### D.79 De­riva­tion of per­tur­ba­tion the­ory

This note de­rives the per­tur­ba­tion the­ory re­sults for the so­lu­tion of the eigen­value prob­lem where is small. The con­sid­er­a­tions for de­gen­er­ate prob­lems use lin­ear al­ge­bra.

First, small is not a valid math­e­mat­i­cal term. There are no small num­bers in math­e­mat­ics, just num­bers that be­come zero in some limit. There­fore, to math­e­mat­i­cally an­a­lyze the prob­lem, the per­tur­ba­tion Hamil­ton­ian will be writ­ten as

where is some cho­sen num­ber that phys­i­cally in­di­cates the mag­ni­tude of the per­tur­ba­tion po­ten­tial. For ex­am­ple, if the per­tur­ba­tion is an ex­ter­nal elec­tric field, could be taken as the ref­er­ence mag­ni­tude of the elec­tric field. In per­tur­ba­tion analy­sis, is as­sumed to be van­ish­ingly small.

The idea is now to start with a good eigen­func­tion of , (where good is still to be de­fined), and cor­rect it so that it be­comes an eigen­func­tion of . To do so, both the de­sired en­ergy eigen­func­tion and its en­ergy eigen­value are ex­panded in a power se­ries in terms of :

If is a small quan­tity, then will be much smaller still, and can prob­a­bly be ig­nored. If not, then surely will be so small that it can be ig­nored. A re­sult that for­gets about pow­ers of higher than one is called first or­der per­tur­ba­tion the­ory. A re­sult that also in­cludes the qua­dratic pow­ers, but for­gets about pow­ers higher than two is called sec­ond or­der per­tur­ba­tion the­ory, etcetera.

Be­fore pro­ceed­ing with the prac­ti­cal ap­pli­ca­tion, a dis­claimer is needed. While it is rel­a­tively easy to see that the eigen­val­ues ex­pand in whole pow­ers of , (note that they must be real whether is pos­i­tive or neg­a­tive), it is much more messy to show that the eigen­func­tions must ex­pand in whole pow­ers. In fact, for de­gen­er­ate en­er­gies they only do if you choose good states . See Rel­lich’s lec­ture notes on Per­tur­ba­tion The­ory [Gor­don & Breach, 1969] for a proof. As a re­sult the prob­lem with de­gen­er­acy be­comes that the good un­per­turbed eigen­func­tion is ini­tially un­known. It leads to lots of messi­ness in the pro­ce­dures for de­gen­er­ate eigen­val­ues de­scribed be­low.

When the above power se­ries are sub­sti­tuted into the eigen­value prob­lem to be solved,

the net co­ef­fi­cient of every power of must be equal in the left and right hand sides. Col­lect­ing these co­ef­fi­cients and re­ar­rang­ing them ap­pro­pri­ately pro­duces:

These are the equa­tions to be solved in suc­ces­sion to give the var­i­ous terms in the ex­pan­sion for the wave func­tion and the en­ergy . The fur­ther you go down the list, the bet­ter your com­bined re­sult should be.

Note that all it takes is to solve prob­lems of the form

The equa­tions for the un­known func­tions are in terms of the un­per­turbed Hamil­ton­ian , with some ad­di­tional but in prin­ci­ple know­able terms.

For dif­fi­cult per­tur­ba­tion prob­lems like you find in en­gi­neer­ing, the use of a small pa­ra­me­ter is es­sen­tial to get the math­e­mat­ics right. But in the sim­ple ap­pli­ca­tions in quan­tum me­chan­ics, it is usu­ally overkill. So most of the time the ex­pan­sions are writ­ten with­out, like

where you are as­sumed to just imag­ine that and are first or­der small, and are “sec­ond or­der small,” etcetera. In those terms, the suc­ces­sive equa­tions to solve are:
 (D.55) (D.56) (D.57) (D.58)

Now con­sider each of these equa­tions in turn. First, (D.55) is just the Hamil­ton­ian eigen­value prob­lem for and is al­ready sat­is­fied by the cho­sen un­per­turbed so­lu­tion and its eigen­value . How­ever, the re­main­ing equa­tions are not triv­ial. To solve them, write their so­lu­tions in terms of the other eigen­func­tions of the un­per­turbed Hamil­ton­ian . In par­tic­u­lar, to solve (D.56), write

where the co­ef­fi­cients are still to be de­ter­mined. The co­ef­fi­cient of is zero on ac­count of the nor­mal­iza­tion re­quire­ment. (And in fact, it is eas­i­est to take the co­ef­fi­cient of also zero for , , ..., even if it means that the re­sult­ing wave func­tion will no longer be nor­mal­ized.)

The prob­lem (D.56) be­comes

where the left hand side was cleaned up us­ing the fact that the are eigen­func­tions of . To get the first or­der en­ergy cor­rec­tion , the trick is now to take an in­ner prod­uct of the en­tire equa­tion with . Be­cause of the fact that the en­ergy eigen­func­tions of are or­tho­nor­mal, this in­ner prod­uct pro­duces zero in the left hand side, and in the right hand side it pro­duces:

And that is ex­actly the first or­der cor­rec­tion to the en­ergy claimed in {A.38.1}; equals the Hamil­ton­ian per­tur­ba­tion co­ef­fi­cient . If the prob­lem is not de­gen­er­ate or is good, that is.

To get the co­ef­fi­cients , so that you know what is the first or­der cor­rec­tion to the wave func­tion, just take an in­ner prod­uct with each of the other eigen­func­tions of in turn. In the left hand side it only leaves the co­ef­fi­cient of the se­lected eigen­func­tion be­cause of or­tho­nor­mal­ity, and for the same rea­son, in the right hand side the fi­nal term drops out. The re­sult is

The co­ef­fi­cients can nor­mally be com­puted from this.

Note how­ever that if the prob­lem is de­gen­er­ate, there will be eigen­func­tions that have the same en­ergy as the eigen­func­tion be­ing cor­rected. For these the left hand side in the equa­tion above is zero, and the equa­tion can­not in gen­eral be sat­is­fied. If so, it means that the as­sump­tion that an eigen­func­tion of the full Hamil­ton­ian ex­pands in a power se­ries in start­ing from is un­true. Eigen­func­tion is bad. And that means that the first or­der en­ergy cor­rec­tion de­rived above is sim­ply wrong. To fix the prob­lem, what needs to be done is to iden­tify the sub­ma­trix of all Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients in which both un­per­turbed eigen­func­tions have the en­ergy , i.e. the sub­ma­trix

The eigen­val­ues of this sub­ma­trix are the cor­rect first or­der en­ergy changes. So, if all you want is the first or­der en­ergy changes, you can stop here. Oth­er­wise, you need to re­place the un­per­turbed eigen­func­tions that have en­ergy . For each or­tho­nor­mal eigen­vec­tor of the sub­ma­trix, there is a cor­re­spond­ing re­place­ment un­per­turbed eigen­func­tion

You will need to rewrite the Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients in terms of these new eigen­func­tions. (Since the re­place­ment eigen­func­tions are lin­ear com­bi­na­tions of the old ones, no new in­te­gra­tions are needed.) You then need to re­s­e­lect the eigen­func­tion whose en­ergy to cor­rect from among these re­place­ment eigen­func­tions. Choose the first or­der en­ergy change (eigen­value of the sub­ma­trix) that is of in­ter­est to you and then choose as the re­place­ment eigen­func­tion cor­re­spond­ing to a cor­re­spond­ing eigen­vec­tor. If the first or­der en­ergy change is not de­gen­er­ate, the eigen­vec­tor is unique, so is now good. If not, the good eigen­func­tion will be some com­bi­na­tion of the re­place­ment eigen­func­tions that have that first or­der en­ergy change, and the good com­bi­na­tion will have to be fig­ured out later in the analy­sis. In any case, the prob­lem with the equa­tion above for the will be fixed, be­cause the new sub­ma­trix will be a di­ag­o­nal one: will be zero when and . The co­ef­fi­cients for which re­main in­de­ter­mi­nate at this stage. They will nor­mally be found at a later stage in the ex­pan­sion.

With the co­ef­fi­cients as found, or not found, the sum for the first or­der per­tur­ba­tion in the wave func­tion be­comes

The en­tire process re­peats for higher or­der. In par­tic­u­lar, to sec­ond or­der (D.57) gives, writ­ing also in terms of the un­per­turbed eigen­func­tions,

To get the sec­ond or­der con­tri­bu­tion to the en­ergy, take again an in­ner prod­uct with . That pro­duces, again us­ing or­tho­nor­mal­ity, (and di­ag­o­nal­ity of the sub­ma­trix dis­cussed above if de­gen­er­ate),

This gives the sec­ond or­der change in the en­ergy stated in {A.38.1}, if is good. Note that since is Her­mit­ian, the prod­uct of the two Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients in the ex­pres­sion is just the square mag­ni­tude of ei­ther.

In the de­gen­er­ate case, when tak­ing an in­ner prod­uct with a for which , the equa­tion can be sat­is­fied through the still in­de­ter­mi­nate pro­vided that the cor­re­spond­ing di­ag­o­nal co­ef­fi­cient of the di­ag­o­nal­ized sub­ma­trix is un­equal to . In other words, pro­vided that the first or­der en­ergy change is not de­gen­er­ate. If that is un­true, the higher or­der sub­ma­trix

will need to be di­ag­o­nal­ized, (the rest of the equa­tion needs to be zero). Its eigen­val­ues give the cor­rect sec­ond or­der en­ergy changes. To pro­ceed to still higher en­ergy, re­s­e­lect the eigen­func­tions fol­low­ing the same gen­eral lines as be­fore. Ob­vi­ously, in the de­gen­er­ate case the en­tire process can be­come very messy. And you may never be­come sure about the good eigen­func­tion.

This prob­lem can of­ten be elim­i­nated or greatly re­duced if the eigen­func­tions of are also eigen­func­tions of an­other op­er­a­tor , and com­mutes with . Then you can arrange the eigen­func­tions into sets that have the same value for the good quan­tum num­ber of . You can an­a­lyze the per­turbed eigen­func­tions in each of these sets while com­pletely ig­nor­ing the ex­is­tence of eigen­func­tions with dif­fer­ent val­ues for quan­tum num­ber .

To see why, con­sider two ex­am­ple eigen­func­tions and of that have dif­fer­ent eigen­val­ues and . Since and both com­mute with , their sum does, so

and since is not zero, must be. Now is the amount of eigen­func­tion pro­duced by ap­ply­ing on . It fol­lows that ap­ply­ing on an eigen­func­tion with an eigen­value does not pro­duce any eigen­func­tions with dif­fer­ent eigen­val­ues . Thus an eigen­func­tion of sat­is­fy­ing

can be re­placed by just , since this by it­self must sat­isfy the eigen­value prob­lem: the Hamil­ton­ian of the sec­ond sum does not pro­duce any amount of eigen­func­tions in the first sum and vice-versa. (There must al­ways be at least one value of for which the first sum at 0 is in­de­pen­dent of the other eigen­func­tions of .) Re­duce every eigen­func­tion of to an eigen­func­tion of in this way. Now the ex­is­tence of eigen­func­tions with dif­fer­ent val­ues of than the one be­ing an­a­lyzed can be ig­nored since the Hamil­ton­ian does not pro­duce them. In terms of lin­ear al­ge­bra, the Hamil­ton­ian has been re­duced to block di­ag­o­nal form, with each block cor­re­spond­ing to a set of eigen­func­tions with a sin­gle value of . If the Hamil­ton­ian also com­mutes with an­other op­er­a­tor that the are eigen­func­tions of, the ar­gu­ment re­peats for the sub­sets with a sin­gle value for .

The Hamil­ton­ian per­tur­ba­tion co­ef­fi­cient is zero when­ever two good quan­tum num­bers and are un­equal. The rea­son is the same as for above. Only per­tur­ba­tion co­ef­fi­cients for which all good quan­tum num­bers are the same can be nonzero.