### 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 , be­ing the place of the term in the or­der­ing.

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

or in in­dex no­ta­tion:
 (5.32)

where is the to­tal count of the cho­sen -​par­ti­cle wave func­tions and the sin­gle counter in re­places a set of in­dices in the de­scrip­tion used in the pre­vi­ous sec­tion. The -​par­ti­cle func­tions 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 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 takes the form:

The trick is now to take the in­ner prod­uct of both sides of this equa­tion with each func­tion in the set of wave func­tions in turn. In other words, take an in­ner prod­uct with to get one equa­tion, then take an in­ner prod­uct with 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,

where

are the ma­trix co­ef­fi­cients, or Hamil­ton­ian co­ef­fi­cients.

This can again be writ­ten more com­pactly in in­dex no­ta­tion:

 (5.33)

which is just a fi­nite-size ma­trix eigen­value prob­lem.

Since the func­tions are known, cho­sen, func­tions, and the Hamil­ton­ian is also known, the ma­trix co­ef­fi­cients can be de­ter­mined. The eigen­val­ues and cor­re­spond­ing eigen­vec­tors can then be found us­ing lin­ear al­ge­bra pro­ce­dures. Each eigen­vec­tor pro­duces a cor­re­spond­ing ap­prox­i­mate eigen­func­tion with an en­ergy equal to the eigen­value .

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

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 2 hy­dro­gen mol­e­cule spa­tial states and . Write the ma­trix eigen­value prob­lem and iden­tify the two eigen­val­ues and eigen­vec­tors. Com­pare with the re­sults of sec­tion 5.3.

As­sume that and have been slightly ad­justed to be or­tho­nor­mal. Then so are and or­tho­nor­mal, since the var­i­ous six-di­men­sion­al in­ner prod­uct in­te­grals, like

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

which is zero if and are or­tho­nor­mal.

Also, do not try to find ac­tual val­ues for , , , and . As sec­tion 5.2 noted, that can only be done nu­mer­i­cally. In­stead just re­fer to as and to as :

Next note that you also have

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 or .
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 4 Slater de­ter­mi­nants can be writ­ten out ex­plic­itly to give the two-par­ti­cle states

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 can be writ­ten out like

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

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.