \langle n|\psi\rangle$$ for every vector $$\psi \in {\cal H}= L^2(\mathbb R,dx)$$. Therefore $$\left|\left|\left(e^{-\beta H} - \sum_{n=0}^N e^{-\beta (n+1/2)}|n\rangle \langle n|\right)\psi\right|\right| = \left|\left|\left(\sum_{n=N+1}^\infty e^{-\beta (n+1/2)}|n\rangle \langle n|\right)\psi\right|\right|\:.$$ Taking the $$\sup$$ over the set of unit vectors in both sides, we also have $$\left|\left|e^{-\beta H}- \sum_{n=0}^N e^{-\beta (n+1/2)}|n\rangle \langle n|\right|\right| = \left|\left| \sum_{n=N+1}^{+\infty} e^{-\beta (n+1/2)}|n\rangle \langle n|\right|\right|\:,$$ but, since $$|| e^{-\beta (n+1/2)}|n\rangle \langle n||| = e^{-\beta (n+1/2)}|| |n\rangle \langle n|||= e^{-\beta (n+1/2)}$$, we also get $$\left|\left|e^{-\beta H}- \sum_{n=0}^N e^{-\beta (n+1/2)}|n\rangle \langle n|\right|\right| \leq \sum_{n=N+1}^{+\infty} \left|\left|e^{-\beta (n+1/2)}|n\rangle \langle n|\right|\right|= \sum_{n=N+1}^{+\infty}e^{-\beta (n+1/2)} = e^{-\beta/2}\sum_{n=N+1}^{+\infty}(e^{-\beta})^n\to 0$$ if $$N\to +\infty$$ because $$\sum_{n=0}^{N}(e^{-\beta})^n \to \frac{1}{1-e^{-\beta}}\quad \mbox{if N \to +\infty}\:.$$ Therefore $$e^{-\beta H}$$ is the limit, with respect to the uniform operator topology, of a sequence of compact operators $$A_N = \sum_{n=0}^N e^{-\beta (n+1/2)}|n\rangle \langle n|$$ $$A_N$$ is compact because it is of finite rank. This is a standard result on compact operators. (See explanation below.)
Since the ideal of compact operators is closed in $$\mathfrak B(\cal H)$$ with respect to that topology, $$e^{-\beta H}$$ is compact as well.
Compactness of finite-rank operators. Compactness for an operator $$T$$, means that it transforms the unit ball into a set whose closure is compact. If $$Ran(T)$$ has finite dimension, the unit ball $$B$$ is sent to a bounded set ($$||T(B)|| \leq ||T|| 1$$) in a closed subspace which can be identified to $$\mathbb C^{\dim(Ran(T))}$$. Since closed bounded sets in $$\mathbb C^n$$ are compact, $$\overline{T(B)}$$ is compact in that space. The abstract properties of compactness (a set is compact in a topological space if and only if it is compact in a sub-topological space containing it) imply that $$\overline{T(B)}$$ is also compact in the whole Hilbert space.
REMARK. I stress that $$\sum_{k=0}^n \frac{(-\beta H)^k}{k!} \not \to e^{-\beta H}\quad \mbox{for n \to +\infty}$$ if the limit is referred to the uniform topology. The limit is true only in the strong operator topology, when restricting the domain of both sides to the span of vectors $$|n\rangle$$,but it is by no means enough to conclude. In particular the operators $$\sum_{k=0}^n \frac{(-\beta H)^k}{k!}$$ are not compact since they are not even bounded! So your idea does not works as it stands, but it has to be modified as I indicated.