How to choose an appropriate value for the regularization $\eta$ in correlation functions used in linear response for a discretized Brillouin zone?

For more context, please see below.

Correlation function in linear response

In linear response we try to find a correlation function, which typically looks like $$ \chi(r-r',t-t') = -i \theta(t-t')[A(r,t-t'),B(r',0)] $$ The $[\cdot,\cdot]$ is a commutator. Moreover if the systems has translation and time-translation symmetry, we can transform the Green's function to Fourier space, i.e., to $\chi(\vec q,\omega)$. The form of which for non-interacting system in the spectral representation reads:

$$ \chi(\vec q,\omega) = \frac{1}{V}\sum_{\vec k} \sum_{\alpha,\beta} \frac{n_F(\epsilon_\alpha(\vec k)) - n_F(\epsilon_\beta(\vec k+\vec q))}{\omega + \epsilon_\alpha(\vec k) - \epsilon_\beta(\vec k+\vec q) + i \eta} \left<\alpha;\vec k\right|A\left|\beta;\vec k+\vec q\right> \left<\beta;\vec k+ \vec q\right|B\left|\alpha;\vec k\right> \label{1}\tag{1} $$ $$ -\mathrm{Im}[\chi(\vec q,\omega)] = \frac{\pi}{V}\sum_{\vec k} \sum_{\alpha,\beta} [n_F(\epsilon_\alpha(\vec k)) - n_F(\epsilon_\beta(\vec k+\vec q))]\delta(\omega + \epsilon_\alpha(\vec k) - \epsilon_\beta(\vec k+\vec q)) \\ \times \left<\alpha;\vec k\right|A\left|\beta;\vec k+\vec q\right> \left<\beta;\vec k+ \vec q\right|B\left|\alpha;\vec k\right> \label{2}\tag{2} $$ Here, $H \left|\alpha;\vec k\right> = \epsilon_{\alpha}(\vec k)\left|\alpha;\vec k\right>$. And $n_F(\epsilon) = 1/(\exp(\epsilon/T)+1)$, the Fermi distribution, $\eta$ is in principle $0^+$. Typically $-\mathrm{Im}[\chi(\vec q,\omega)]$ is related to some observable (if $A$ and $B$ are current operators, then this is proportional to conductivity $\sigma(\vec q,\omega)$)

Discretization of the Brillouin zone (BZ)

Say we have a 2D BZ, the reciprocal lattice vectors are $G_1$ and $G_2$. We can discretize the BZ in an $N\times N$ grid. The smallest Grids of BZ are given by $\mathrm{d}G_1 = G_1/N$ and $\mathrm{d}G_2 = G_2/N$. The Eigenvalues and the Eigenvectors of $H$ is obtained in this discretized BZ.

Choice of $\eta$

$\eta$ is ideally $0^+$ but in my experience very small $\eta$ doesn't give smooth results. The reason being that in (\ref{2}) the delta function can approximated as a Lorentzian. If the Lorentzian is very sharp then $\mathrm{Im}[\chi(\vec q,\omega)]$ is also very sharply peaked and the result is not smooth anymore. On the other hand A thick Lorentzian gives smooth plot. Now my question is how should be the $\eta$ needs to be choose? Making it smaller makes the $\mathrm{Im}[\chi(\vec q,\omega)]$ spiky, larger $\eta$ washes away all the details. What should be a guiding principle of choosing $\eta$. A brute-force way is to perform the calculations for different $\eta$ and see for yourself, which is not always possible for long numerical calculations.

I think $\eta$ depends on

  • The grid-size, i.e. the length of $(\mathrm dG_1,\mathrm dG_2)$.
  • The Energy scale of the problem.
  • It should be same as how the energy changes over a grid on average.
  • It should be small than the $\omega$ grid.

What are the guiding principles to choose an appropriate value of $\eta$?


1 Answer 1


A crude prescription is that $\eta$ should be (at least several times) bigger than the grid size, but much smaller than the other energy scales in the problem. If this cannot be achieved, than the problem is with the grid size, - not the choice of $\eta$.

Physics: Determining the relevant scales of the problem requires some analytical investigation - this is the art of physics, and a cannot be avoided by recourse to numerics alone.

Statistics How small should be $\eta$ is ultimately determined by the required precision, i.e., statistics.

Numerics Choosing the grid size just small enough to assure the smooth results and (more importantly!) the convergence of the numerical method, while keeping computations manageable in terms of time and cost is a matter of numerical calculations - covered in many textbooks and numerical courses (starting with Numerical recipies.) These are unfortunately less appreciated nowadays, since the available computational power and ready packages of solvers/simulators made numerics look easy and accessible to everyone at redundant precision... but only for simple problems.


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