### D.38 Time-de­pen­dent per­tur­ba­tion the­ory

The equa­tions to be solved are

To sim­plify the use of per­tur­ba­tion the­ory, it is con­ve­nient to use a trick that gets rid of half the terms in these equa­tions. The trick is to de­fine new co­ef­fi­cients and by

 (D.22)

The new co­ef­fi­cients and are phys­i­cally just as good as and . For one, the prob­a­bil­i­ties are given by the square mag­ni­tudes of the co­ef­fi­cients, and the square mag­ni­tudes of and are ex­actly the same as those of and . That is be­cause the ex­po­nen­tials have mag­ni­tude one. Also, the ini­tial con­di­tions are un­changed, as­sum­ing that you choose the in­te­gra­tion con­stants so that the in­te­grals are ini­tially zero.

The evo­lu­tion equa­tions for and are

 (D.23)

with . Ef­fec­tively, the two en­ergy ex­pec­ta­tion val­ues have been turned into zero. How­ever, the ma­trix el­e­ment is now time-de­pen­dent, if it was not al­ready. To check the above evo­lu­tion equa­tions, just plug in the de­f­i­n­i­tion of the co­ef­fi­cients.

It will from now on be as­sumed that the orig­i­nal Hamil­ton­ian co­ef­fi­cients are in­de­pen­dent of time. That makes the dif­fer­ence in ex­pec­ta­tion en­er­gies con­stant too.

Now the for­mal way to per­form time-de­pen­dent per­tur­ba­tion the­ory is to as­sume that the ma­trix el­e­ment is small. Write as where is a scale fac­tor. Then you can find the be­hav­ior of the so­lu­tion in the lim­it­ing process by ex­pand­ing the so­lu­tion in pow­ers of . The de­f­i­n­i­tion of the scale fac­tor is not im­por­tant. You might iden­tify it with a small phys­i­cal pa­ra­me­ter in the ma­trix el­e­ment. But in fact you can take the same as and as an ad­di­tional math­e­mat­i­cal pa­ra­me­ter with no mean­ing for the phys­i­cal prob­lem. In that ap­proach, dis­ap­pears when you take it to be 1 in the fi­nal an­swer.

But be­cause the prob­lem here is so triv­ial, there is re­ally no need for a for­mal time-de­pen­dent per­tur­ba­tion ex­pan­sion. In par­tic­u­lar, by as­sump­tion the sys­tem stays close to state , so the co­ef­fi­cient must re­main small. Then the evo­lu­tion equa­tions above show that will hardly change. That al­lows it to be treated as a con­stant in the evo­lu­tion equa­tion for . That then al­lows to be found by sim­ple in­te­gra­tion. The in­te­gra­tion con­stant fol­lows from the con­di­tion that is zero at the ini­tial time. That then gives the re­sult cited in the text.

It may be noted that for the analy­sis to be valid, must be small. That en­sures that is cor­re­spond­ingly small ac­cord­ing to its evo­lu­tion equa­tion. And then the change in from its orig­i­nal value is small of or­der ac­cord­ing to its evo­lu­tion equa­tion. So the as­sump­tion that it is about con­stant in the equa­tion for is ver­i­fied. The er­ror will be of or­der .

To be sure, this does not ver­ify that this er­ror in de­cays to zero when tends to in­fin­ity. But it does, as can be seen from the ex­act so­lu­tion,

By split­ting it up into ranges no larger than and no larger than 1, you can see that the er­ror is never larger than or­der for no larger than 1. And it is of or­der out­side that range.

Fi­nally, con­sider the case that the state can­not just tran­si­tion to one state but to a large num­ber of them, each with its own co­ef­fi­cient . In that case, the in­di­vid­ual con­tri­bu­tions of all these states add up to change . And must def­i­nitely stay ap­prox­i­mately con­stant for the above analy­sis to be valid. For­tu­nately, if you plug the ap­prox­i­mate ex­pres­sions for the into the evo­lu­tion equa­tion for , you can see that stays ap­prox­i­mately con­stant as long as the sum of all the tran­si­tion prob­a­bil­i­ties does. So as long as there is lit­tle prob­a­bil­ity of any tran­si­tion at time , time-de­pen­dent per­tur­ba­tion the­ory should be OK.