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

[tex]\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}[/tex]

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

[tex]\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)}[/tex]

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

Thanks for your help!

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# Orthogonality of Matsubara Plane Waves

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