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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!

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  • $\begingroup$ Can someone comment to explain the downvote? $\endgroup$
    – dllahr
    Jan 5 '20 at 14:59
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    $\begingroup$ 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 $v$? $\endgroup$ Jan 8 '20 at 11:12
  • $\begingroup$ Thank you for looking, I edited to add definitions of the variables to the best of my understanding $\endgroup$
    – dllahr
    Jan 8 '20 at 12:38
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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).

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  • $\begingroup$ Thank you, you've added a ton! I'd like to try it out and think about it a bit, then I'll upvote and accept. Is another way of thinking about the U(K=0) = 0 that if I were doing a self consistent field calculation and I chose an arbitrary initial distribution for the electron(s) I could then calculate a converged / well-behaved integral for K=0? $\endgroup$
    – dllahr
    Jan 14 '20 at 15:38
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    $\begingroup$ 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. $\endgroup$ Jan 14 '20 at 22:13

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