5.8 Ma­trix For­mu­la­tion

When the num­ber of un­knowns in a quan­tum me­chan­i­cal prob­lem has been re­duced to a fi­nite num­ber, the prob­lem can be re­duced to a lin­ear al­ge­bra one. This al­lows the prob­lem to be solved us­ing stan­dard an­a­lyt­i­cal or nu­mer­i­cal tech­niques. This sec­tion de­scribes how the lin­ear al­ge­bra prob­lem can be ob­tained.

Typ­i­cally, quan­tum me­chan­i­cal prob­lems can be re­duced to a fi­nite num­ber of un­knowns us­ing some fi­nite set of cho­sen wave func­tions, as in the pre­vi­ous sec­tion. There are other ways to make the prob­lems fi­nite, it does not re­ally make a dif­fer­ence here. But in gen­eral some sim­pli­fi­ca­tion will still be needed af­ter­wards. A mul­ti­ple sum like equa­tion (5.30) for dis­tin­guish­able par­ti­cles is awk­ward to work with, and when var­i­ous co­ef­fi­cients drop out for iden­ti­cal par­ti­cles, its gets even messier. So as a first step, it is best to or­der the terms in­volved in some way; any or­der­ing will in prin­ci­ple do. Or­der­ing al­lows each term to be in­dexed by a sin­gle counter $q$, be­ing the place of the term in the or­der­ing.

Us­ing an or­der­ing, the wave func­tion for a to­tal of $I$ par­ti­cles can be writ­ten more sim­ply as

\begin{displaymath}
\Psi =
a_1 \psi^{\rm S}_1({\skew0\vec r}_1,S_{z1}, {\skew0...
...w0\vec r}_2,S_{z2}, \ldots, {\skew0\vec r}_I,S_{zI})
+ \ldots
\end{displaymath}

or in in­dex no­ta­tion:
\begin{displaymath}
\Psi = \sum_{q=1}^Q a_q
\psi^{\rm S}_q({\skew0\vec r}_1,S_{z1}, {\skew0\vec r}_2,S_{z2}, \ldots, {\skew0\vec r}_I,S_{zI}).
\end{displaymath} (5.32)

where $Q$ is the to­tal count of the cho­sen $I$-​par­ti­cle wave func­tions and the sin­gle counter $q$ in $a_q$ re­places a set of $I$ in­dices in the de­scrip­tion used in the pre­vi­ous sec­tion. The $I$-​par­ti­cle func­tions $\psi^{\rm S}_q$ are al­lowed to be any­thing; in­di­vid­ual (Hartree) prod­ucts of sin­gle-par­ti­cle wave func­tions for dis­tin­guish­able par­ti­cles as in (5.30), Slater de­ter­mi­nants for iden­ti­cal fermi­ons, per­ma­nents for iden­ti­cal bosons, or what­ever. The only thing that will be as­sumed is that they are mu­tu­ally or­tho­nor­mal. (Which means that any un­der­ly­ing set of sin­gle-par­ti­cle func­tions $\pp{n}/{\skew0\vec r}///$ as de­scribed in the pre­vi­ous sec­tion should be or­tho­nor­mal. If they are not, there are pro­ce­dures like Gram-Schmidt to make them so. Or you can just put in some cor­rec­tion terms.)

Un­der those con­di­tions, the en­ergy eigen­value prob­lem $H\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$ takes the form:

\begin{displaymath}
\sum_{q=1}^Q H a_q \psi^{\rm S}_q = \sum_{q=1}^Q E a_q \psi^{\rm S}_q
\end{displaymath}

The trick is now to take the in­ner prod­uct of both sides of this equa­tion with each func­tion $\psi^{\rm S}_{\underline q}$ in the set of wave func­tions in turn. In other words, take an in­ner prod­uct with $\langle\psi^{\rm S}_1\vert$ to get one equa­tion, then take an in­ner prod­uct with $\langle\psi^{\rm S}_2\vert$ to get a sec­ond equa­tion, and so on. This pro­duces, us­ing the fact that the func­tions are or­tho­nor­mal to clean up the right-hand side,

\begin{displaymath}
\begin{array}{ccccccccl}
H_{11} a_1 & + & H_{12} a_2 & + &...
...{Q2} a_2 & + & \ldots & + & H_{QQ} a_Q & = & E a_Q
\end{array}\end{displaymath}

where

\begin{displaymath}
H_{11} = \langle \psi^{\rm S}_1\vert H \psi^{\rm S}_1\rangl...
... H_{QQ} = \langle \psi^{\rm S}_Q\vert H \psi^{\rm S}_Q\rangle.
\end{displaymath}

are the ma­trix co­ef­fi­cients, or Hamil­ton­ian co­ef­fi­cients.

This can again be writ­ten more com­pactly in in­dex no­ta­tion:

\begin{displaymath}
\sum_{q=1}^Q H_{{\underline q}q} a_q = E a_{\underline q}
...
...angle \psi^{\rm S}_{\underline q}\vert H \psi^{\rm S}_q\rangle
\end{displaymath} (5.33)

which is just a fi­nite-size ma­trix eigen­value prob­lem.

Since the func­tions $\psi^{\rm S}_q$ are known, cho­sen, func­tions, and the Hamil­ton­ian $H$ is also known, the ma­trix co­ef­fi­cients $H_{{{\underline q}}q}$ can be de­ter­mined. The eigen­val­ues $E$ and cor­re­spond­ing eigen­vec­tors $(a_1,a_2,\ldots)$ can then be found us­ing lin­ear al­ge­bra pro­ce­dures. Each eigen­vec­tor pro­duces a cor­re­spond­ing ap­prox­i­mate eigen­func­tion $a_1\psi^{\rm S}_1+a_2\psi^{\rm S}_2+\ldots$ with an en­ergy equal to the eigen­value $E$.


Key Points
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Op­er­a­tor eigen­value prob­lems can be ap­prox­i­mated by the ma­trix eigen­value prob­lems of lin­ear al­ge­bra.

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That al­lows stan­dard an­a­lyt­i­cal or nu­mer­i­cal tech­niques to be used in their so­lu­tion.

5.8 Re­view Ques­tions
1.

As a rel­a­tively sim­ple ex­am­ple, work out the above ideas for the $Q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 hy­dro­gen mol­e­cule spa­tial states $\psi^{\rm S}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {l}}\psi_{\rm {r}}$ and $\psi^{\rm S}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {l}}\psi_{\rm {r}}$. Write the ma­trix eigen­value prob­lem and iden­tify the two eigen­val­ues and eigen­vec­tors. Com­pare with the re­sults of sec­tion 5.3.

As­sume that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ have been slightly ad­justed to be or­tho­nor­mal. Then so are $\psi^{\rm S}_1$ and $\psi^{\rm S}_2$ or­tho­nor­mal, since the var­i­ous six-di­men­sion­al in­ner prod­uct in­te­grals, like

\begin{eqnarray*}\lefteqn{ \langle\psi^{\rm S}_1\vert\psi^{\rm S}_2\rangle\equiv...
... r}_2) {\,\rm d}^3 {\skew0\vec r}_1 {\,\rm d}^3 {\skew0\vec r}_2
\end{eqnarray*}

can ac­cord­ing to the rules of cal­cu­lus be fac­tored into three-di­men­sion­al in­te­grals as

\begin{eqnarray*}\lefteqn{\langle\psi^{\rm S}_1\vert\psi^{\rm S}_2\rangle} \\ &&...
..._{\rm {r}}\rangle\langle\psi_{\rm {r}}\vert\psi_{\rm {l}}\rangle
\end{eqnarray*}

which is zero if $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are or­tho­nor­mal.

Also, do not try to find ac­tual val­ues for $H_{11}$, $H_{12}$, $H_{21}$, and $H_{22}$. As sec­tion 5.2 noted, that can only be done nu­mer­i­cally. In­stead just re­fer to $H_{11}$ as $J$ and to $H_{12}$ as $\vphantom0\raisebox{1.5pt}{$-$}$$L$:

\begin{eqnarray*}& H_{11} \equiv\langle\psi^{\rm S}_1\vert H\psi^{\rm S}_1\rangl...
...i_{\rm {r}}\vert H\psi_{\rm {r}}\psi_{\rm {l}}\rangle\equiv - L.
\end{eqnarray*}

Next note that you also have

\begin{eqnarray*}& H_{22} \equiv\langle\psi^{\rm S}_2\vert H\psi^{\rm S}_2\rangl...
...}}\psi_{\rm {l}}\vert H\psi_{\rm {l}}\psi_{\rm {r}}\rangle = - L
\end{eqnarray*}

be­cause they are the ex­act same in­ner prod­uct in­te­grals; the dif­fer­ence is just which elec­tron you num­ber 1 and which one you num­ber 2 that de­ter­mines whether the wave func­tions are listed as $\psi_{\rm {l}}\psi_{\rm {r}}$ or $\psi_{\rm {r}}\psi_{\rm {l}}$.

So­lu­tion mat­for-a

2.

Find the eigen­states for the same prob­lem, but now in­clud­ing spin.

As sec­tion 5.7 showed, the an­ti­sym­met­ric wave func­tion with spin con­sists of a sum of six Slater de­ter­mi­nants. Ig­nor­ing the highly ex­cited first and sixth de­ter­mi­nants that have the elec­trons around the same nu­cleus, the re­main­ing $C$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4 Slater de­ter­mi­nants can be writ­ten out ex­plic­itly to give the two-par­ti­cle states

\begin{eqnarray*}\psi^{\rm S}_1 = \frac{\psi_{\rm {l}}\psi_{\rm {r}}{\uparrow}{\...
... \psi_{\rm {r}}\psi_{\rm {l}}{\downarrow}{\downarrow}}{\sqrt{2}}
\end{eqnarray*}

Note that the Hamil­ton­ian does not in­volve spin, to the ap­prox­i­ma­tion used in most of this book, so that, fol­low­ing the tech­niques of sec­tion 5.5, an in­ner prod­uct like $H_{23}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\langle\psi^{\rm S}_2\vert H\psi^{\rm S}_3\rangle$ can be writ­ten out like

\begin{eqnarray*}H_{23} & = & \frac 12\langle\psi_{\rm {l}}\psi_{\rm {r}}{\uparr...
...w}- (H\psi_{\rm {r}}\psi_{\rm {l}}){\uparrow}{\downarrow}\rangle
\end{eqnarray*}

and then mul­ti­plied out into in­ner prod­ucts of match­ing spin com­po­nents to give

\begin{displaymath}
H_{23} = -\frac 12\langle\psi_{\rm {l}}\psi_{\rm {r}}\vert H...
...}\psi_{\rm {l}}\vert H\psi_{\rm {l}}\psi_{\rm {r}}\rangle = L.
\end{displaymath}

The other 15 ma­trix co­ef­fi­cients can be found sim­i­larly, and most will be zero.

If you do not have ex­pe­ri­ence with lin­ear al­ge­bra, you may want to skip this ques­tion, or bet­ter, just read the so­lu­tion. How­ever, the four eigen­vec­tors are not that hard to guess; maybe eas­ier to guess than cor­rectly de­rive.

So­lu­tion mat­for-b