Matsubara frequency

Remarks

Check: $e^{-i \omega_n (\tau + \beta)} = e^{-i\omega_n \tau} e^{-i \omega_n \beta}$, and $e^{-i\omega_n \beta}$ equals $e^{-2\pi i n} = 1$ for bosonic and $e^{-(2n+1)\pi i} = -1$ for fermionic frequencies. Example: for $\varepsilon > 0$, the classical sum $\sum_{n \in \Z} 1/(n^2 + a^2) = (\pi / a) \coth(\pi a)$ gives $$\frac{1}{\beta} \sum_{n \in \Z} \frac{1}{\omega_n^2 + \varepsilon^2} = \frac{1}{2\varepsilon} \coth\frac{\beta\varepsilon}{2} = \frac{1}{2\varepsilon}\big(1 + 2 n_B(\varepsilon)\big)$$ for bosonic $\omega_n$, where $n_B(\varepsilon) = 1/(e^{\beta\varepsilon} - 1)$ is the Bose–Einstein factor. For fermionic $\omega_n$ the same sum equals $\frac{1}{2\varepsilon} \tanh\frac{\beta\varepsilon}{2} = \frac{1}{2\varepsilon}\big(1 - 2 n_F(\varepsilon)\big)$, with the Fermi–Dirac factor $n_F(\varepsilon) = 1/(e^{\beta\varepsilon} + 1)$. Such sums are computed in general by the residue theorem, using a function such as $\beta/(e^{\beta z} - 1)$ whose poles sit at $z = i\omega_n$. Non-example: the frequency $\pi/(2\beta)$ is neither bosonic nor fermionic, since $e^{-i\pi/2} = -i$, so $e^{-i\omega\tau}$ is neither periodic nor antiperiodic with period $\beta$.