Disclaimer: If this is the wrong place for this, I apologise, this probably comes somewhere between QM, Atomic, Linear algebra and a spoonful of Quantum chemistry for good measure.(adsbygoogle = window.adsbygoogle || []).push({});

Anyway, for a group of non interacting (mean field) electrons, moving in a potential generated by nuclei and the average interactions with other electrons, the single particle, non relativistic Hamiltonian can be solved numerically quite easily. Then the ground state of the system is first estimated by the Slater determinant of the single electron wavefunctions, building up from the lowest energy wavefunctions to the highest required by the number of electrons.

[itex]\psi(x_1,\ldots,x_N) = \left|\begin{array}{ccc}

\phi_{1}(x_{1}) & \ldots & \phi_{N}(x_{1})\\

\vdots & \ddots\\

\phi_{1}(x_{N}) & & \phi_{N}(x_{N})

\end{array}\right|[/itex]

Where the variables include position and spin. Now, for small numbers of electrons, It is reasonable to just take the determinant, mod square it get the probability density function and then integrate over the required region, to generate the potential, then essentially go through the Hartree-Foch self consistency method. The problem with this is that for large N (say, 40 or 50), the determinant has enormous numbers of terms depending on 40 or 50 variables, which obviously isn't particularly practical for actually computing anything of any interest.

For the issue at hand (i.e. atomic systems), on spherical symmetry and non relativistic (ignoring the LS couping), the states are degenerate in all generality in the projected spin [itex]s_z[/itex] and the projected orbital angular momentum [itex]l_z[/itex]. Furthermore, looking just for a spherical potential means that all the angular integrals can be done ahead of time with the orthogonality relations of the spherical harmonics, so just looking the radial cumulative probability (which is the thing of interest for generating the potential), will look something like;

[itex] P(R) = \int dx_1 ... \int dx_N \psi^*\psi [/itex]

Is there some general rule for splitting up this integral into the non orthogonal bits? For Helium and Lithium it seems that the states separate straightforwardly to (within normalisation, which has been played fast and loose with) the products of the determinants of the bits which are non-orthogonal, or for example, in Lithium, the effective state looks something like

[itex] \left|1s\downarrow;\, x_{1}\right\rangle \otimes\left|\begin{array}{cc}

\left|1s\uparrow;\, x_{2}\right\rangle & \left|2s\uparrow;\, x_{2}\right\rangle \\

\left|1s\uparrow;\, x_{3}\right\rangle & \left|2s\uparrow;\, x_{3}\right\rangle

\end{array}\right| [/itex]

And then further states added in the same manner? (i.e. in seperate determinants for the different angular momentum and spin quantum numbers)

I've probably not been particularly clear for which I apologise, but any help would be much appreciated!

(n.b. sorry for mixing wavefunction and Dirac notation)

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# Slater Determinants for Large numbers of electrons

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