If $\mathcal{O}$ is an hermitian operator in a system given by Hamiltonian $H$ and inverse temperature $\beta$, is $$\langle \mathcal{O} \mathcal{O} \rangle = Tr (e^{-\beta H} \mathcal{O} \mathcal{O})$$ always convergent, or do we have to regulate it ? I am assuming that the system is a very general quantum field theory and the hamiltonian is bounded from below.
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$\begingroup$ Well it's simultaneously diagonal. So do a basis expansion and see.. $\endgroup$– user18764May 28, 2018 at 17:27
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3 Answers
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No.
Consider for example observables of the form $\mathcal O = e^{\beta H/2}$; the value exponentially increases as the probability exponentially decreases and you can get what's in principle an infinite sum of constant terms.
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$\begingroup$ Your counter example is not really illustrative, because taking $\mathcal{O}$ to be a function of Hamiltonian itself, you often run into problems, and in practice $\mathcal{O}$ operator has a finite support over energy eigen basis. For instance Eigen-state thermalization hypothesis fails when the operator itself is considered as the Hamiltonian. $\endgroup$– JaswinMay 29, 2018 at 15:48
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In general, the expectation value $tr(A e^{-\beta H})$ is not defined for a generic selfadjoint operator $A$ (of the form ${\cal O}^2$ or not) if it is unbounded as it is the standard situation in QFT.
$tr(A e^{-\beta H})$ however converges if (with the written order!) the range of $e^{-\beta H}$ belongs to the domain of $A$ and the composition is trace class.
As an example, the hypotheses are valid if $H$ is selfadjoint and bounded below, with pure point spectrum and the dimension of eigenspaces is finite and bounded. In this case, for instance, $A= H^n$ for $n=0,1,2, \ldots$ satisfies all conditions.
When the expectation value is not defined with the formula above, one may try to regularize taking some thermidynamical limit or directly use a more advanced formalism like the algebraic one, together with the KMS condition to characterize thermal states.
In QFT in Minkowski spacetime, for a free theory, $e^{-\beta H}$ though bounded is never trace class, since its spectrum is not a pure point spectrum. Therefore you always must regularize, independently from the choice of $A$, barring forced unphysical choices ($A$ of trace class).
answered May 28, 2018 at 18:31
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Now I have a clear reason why thermal expectation value of a square of an operator is not convergent. In any generic field theory I can express the expectation value as $$ Tr (e^{-\beta H} O O) = \sum_{i,k} e^{-\beta e_{i} }O_{ik} O^*_{ik}$$ Since there is no damping factor in the summation k, the sum is not convergent for any theory and operator.