# electronic band structure calculation using plane wave expansion: what are the diagonal matrix elements for a coulomb potential?

I'm trying to do a plane wave basis expansion calculation for the band structure / wavefunctions of electrons in a periodic solid, using a coulomb potential for the nuclei. I'm working from Ashcroft and Mermin, mainly. Equation 11.3 on page 194 gives the potential energy term:

$$U_K \approx -\left( \frac{4\pi Z_a e^2}{K^2} \right) \frac{1}{v}$$

$$K$$ (my understanding) is the length of the difference between any 2 reciprocal lattice vectors (NB of course the difference between any 2 reciprocal lattice vectors is itself a reciprocal lattice vector)

$$v$$ is the volume of the unit cell

$$Z_a$$ is the charge on the nuclei

$$e$$ is the charge of an electron

For off-diagonal elements $$K$$ is non-zero, however for the diagonal it is $$0$$ and hence the above goes to negative infinity. What is the value for the potential for the diagonal elements?

One idea I had was to integrate just the Coulomb potential over the unit cell, but that seems to contradict Ashcroft & Mermin eqn. 9.34, p. 167, which they indicate is where they get the above from, and which indicates the integral should be over all space:

$$\phi(K) = \int_{all\ space} dr\ e^{-iKr}\phi(r)$$

$$\phi(r)$$ is the potential produced by an individual ion/nuclei in the lattice

Any help much appreciated!

• Can someone comment to explain the downvote? Jan 5, 2020 at 14:59
• I don't have Ashcroft and Mermin to hand, so could you give some detail on the symbols? In $U_K$ (your first equation) I guess that $K$ is the momentum, but what is the $v$? Jan 8, 2020 at 11:12
• Thank you for looking, I edited to add definitions of the variables to the best of my understanding Jan 8, 2020 at 12:38

I do not know if I can add much, but I can try to explain how these matrix elements arise. The integral should indeed be over all space, but the lattice structure manifests itself in that the momentum $$K$$ only takes discrete values: the reciprocal lattice vectors. If you make a Fourier transform of the Coulomb potential, you indeed encounter a problem of the singularity at $$r=0$$. This can be addressed by modifying the Coulomb potential by multiplying it by a factor $$e^{- \lambda r}$$, performing the Fourier transform, and then letting the parameter $$\lambda \rightarrow 0$$ (other forms of obtaining the result are also available).
The result that you get is the $$U(K)$$ (Eq. 113) that you quote above, where $$K$$ is a reciprocal lattice vector. This is well-behaved at all points except $$K=0$$. You can set the $$U(K=0) = 0$$ by hand. If this seems rather arbitrary, the $$K=0$$ value is related to the mean charge of the unit cell, which is indeed zero for physical cases (otherwise the electrostatic energy of the crystal would indeed diverge).
• The $K=0$ divergence relates to the large-scale behavior of the density (small momentum $\leftrightarrow$ large distance, and vice-versa). I would think that any physical distribution must have a net charge of zero, and so the divergence will not manifest, and so the integral is well-behaved. Jan 14, 2020 at 22:13