window.onload=function() {document.getElementById("pageHeader").style.cursor = "hand"; document.getElementById("pageHeader").style.cursor = "pointer"; document.getElementById("pageHeader").onclick = function() { window.location.href="http://www.cam.ac.uk"; }; }

Partition function for N identical particles

The purpose of this post is to provide some of the details that go into the derivation of the partition function for a system of N identical particles given in Eq. (5.46) in the handout. Although I gave a more detailed derivation in Thursday’s lectures, there are perhaps two points that merit emphasising:

• The many particle states $$|\mathbf{r}_{1},\ldots,\mathbf{r}_{N}\rangle_{S/A} = \frac{1}{\sqrt{N!}}\sum_{P} \eta_{S/A}(P) |\mathbf{r}_{P1},\ldots,\mathbf{r}_{PN}\rangle,$$ only depend on the set of points $$\{\mathbf{r}_i\}$$ and not on their labelling. Thus, for two bosons $$|\mathbf{r}_1,\mathbf{r}_2\rangle_S=\left(|\mathbf{r}_1,\mathbf{r}_2\rangle+|\mathbf{r}_2,\mathbf{r}_1\rangle\right)/\sqrt{2}=|\mathbf{r}_2,\mathbf{r}_1\rangle_S.$$ This means that when we take a trace over these N particle states, we have to include a factor $$1/N!$$ in order to avoid over counting. For example $$\text{tr}\,\rho_2 = \frac{1}{2}\int d\mathbf{r}_1 d\mathbf{r}_2\,\rho_2(\mathbf{r}_1,\mathbf{r}_2),$$ where $$\rho_2(\mathbf{r}_1,\mathbf{r}_2)=\langle\mathbf{r}_1,\mathbf{r}_2|\exp(-\beta H_2)|\mathbf{r}_1,\mathbf{r}_2\rangle_S$$ is the symmetric two particle density matrix at thermal equilibrium, and $$H_2 = -\frac{\hbar^2}{2m}\left[\nabla_{\mathbf{r}_1}^2+\nabla_{\mathbf{r}_2}^2\right]$$ is the two-particle free Hamiltonian.
• After substituting the expression for the single particle density matrix, this gives rise to the expression for the partition function involving a double sum over permutations, one for the bra and one for the ket, with an additional $$1/N!$$ for over counting $$Z_{N} = \frac{1}{(N!)^2\lambda_{\text{dB}}^{3N}} \sum_{P,Q}\int d\mathbf{r}_{1}\cdots d\mathbf{r}_{N}\,\eta_{S/A}(P)\eta_{S/A}(Q)\exp\left(-\frac{\pi}{\lambda_{\text{dB}}^{2}}\sum_{j=1}^{N}|\mathbf{r}_{Pj}-\mathbf{r}_{Qj}|^{2}\right).$$ Now, we can do one of the sums (that labelled by Q, for example) by observing that, for each permutation Q, the sum over permutations P contains exactly the same terms. The only difference is that they are permuted, and the sum over Q is just $$N!$$ copies of the same sum. This gives the result given in Eq. (5.46) $$Z_{N} = \frac{1}{N!\lambda_{\text{dB}}^{3N}} \sum_{P}\int d\mathbf{r}_{1}\cdots d\mathbf{r}_{N}\,\eta_{S/A}(P)\exp\left(-\frac{\pi}{\lambda_{\text{dB}}^{2}}\sum_{j=1}^{N}|\mathbf{r}_{j}-\mathbf{r}_{Pj}|^{2}\right).$$ Since the product of the $$\eta$$’s reflects the number of transpositions to go from permutation P to permutation Q, the signs are unaffected by this relabelling.
blog comments powered by Disqus