5.8.2 So­lu­tion mat­for-b

Ques­tion:

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.

An­swer:

Eval­u­at­ing all 16 in­ner prod­ucts as above, the ma­trix eigen­value prob­lem is found to be

Af­ter bring­ing every­thing to the left hand side, the de­ter­mi­nant of the re­sult­ing ma­trix can again be set to zero, and the pos­si­ble en­ergy val­ues found. From those the eigen­vec­tors can be de­duced. It turns out that the larger eigen­value is triply de­gen­er­ate, so the three cor­re­spond­ing eigen­vec­tors are not unique; more than one ac­cept­able choice ex­ists for them. How­ever, you do have to nor­mal­ize them to unit length and en­sure that they are mu­tu­ally or­thog­o­nal.

Rather than go­ing through that math, it is quicker just to guess the eigen­vec­tors. One guess that works is

Just sub­sti­tute 1 and 0 into the equa­tions above and see that they are sat­is­fied pro­vided that the en­ergy has the ex­cited value . The cor­re­spond­ing eigen­func­tion has, ac­cord­ing to its de­f­i­n­i­tion above, both elec­trons spin-up and in the ex­cited an­ti­sym­met­ric spa­tial state .

A sim­i­lar guess that works is

This cor­re­sponds to the eigen­func­tion in which both elec­trons are spin-down, and again in the an­ti­sym­met­ric spa­tial state. The third one that works is

where the scale fac­tor is only needed to en­sure that the vec­tor is of unit length. The eigen­state cor­re­sponds, ac­cord­ing to the de­f­i­n­i­tions of and above, to the elec­trons be­ing in the an­ti­sym­met­ric spa­tial state times the triplet state of sec­tion 5.5.6. The en­ergy is again the el­e­vated value .

The fi­nal eigen­vec­tor

gives the eigen­state . This cor­re­sponds to the elec­trons be­ing in the sym­met­ric spa­tial state times the sin­glet spin state. The en­ergy is the ground state value .