In his lecture, E. Fradkin performs a Matsubara sum to show that the finite temperature contribution to the thermal propagator of the free scalar field contains the Bose-Einstein factor (see 5.209 - 218 in the lecture) $$ n(\mathbf{p},T)=\left[ \exp\left( \frac{\sqrt{\mathbf{p}^2+m^2}}{T} \right) -1 \right]^{-1}. $$ However, the author's result (5.215) of the frequency sum seems incorrect to me, which I discussed in a Math StackExchange post. If the conclusion there is correct, (5.215) should be rewritten as (with $\omega_{\mathbf{p}}=\sqrt{\mathbf{p}^2+m^2}$) $$ G_T(\mathbf{x},\tau) = \int \frac{d^{d}p}{(2\pi)^d} \left\{ \frac{\coth\left(\frac{\omega_{\mathbf{p}}}{2T}\right)}{2\omega_{\mathbf{p}}}\cosh{\left(\omega_{\mathbf{p}}\tau\right)}-\sinh{\left(\omega_{\mathbf{p}}\tau\right)} \right\} e^{i\mathbf{p}\cdot\mathbf{x}}, $$ which reduces for small $\tau$ to $$ G_T(\mathbf{x},\tau) \simeq \int \frac{d^{d}p}{(2\pi)^d} \frac{\coth\left(\frac{\omega_{\mathbf{p}}}{2T}\right)}{2\omega_{\mathbf{p}}} \frac{1}{2} \left[ e^{\omega_{\mathbf{p}}\tau}+e^{-\omega_{\mathbf{p}}\tau} \right] e^{i\mathbf{p}\cdot\mathbf{x}}. $$ But then the real-time propagator (5.218) would be $$ G^{(0)}(\mathbf{x};T)=G_M^{(0)}(\mathbf{x}) + \int \frac{d^{d}p}{(2\pi)^d} \frac{n(\mathbf{p},T)}{\omega_{\mathbf{p}}} \frac{1}{2} \left[ e^{i\omega_{\mathbf{p}}x_0}+e^{-i\omega_{\mathbf{p}}x_0} \right] e^{i\mathbf{p}\cdot\mathbf{x}}, $$ where $$ G_M^{(0)}(\mathbf{x}) = \int \frac{d^{d}p}{(2\pi)^d 2\omega_{\mathbf{p}}} \frac{1}{2} \left[ e^{i\omega_{\mathbf{p}}x_0}+e^{-i\omega_{\mathbf{p}}x_0} \right] e^{i\mathbf{p}\cdot\mathbf{x}} \neq \int \frac{d^{d}p}{(2\pi)^d 2\omega_{\mathbf{p}}} e^{-i\omega_{\mathbf{p}}|x_0|} e^{i\mathbf{p}\cdot\mathbf{x}} $$ is not the real-time propagator at $T=0$ (the correct expression for propagator can be found, for example, on page 24, Zee, [Nutshell]).

How do I fix this apparent inconsistency?


1 Answer 1


You can also derive @Svyatoslav's correct expression by Poisson Summation: $$ \frac 1{2\pi} \sum_{n=-\infty}^\infty \frac{e^{in\tau}}{n^2+M^2}= \sum_{n=-\infty}^\infty \frac 1{2|M |} e^{-|M||\tau+2\pi n|}, \quad \hbox{(Poisson Summation)}\\ = \frac 1 {2M} \frac{\cosh(\pi -\tau)M}{\sinh \pi M}, \quad 0<\tau<2\pi,\nonumber $$ The first line come from applying Poisson summation to the zero temperature expression
$$ \int_{-\infty}^{\infty} \frac{dk}{2\pi}\frac{e^{ik\tau}} {k^2+M^2}=\frac 1 {2|M|}e^{-|\tau||M|} $$ and has the physical interpretation as the method-of-images sum over the $n$-fold winding of the particle trajectory around the periodic imaginary time direction. The passage from the first to second lines is just summing the two geometric series from $n=0$ to $ \infty$ and $n=-\infty$ to $-1$.

I saw Eduardo while out walking yesterday, and informed him of the problem. I expect that he will fix his notes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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