Developers:Separation of the pseudopotential

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We will separate the local part of the pseudpotential in a long range and a short range parts. The long range part will be the Coulombian potential associated to a radial gaussian charge distribution:

\rho(r)=\frac{Z_{val}}{\left(\sqrt{2\pi}\sigma\right)^3}e^{-\frac{r^2}{\sqrt(2\pi)\sigma}}\ .


The potential is

\Phi(r) = \frac{Z_{val}}{r}\mbox{erf}\left(\frac{r}{\sqrt{2}\sigma}\right)\ ,

\Phi(r\rightarrow0) \longrightarrow \frac{2Z_{val}}{\sqrt{2\pi}\sigma}\ ,

\Phi(r\rightarrow\infty) \longrightarrow -\frac{Z_{val}}{r}\ .

In fourier space:

\rho(q)=\frac{Z_{val}}{2\pi\sigma}e^{-\frac12{\sigma^2q^2}}\ .


We want the potential generated by this charge to be smooth so we consider the associated potential (for simplicity we consider the 1D poisson equation)

\Phi(q) = \frac1{q^2}\left(-4\pi\rho(q)\right) = -\frac{2Z_{val}}{q^2\sigma^2}e^{-\frac12\sigma^2q^2}


What we want is that \left|\Phi(q_{max})\right| must be smaller than a certain threshold:

\frac{\Phi(q_{max})}{Z_{val}} = \frac{e^{-x}}x

with x = \frac12\sigma^2q_{max}^2.

Once we get x, we can get the optimum value of σ:

\sigma = \frac{\sqrt{2x}}{q_{max}} = \frac{\sqrt{2x}}{\pi}h\ .

For \frac{e^{-x}}x = 0.001 we get x=5.2496, which implies

\sigma = 1.03 h\ .

This value will be used in the code, in principle there is a dependency on Zval, but solving a trascendent equation in the code is not very reliable.

Note: this has been taken from the code, in case of discrepancies the code is more likely to be correct.

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