Prefactors for Fourier Transforms in Altland Simons

I am trying to understand how to do the Fourier Transform on pages 184 - 185 in Altland Simons(2nd ed). In particular, we are told in the problem statement part b:

$$S[\theta] = \frac{1}{2 c} \int dx d \tau [(\partial_\tau \theta)^2 + (\partial_x \theta)^2]$$

Writing in terms of a Fourier series:

$$\partial_\tau \theta = \frac{1}{\sqrt{\beta L}} \sum_{\omega_{\, n} \, , \, p} \theta_{n, p} \; e^{i (px - \omega_{\, n} \; \tau)} \; \; (-i \omega_n) \;$$ $$\partial_x \theta = \frac{1}{\sqrt{\beta L}} \sum_{\omega_{\, n} \, , \, p} \theta_{n, p} \; e^{i (px - \omega_{\, n} \; \tau)} \; \; (ip) \;$$

Now using the formulas:

$$\int_0^\beta d \tau e^{- i \omega_{\, n} \; \tau} = \beta \delta_{\omega_{\, n} \, , \, 0} \;$$ $$\frac{1}{L} \int dx e^{-i (k + k') x} = \delta_{k + k', 0}\;$$ $$\beta \equiv \frac{1}{T} \;$$

the first from page 168 (right below eq. 4.32), the second from the first line of page 21, and the third from page 166 (right below eq. 4.23), we find:

$$S[\theta] = \frac{1}{2c} \sum_{p, n} |\theta_{p, n} \, |^2 \, (p^2 + \omega_n^2 \, )$$

since the prefactor $$\frac{1}{\beta L}$$ arising from $$(\partial_\tau \theta)^2, (\partial_x \theta)^2$$ is canceled by the $$\beta, L$$ from the first two formulas under "now using the formulas".

However, on page 185 in the solution section AS gives the answer as

$$S[\theta] = \frac{L}{2cT} \sum_{p, n} |\theta_{p, n} \, |^2 \, (p^2 + \omega_n^2 \, ),$$

which has an additional factor of $$\frac{L}{T}$$.

• Could you add the edition of the book and the numbers of equations? Aug 5, 2021 at 11:46
• Sure, will do that now. Aug 6, 2021 at 16:44
• A list of errors for the first edition (I think) can be found here. You could check whether your concerns are listed there. Of course, even if not listed there, it could be an error. Aug 6, 2021 at 17:53

The answer is simply that they use a weird normalization of their Fourier modes. To get the same action that they use, define $$\theta_{p,n}$$ via $$\theta_{p,n} = \frac{1}{\beta L} \int_0^{\beta}\! d\tau \int_0^L \! dx \ e^{i (\omega_n \tau - kx)} \theta(\tau,x), \quad \theta(\tau,x) = \sum_{k,n} e^{-i(\omega_n \tau - kx)} \theta_{p,n}$$ This is essentially changing the normalization of $$\theta_{p,n}$$ by a factor $$\sqrt{\beta L}$$. This normalization is a little bit silly (if anything, I would put the factor of $$\beta L$$ on the Fourier sum instead of the integral for ease of taking $$L,\beta \rightarrow \infty$$), but there's no reason you can't do it. Note that the result of the Gaussian integral is independent of your choice of Fourier decomposition, as it must be: if the factor of $$T/L$$ didnt come from the prefactor in the action, it would instead come from expanding $$\theta(x,t)$$ in Fourier modes inside the correlation function.