#### 5.8.1 So­lu­tion mat­for-a

Ques­tion:

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 .

An­swer:

Us­ing the ab­bre­vi­a­tions and , the ma­trix eigen­value prob­lem be­comes

or tak­ing every­thing to the left hand side,

For this ho­mo­ge­neous sys­tem of equa­tions to have a so­lu­tion other than the triv­ial one 0, the de­ter­mi­nant of the ma­trix must be zero:

which al­lows for the two pos­si­bil­i­ties

So there are two en­ergy eigen­val­ues:

In the first case, since ac­cord­ing to the equa­tions above

it fol­lows that and must be equal, pro­duc­ing the eigen­func­tion

and nor­mal­iza­tion shows that 1/, as­sum­ing you take it real and pos­i­tive, (and as­sum­ing that and are re­ally ad­justed to be or­tho­nor­mal as as­sumed in this par­tic­u­lar ques­tion).

So the first en­ergy eigen­state is

Sim­i­larly the sec­ond en­ergy eigen­state is

Com­par­ing with sec­tion 5.3, the first eigen­state can be rec­og­nized as the ground state in which the nu­clei share the elec­trons sym­met­ri­cally. The sec­ond eigen­state is the ex­cited-en­ergy an­ti­sym­met­ric state in which the elec­trons share the elec­trons an­ti­sym­met­ri­cally. Other states of this type, like sim­ply , say, have ex­pec­ta­tion en­ergy some­where in be­tween and , but they are not eigen­states and do not have def­i­nite en­ergy.

Of course, the analy­sis here is ap­prox­i­mate. But as dis­cussed more in the ear­lier sec­tion 5.2, the true ground state is re­ally sym­met­ric, and the ex­cited en­ergy eigen­state re­ally is an­ti­sym­met­ric. Af­ter all, the Hamil­ton­ian com­mutes with the op­er­a­tion of swap­ping the elec­trons, (swap­ping the elec­trons does not do any­thing phys­i­cally,) so the en­ergy eigen­states must also be eigen­states of the swap­ping op­er­a­tor. The sym­met­ric state is an eigen­state of the swap­ping op­er­a­tor with eigen­value 1, it stays the same, and the an­ti­sym­met­ric state is an eigen­state with eigen­value 1; it changes sign un­der the swap.