N.17 Gen­er­al­ized or­bitals

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

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

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

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

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

$$\pp n//// = c_{n+} \pe n+,0//u// + c_{n-} \pe n-,0//d//,$$

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

In that case, the ex­pec­ta­tion en­ergy mul­ti­plies out to

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

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 min­i­mum 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.