# E. Fradkin on thermal propagator of free scalar field

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?

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 $$-1$$.