Considering a single free bosonic mode with Hamiltonian $H=\epsilon a^{\dagger}a$, we can use (bosonic) coherent states to write the corresponding partition function $\mathcal{Z}=\mathrm{tr} (e^{-\beta H})$ as: \begin{equation} \mathcal{Z}=\int d(\bar{\psi},\psi)e^{-\bar{\psi}\psi}\langle\psi|e^{-\beta H}|\psi\rangle, \end{equation} where we integrate over the complex variables $\bar{\psi}$ and $\psi$. To solve this using a path integral we apply the usual procedure, partitioning the imaginary time in $N$ intervals and subsequently inserting resolutions of identity. This simply yields the usual path integral expression (cf. eq. (4.27) in A. Altland's "Condensed matter field theory"): \begin{equation} \mathcal{Z}=\int \prod_i d(\bar{\psi}_i,\psi_i)e^{\sum_i-\bar{\psi}_i\psi_i + \bar{\psi}_{i+1}\psi_i-\delta\epsilon\bar{\psi}_{i+1}\psi_i}, \end{equation} where $\delta\equiv \beta/N$. Usually, one would take the continuum limit at this point, but instead I'd like to first evaluate this integral for large $N$, and afterwards take $N\rightarrow \infty$ (since the integral is quadratic, this should be fine). This being a homework problem, there is a hint that says: \begin{equation} 1+a+a^2+a^3+...=\frac{1}{1-a}, \end{equation} which seems to indicate that the $\mathcal{Z}$ above should be evaluated at each seperate timeslice, resulting in \begin{equation} \mathcal{Z}=\sum_n e^{-\beta n\epsilon}, \end{equation} since we know from quantum statistical mechanics that a single bosonic mode has partition function \begin{equation} \frac{1}{1-e^{-\beta \epsilon}}. \end{equation} In trying to carry out this program, I have tried to use the standard complex Gaussian, but in no way do I seem to get the series mentioned in the hint. Therefore, my question is: how do I evaluate $\mathcal Z$ above for large $N$?


To begin with, the result of the multi-dimensional complex Gaussian integral is \begin{equation} I(A) := \int d(z_1, z_1^\ast) \cdots d(z_n, z_n^\ast)\, \exp\big(-\vec{z}^{\dagger} A \vec{z}\big) = \det(A)^{-1}. \end{equation} Here, $A$ is an $n\times n$ matrix, and $\vec{z} = (z_1, \ldots, z_n)^T$. The integration measure $d(z, z^\ast)$ is defined by \begin{equation} d(z, z^\ast) := \frac{d\mathrm{Re}(z)\, d\mathrm{Im}(z)}{\pi}. \end{equation}

Then, the partition function of a single free bosonic mode with $H = \epsilon \,a^\dagger a$ is simply \begin{equation} \mathcal{Z} = \lim_{n \to \infty} I(M_n) = \lim_{n \to \infty}\det(M_n)^{-1} \end{equation} with \begin{equation} M_n := \begin{pmatrix}1&&&&-1+x_n\\ -1+x_n&1&&&\\& -1+x_n&1&&\\ &&\ddots&\ddots&\\ &&&-1+x_n&1\end{pmatrix}, \end{equation} where $x_n := \beta\epsilon/n$. From the determinant formula \begin{equation} \det(A) = \varepsilon_{j_1 \ldots j_n} A_{1j_1}A_{2j_2} \ldots A_{nj_n}, \end{equation} it is easy to see that $\det(M_n) = 1 - (1-x_n)^n$. Then, we have \begin{equation} \lim_{n\to \infty} (1 - x_n)^n = \left(1-\frac{\beta\epsilon}{n}\right)^n = e^{-\beta\epsilon}, \end{equation} from which it follows that \begin{equation} \mathcal{Z} = \lim_{n \to \infty}\det(M_n)^{-1} = (1 - e^{-\beta\epsilon})^{-1}. \end{equation}

| cite | improve this answer | |

Do you need to use the coherent states to calculate the partition function? You might be overthinking it a bit.

A partition function is the sum of Boltzmann factors ($e^{-\beta E}$) over the Fock space. Your Fock space is defined by the occupation numbers: $|0\rangle$, $|1\rangle$, etc. For you, $\hat H|n\rangle = n\epsilon|n\rangle$. So you need to perform the sum as follows:

$\mathcal{Z} = \sum_n \langle n|e^{-\beta \hat H}|n\rangle = \sum_n e^{-\beta n\epsilon}$,

which is exactly the formula from the hint.

If you insist on using QFT, start by writing down the action for your system using Matsubara frequencies:

$S[\bar\psi,\psi] = \sum_n \bar\psi_n\left[-i\omega_n + \epsilon\right]\psi_n$

Your partition function is

$\mathcal{Z} = \int D(\bar\psi,\psi) e^{-S[\bar\psi,\psi]}$.

From your question, I assume you have Altland's book, so to proceed see Eq.(4.33)

| cite | improve this answer | |
  • $\begingroup$ Thanks for answering. However, the idea behind this question is to stray away from the conventional methods you mentioned, and instead take this unusual route. $\endgroup$ – temperature-dependent Sep 30 '17 at 14:05

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.