I am trying to calculate $$Z = \int\limits_{\phi(\beta) = \phi(0) =0} D \phi\ e^{-\frac{1}{2} \int_0^{\beta} d\tau \dot{\phi}^2}$$ without transforming it to the Matsubara frequency space, I can show that $Z = \sqrt{\frac{1}{2\pi \beta}}$. However, I have a problem in obtaining the same result in the Matsubara frequency space: \begin{equation} \phi (\tau) = \frac{1}{\sqrt{\beta}} \left( \sum_{n} \phi_n \ e^{i\omega_n\tau} \right), \end{equation} with $\sum_n \phi_n =0, \omega_n = \frac{2\pi n}{\beta}$. And \begin{equation} Z = \int \prod_n D\phi_n\ \delta\left(\sum_n \phi_n\right)\ e^{-\frac{1}{2} \sum_n \phi_n \phi_{-n} \omega_n^2 } \end{equation} which, I think, vanishes.

I guess the problem lies in the measure. Any comments?

Info: I write the Schulman's derivation in imaginary time here. \begin{eqnarray} Z &=& \int\limits_{\phi(0) =\phi(\beta) = 0} D\phi(\tau) e^{-\frac{1}{2}\int_0^{\beta}d\tau\dot{\phi}^2}\\ &=& \text{lim}_{N \rightarrow \infty} (\frac{1}{2\pi \epsilon})^{(N+1)/2} \int d\phi_1 \dots d\phi_N e^{-\frac{1}{2\epsilon} \sum_{i =0}^N (\phi_{i+1} -\phi_i)^2} \end{eqnarray}

Then, we can use the identity \begin{equation} \int_{-\infty}^{\infty} du \sqrt{\frac{a}{\pi}} e^{-a(x-u)^2}\sqrt{\frac{b}{\pi}} e^{-b(u -y)^2} = \sqrt{\frac{ab}{\pi(a+b)}} e^{-\frac{ab}{a+b}(x-y)^2} \end{equation} to evaluate the sum to be \begin{equation} Z = \sqrt{\frac{1}{2\pi \beta}}. \end{equation}


1 Answer 1


I) The Euclidean path integral with $\hbar=1$ reads

$$ Z~=~\int_{DBC} \!{\cal D}x ~e^{-S},\tag{1}$$

with Dirichlet boundary conditions (DBC)

$$ x(0)~=~0~=~x(T).\tag{2}$$

We expand the real periodic variable $x\in\mathbb{R}$ in Fourier series$^1$

$$ \begin{align} x(t) ~=~&\frac{a_0}{2}+\sum_{n\in\mathbb{N}} \left\{a_n \cos(\omega_n t) + b_n \sin(\omega_n t)\right\}\cr ~=~& \sum_{n\in\mathbb{Z}} c_n e^{i\omega_n t}, \cr \omega_n~:=~&\frac{2\pi n}{T},\qquad a_n,b_n~\in~ \mathbb{R},\cr c_n~\in~& \mathbb{C}, \qquad c_n^{\ast}~=~c_{-n}. \end{align} \tag{3} $$

The DBC (2) becomes

$$ \sum_{n\in\mathbb{Z}} c_n~=~0\qquad\Leftrightarrow\qquad c_0 ~=~ -2{\rm Re}\sum_{n\in\mathbb{N}}c_n .\tag{4}$$

The action for a free non-relativistic point particle with mass $m=1$ reads:

$$ S~=~\frac{1}{2} \int_0^T \!dt~ \dot{x}^2~=~T\sum_{n\in\mathbb{N}}\omega_n^2 |c_n|^2 . \tag{5} $$

II) We know that the proper normalization of the path integral (1) is

$$ Z~=~\frac{1}{\sqrt{2 \pi T}}.\tag{6} $$

This can e.g. be deduced (without introducing fudge factors!) from the (semi)group property of Feynman path integrals, cf. this Phys.SE post and links therein. Up till now we have basically just restated what OP wrote in his question.

III) Now we would like to repeat the same calculation using Fourier series, i.e. work with the Matsubara frequencies. In this answer, we will not explore the (semi)group property, but just do a quick and dirty calculation using various fudge factors, and see what we get. Since this is homework, the explanation will be somewhat brief.

To make heuristic sense of the path integral (1), we will use the following zeta function regularization rules:

$$ \prod_{n\in\mathbb{N}} a ~=~\frac{1}{\sqrt{a}} \quad\text{and}\quad \prod_{n\in\mathbb{N}} n ~=~\sqrt{2\pi}, \tag{7}$$

stemming from the zeta function values

$$ \zeta(0)~=~-\frac{1}{2} \quad\text{and}\quad \zeta^{\prime}(0)~=~-\ln\sqrt{2\pi} , \tag{8} $$ respectively. Now let the path integral measure be

$$ \begin{align} {\cal D}x~:=~&\delta\left(B\sum_{n\in\mathbb{Z}} c_n\right) A\mathrm{d} c_0 \prod_{n\in\mathbb{N}} A^2 \mathrm{d}^2c_n \cr ~\stackrel{(7)}{=}~&\frac{1}{B}\delta\left(\sum_{n\in\mathbb{Z}} c_n\right) \mathrm{d} c_0 \prod_{n\in\mathbb{N}} \mathrm{d}^2c_n , \end{align}\tag{9} $$

where $A$, $B$ are fudge factors. Interestingly, eq. (9) is independent of the $A$-fudge factor! After performing the delta function integration and the Gaussian integrals, we find

$$ \begin{align} Z~=~&\frac{1}{B} \prod_{n\in\mathbb{N}} \frac{\pi}{T\omega_n^2}\cr ~=~&\frac{1}{B} \prod_{n\in\mathbb{N}} \frac{T}{4\pi n^2}\cr ~\stackrel{(7)}{=}~&\frac{1}{B\sqrt{\pi T}}. \end{align}\tag{10}$$

Apparently we should chose the fudge factor $B=\sqrt{2}$ in order to achieve the correct normalization (6).


$^1$ Note that the sine (cosine) modes (3) correspond trivially (non-trivially) to the even (odd) modes in my Phys.SE answer here, respectively.

  • $\begingroup$ emm, this won't address op's question in the comment: if $\prod_{n\in\mathbb{N}} \frac{T}{4\pi n^2}$ is already convergent(to 0), why do we still need to regularize? $\endgroup$
    – Jia Yiyang
    Jul 13, 2014 at 8:14
  • $\begingroup$ In statistical physics, the partition function $0<Z<\infty$ is a manifestly positive quantity. In other words, the logarithm $\ln Z\in\mathbb{R}$ should be a finite quantity. (In particular, getting $Z=0$ is an unphysical wrong result, and is in some sense just as bad as getting $Z=\infty$, and a sure sign that regularization (& possibly renormalization) are needed.) $\endgroup$
    – Qmechanic
    Jul 13, 2014 at 8:31
  • $\begingroup$ The other thing is, I think, interpreting $\prod_{n\in\mathbb{N}} \frac{T}{4\pi n^2}$ as $\lim_{N\to\infty}\prod_{n=1}^N \frac{T}{4\pi n^2}$ is already a regularization, i.e. lattice regularization, $\omega_n=\frac{2\pi n}{\beta}, n=0, \pm 1, \pm2,\ldots\pm \frac{N}{2}$, where $N=\frac{\beta}{\epsilon}$, and in such regularization $Z$ literally equals $\lim_{N\to\infty}\prod_{n=1}^N \frac{T}{4\pi n^2}=0$ $\endgroup$
    – Jia Yiyang
    Jul 13, 2014 at 9:07
  • $\begingroup$ @Qmechanic I read your answer. Thank you. As I am not familiar with the method you use, I can only say it might be able to solve the problem. But I did not expect this problem to be a tough academic question. $\endgroup$
    – DKS
    Jul 13, 2014 at 14:29
  • $\begingroup$ The cut-off regularization $\prod_{n=1}^N \frac{T}{4\pi n^2}$ also needs renormalization, i.e. insertion of counter-terms. $\endgroup$
    – Qmechanic
    Jan 14, 2016 at 18:36

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