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In appendix C of Quantum Physics in One Dimension of Thierry Giamarchi, it is claimed that (See (C.22)) after performing the Matsubara sum over the bosonic frequencies $\omega_n=2\pi n/\beta$ in $$\frac{1}{\beta\Omega}\sum_k\sum_n e^{i(kx-\omega_n\tau)}\times \frac{-i2\pi\omega_n/k}{\omega_n^2+u^2k^2}, $$ and taking the $\beta\to\infty$ limit we are left with (see (C.25)) $$-i\int_0^\infty\frac{dk}{k}e^{-\alpha k}e^{-u\cdot\text{sign}(\tau)\tau}\sin(k\cdot\text{sign}(\tau)x),$$ where the factor $e^{-\alpha k}$ is introduced to ensure convergence. The "naive" result would be (see (C.26)) $$-i\text{sign}(\tau)\arctan\left[\frac{x}{u|\tau|+\alpha}\right]$$ I have had troubles when trying to reproduce these results. First I do not get (C.25) when computing the Matsubara sum. I have used the weighting function $g(z)=\theta(\tau)(-\beta f_B(-z))+\theta(-\tau)f_B(z)$, where $f_B(z)$ is the Bose-Einstein distribution (see Matsubara sums). On the other hand, if I take the limit from the beginning and compute the sum as the integral $$\sum_n \to \int d\omega \frac{\beta}{2\pi},$$ using complex integration then I still get something slightly different, namely, $$-\frac{1}{2}\int_0^\infty \text{sign}(\tau)\frac{e^{-\alpha k}}{k}(e^{ikx}e^{-uk|\tau|}-e^{-ikx}e^{uk|\tau|}).$$

Where I messed up?

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    $\begingroup$ Can you make the question self-contained, so that we do not need to look for a copy of the book to find out whether these are Bose on Fermi $\omega_n$'s. $\endgroup$ Commented Dec 27, 2022 at 17:08
  • $\begingroup$ @mikestone Yeah, I forgot to point that out, thanks. $\endgroup$ Commented Dec 27, 2022 at 17:39

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I'm not sure that this will help, but I often use the following dedimensionalized algebra to do the Bose sums:

$$ \frac 1{2\pi} \sum_{n=-\infty}^\infty \frac{e^{in\tau}}{n^2+M^2}= \sum_{n=-\infty}^\infty \frac 1{2|M |} e^{-|M||\tau+2\pi n|}, \quad \hbox{(Poisson Summation)}\nonumber\\ = \frac 1 {2|M|} \frac{\cosh(\pi -\tau)M}{\sinh \pi |M|}, \quad 0<\tau<2\pi,\nonumber\\ = \frac 1{2|M|} e^{-M\tau} +\frac 1 {|M|}\frac{ \cosh M\tau}{(e^{2\pi |M|}-1)}\quad 0<\tau<2\pi.\nonumber\\ %= \frac 1{2M} (\coth \pi M\cosh M \tau- \sinh M\tau) \nonumber $$ 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 $-1$.

You can get the extra factor of $\omega_n=n$ by differentiating wrt $\tau$

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  • $\begingroup$ May you clarify how in the first line you performed Poisson summation but there is still a sum over $n$? Are you integrating over $k$ first even though it does not appear at the initial expression? $\endgroup$ Commented Dec 28, 2022 at 22:11
  • $\begingroup$ I am not touching $k$. The $n$ on the left and the $n$ or the right are independent dummy summation labels. I could have called then RHS $n$ anything, but I am lazy. so I reused $n$. Poisson says that $\sum_{n=-\infty}^\infty F(n) =\sum_{n=-\infty}^\infty \tilde F(n)$ where $\tilde F$ is the Fourier transform of $F(n)$. $\endgroup$ Commented Dec 29, 2022 at 1:12
  • $\begingroup$ Got it, thanks. $\endgroup$ Commented Dec 29, 2022 at 23:53

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