# When is $e^{(-Ht)}$ trace-class?

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Hi

often, when you are doing for example Statistical Physics you need that the operator  $e^{(-tH)}$ is trace-class as you otherwise run into mathematical difficulties. Now, consider a one-dimensional Hamiltonian $H = -d^2_x + V$. Are there any known conditions for $V$ such that this operator is trace-class? Does anybody know a good reference?

asked May 28, 2014

## 2 Answers

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A necessary condition is that your Hamiltonian $H$ (I assume self-adjoint) is such that

(1) it has pure point spectrum,

(2) it has spectrum bounded below,

(3) every eigenspace has finite dimension.

This is because if $e^{-tH}$ (I assume $t>0$) is trace class, it must be compact and compact operators have bounded spectrum made of proper eigenvalues, with $0$ as the only possible accumulation point and only possible point of the continuous spectrum.

If you work in $L^2(\mathbb R)$, Hamiltonians like $-\frac{d^2}{dx^2} + V(x)$ generally have a continuous part unless $V$ diverges as $|x| \to +\infty$. In the former  case $e^{-tH}$ cannot be trace class, the latter has to by studied case by  case.

If, instead, you work in compact regions (with suitable boundary conditions), as in the case you are considering,  the spectrum is well behaved (in view of known properties of the heat kernel of the Laplace-Beltrami operator corrected with a regular potential).  Thus you only have to check if $\sum_{n=0}^{+\infty} e^{-t E_n}$ converges. To this end there is the celebrated Weyl's asymptotic formula $$E_n^{D/2} \sim C n\:,$$ where $D$ is the dimension of the manifold $D=1$ in our case, $C>0$ is a constant and $E_{k+1} \geq E_k$.

Thus $E_n \sim C n^2$ and $$\sum_{n} e^{-t E_n} \sim \sum_n e^{-tCn^2}$$ always converges for $t>0$.

The final answer is: Yes $e^{-tH}$, for $t>0$, is trace class for the potential $\cos x$ in $L^2([0,2\pi])$ with periodic boundary conditions.

Compactness of the region plays a crucial role here. This is the reason why a quantum gas  in a finite box can be treated by means of the standard QM (states are positive trace class operators with unit trace) whereas infinitely extend systems need the algebraic formalism and the KMS condition to be mathematically rigorously handled  (if one is not content with the standard thermodynamic limit procedure).

answered May 29, 2014 by (2,085 points)
edited May 29, 2014
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A Hermitian operator is trace class iff its sum of absolute values of eigenvalues converges. For $e^{-tH}$ this means that $\sum_k e^{-tE_k}$ converges, where the $E_k$ are the eigenvalues of $H$, counted with multiplicity. By the root criterion, a necessary condition is that the inferior limit of $E_k/k$ is nonnegative, and a sufficient condition is that the inferior limit of $E_k/k$ is positive, i.e., $E_k\ge Ck$ for sufficiently large $k$ and some $C>0$. This is the case for the harmonic oscillator and (by majorization) for any potential that can be underestimated by a harmonic potential.

On the other hand, much of statistical mechanics is about Hamiltonians that do not satisfy this criterion. Then a thermodynamic limit is needed to make everything work, and the right setting is $C^*$-algebras and KMS states.

answered May 28, 2014 by (15,518 points)
edited Jun 2, 2014

right, thank you for this answer, but I was more interested in the case when you don't know the spectrum explicitely and the reasoning is just based on the Hamiltonian. so, let's consider the Schrödinger operator $H = -d_x^2+\cos(x)$ on $L^2[0,2\pi]$ for example. This is Mathieu's equation, but there is no explicit representation for the eigenvalues available. Can I say anything in this case?

@Lefschetz: You need to underestimate the potential of interest by a potential for which you can solve the eigenvalue problem. The k-th eigenvalue varies monotonically with the potential, thus you may be able to verify the test given. On the other hand, if you overestimate instead, you may be able to show that the condition is violated.

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