N.17 Generalized orbitals

This note has a brief look at generalized orbitals of the form

\begin{displaymath}
\pp n/{\skew0\vec r}/// = \pe n+/{\skew0\vec r}/u/z/+\pe n-/{\skew0\vec r}/d/z/.
\end{displaymath}

For such orbitals, the expectation energy can be worked out in exactly the same way as in {D.52}, except without simplifying the spin terms. The energy is

\begin{eqnarray*}
\big\langle E\big\rangle & = &
\sum_{n=1}^I \Big\langle\pp...
...
v^{\rm ee}\Big\vert\pp{\underline n}////\pp n////\Big\rangle
\end{eqnarray*}

To multiply out to the individual spin terms, it is convenient to normalize the spatial functions, and write

\begin{displaymath}
\pp n//// = c_{n+} \pe n+,0//u// + c_{n-} \pe n-,0//d//,
\end{displaymath}


\begin{displaymath}
\langle\pe n+,0////\vert\pe n+,0////\rangle=
\langle\pe ...
...ngle=1,
\quad \vert c_{n+}\vert^2 + \vert c_{n-}\vert^2 = 1
\end{displaymath}

In that case, the expectation energy multiplies out to

\begin{eqnarray*}
\big\langle E\big\rangle & = &
\sum_{n=1}^I
\Big\langle\...
...n+}^* c_{n-} c_{{\underline n}-}^* c_{{\underline n}+}
\bigg)
\end{eqnarray*}

where $\Re$ stands for the real part of its argument.

Now assume you have a normal unrestricted Hartree-Fock solution, and you try to lower its ground-state energy by selecting, for example, a spin-up orbital $\pe{m}//u//$ $\vphantom0\raisebox{1.5pt}{$\equiv$}$ $\pe{m+,0}//u//$ and adding some amount of spin down to it. First note then that the final sum above is zero, since at least one of $c_{n+}$, $c_{n-}$, $c_{{\underline n}-}$, and $c_{{\underline n}+}$ must be zero: all states except $m$ are still either spin-up or spin-down, and $m$ cannot be both $n$ and ${\underline n}$ $\raisebox{.2pt}{$\ne$}$ $n$. With the final sum gone, the energy is a linear function of $\vert c_{m-}\vert^2$, with $\vert c_{m+}\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1-\vert c_{m-}\vert^2$. The maximum energy must therefore occur for either $\vert c_{m-}\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the original purely spin up orbital, or for $\vert c_{m-}\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. (The latter case means that the unrestricted solution with the opposite spin for orbital $m$ must have less energy, so that the spin of orbital $m$ was incorrectly selected.) It follows from this argument that for correctly selected spin states, the energy cannot be lowered by replacing a single orbital with a generalized one.

Also note that for small changes, $\vert c_{m-}\vert^2$ is quadratically small and can be ignored. So the variational condition of zero change in energy is satisfied for all small changes in orbitals, even those that change their spin states. In other words, the unrestricted solutions are solutions to the full variational problem $\delta\langle{E}\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 for generalized orbitals as well.

Since these simple arguments do not cover finite changes in the spin state of more than one orbital, they do not seem to exclude the possibility that there might be additional solutions in which two or more orbitals are of mixed spin. But since either way the error in Hartree-Fock would be finite, there may not be much justification for dealing with the messy problem of generalized orbitals with dubious hopes of improvement. Procedures already exist that guarantee improvements on standard Hartree-Fock results.