Deriving the path integral for periodic boundary conditions

Question

I'm thinking about path integrals with the Euclidean time formalism, where I have partition function $Z=\operatorname{Tr} e^{-\beta \hat H}$. I'm used to the following derivation of the path integral:

\begin{align*} Z&=\int_{\Omega} \langle x_0|e^{-\beta \hat H}|x_0\rangle dx_0\\ &=\iiint_{\Omega^N} dx_0\cdots dx_{N-1} \langle x_0|e^{-\Delta\tau\hat H}|x_1\rangle\langle x_1| (\ldots) |x_{N-1}\rangle\langle x_{N-1}|e^{-\Delta \tau \hat H}|x_0\rangle \\ &=\vdots\\ &=\int [Dx] e^{-\int_0^\beta Ld\tau} \\ \end{align*}

This relies on the repeated application of the identity ${\bf 1}=\int_{\Omega}|x\rangle\langle x|dx$ ($\Omega$ being the volume in question), and the trace enforces periodicity in Euclidean time $x(\tau)=x(0)$. My question is: how can I save this derivation if I put periodic boundary condition on space?

For concreteness I consider a particle in a box of size $2\pi$. If the box has Dirichlet boundary conditions, the proper basis of eigenstates is $\psi_n^d(\theta)=\frac{1}{\sqrt{\pi}}\sin(\frac{n\theta}{2})$, the orthogonality relation $$\int_0^{2\pi}\psi_n^d (\theta)^*\psi_m^d(\theta)d\theta=\delta_{nm},$$ and the completeness relation $$\sum_{n=1}^\infty \frac{1}{\pi}\sin(\frac{n \theta}{2})\sin(\frac{n \theta'}{2})=\delta(\theta-\theta').$$ Each time we insert the identity or take the trace, we're really using this resolution. We should get the following partition function, where paths are constrained to have $\theta(\tau)\in [0,2\pi]$: \begin{align*} Z&=\sum_{n=1}^\infty e^{-\beta \frac{\hbar^2}{2m}\left(\frac{n}{2}\right)^2}\\ &=\int [D\theta] e^{-\int_0^\beta Ld\tau} \end{align*}

If we have periodic boundary conditions, we have $\psi^p_k(\theta)=\frac{1}{\sqrt{2\pi}}e^{ik\theta}$ for any integer $k$, and completeness relation $$\sum_{k\in\mathbb{Z}} \frac{1}{2\pi}e^{ik(\theta-\theta')}=\delta(\theta-\theta').$$ We should get the following path integral which includes a sum over winding numbers and allows $\theta(\tau)$ to step outside the region $[0,2\pi]$.

\begin{align*} Z&=1+\sum_{k=1}^\infty 2e^{-\beta \frac{\hbar^2}{2m}k^2}\\ &=\int [D\theta] e^{-\int_0^\beta Ld\tau} \end{align*}

In the second case, how can I derive this? It seems like the key must be in the two different resolutions of the identity.

I don't have a problem with the intuition, just the mathematics of resolving the identity with these two different boundary conditions.

Answer 1

As Sunyam pointed out in the comment, p. 578 of Kleinert "Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets" covers the derivation with basically the same terminology I'm using. The key is indeed in the different ways to write $|k\rangle$.

Periodic boundary conditions.

Denote $\Omega=[0,2\pi]$. We have a general hamiltonian $H(-i\partial_x,\hat{x})$. We want to sum over the following energy eigenstates (= momentum eigenstates): $$ |k\rangle=\int\frac{1}{\sqrt{2\pi}}e^{ikx}|x\rangle dx$$

We can still start with evaluating $Z$ has a sum over position eigenstates:

\begin{align*} Z&=\int dx_0\langle x_0 | e^{-\beta H}|x_0\rangle\\ &=\int dx_0\cdots dx_{N-1}\langle x_0 | e^{-\Delta \tau H}|x_1\rangle\langle x_1|\cdots | x_{N-1}\rangle\langle x_{N-1}|e^{-\Delta \tau H}|x_0\rangle \end{align*}

The key, from Kleinert, is to evaluate the inner product in a clever way, by making note that:

\begin{align*} \langle x_{n-1}|x_n\rangle&=\sum_k\langle x_{n-1}|k\rangle\langle k | x_n\rangle \\ &=\sum_k \frac{1}{2\pi}e^{ik(x_n-x_{n-1})}\\ &=\sum_\ell \delta(x_n-x_{n-1}+2\pi\ell)\\ &=\sum_{\ell_n} \int\frac{dp_n}{2\pi}\exp\left( i p_n(x_n-x_{n-1}+2\pi\ell_n) \right) \end{align*}

So at this point, we have:

\begin{align*} Z&=\sum_{\ell_0\cdots\ell_{N-1}}\int_0^{2\pi} dx_0\cdots dx_{N-1} \int_\mathbb{R}\frac{dp_0\cdots dp_{N-1}}{(2\pi)^N}\exp\sum_n\left( i p_n(x_n-x_{n-1}+2\pi\ell_n)-\Delta\tau H(p_n,x_n)\right) \end{align*}

The following manipulation can get rid of the summation over $\ell$:

$$\int_0^{2\pi} dx\sum_\ell f(x+2\pi\ell)=\int_{-\infty}^\infty dx f(x)$$

We do this trick on $x_1$, then $x_2$, etc. until we get to the end of the chain.

\begin{align*} Z&=\sum_{\ell_0}\int_0^{2\pi} dx_0 \int_\mathbb{R^{2N-1}}\frac{dx_1\cdots dx_{N-1} dp_0\cdots dp_{N-1}}{(2\pi)^N}\\ &\cdot\exp\left(\sum_{n=1}^{N-1} (i p_n(x_n-x_{n-1})-\Delta\tau H(p_n,x_n))+i p_0(x_0-x_{N-1}+2\pi\ell)-\Delta\tau H(p_0,x_0)\right) \end{align*}

This can't be put into the form $\sum_\ell f(x_0+2\pi\ell)$ (because of the factor of $\exp-ip_1x_0$), and so we're left with a sum over winding numbers and our phase space path integral:

\begin{align*} Z&=\sum_{\ell}\int_{x(0)=x(\tau)+2\pi\ell} \mathcal{D}[x]\mathcal{D}[p]\exp(\int_0^\beta d\tau\left( i p\partial_\tau x-H(p,x)\right)) \end{align*}

You could also plug in $H=p^2/(2m)+V(x)$ and evaluate the $p$ integrals to find the non phase space version:

\begin{align*} Z&=\sum_{\ell}C\int_{x(0)=x(\tau)+2\pi\ell} \mathcal{D}[x]\exp(\int_0^\beta d\tau\left( - \frac{m}{2}\dot{x}^2-V(x)\right) \end{align*}

(where $C=\sqrt{\frac{m}{2\pi\Delta\tau}}^N$ is from evaluating the $p$ integrals)