No subject

\begin{equation}
\sum_{ \k,v } \epsilon_{ \k,v } =
- {\hbar^2 \over 2m}  \sum_{ \k,v } \int \psi^*_{ \k,v }(\r) 
   \nabla ^2 \psi_{ \k,v }(\r) d\r
+ \sum_{ \k,v,\mu,\R} \int \psi^*_{ \k,v }(\r) 
   \hat V^\mu(\r-\d_\mu-\R) \psi_{ \k,v }(\r) 
+ e^2 \int { n(\r) n(\r') \over \mid \r-\r' \mid } 
           d\r d\r'
+ \int n(\r) V_{xc}(\r) d\r 
\end{equation}
and we can give an alternate formula for the total energy of a crystal:
\begin{equation}
E_{tot} =  \sum_{ \k,v } \epsilon_{ \k,v }
- {e^2 \over 2} \int {n(\r) n(\r') \over
            \mid \r-\r' \mid} d\r d\r'
+  \int n(\r)\left(\epsilon_{xc}(\r) - V_{xc}(\r)\right) d\r 
+ {e^2 \over 2} \sum'_{\mu,\nu,\R,\R'} {Z_\mu Z_\nu \over 
\mid \d_\mu+\R-\d_\nu-\R' \mid }
\end{equation}

\newpage
{\bf Plane-wave -- Pseudopotential formalism}
\par
Let us consider the \G-space representation
of the wavefunctions:
\begin{equation} |\psi_\k> = \sum_\G \Psi(\k+\G) |\k+\G>, \qquad
   \Psi ({\bf k+G}) = <{\bf k+G} \mid \psi_\k>,\qquad
   \mid {\bf k+G} > = {1 \over \sqrt V} e^{i {\bf (k+G) r}},\end{equation}
where $ V = N \Omega $ is the volume of the crystal. With these
definitions, the normalizations are:
\begin{equation}
<\k+\G|\k+\G>'=\delta_{\G,\G'},
\qquad<\psi_\k |\psi_\k>=1 \quad\mbox{if}\quad\sum_\G |\Psi(\k+\G)|^2 = 1.
\end{equation}
Let us define the Fourier trasform for a periodic function 
$ F(\r)=\sum_\R f(\r-\R) $ as:
\begin{eqnarray}
F(\G) & = & {1\over N\Omega} \int d\r F(\r) e^{-i{\bf G r}} 
= {1 \over \Omega} \int d\r f(\r) e^{-i{\bf G r}}
= <{\bf k+G}_1 \mid F(\r) \mid {\bf k+G}_2> ~,
\qquad {\bf G = G}_1-\G_2 \\
F(\r) & = &\sum_\G F(\G) e^{i{\bf G r}}.
\end{eqnarray}
We assume non local pseudopotential of general form 
$\hat V^\mu = V_\mu(r)+\sum_i V_{\mu,i}(\r,\r')$.
The total energy per unit cell in reciprocal space is:
\begin{eqnarray}
{ E_{tot} \over N } & = & {1 \over N} {\hbar^2 \over 2m} \sum_{\k,v} 
       \sum_\G {\mid \Psi_v({\bf k+G}) \mid}^2 ({\bf k+G})^2
+ \Omega \sum_\G n^*(\G) \sum_\mu S_\mu(\G) V_\mu(\G)
+ {1 \over N} \sum_{\k,v} \sum_{\mu,i} \sum_{{\bf G,G'}} 
   S_\mu({\bf G - G'}) \times \nonumber \\ & & \mbox{} \times
    \Psi_v^*({\bf k+G}) \Psi_v({\bf k+G'}) V_{\mu,i}({\bf k+G},{\bf k+G'})
+ { \Omega \over 2} \sum_\G n^*(\G) V_{Hartree}(\G)
+ \Omega\sum_\G n^*(\G)\epsilon_{xc}(\G)d\r
+ {e^2 \over 2} \sum'_{\mu,\nu,\R} 
{Z_\mu Z_\nu \over \mid \d_\mu-\d_\nu-\R\mid }
\end{eqnarray}
where $ S_\mu(\G) = \sum_\mu e^{-i{\bf G d_\mu}}$ is the structure factor, and
\begin{equation} V_{Hartree}(\G) = 4\pi e^2 { n({\bf G)} \over \G^2 }, \qquad
   V_\mu(\G)  = {1 \over \Omega} \int V_\mu(\r) 
                    e^{-i{\bf G r}} d\r, \qquad
   V_{\mu,i}(\k_1,\k_2) = {1 \over \Omega} \int
 e^{-i\k_1\r} V_{\mu,i}(\r,\r') e^{i\k_2\r'} d\r d\r' . 
\end{equation}
Note that we have assumed one atom of each kind. The generalization
is straightforward: the structure factor becomes 
$ S_\mu(\G) = \sum_{i_\mu} e^{-i{\bf G d_{i\mu}}}$
where $i_\mu$ runs over atoms of the same kind $\mu$.

Using eigenvalues sum, the total energy per unit cell is
\begin{equation}
{ E_{tot} \over N } = {1 \over N} \sum_{\k,v} \epsilon_{\k,v}
- { \Omega \over 2} \sum_\G n^*(\G) V_{Hartree}(\G)
+ \Omega \sum_\G n^*(\G) \left(\epsilon_{xc}(\G)-V_{xc}(\G)\right)
+ {e^2 \over 2} \sum'_{\mu,\nu,\R} 
  {Z_\mu Z_\nu \over \mid \d_\mu-\d_\nu-\R \mid }.
\end{equation}
In the plane-wave representation the Kohn-Sham equation becomes
\begin{equation}
 \sum_{\G'} <{\bf k+G} \mid H-\epsilon \mid {\bf k+G'}> \Psi ({\bf k+G'}) = 0,
\qquad\mbox{or} \qquad
 \sum_{\G'} <{\bf k+G} \mid H \mid {\bf k+G'}> \Psi ({\bf k+G'}) =
 \epsilon \Psi ({\bf k+G})
\end{equation}
The matrix elements of the hamiltonian are
\begin{eqnarray} <{\bf k+G} \mid H-\epsilon \mid {\bf k+G'}> & = &
 \biggl ( - {\hbar^2 \over 2m} ({\bf k+G})^2 -\epsilon \biggr ) \delta_{\bf GG'}
+ \sum_\mu S_\mu(\G-\G') \biggl (  V_\mu(\G-\G')
+ \sum_i V_{\mu,i}({\bf k+G},{\bf k+G'}) \biggr ) \nonumber \\
& + & V_{Hartree}(\G-\G') + V_{xc}(\G-\G') .
\end{eqnarray}

{\em Divergent Terms in the potential}
\par
The Hartree term, $V_{Hartree}(0)$, and local potential term,
$\sum_\mu S_\mu(0)V_\mu(0)$, are separately divergent and must 
be treated in a special way.
Let us consider their sum $\widetilde V(\r)=V_{loc}(\r)+V_{Hartree}(\r)$.
Its $\G=0$ term is not divergent:
\begin{equation}
    \widetilde V(\G=0) ={1\over\Omega} \int d\r\left(\sum_\mu V_\mu(\r-\d_\mu) 
        + {1\over N} e^2 \int {n(\r')\over |\r-\r'|} d\r'\right)=
       {1\over\Omega} \sum_\mu \int d\r \left( V_\mu(r) 
       + {Z_\mu e^2\over r} \right ) = {1\over\Omega} \sum_\mu \alpha_\mu
\end{equation}
where we used
\begin{equation}
    V_\mu(r) \sim -{Z_\mu e^2\over r}\quad\mbox{for large~} r,\qquad
    {1\over N} \int n(\r) = \sum_\mu Z_\mu.
\end{equation}

The $\alpha_\mu$ are parameters depending only on the pseudopotential.

\newpage
{\em Divergent Terms in the energy}

\par
The $\G=0$ terms of the ion-ion, Hartree, and local
pseudopotential terms in the total energy are separately divergent
and must be treated in a special way. Let us call $E_{div}$ the
sum of all divergent terms.

First Step: split $E_{div} = E_{div}^{(1)} + E_{div}^{(2)}$, with
\begin{equation}
E_{div}^{(1)} = \int n(\r) \sum_\mu V_\mu(\r-\d_\mu) d\r
        + {1\over N} e^2 \int {n(\r)n(\r')\over |\r-\r'|} d\r d\r'
\end{equation}
\begin{equation}
E_{div}^{(2)} = {e^2\over 2} 
      \sum'_{\mu,\nu,\R}{Z_\mu Z_\nu\over|\d_\mu-\d_\nu-\R|}
    - {1\over N} {e^2\over 2} \int {n(\r)n(\r')\over |\r-\r'|} d\r d\r'
\end{equation}
Using the previous definition of $\widetilde V(\r)$, 
the first divergent term can be written as
\begin{equation}
E_{div}^{(1)}= \int n(\r) \widetilde V(\r) d\r.
\end{equation}
The $\G=0$ term of $\widetilde V(\G)$ is not divergent and has been
previously calculated:
\begin{equation}
 \widetilde V(\G=0) = {1\over\Omega} \sum_\mu \alpha_\mu,\qquad
 n(\G=0)=\sum_\mu {Z_\mu \over \Omega}.
\end{equation}
We finally get for the $\G=0$ contribution what is usually called
``$\alpha Z$ term'':
\begin{equation}
E_{div}^{(1)} = \Omega\sum_{\G\neq 0} n^*(\G) \widetilde V(\G)
              + {1\over\Omega}(\sum_\mu Z_\mu)(\sum_\mu \alpha_\mu)
              = \Omega\sum_\G n^*(\G) \widetilde V_{loc}(\G)
              + 2 \widetilde E_{Hartree}
\end{equation}
where $\widetilde V_{loc}$ is the local potential for $\G\neq 0$,
contains the $\alpha Z$ term in $\G=0$ component, and
\begin{equation}
\widetilde E_{Hartree} = {\Omega\over 2}\sum_{\G\neq 0} n^*(\G)V_{Hartree}(\G).
\end{equation}
Second step: write
$E_{div}^{(2)}= E_{Ewald}^{(1)} + E_{Ewald}^{(2)}-E_{Hartree}$,
with
\begin{equation}
E_{Ewald}^{(1)} = 
  {e^2\over 2} \sum'_{\mu,\nu,\R}{Z_\mu Z_\nu\over|\d_\mu-\d_\nu-\R|}
  \mbox{erfc}(\sqrt{\eta}|\d_\mu-\d_\nu-\R|)~,\qquad
E_{Ewald}^{(2)} = 
   {e^2\over 2} \sum_{\mu,\nu,\R}{Z_\mu Z_\nu\over|\d_\mu-\d_\nu-\R|}
   \mbox{erf} (\sqrt{\eta}|\d_\mu-\d_\nu-\R|)
   - e^2 \sqrt{\eta\over\pi}\sum_\mu Z_\mu^2.
\end{equation}
This identity is verified for any value of $\eta$.
The sum in $E_{Ewald}^{(2)}$ includes the term with $\d_\mu-\d_\nu-\R=0$
(note the missing prime), that is subtracted back in the second term of 
$E_{Ewald}^{(2)}$ (note that 
$\mbox{erf}(x)\rightarrow 2x/\sqrt\pi$ for small $x$).

The first Ewald term $E_{Ewald}^{(1)}$ is rapidly convergent 
in real space for any reasonable values of $\eta$.

The sum in $E_{Ewald}^{(2)}$ can be written
as the interaction energy between point charges $n_c(\r)$ 
and the potential $V_g(\r)$ produced by a gaussian distribution
of charges:
\begin{equation}
E_{Ewald}^{(2)}= {1\over 2}\int n_c(\r)V_g(\r)d\r
   - e^2 \sqrt{\eta\over\pi}\sum_\mu Z_\mu^2~,\qquad
n_c(\r)=\sum_\mu Z_\mu\delta(\r-\d_\mu)~,
\qquad V_g(\r)=e^2\sum_{\mu,\R} 
{Z_\mu\mbox{erf}(\sqrt{\eta}|\r-\d_\mu-\R|) \over |\r-\d_\mu-\R|}
\end{equation}
In reciprocal space, by using the Fourier transform
\begin{equation}
   {1\over r'} \mbox{erf} (\sqrt{\eta}r')
= \left({\eta\over \pi}\right)^{3/2} \int {e^{-\eta r^2}\over |\r-\r'|} d\r
 = \int {4\pi e^{-G^2/4\eta}\over G^2}e^{i\G\cdot\r'} d\G
\end{equation}
one obtains (forgetting for the moment the divergence of 
$V_g(\G=0)$):
\begin{equation}
E_{Ewald}^{(2)}= {\Omega\over 2}\sum_{\G} n_c^*(\G)V_g(\G)
   - e^2 \sqrt{\eta\over\pi}\sum_\mu Z_\mu^2~,\qquad
n_c(\G)={1\over\Omega}\sum_\mu Z_\mu e^{i\G\cdot\d_\mu}~,
\qquad V_g(\G) =  {4\pi e^2\over\Omega}
\sum_\mu Z_\mu e^{i\G\cdot\d_\mu} {e^{-G^2/4\eta}\over G^2}
\end{equation}
The $\G = 0$ contribution to $E_{Ewald}^{(2)}-E_{Hartree}$:
\begin{equation}
E_0 = {\Omega\over 2} \left(n_c(0)V_g(0)-n(0)V_{Hartree}(0)\right)
\end{equation}
is no longer divergent, because $n(0)=n_c(0)=\sum_\mu Z_\mu/\Omega$ 
due to the neutrality of the system:
\begin{eqnarray}
(V_g-V_{Hartree})(\G=0) & = & {e^2\over N \Omega}
 \int\left( \sum_{\mu,\R} Z_\mu
{\mbox{erf}(\sqrt{\eta}|\r-\d_\mu-\R)|)\over |\r-\d_\mu-\R)|}
- \int {n(\r')\over |\r-\r'|}d\r'\right)d\r \nonumber \\
& = &  {e^2\over\Omega}\left(\sum_\mu Z_\mu\right)
  \int {\mbox{erf}(\sqrt{\eta}r)-1\over r}d\r 
  = {e^2\over \Omega} \left(\sum_\mu Z_\mu\right) {\pi\over\eta}
\end{eqnarray}
The integral appearing in the last expression can be found
in tables:
\begin{equation}
  \int {\mbox{erf}(\sqrt{\eta}r)-1\over r}d\r = 
  4\pi \int (\mbox{erf}(\sqrt{\eta}r)-1)rdr = 4\pi{1 \over 4\eta}.
\end{equation}
Putting all pieces together, one obtains for $E^{(2)}_{div}$:
\begin{eqnarray}
E^{(2)}_{div} = -\widetilde E_{hartree} + E_{Ewald}
  & = & - {\Omega\over 2} \sum_{\G\neq0} n^*(\G)V_{Hartree}(\G)
                + {4\pi\over\Omega} {e^2\over 2}\sum_{\G\neq 0}
                  \left |\sum_\mu Z_\mu e^{i\G\d_\mu}\right |^2
                  {e^{-G^2/4\eta}\over G^2} \nonumber \\ & & \mbox{}
                + {e^2\over 2} \sum'_{\mu,\nu,\R}
                  {Z_\mu Z_\nu\over|\d_\mu-\d_\nu-\R|}
                  \mbox{erfc}(\sqrt{\eta}|\d_\mu-\d_\nu-\R|)
                - e^2 \sqrt{\eta\over\pi}\sum_\mu Z_\mu^2
                - {4\pi\over\Omega} {e^2\over 2} {1\over 4\eta}
                  \left(\sum_\mu Z_\mu\right)^2
\end{eqnarray}
and for the total energy:
\begin{eqnarray}
{ E_{tot} \over N } & = & {1 \over N} {\hbar^2 \over 2m} \sum_{\k,v} 
       \sum_\G {\mid \Psi_v({\bf k+G}) \mid}^2 ({\bf k+G})^2
+ {1 \over N} \sum_{\k,v} \sum_{\mu,i} \sum_{{\bf G,G'}} 
   S_\mu({\bf G - G'}) \Psi_v^*({\bf k+G}) \Psi_v({\bf k+G'}) 
   V_{\mu,i}({\bf k+G},{\bf k+G'}) \nonumber \\ & & \mbox{} 
+ \Omega \sum_\G n^*(\G) \epsilon_{xc}(\G)
+ \Omega \sum_\G n^*(\G) \widetilde V_{loc}(\G)
+ \widetilde E_{Hartree} + E_{Ewald}.
\end{eqnarray}
One can use the sum of the eigenvalues to calculate the total energy: 
the kinetic, non local, and local (including the $\alpha Z$ term) 
contributions disappear and the expression of the total energy
becomes: 
\begin{eqnarray}
{ E_{tot} \over N } = {1 \over N} \sum_{\k,v} \epsilon_{\k,v}
               + \Omega \sum_\G n^*(\G)\left(\epsilon_{xc}(\G)-V_{xc}(\G)\right)
               - \tilde E_{Hartree} + E_{Ewald}.
\end{eqnarray}

{\em Calculation of energy in CP}

\par
Let us define $n_g$ as the sum of gaussian charges centered at 
atomic sites:
\begin{equation}
n_g(\r) = \sum_\mu Z_\mu \left ({\eta\over\pi}\right)^{3/2}
                         e^{-\eta(\r-\d_\mu)^2}.
\end{equation}
The divergent terms $E_{div}$ of the energy can be rewritten as
$E_{div} = E_{div}^{(1)} + E_{div}^{(2)}$, with
\begin{equation}
E_{div}^{(1)} = \int n(\r) \sum_\mu V_\mu(\r-\d_\mu) d\r
        - {1\over N} e^2 \int {n(\r)n_g(\r')\over |\r-\r'|} d\r d\r'
\end{equation}
\begin{equation}
E_{div}^{(2)} = {e^2\over 2} 
      \sum'_{\mu,\nu,\R}{Z_\mu Z_\nu\over|\d_\mu-\d_\nu-\R|}
    + {1\over N} {e^2\over 2} \int {n(\r)n(\r')\over |\r-\r'|} d\r d\r'
    + {1\over N} e^2 \int {n  (\r)n_g(\r')\over |\r-\r'|} d\r d\r'
\end{equation}
$E_{div}^{(1)}$ can be rewritten as
\begin{equation}
E_{div}^{(1)}= \int n(\r) V_{loc+g}(\r)d\r,\qquad
 V_{loc+g}(\r) = \left ( \sum_\mu V_\mu(\r-\d_\mu)  + V_g(\r) \right ),
\end{equation}
where $V_g(\r)$ is the potential generated by the gaussians.
The singularity at $\G=0$ disappears :
\begin{equation}
 V_{loc+g}(\G=0) = {1\over\Omega} \sum_\mu \int d\r 
    \left(V_\mu(r) + {Z_\mu e^2\mbox{erf}(\sqrt{\eta}r)\over r}\right)
   \equiv  {1\over\Omega} \sum_\mu \alpha'_\mu.
\end{equation}
Note that this term is similar to, but not equal to, the $\alpha Z$ term.
$E_{div}^{(1)}$ can be rewritten as:
\begin{equation}
E_{div}^{(1)} =  {1\over\Omega} \left(\sum_\mu Z_\mu\right)
                                \left(\sum_\mu \alpha'_\mu\right)
        + \Omega \sum_{\G\neq 0} n^*(\G)V_{loc+g}(\G) .
\end{equation}

$E_{div}^{(2)}$ can be rewritten as $E_{div}^{(2)}= E_{Ewald} + E_{H+g}$,
where $E_{Ewald}$ is the well-known Ewald sum,
\begin{equation}
E_{Ewald} = {e^2\over 2} 
      \sum'_{\mu,\nu,\R}{Z_\mu Z_\nu\over|\d_\mu-\d_\nu-\R|}
    - {1\over N} {e^2\over 2} \int {n_g(\r)n_g(\r')\over |\r-\r'|} d\r d\r',
\end{equation}
while $E_{H+g}$ is the electrostatic energy of a system of electrons
and ions with a gaussian charge distribution:
\begin{equation}
E_{H+g} = {1\over N} {e^2\over 2} 
         \int {(n(\r)+n_g(\r))(n(\r')+n_g(\r'))\over |\r-\r'|} d\r d\r'.
\end{equation}
Both terms are regular at $\G=0$ because they are the electrostatic energy 
of neutral systems. $E_{H+g}$ can be directly calculated:
\begin{equation}
E_{H+g} = \Omega \sum_{\G\neq 0} (n^*(\G)+n_g^*(\G)) V_{H+g}(\G), \quad
V_{H+g}(\G) = {4\pi e^2\over G^2} \left(n(\G)+n_g(\G) \right) .
\end{equation}

$E_{Ewald}$ can be easily computed using the same technique used before:
\begin{equation}
E_{Ewald} = E^{(1)}_{Ewald} + E^{(2)}_{Ewald}, 
\end{equation}
where (note the sum over all vectors including $\d_\mu-\d_\nu-\R=0$):
\begin{equation}
E^{(1)}_{Ewald} = 
  {e^2\over 2} \sum_{\mu,\nu,\R}{Z_\mu Z_\nu\over|\d_\mu-\d_\nu-\R|}
  \mbox{erf}(\sqrt{\zeta}|\d_\mu-\d_\nu-\R|)
    - {1\over N} {e^2\over 2} \int {n_g(\r)n_g(\r')\over |\r-\r'|} d\r d\r'
\end{equation}
and (note the self-interaction term compensating the $\d_\mu-\d_\nu-\R=0$
contribution of the former term):
\begin{equation}
E^{(2)}_{Ewald} = 
  {e^2\over 2} \sum'_{\mu,\nu,\R}{Z_\mu Z_\nu\over|\d_\mu-\d_\nu-\R|}
  \mbox{erfc}(\sqrt{\zeta}|\d_\mu-\d_\nu-\R|)
   - e^2 \sqrt{\zeta\over\pi}\sum_\mu Z_\mu^2.
\end{equation}
for an arbitrary value of $\zeta$.
$E^{(1)}_{Ewald}$ can be made to vanish, because both terms appearing 
in it can be written in reciprocal space as an Ewald sum. Leaving apart
the $\G=0$ contribution,
\begin{equation}
  {1\over N} {e^2\over 2} \int {n_g(\r)n_g(\r')\over |\r-\r'|} d\r d\r'
 = {\Omega \over 2}\sum_{\G\neq 0} { n_g^*(\G)V_g(\G)}
 = {4\pi\over\Omega}{e^2\over 2}\sum_{\G\neq 0}
    \left|\sum_\mu Z_\mu e^{i\G\cdot\d_\mu}\right|^2 {e^{-G^2/2\eta}\over G^2}.
\end{equation}
This expression comes from the Fourier transform of $n_g(\r)$ and $V_g(\r)$:
\begin{equation}
n_g(\G) = {1\over \Omega} 
\sum_\mu Z_\mu e^{i\G\cdot\d_\mu} {e^{-G^2/4\eta}},\quad
V_g(\G) = {4\pi e^2 n_g(\G) \over G^2}.
\end{equation}
By setting $2\zeta=\eta$, one finds exactly the reciprocal space
expression for the first term:
\begin{equation}
  {e^2\over 2} \sum_{\mu,\nu,\R}{Z_\mu Z_\nu\over|\d_\mu-\d_\nu-\R|}
  \mbox{erf}(\sqrt{\zeta}|\d_\mu-\d_\nu-\R|)
 = {4\pi\over\Omega}{e^2\over 2}\sum_{\G\neq 0}
    \left|\sum_\mu Z_\mu e^{i\G\cdot\d_\mu}\right|^2 {e^{-G^2/4\zeta}\over G^2}.
\end{equation}
$E^{(2)}_{Ewald}$ is a rapidly convergent sum in real space,
plus a term coming from the self-interaction of gaussians.

{\em Miscellaneous}

\par
When a set of special points $\{ \k_i \} $, with weights 
$ w_i$, $ \sum_i w_i = 1$, 
is used to sample the Brillouin Zone, one has:
\begin{equation}
{1 \over N} \sum_\k f(\k) \Longrightarrow 
\sum_i w_i f(\k_i).
\end{equation}

The $\psi(\r)$ as defined above are vanishingly small in order 
to be normalized. What is actually calculated, and used in the 
Fast Fourier Transform algorithm, is $\sqrt{N} \psi(\r)$:
$\Psi(\k+\G) \stackrel{FFT}{\longleftrightarrow}\sqrt{N}\psi(\r)$.
This ensures the correct normalization of the charge density.

\end{document}	

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