Many BZ integrals of interest are related to the spectral function $$ G(\omega) = \int_{BZ}\frac{F(\mathbf{k})d\mathbf{k}}{\omega - \omega(\mathbf{k}) + i\eta} \tag{1} $$ where $\omega(\mathbf{k})$ are the eigenvalues, $F(\mathbf{k})$ some matrix elements. It is shown that the imaginary part of $G(\omega)$ is $$ \Im(G) = -\pi \int d\mathbf{k} F(\mathbf{k}) \delta(\omega - \omega(\mathbf{k})) \tag{2} $$ In practice, the $\delta$ function can be replaced by some broadening functions (gaussian, lorentzian) or using the tetrahedron method. The real part can then be obtained by Kramers-Kronig transformation of the imaginary part. Now, I would like to evaluate $$ G^\prime(\omega) = \int_{BZ}\frac{F(\mathbf{k})d\mathbf{k}}{(\omega - \omega(\mathbf{k}) + i\eta)^2} \tag{3} $$ I'm wondering if the imaginary part of $G^\prime$ can be expressed as $\delta$ function, then I can evaluate it with similar recipe?


2 Answers 2


I think in this case one should realize what the imaginary constant $i\eta$ stands for.

Ideally this is a non-physical constant that is used to remove the possibility of hitting the singular point $\omega = \omega(\mathbf k)$. As such it is a user-defined value that is used to ensure that the inverse is not infinite, i.e. can be calculated.

In Green function methods the imaginary part of the Green function as given in your Eq. (1) is directly related to the Lorentzian function with broadening $\eta$. So here there is a clear connection between the $\eta$ value and a temperature.

The same will apply when you are dealing with the squared $\delta \omega$. So the $\eta$ will still be a non-physical constant used to remove infinities.

In this case I would instead do:

$$\tag{1} G^\prime(\omega) = \int_{BZ}\frac{F(\mathbf{k})d\mathbf{k}}{(\omega - \omega(\mathbf{k}))^2 + i\eta^2} $$ The purpose here is to make the broadening $\eta$ on the same order as your $\delta\omega$ (and with same units).

A good example is the Green function for electrons (denominator is $E-E'$) and phonons (denominator $(\omega - \omega')^2$). And to make the imaginary number a relatable physical quantity one will still use the same energy unit for $\eta$ and just follow the power of the denominator.

This will give you the same Lorentzian $\delta$ function for any power of the Green function as this, note that the broadening is changed by the same power: $$\tag{2} \mathcal L(\omega) = \frac{\eta^p}{\delta\omega^{2p} + \eta^{2p}} $$

Generally books and texts are not focusing too many details on where the parenthesis is placed for one obvious reason: $\eta$ should always be chosen to be so small that $\delta\omega\eta\sim 0$. If this is not the case then $\eta$ is probably too large to gather any physical meaning of the calculation.

  • $\begingroup$ Thanks for the detailed explanation. So, I have to evaluate $\delta[(\omega - \omega(\mathbf{k}))^2]$. For broadening function, one need change the power as your second equation. My question is what about for tetrahedron method? Basically, how to evaluate $\delta[(\omega - \omega(\mathbf{k}))^2]$ with tetrahedron method? $\endgroup$ Feb 10, 2022 at 16:52
  • 1
    $\begingroup$ I think you are misunderstanding what the tethrahedron method (TM) does? The TM method chooses suitable weighting coefficients for specific k-points and simply evaluates the summation of the $F(\omega)$ function. Generally any $\delta[(\omega -\omega(\mathbf k))^2]$ function may be approximated by a Gaussian, Lorentzian etc function. I am not entirely sure whether in the tetrahedron method you'll just have to rederive eq. 2.2 in said paper with respect to the power 2. Probably it will be just be a power 2, but I am not 100% sure about that. $\endgroup$
    – nickpapior
    Feb 10, 2022 at 19:20

The short answer is No.

From your Eq.(1) to Eq.(2), the following identity is used: \begin{align}\tag{1} \dfrac{1}{x-i\eta} = \dfrac{x+i\eta}{x^2+\eta^2}=\dfrac{P}{x}+i\pi\delta(x) \end{align} where $P$ represents the Cauchy principal value.

You can't do the same thing for your Eq.(3).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .