I am trying to understand the solution to a problem in Altland & Simons, chapter 4, p. 183. As a demonstration of the finite temperature path integral, the problem asks to calculate the partition function of a single harmonic oscillator. The coherent state path integral is $$ \mathcal{Z} = \int D(\overline{\phi},\phi) \exp \Big[ -\int_0^{\beta} d\tau \, \overline{\phi} (\partial_{\tau} + \omega) \phi \Big] \sim [ \det(\partial_{\tau} + \omega) ]^{-1} \tag{4.53}$$ where the $\sim$ follows from simply treating the path integral as if it were an ordinary Gaussian integral. Using the fact that $\phi(\tau)$ must be periodic, we can expand $\phi$ in a Fourier series and find that the eigenvalues of $\tau$ are $\omega_n = 2\pi n / \beta$, from which we obtain the expression $$ \mathcal{Z} \sim \prod_{\omega_n} (-i \omega_n + \omega)^{-1} = \prod_{n = 1}^{\infty} \Big[ \Big( \frac{2\pi n}{\beta} \Big)^2 + \omega^2 \Big]^{-1}. $$ We obtain the latter expression by pairing each $n$th term with the $-n$th term.

Now, here comes the question: to compute this infinite product, Altland & Simons perform the following steps: $$ \prod_{n = 1}^{\infty} \Big[ \Big( \frac{2\pi n}{\beta} \Big)^2 + \omega^2 \Big]^{-1} \sim \prod_{n = 1}^{\infty} \Big[ 1 + \Big( \frac{\beta \omega}{2\pi n} \Big)^2 \Big]^{-1} \sim \frac{1}{\sinh(\beta \omega / 2)}. $$ It seems to me that to get from the first to the second expression, they are multiplying and dividing by $\prod_{n = 1}^{\infty} (\beta / 2\pi n)^2 $, so as to use the formula $x/ \sin x = \prod_{n = 1}^{\infty} (1-x^2 / (\pi n)^2 )^{-1} $. This seems completely unjustified to me -- not only are you dropping temperature dependence in the $\sim$, but you're effectively multiplying and dividing by zero! Not to mention that the final $\sim$ conveniently ignores a factor of $\beta$ in the numerator in order to get the correct final answer.

Is there something I'm missing, or is this calculation completely bogus? And what is the correct means to get the right answer?

  • $\begingroup$ This might be beside the point of your question, but you are missing the zero mode in your middle expression, aren't you? That gives you a factor of $\omega$. $\endgroup$ – octonion Jun 23 '20 at 1:07
  • $\begingroup$ Yep! I forgot about it because A/S doesn't include it (presumably because it's just a constant), but it should certainly be there. $\endgroup$ – Zack Jun 23 '20 at 3:06

The key is that if you treat the measure of the path integral properly $\mathcal{Z}$ is unitless. It is just a sum of Boltzmann factors. When you write $$\mathcal{Z} \sim \prod_{\omega_n} (-i \omega_n + \omega)^{-1}$$ This is an infinite product of dimensionful quantities. Since $\beta$ is the only dimensionful quantity involved in the definition of a path integral measure ($\omega$ is something depending on the dynamics) you can immediately guess that if you were careful about the definition of the path integral measure you would get a factor of $\prod_{n} \beta^{-1}$. I'm not going to actually show this here, just point out that due to dimensional analysis there is really only one thing it could be.

That answers why Atland/Simons are justified in multiplying by that factor involving an infinite product of $\beta$ that seemed completely ad hoc. The $\beta$ dependence is really coming from a careful treatment of the measure. Note that the one extra missing $\beta$ you point out is exactly what is needed to match with the $\omega$ you missed from the zero mode, as I pointed out in the comments.

To answer your other question about multiplying by a factor that is formally zero. There is what seems to me a more satisfying way to treat it. Let me begin with your middle expression, including the zero mode and factors of temperature coming from the measure. $$\mathcal{Z} \sim \beta\omega\prod_{n = 1}^{\infty} \Big[ ( 2\pi n)^2 + (\beta\omega)^2 \Big]^{-1}=\prod_{n \in Z} \Big[ ( 2\pi n)^2 + (\beta\omega)^2 \Big]^{-\frac{1}{2}}$$ Now let's turn this into a sum by taking the log, and also take the derivative by the quantity $\beta^2\omega^2$. Taking the log and derivative sweeps those divergent constant factors under the rug. $$\frac{d}{d(\beta^2\omega^2)}\log \mathcal{Z} = -\frac{1}{2}\sum_{n\in Z} \frac{1}{( 2\pi n)^2 + \beta^2\omega^2}$$ This sum is regular and you can treat it with the Matsubara sum trick, which is a useful thing to learn, but I won't go over it here. The result is $$\sum_{n\in Z} \frac{1}{( 2\pi n)^2 + \beta^2\omega^2}=\frac{1}{2\beta\omega}\frac{\cosh\left(\beta\omega/2\right)}{\sinh\left(\beta\omega/2\right)}$$ So $$\frac{d}{d(\beta^2\omega^2)}\log \mathcal{Z} =\frac{d}{d(\beta^2\omega^2)}\log \left(\frac{1}{\sinh\left(\beta\omega/2\right)}\right)$$

  • $\begingroup$ Fantastic answer, thank you! It's nice that your solution avoids the "multiplication by zero", but what's the idea behind the divergent constants? I imagine that if we worked with a properly discretized path integral, these terms would not be divergent, and they would cancel with some factors in the measure which are also nominally divergent. Is this correct? $\endgroup$ – Zack Jun 23 '20 at 3:17
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    $\begingroup$ Yes if you go into how the path integral was derived from the quantum mechanics and take the limit carefully everything should be finite at each step. It should be an exercise that is simple enough to do for the harmonic oscillator if you're curious. Really in practice people just trust all this (like how I'm waving my hands here) and just don't worry about divergent constant factors. $\endgroup$ – octonion Jun 23 '20 at 3:35
  • $\begingroup$ I see, thank you for the help, this had me stumped all day! $\endgroup$ – Zack Jun 23 '20 at 3:51
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    $\begingroup$ No problem. I graded a problem like this for a class once, and I wasn't satisfied with the instructor's solution so I spent a day thinking about it then! $\endgroup$ – octonion Jun 23 '20 at 3:53

OP's partition function for the harmonic oscillator

$$\begin{align}Z^{-1} ~=~&\prod_{n\in \mathbb{Z}}\left[ -\frac{2\pi i n}{\beta} + \omega\right] \cr ~=~&\omega\prod_{n\in \mathbb{N}}\left[\left( \frac{2\pi n}{\beta} \right)^2 + \omega^2\right] \cr ~=~&\omega\left[ \prod_{n\in \mathbb{N}}\frac{2\pi }{\beta}\right]^2\left[ \prod_{n\in \mathbb{N}}n\right]^2 \prod_{n\in \mathbb{N}}\left[1 + \left( \frac{\beta \omega}{2\pi n} \right)^2 \right] \cr ~\stackrel{(2)}{=}~&\omega\cdot \frac{\beta}{2\pi }\cdot 2\pi \cdot\frac{\sinh\frac{\beta\omega}{2}}{\frac{\beta\omega}{2}}\cr ~=~&2\sinh\frac{\beta\omega}{2}\cr ~=~&\left(\sum_{n\in\mathbb{N}_0}e^{-(n+1/2)\beta\omega}\right)^{-1} \end{align}\tag{1}$$

can be understood via the following zeta function regularization rules:

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

stemming from the zeta function values

$$ \zeta(0)~=~-\frac{1}{2} \quad\text{and}\quad \zeta^{\prime}(0)~=~-\ln\sqrt{2\pi} ,\tag{3} $$

respectively. See also e.g. this & this related Phys.SE posts.

  • $\begingroup$ I see, very interesting! But this approach somehow clashes with my previous understanding -- it was my original impression that the nominally divergent quantities you've handled with zeta function regularization cancel with terms we've neglected in the measure of the integral (for example, as octonion pointed out, each integral over the properly discretized field $\phi_i$ ought to carry a factor of $\sqrt{\beta}$ so that the partition function remains unitless). Your solution seems to imply that we haven't neglected any factors. Is my previous thinking incorrect? $\endgroup$ – Zack Jun 23 '20 at 16:48
  • $\begingroup$ $\prod_{n\in\mathbb{Z}} \beta~=~1$ in zeta function regularization. $\endgroup$ – Qmechanic Jun 23 '20 at 17:56

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