I would like to derive the partition function for the quantum Harmonic oscillator from scratch:
$$\tag{1} Z = \int dp \, dx\, e^{-\beta H}.$$
The free particle appears in many textbooks. $H = p^2$ so it is a Gaussian integral
$$\tag{2} Z = \int dp \, dx \, e^{-\beta p^2} = L \int dp \, e^{-\beta p^2}
= L \sqrt{2\pi/\beta}.$$
I wanted to do the same calculation for the Harmonic oscillator I get could take advantage that
$$\tag{3} Z = \int dp \, dx \, e^{-\beta (p^2 + x^2)} =
\int dx \, e^{-\beta x^2} \cdot \int dp \, e^{-\beta p^2 } =
\sqrt{2\pi} \cdot \sqrt{2\pi} = 2\pi .$$
Or I could integrate over the circles
$$\tag{4} H = p^2 + x^2 = E$$
take advantage the Hamiltonian is symmetric on rotation in phase space
$$\tag{5} Z = \int dp \, dx \, e^{-\beta (p^2 + x^2)} =
\int dE \, d\theta \, E \, e^{-\beta E} = 2\pi. $$
It seems I have forgotten that energy levels are quantized, so I should integrate over the circles
$$\tag{6} H = E_n = \hbar \omega (n + \tfrac{1}{2}).$$
Where are the $E_n$ wavefunctions $\psi_n$ located in phase space? By Heisenberg uncertainty, we can't specify both $(p,x)$ in the phase plane. Are they equidistributed on the circle $p^2 + x^2 = E$?
Best Answer
(Not really an answer, but as one should not state such things in comments, I'm putting it here)
You commented: "This seems to boil down to the relationship between the phase space and the Hilbert space."
That's a deep question. I recommend reading Urs Schreiber's excellent post on how one gets from the phase space to the operators on a Hilbert space in a natural fashion. I'm not certain how the Wigner/Moyal picture of QM relates to quantum statistical mechanics, since we define the quantum canonical partition function to be $Z(\beta) := \mathrm{Tr}(\mathrm{e}^{\beta H})$ on the Hilbert space of states, as we basically draw the analogy that the classical phase space is the "space of states" for our classical theory, and the integral the trace over it, and generalize that to the quantum theory.
Also note that, in a quantum world, $\int\mathrm{d}x\mathrm{d}p\mathrm{e}^{-\beta H}$ is a bit of a non-sensical expression, since $H$ is an operator - the result of this would not be a number, which the partition function certainly should be.