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A..4 Numerical solution

The radial (scalar-relativistic) KS equation is integrated on a radial grid. It is convenient to have a denser grid close to the nucleus and a coarser one far away. Traditionally a logarithmic grid is used: ri = r0exp(i$ \Delta$x). With this grid, one has

$\displaystyle \int_{0}^{\infty}$f (r)dr = $\displaystyle \int_{0}^{\infty}$f (x)r(x)dx (15)
and

$\displaystyle {d f(r)\over d r}$ = $\displaystyle {1\over r}$$\displaystyle {d f(x)\over d x}$,        $\displaystyle {d^2 f(r)\over d r^2}$ = - $\displaystyle {1\over r^2}$$\displaystyle {d f(x)\over d x}$ + $\displaystyle {1\over r^2}$$\displaystyle {d^2 f(x)\over d x^2}$. (16)
We start with a given self-consistent potential V and a trial eigenvalue $ \epsilon$. The equation is integrated from r = 0 outwards to rt, the outermost classical (nonrelativistic for simplicity) turning point, defined by l (l+1)/rt2 + $ \left(\vphantom{V(r_t)-\epsilon}\right.$V(rt) - $ \epsilon$$ \left.\vphantom{V(r_t)-\epsilon}\right)$ = 0. In a logarithmic grid (see above) the equation to solve becomes:
$\displaystyle {1\over r^2}$$\displaystyle {d^2 R_{nl}(x)\over d x^2}$ = $\displaystyle {1\over r^2}$$\displaystyle {d R_{nl}(x)\over d x}$ + $\displaystyle \left(\vphantom{ {l(l+1)\over r^2} + M(r)\left(V(r)-\epsilon\right)
}\right.$$\displaystyle {l(l+1)\over r^2}$ + M(r)$\displaystyle \left(\vphantom{V(r)-\epsilon}\right.$V(r) - $\displaystyle \epsilon$$\displaystyle \left.\vphantom{V(r)-\epsilon}\right)$$\displaystyle \left.\vphantom{ {l(l+1)\over r^2} + M(r)\left(V(r)-\epsilon\right)
}\right)$Rnl(r)  
      - $\displaystyle {\alpha^2\over 4M(r)}$$\displaystyle {dV(r)\over dr}$$\displaystyle \left(\vphantom{{1\over r} {dR_{nl}(x)\over dx} +
\langle\kappa\rangle {R_{nl}(r)\over r}}\right.$$\displaystyle {1\over r}$$\displaystyle {d R_{nl}(x)\over d x}$ + $\displaystyle \langle$$\displaystyle \kappa$$\displaystyle \rangle$$\displaystyle {R_{nl}(r)\over r}$$\displaystyle \left.\vphantom{{1\over r} {dR_{nl}(x)\over dx} +
\langle\kappa\rangle {R_{nl}(r)\over r}}\right)$. (17)

This determines d2Rnl(x)/dx2 which is used to determine dRnl(x)/dx which in turn is used to determine Rnl(r), using predictor-corrector or whatever classical integration method. dV(r)/dr is evaluated numerically from any finite difference method. The series is started using the known (?) asymptotic behavior of Rnl(r) close to the nucleus (with ionic charge Z)

Rnl(r) $\displaystyle \simeq$ r$\scriptstyle \gamma$,        $\displaystyle \gamma$ = $\displaystyle {l\sqrt{l^2-\alpha^2 Z^2}+
(l+1)\sqrt{(l+1)^2-\alpha^2 Z^2}\over 2l+1}$. (18)
The number of nodes is counted. If there are too few (many) nodes, the trial eigenvalue is increased (decreased) and the procedure is restarted until the correct number n - l - 1 of nodes is reached. Then a second integration is started inward, starting from a suitably large r $ \sim$ 10rt down to rt, using as a starting point the asymptotic behavior of Rnl(r) at large r:

Rnl(r) $\displaystyle \simeq$ e-k(r)r,        k(r) = $\displaystyle \sqrt{{{l(l+1)\over r^2} + \left(V(r)-\epsilon\right)}}$. (19)
The two pieces are continuously joined at rt and a correction to the trial eigenvalue is estimated using perturbation theory (see below). The procedure is iterated to self-consistency.

The perturbative estimate of correction to trial eigenvalues is described in the following for the nonrelativistic case (it is not worth to make relativistic corrections on top of a correction). The trial eigenvector Rnl(r) will have a cusp at rt if the trial eigenvalue is not a true eigenvalue:

A = $\displaystyle {d R_{nl}(r_t^+)\over dr}$ - $\displaystyle {d R_{nl}(r_t^-)\over dr}$$\displaystyle \ne$0. (20)
Such discontinuity in the first derivative translates into a $ \delta$(rt) in the second derivative:

$\displaystyle {d^2 R_{nl}(r)\over dr^2}$ = $\displaystyle {d^2 \widetilde R_{nl}(r)\over dr^2}$ + A$\displaystyle \delta$(r - rt) (21)
where the tilde denotes the function obtained by matching the second derivatives in the r < rt and r > rt regions. This means that we are actually solving a different problem in which V(r) is replaced by V(r) + $ \Delta$V(r), given by

$\displaystyle \Delta$V(r) = - $\displaystyle {\hbar^2\over 2m}$$\displaystyle {A\over R_{nl}(r_t)}$$\displaystyle \delta$(r - rt). (22)
The energy difference between the solution to such fictitious potential and the solution to the real potential can be estimated from perturbation theory:

$\displaystyle \Delta$$\displaystyle \epsilon_{{nl}}^{}$ = - $\displaystyle \langle$$\displaystyle \psi$|$\displaystyle \Delta$V|$\displaystyle \psi$$\displaystyle \rangle$ = $\displaystyle {\hbar^2\over 2m}$Rnl(rt)A. (23)


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Next: B. Equations for the Up: A. Atomic Calculations Previous: A..3 Scalar-relativistic case   Contents
Filippo Spiga 2016-04-24