Relationship between the partition function and the Hamiltonian

Question

The quantum Hamiltonian $H$ is related to the partition function $Z$ of statistical mechanics by $$ Z = \textrm{Tr} \; e^{-\beta H} $$

But this wikipedia page uses Feynman's path integral formulation of QM to derive that $$ Z = \textrm{Tr} \; e^{-HT / \hbar} $$

It's very counterintuitive to me that these two expressions would both give the same result, given that $\beta = (k_BT)^{-1}$. Is the second formula correct? If so, is there a simple way to see that the two expressions are equal?

Answer 1

The second formula must be handle with care. It is possible to show that the thermodynamic partition function of a system described by an Hamiltonian $H$ may correspond to the path integral under some assumptions (see for example here or look for a textbook as that) and steps

1. The Hamiltonian contains a term that is quadratic in the impulses (there is a gaussian integration over them, but I do not know if this step can be avoided in some way)
2. The theory is written the imaginary time using a Wick rotation $t\to-i\tau$
3. Imaginary-time direction is compactified with a certain finite time length $\beta$
4. Dynamical variables (or fields) must follow appropriate boundary conditions (periodic or anti-periodic) when winding over the imaginary time direction (for example, anti-periodic for fermions and periodic for bosons).

If one follows this recipe, then formally it is possible to say that the path integral representation coincides with the thermodynamic partition function of the system with temperature given by $$ \beta = 1/kT $$ where $\beta$ is the length of the imaginary time interval.