D.15 The hy­dro­gen ra­dial wave func­tions

This will be child’s play for har­monic os­cil­la­tor, {D.12}, and spher­i­cal har­mon­ics, {D.14}, vet­er­ans. If you re­place the an­gu­lar terms in (4.33) by , and then di­vide the en­tire equa­tion by , you get

Since is nondi­men­sion­al, all terms in this equa­tion must be. In par­tic­u­lar, the ra­tio in the third term must be the rec­i­p­ro­cal of a con­stant with the di­men­sions of length; so, de­fine the con­stant to be the Bohr ra­dius . It is con­ve­nient to also de­fine a cor­re­spond­ingly nondi­men­sion­al­ized ra­dial co­or­di­nate as . The fi­nal term in the equa­tion must be nondi­men­sion­al too, and that means that the en­ergy must take the form , where is a nondi­men­sion­al en­ergy. In terms of these scaled co­or­di­nates you get

or writ­ten out

where the primes de­note de­riv­a­tives with re­spect to .

Sim­i­lar to the case of the har­monic os­cil­la­tor, you must have so­lu­tions that be­come zero at large dis­tances from the nu­cleus: gives the prob­a­bil­ity of find­ing the par­ti­cle in­te­grated over all pos­si­ble po­si­tions, and if does not be­come zero suf­fi­ciently rapidly at large , this in­te­gral would be­come in­fi­nite, rather than one (cer­tainty) as it should. Now the ODE above be­comes for large ap­prox­i­mately 0, which has so­lu­tions of the rough form for pos­i­tive that do not have the re­quired de­cay to zero. Zero scaled en­ergy is still too much, as can be checked by solv­ing in terms of Bessel func­tions, so you must have that is neg­a­tive. In clas­si­cal terms, the earth can only hold onto the moon since the moon’s to­tal en­ergy is less than the po­ten­tial en­ergy far from the earth; if it was not, the moon would es­cape.

Any­way, for bound states, you must have the scaled en­ergy neg­a­tive. In that case, the so­lu­tion at large takes the ap­prox­i­mate form . Only the neg­a­tive sign is ac­cept­able. You can make things a lot eas­ier for your­self if you peek at the fi­nal so­lu­tion and rewrite as be­ing 1 (that is not re­ally cheat­ing, since you are not at this time claim­ing that is an in­te­ger, just a pos­i­tive num­ber.) In that case, the ac­cept­able ex­po­nen­tial be­hav­ior at large dis­tance takes the form where . Split off this ex­po­nen­tial part by writ­ing where must re­main bounded at large . Sub­sti­tut­ing these new vari­ables, the ODE be­comes

where the primes in­di­cate de­riv­a­tives with re­spect to .

If you do a power se­ries so­lu­tion of this ODE, you see that it must start with ei­ther power or with power . The lat­ter is not ac­cept­able, since it would cor­re­spond to an in­fi­nite ex­pec­ta­tion value of en­ergy. You could now ex­pand the so­lu­tion fur­ther in pow­ers of , but the prob­lem is that tab­u­lated poly­no­mi­als usu­ally do not start with a power but with power zero or one. So you would not eas­ily rec­og­nize the poly­no­mial you get. There­fore it is best to split off the lead­ing power by defin­ing , which turns the ODE into

Sub­sti­tut­ing in a power se­ries , you get

The ac­cept­able low­est power of is now zero. Again the se­ries must ter­mi­nate, oth­er­wise the so­lu­tion would be­have as at large dis­tance, which is un­ac­cept­able. Ter­mi­na­tion at a high­est power re­quires that equals . Since and are in­te­gers, so must be , and since the fi­nal power is at least zero, is at least . The cor­rect scaled en­ergy 1 with has been ob­tained.

With iden­ti­fied, you can iden­tify the ODE as La­guerre's as­so­ci­ated dif­fer­en­tial equa­tion, e.g. [41, 30.26], the -​th de­riv­a­tive of La­guerre's dif­fer­en­tial equa­tion, e.g. [41, 30.1], and the poly­no­mial so­lu­tions as the as­so­ci­ated La­guerre poly­no­mi­als , e.g. [41, 30.27], the -​th de­riv­a­tives of the La­guerre's poly­no­mi­als , e.g. [41, 30.2]. To nor­mal­ize the wave func­tion use an in­te­gral from a ta­ble book, e.g. [41, 30.46].

Putting it all to­gether, the generic ex­pres­sion for hy­dro­gen eigen­func­tions are, drums please:

 (D.8)

The prop­er­ties of the as­so­ci­ated La­guerre poly­no­mi­als are in ta­ble books like [41, pp. 169-172], and the spher­i­cal har­mon­ics were given ear­lier in chap­ter 4.2.3 and in de­riva­tion {D.14}, (D.5).

Do keep in mind that dif­fer­ent ref­er­ences have con­tra­dic­tory de­f­i­n­i­tions of the as­so­ci­ated La­guerre poly­no­mi­als. This book fol­lows the no­ta­tions of [41, pp. 169-172], who de­fine

In other words, is sim­ply the -​th de­riv­a­tive of , which cer­tainly tends to sim­plify things. Ac­cord­ing to [25, p. 152], the most nearly stan­dard no­ta­tion de­fines

Com­bine the messy de­f­i­n­i­tion of the spher­i­cal har­mon­ics (D.5) with the un­cer­tain de­f­i­n­i­tion of the La­guerre poly­no­mi­als in the for­mu­lae (D.8) for the hy­dro­gen en­ergy eigen­func­tions above, and there is of course al­ways a pos­si­bil­ity of get­ting an eigen­func­tion wrong if you are not care­ful.

Some­times the value of the wave func­tions at the ori­gin is needed. Now from the above so­lu­tion (D.8), it is seen that

 (D.9)

so only the eigen­func­tions are nonzero at the ori­gin. To find the value re­quires where is the de­riv­a­tive of the La­guerre poly­no­mial . Skim­ming through ta­ble books, you can find that , [41, 30.19], while the dif­fer­en­tial equa­tion for these func­tion im­plies that . There­fore:
 (D.10)