1
$\begingroup$

In Kerson Huang's Statistical Mechanics (2nd ed.), it is claimed that the phase space volume occupied by the canonical ensemble is the partition function: $$ Q_N (V, T) \equiv \int \frac{dp dq}{N! h^{3N}} e^{-\beta \mathcal{H}(p, q)}, \tag{1} $$ where $N$ is the number of particles and $\mathcal{H}(p, q)$ is the Hamiltonian that describes the macroscopic system. I cannot understand why this is the case. Shouldn't it be something like the following: $$ V_{\text{phase space}} = \int_{\Sigma} dp dq, \tag{2} $$ where $\Sigma$ is the region "under" the hyper-surface of constant energy $E$, which would be the maximum energy of the system + heat bath. Now, I can understand why Eq. $(1)$ is the sum over all states, but I fail to see how $(1)$ and $(2)$ are the same. What am I missing here?

$\endgroup$
  • 1
    $\begingroup$ (1) and (2) are indeed not the same. The wording in Huang's book is pretty bad. The partition function is indeed (1), but it is not equal to the volume of the phase space (one could say that it is the measure of the phase space according to $e^{-\beta\mathcal{H}}/(N!h^{3N})$, but that would not convey any additional information, so I don't see the point)$. $\endgroup$ – Yvan Velenik May 6 at 13:13
  • $\begingroup$ @YvanVelenik thanks for your reply. Could you please elaborate on what you mean by "measure of the phase space according to $e^{-\beta \mathcal{H}}/(N! h^{3N})$ " Also, is there any reason why Huang calls the partition function the phase space volume occupied by the canonical ensemble? $\endgroup$ – grav.field May 6 at 19:15
  • $\begingroup$ I don't know what he has in mind when he uses this terminology. I am using measure in its usual mathematical sense. $\endgroup$ – Yvan Velenik May 7 at 6:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.