2.6.9 So­lu­tion herm-i

Ques­tion:

A com­plete set of or­tho­nor­mal eigen­func­tions of the op­er­a­tor that are pe­ri­odic on the in­ter­val 0 are the in­fi­nite set of func­tions

Check that these func­tions are in­deed pe­ri­odic, or­tho­nor­mal, and that they are eigen­func­tions of with the real eigen­val­ues

Com­plete­ness is a much more dif­fi­cult thing to prove, but they are. The com­plete­ness proof in the notes cov­ers this case.

An­swer:

Any eigen­func­tion of the above list can be writ­ten in the generic form where is a whole num­ber, in other words where is an in­te­ger, one of ..., 3, 2, 1, 0, 1, 2, 3, ... If you show that the stated prop­er­ties are true for this generic form, it means that they are true for every eigen­func­tion.

Now pe­ri­od­ic­ity re­quires that , and the Euler for­mula ver­i­fies this: sines and cosines are the same if the an­gle changes by a whole mul­ti­ple of . (For ex­am­ple, , , , etcetera are phys­i­cally all equiv­a­lent to a zero an­gle.)

The de­riv­a­tive of with re­spect to is , and mul­ti­ply­ing by you get , so is an eigen­func­tion of with eigen­value .

To see that is nor­mal­ized, check that its norm is unity:

To ver­ify that is or­thog­o­nal to every other eigen­func­tion, take the generic other eigen­func­tion to be with an in­te­ger dif­fer­ent from . You must then show that the in­ner prod­uct of these two eigen­func­tions is zero. Since the nor­mal­iza­tion con­stants do not make any dif­fer­ence here, you can just show that is zero. You get

since 1. So dif­fer­ent eigen­func­tions are or­thog­o­nal, their in­ner prod­uct is zero.