We define the thermal density operator as

$$\tau(\beta) = \frac{e^{-\beta H}}{\mathrm{Tr}(e^{-\beta H})}$$

where $H$ is the systems Hamiltonian.

Today I was told that the eigenvalues of the thermal state density operator are the same as the classical thermal distribution.

I've been looking for sources to confirm this but was not able to find any. Can maybe someone explain this relation to me?


1 Answer 1


Let's consider the basis of the eigenvectors of $H$, where such operator is diagonal. The argument is valid for any basis, as both the eigenvalues and the trace are base invariant.

Let's consider that $\{\epsilon_i\}_{i = 1,...,N}$ are the eigenvalues of $H$. In the chosen base, your operator reads:

\begin{equation} \tau(\beta)_{ij} = \frac{e^{-\beta \epsilon_i}}{\sum^{N}_{k = 1}e^{-\beta \epsilon_k}} \delta_{ij} \end{equation}

where $\delta_{ij}$ is the Kronekcer delta. Indeed, it is a diagonal matrix whose eigenvalues are $\{\frac{e^{-\beta \epsilon_i}}{\sum^{N}_{k = 1}e^{-\beta \epsilon_k}}\}_{i = 1,...,N}$. This corresponds to the probability of the $i^{th}$ state in the canonical ensemble, according to the Boltzmann distribution. (The denominator of the eigenvalues is actually the $Z$ partition function).

  • $\begingroup$ I understood everything except for the introduction part regarding different bases. Are you saying that $\tau(\beta)$ will always be diagonal ? Which eigenvalues are base independent ? $\endgroup$
    – CatoMaths
    Commented Jan 26, 2019 at 18:14
  • $\begingroup$ $\tau(\beta)$ will be diagonal in the basis of its eigenvectors, which happen to be the same as those of $H$, but of course it won't be diagonal in any arbitrary basis. With respect to the invariance of eigenvalues of $H$ and $\tau(\beta)$, I mean that they will remain the same in every basis and the same goes for the trace (this is a general property). This is to convince you of the generality of the argument: even if the calculations were perfomed in the basis of eigenvectors, the result is valid in any basis, since it only depends on the eigenvalues. $\endgroup$
    – Mat
    Commented Jan 26, 2019 at 18:24
  • $\begingroup$ The only basis I ever have learned about is the basis of the eigenvectors of $H$. Thanks to you now I know that the basis of the eigenvectors of $\tau(\beta)$ is the same as the one of $H$. In this case in makes perfect sense that both are diagonal with their respective eigenvalues as their entries. Now is there any other basis which I am missing or is that the point which you were trying to explain to me? $\endgroup$
    – CatoMaths
    Commented Jan 26, 2019 at 18:43
  • $\begingroup$ Well, there is an infinite number of bases that define the Hilbert space you are working with. In this particular case and with this little information, I can't think of any other relevant basis, but just you to know, you might encounter other problems where the Hamiltonian is not diagonal in the relevant basis to work with. $\endgroup$
    – Mat
    Commented Jan 26, 2019 at 19:13

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