D.79 De­riva­tion of per­tur­ba­tion the­ory

This note de­rives the per­tur­ba­tion the­ory re­sults for the so­lu­tion of the eigen­value prob­lem $(H_0+H_1)\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$ where $H_1$ 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

$$H_1 \equiv \varepsilon H_{\varepsilon}$$

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

The idea is now to start with a good eigen­func­tion $\psi_{{\vec n},0}$ of $H_0$, (where good is still to be de­fined), and cor­rect it so that it be­comes an eigen­func­tion of $H$ $\vphantom0\raisebox{1.5pt}{$=$}$ $H_0+H_1$. To do so, both the de­sired en­ergy eigen­func­tion and its en­ergy eigen­value are ex­panded in a power se­ries in terms of $\varepsilon$:

$$\begin{eqnarray*} && \psi_{\vec n}= \psi_{{\vec n},0} + \varepsilon \psi_{{\v... ...arepsilon} + \varepsilon^2 E_{{\vec n},\varepsilon^2} + \ldots \end{eqnarray*}$$

If $\varepsilon$ is a small quan­tity, then $\varepsilon^2$ will be much smaller still, and can prob­a­bly be ig­nored. If not, then surely $\varepsilon^3$ will be so small that it can be ig­nored. A re­sult that for­gets about pow­ers of $\varepsilon$ 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 $\varepsilon$, (note that they must be real whether $\varepsilon$ is pos­i­tive or neg­a­tive), it is much more messy to show that the eigen­func­tions must ex­pand in whole pow­ers. In fact, for de­gen­er­ate en­er­gies $E_{{\vec n},0}$ they only do if you choose good states $\psi_{{\vec n},0}$. 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 $\psi_{{\vec n},0}$ 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,

$$\left(H_0+\varepsilon H_\varepsilon\right)\psi_{\vec n} = E_{\vec n}\psi_{\vec n}$$

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

$$\begin{eqnarray*} &\varepsilon^0:& (H_0-E_{{\vec n},0})\psi_{{\vec n},0} = 0 \\... ...E_{{\vec n},\varepsilon^3}\psi_{{\vec n},0} \\ &\vdots& \cdots \end{eqnarray*}$$

These are the equa­tions to be solved in suc­ces­sion to give the var­i­ous terms in the ex­pan­sion for the wave func­tion $\psi_{\vec n}$ and the en­ergy $E_{\vec n}$. 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

$$(H_0-E_{{\vec n},0})\psi_{{\vec n},\ldots} = \ldots$$

The equa­tions for the un­known func­tions are in terms of the un­per­turbed Hamil­ton­ian $H_0$, with some ad­di­tional but in prin­ci­ple know­able terms.

For dif­fi­cult per­tur­ba­tion prob­lems like you find in en­gi­neer­ing, the use of a small pa­ra­me­ter $\varepsilon$ 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

$$\begin{eqnarray*} && \psi_{\vec n}= \psi_{{\vec n},0} + \psi_{{\vec n},1} + \p... ...ec n}= E_{{\vec n},0} + E_{{\vec n},1} + E_{{\vec n},2} + \ldots \end{eqnarray*}$$

where you are as­sumed to just imag­ine that $\psi_{{\vec n},1}$ and $E_{{\vec n},1}$ are first or­der small, $\psi_{{\vec n},2}$ and $E_{{\vec n},2}$ are “sec­ond or­der small,” etcetera. In those terms, the suc­ces­sive equa­tions to solve are:

$$(H_0-E_{{\vec n},0})\psi_{{\vec n},0} = 0$$ (D.55)

$$(H_0-E_{{\vec n},0})\psi_{{\vec n},1} = - H_1\psi_{{\vec n},0} + E_{{\vec n},1}\psi_{{\vec n},0}$$ (D.56)

$$(H_0-E_{{\vec n},0})\psi_{{\vec n},2} = - H_1\psi_{{\vec n},1} + E_{{\vec n},1}\psi_{{\vec n},1} + E_{{\vec n},2}\psi_{{\vec n},0}$$ (D.57)

$$(H_0-E_{{\vec n},0})\psi_{{\vec n},3} = - H_{1}\psi_{{\vec n},2} ... ...{\vec n},2} + E_{{\vec n},2}\psi_{{\vec n},1} + E_{{\vec n},3}\psi_{{\vec n},0}$$ (D.58)

$$\cdots$$

Now con­sider each of these equa­tions in turn. First, (D.55) is just the Hamil­ton­ian eigen­value prob­lem for $H_0$ and is al­ready sat­is­fied by the cho­sen un­per­turbed so­lu­tion $\psi_{{\vec n},0}$ and its eigen­value $E_{{\vec n},0}$. How­ever, the re­main­ing equa­tions are not triv­ial. To solve them, write their so­lu­tions in terms of the other eigen­func­tions $\psi_{\underline{\vec n},0}$ of the un­per­turbed Hamil­ton­ian $H_0$. In par­tic­u­lar, to solve (D.56), write

$$\psi_{{\vec n},1} = \sum_{\underline{\vec n}\ne{\vec n}} c_{\underline{\vec n},1} \psi_{\underline{\vec n},0}$$

where the co­ef­fi­cients $c_{\underline{\vec n},1}$ are still to be de­ter­mined. The co­ef­fi­cient of $\psi_{{\vec n},0}$ is zero on ac­count of the nor­mal­iza­tion re­quire­ment. (And in fact, it is eas­i­est to take the co­ef­fi­cient of $\psi_{{\vec n},0}$ also zero for $\psi_{{\vec n},2}$, $\psi_{{\vec n},3}$, ..., even if it means that the re­sult­ing wave func­tion will no longer be nor­mal­ized.)

The prob­lem (D.56) be­comes

$$\sum_{\underline{\vec n}\ne{\vec n}} c_{\underline{\vec n}... ...} = - H_1\psi_{{\vec n},0} + E_{{\vec n},1}\psi_{{\vec n},0}$$

where the left hand side was cleaned up us­ing the fact that the $\psi_{\underline{\vec n},0}$ are eigen­func­tions of $H_0$. To get the first or­der en­ergy cor­rec­tion $E_{{\vec n},1}$, the trick is now to take an in­ner prod­uct of the en­tire equa­tion with $\langle\psi_{{\vec n},0}\vert$. Be­cause of the fact that the en­ergy eigen­func­tions of $H_0$ are or­tho­nor­mal, this in­ner prod­uct pro­duces zero in the left hand side, and in the right hand side it pro­duces:

$$0 = - H_{{\vec n}{\vec n},1} + E_{{\vec n},1} \qquad H_{{... ...1} = \langle\psi_{{\vec n},0}\vert H_1\psi_{{\vec n},0}\rangle$$

And that is ex­actly the first or­der cor­rec­tion to the en­ergy claimed in {A.38.1}; $E_{{\vec n},1}$ equals the Hamil­ton­ian per­tur­ba­tion co­ef­fi­cient $H_{{\vec n}{\vec n},1}$. If the prob­lem is not de­gen­er­ate or $\psi_{{\vec n},0}$ is good, that is.

To get the co­ef­fi­cients $c_{\underline{\vec n},1}$, so that you know what is the first or­der cor­rec­tion $\psi_{{\vec n},1}$ to the wave func­tion, just take an in­ner prod­uct with each of the other eigen­func­tions $\langle\psi_{\underline{\vec n},0}\vert$ of $H_0$ 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

$$c_{\underline{\vec n},1} (E_{\underline{\vec n},0} - E_{{\v... ... \psi_{\underline{\vec n},0} \vert H_1\psi_{{\vec n},0}\rangle$$

The co­ef­fi­cients $c_{\underline{\vec n},1}$ can nor­mally be com­puted from this.

Note how­ever that if the prob­lem is de­gen­er­ate, there will be eigen­func­tions $\psi_{\underline{\vec n},0}$ that have the same en­ergy $E_{{\vec n},0}$ as the eigen­func­tion $\psi_{{\vec n},0}$ 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 $\psi_{\vec n}$ of the full Hamil­ton­ian ex­pands in a power se­ries in $\varepsilon$ start­ing from $\psi_{{\vec n},0}$ is un­true. Eigen­func­tion $\psi_{{\vec n},0}$ 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 $E_{{\vec n},0}$, i.e. the sub­ma­trix

$$\mbox{all}\quad H_{{\vec n}_i{\vec n}_j,1} \quad\mbox{with}\quad E_{{\vec n}_i,0}=E_{{\vec n}_j,0}=E_{{\vec n},0}$$

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 $E_{{\vec n},0}$. For each or­tho­nor­mal eigen­vec­tor $(c_1,c_2,\ldots)$ of the sub­ma­trix, there is a cor­re­spond­ing re­place­ment un­per­turbed eigen­func­tion

$$c_1 \psi_{{\vec n}_1,0,{\rm old}} + c_2 \psi_{{\vec n}_2,0,{\rm old}} + \ldots$$

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 $\psi_{{\vec n},0}$ whose en­ergy to cor­rect from among these re­place­ment eigen­func­tions. Choose the first or­der en­ergy change (eigen­value of the sub­ma­trix) $E_{{\vec n},1}$ that is of in­ter­est to you and then choose $\psi_{{\vec n},0}$ as the re­place­ment eigen­func­tion cor­re­spond­ing to a cor­re­spond­ing eigen­vec­tor. If the first or­der en­ergy change $E_{{\vec n},1}$ is not de­gen­er­ate, the eigen­vec­tor is unique, so $\psi_{{\vec n},0}$ 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 $c_{\underline{\vec n},1}$ will be fixed, be­cause the new sub­ma­trix will be a di­ag­o­nal one: $H_{\underline{\vec n}{\vec n},1}$ will be zero when $E_{\underline{\vec n},0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{{\vec n},0}$ and $\underline{\vec n}$ $\raisebox{.2pt}{$\ne$}$ ${\vec n}$. The co­ef­fi­cients $c_{\underline{\vec n},1}$ for which $E_{\underline{\vec n},0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{{\vec n},0}$ re­main in­de­ter­mi­nate at this stage. They will nor­mally be found at a later stage in the ex­pan­sion.

With the co­ef­fi­cients $c_{\underline{\vec n},1}$ as found, or not found, the sum for the first or­der per­tur­ba­tion $\psi_{{\vec n},1}$ in the wave func­tion be­comes

$$\psi_{{\vec n},1} = - \hspace{-5pt} \sum_{E_{\underline{\ve... ...vec n}}} c_{\underline{\vec n},1} \psi_{\underline{\vec n},0}$$

The en­tire process re­peats for higher or­der. In par­tic­u­lar, to sec­ond or­der (D.57) gives, writ­ing $\psi_{{\vec n},2}$ also in terms of the un­per­turbed eigen­func­tions,

$$\begin{eqnarray*} \sum_{\underline{\vec n}} c_{\underline{\vec n},2} (E_{\und... ...) \psi_{\underline{\vec n},0} + E_{{\vec n},2}\psi_{{\vec n},0} \end{eqnarray*}$$

To get the sec­ond or­der con­tri­bu­tion to the en­ergy, take again an in­ner prod­uct with $\langle\psi_{{\vec n},0}\vert$. That pro­duces, again us­ing or­tho­nor­mal­ity, (and di­ag­o­nal­ity of the sub­ma­trix dis­cussed above if de­gen­er­ate),

$$0 = \sum_{E_{\underline{\vec n},0}\ne E_{{\vec n},0}} \fr... ... {E_{\underline{\vec n},0} - E_{{\vec n},0}} + E_{{\vec n},2}$$

This gives the sec­ond or­der change in the en­ergy stated in {A.38.1}, if $\psi_{{\vec n},0}$ is good. Note that since $H_1$ 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 $\langle\psi_{\underline{\vec n},0}\vert$ for which $E_{\underline{\vec n},0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{{\vec n},0}$, the equa­tion can be sat­is­fied through the still in­de­ter­mi­nate $c_{\underline{\vec n},1}$ pro­vided that the cor­re­spond­ing di­ag­o­nal co­ef­fi­cient $H_{\underline{\vec n}\underline{\vec n},1}$ of the di­ag­o­nal­ized sub­ma­trix is un­equal to $E_{{\vec n},1}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $H_{{\vec n}{\vec n},1}$. In other words, pro­vided that the first or­der en­ergy change is not de­gen­er­ate. If that is un­true, the higher or­der sub­ma­trix

$$\mbox{all } \sum_{E_{\underline{\vec n},0}\ne E_{{\vec n},... ...n},0} \quad E_{{\vec n}_i,1}=E_{{\vec n}_j,1}=E_{{\vec n},1}$$

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 $H_0$ are also eigen­func­tions of an­other op­er­a­tor $A$, and $H_1$ com­mutes with $A$. Then you can arrange the eigen­func­tions $\psi_{\underline{\vec n},0}$ into sets that have the same value for the good quan­tum num­ber $a$ of $A$. You can an­a­lyze the per­turbed eigen­func­tions in each of these sets while com­pletely ig­nor­ing the ex­is­tence of eigen­func­tions with dif­fer­ent val­ues for quan­tum num­ber $a$.

To see why, con­sider two ex­am­ple eigen­func­tions $\psi_1$ and $\psi_2$ of $A$ that have dif­fer­ent eigen­val­ues $a_1$ and $a_2$. Since $H_0$ and $H_1$ both com­mute with $A$, their sum $H$ does, so

$$0 = \langle\psi_2\vert(H A - A H)\psi_1\rangle = \langle\p... ...si_1\rangle = (a_1-a_2)\langle\psi_2\vert H\vert\psi_1\rangle$$

and since $a_1-a_2$ is not zero, $\langle\psi_2\vert H\vert\psi_1\rangle$ must be. Now $\langle\psi_2\vert H\vert\psi_1\rangle$ is the amount of eigen­func­tion $\psi_2$ pro­duced by ap­ply­ing $H$ on $\psi_1$. It fol­lows that ap­ply­ing $H$ on an eigen­func­tion with an eigen­value $a_1$ does not pro­duce any eigen­func­tions with dif­fer­ent eigen­val­ues $a$. Thus an eigen­func­tion of $H$ sat­is­fy­ing

$$H \left(\sum_{a=a_1}c_{\vec n}\psi_{{\vec n},0} + \sum_{a\... ...vec n},0} + \sum_{a\ne a_1}c_{\vec n}\psi_{{\vec n},0}\right)$$

can be re­placed by just $\sum_{a=a_1}c_{\vec n}\psi_{{\vec n},0}$, 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 $a_1$ for which the first sum at $\varepsilon$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is in­de­pen­dent of the other eigen­func­tions of $H$.) Re­duce every eigen­func­tion of $H$ to an eigen­func­tion of $A$ in this way. Now the ex­is­tence of eigen­func­tions with dif­fer­ent val­ues of $a$ 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 $a$. If the Hamil­ton­ian also com­mutes with an­other op­er­a­tor $B$ that the $\psi_{{\vec n},0}$ are eigen­func­tions of, the ar­gu­ment re­peats for the sub­sets with a sin­gle value for $b$.

The Hamil­ton­ian per­tur­ba­tion co­ef­fi­cient $\langle\psi_2\vert H_1\vert\psi_1\rangle$ is zero when­ever two good quan­tum num­bers $a_1$ and $a_2$ are un­equal. The rea­son is the same as for $\langle\psi_2\vert H\vert\psi_1\rangle$ above. Only per­tur­ba­tion co­ef­fi­cients for which all good quan­tum num­bers are the same can be nonzero.