Orthogonality of Matsubara Plane Waves

[leastaction]

Hi there!

In thermal field theory, the Matsubara frequencies are defined by \nu_n = \frac{2n\pi}{\beta} for bosons and \omega_n = \frac{(2n+1)\pi}{\beta} for fermions. Assuming discrete imaginary time with time indices k=0,\hdots,N, it is easy to obtain the following orthogonality relation for bosons, just by using the standard formula for the geometric series (\beta is the inverse temperature),

\frac{\beta}{N} \sum_{k=0}^{N-1} \mathrm{e}^{\mathrm{i} \frac{\beta}{N} (\nu_n+\nu_m)k} = \begin{cases} \frac{\beta}{N} \sum_{k=0}^{N-1} 1 = \beta & \mathrm{for}\ n=-m \\ \frac{1-\mathrm{e}^{\mathrm{i} \beta (\nu_n+\nu_m)}}{1-\mathrm{e}^{\mathrm{i} \frac{\beta}{N} (\nu_n+\nu_m)}} = 0 & \mathrm{for}\ n\neq -m \end{cases} = \beta\delta_{n,-m}

The second line holds because \beta (\nu_n+\nu_m) is an integer multiple of 2\pi and thus the numerator vanishes. But in the case of fermions, I obtain

\frac{\beta}{N} \sum_{k=0}^{N-1} \mathrm{e}^{\mathrm{i} \frac{\beta}{N} (\omega_n+\omega_m)k} = \begin{cases} \frac{\beta}{N} \sum_{k=0}^{N-1} 1 = \beta & \mathrm{for}\ n=-(m+1) \\ \frac{1-\mathrm{e}^{\mathrm{i} \beta (\omega_n+\omega_m)}}{1-\mathrm{e}^{\mathrm{i}\frac{\beta}{N} (\omega_n+\omega_m)}} = 0 & \mathrm{for}\ n \neq -(m+1)} \end{cases} = \beta\delta_{n,-(m+1)}

Is this true? The Kronecker delta with n,-(m+1) looks rather strange!

Thanks for your help!

 

[kanato]

It's not that strange, but it's a consequence of writing your exponential with a sum of frequencies instead of a difference. Consider the values around n = 0, \omega_{-1} = -T\pi, \omega_0 = T\pi, so for your orthogonality to hold, you need n = 0, m = -1, or n = -1, m = 0. If you choose to write the orthogonality with a difference, you should get n = m in both cases.

 

[leastaction]

Thanks very much! I ask because I've encountered a sum of the above type (with a plus sign) while computing a two-point Green's function and I was wondering if the energy/frequency was conserved...