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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 operators is a Banach space, we can conclude. I cannot figure out how to prove this.

DanielC
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th_phys
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1 Answers1

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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.

Valter Moretti
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