#### 5.3.2 So­lu­tion 2state-b

Ques­tion:

Show that it does not have an ef­fect on the so­lu­tion whether or not the ba­sic states and are nor­mal­ized, like in the pre­vi­ous ques­tion, be­fore the state of low­est en­ergy is found.

This re­quires no de­tailed analy­sis; just check that the same so­lu­tion can be de­scribed us­ing the nonorthog­o­nal and or­thog­o­nal ba­sis states. It is how­ever an im­por­tant ob­ser­va­tion for var­i­ous nu­mer­i­cal so­lu­tion pro­ce­dures: your set of ba­sis func­tions can be cleaned up and sim­pli­fied with­out af­fect­ing the so­lu­tion you get.

An­swer:

Us­ing the orig­i­nal ba­sis states, the so­lu­tion, say the ground state of low­est en­ergy, can be writ­ten in the form

for some val­ues of the con­stants and . Now the ex­pres­sion for the or­thog­o­nal­ized func­tions,

can for given and be thought of as two equa­tions for and that can be solved. In par­tic­u­lar, adding times the sec­ond equa­tion to the first gives

Sim­i­larly, adding times the first equa­tion to the sec­ond gives

If this is plugged into the ex­pres­sion for the so­lu­tion, , it takes the form

where

So, while the con­stants and are dif­fer­ent from and , the same so­lu­tion can be found equally well in terms of and as in terms of and .

In the terms of lin­ear al­ge­bra, and span the same func­tion space as and : any wave func­tion that can be de­scribed as a com­bi­na­tion of and can also be de­scribed in terms of and , al­though with dif­fer­ent con­stants. This is true as long as the de­f­i­n­i­tions of the new func­tions can be solved for the old func­tions as above. The ma­trix of co­ef­fi­cients, here

must have a nonzero de­ter­mi­nant.