A.38 Per­tur­ba­tion The­ory

Most of the time in quan­tum me­chan­ics, ex­act so­lu­tion of the Hamil­ton­ian eigen­value prob­lem of in­ter­est is not pos­si­ble. To deal with that, ap­prox­i­ma­tions are made.

Per­tur­ba­tion the­ory can be used when the Hamil­ton­ian $H$ con­sists of two parts $H_0$ and $H_1$, where the prob­lem for $H_0$ can be solved and where $H_1$ is small. The idea is then to ad­just the found so­lu­tions for the un­per­turbed Hamil­ton­ian $H_0$ so that they be­come ap­prox­i­mately cor­rect for $H_0+H_1$.

This ad­den­dum ex­plains how per­tur­ba­tion the­ory works. It also gives a few sim­ple but im­por­tant ex­am­ples: the he­lium atom and the Zee­man and Stark ef­fects. Ad­den­dum,{A.39} will use the ap­proach to study rel­a­tivis­tic ef­fects on the hy­dro­gen atom.

A.38.1 Ba­sic per­tur­ba­tion the­ory

To use per­tur­ba­tion the­ory, the eigen­func­tions and eigen­val­ues of the un­per­turbed Hamil­ton­ian $H_0$ must be known. These eigen­func­tions will here be in­di­cated as $\psi_{{\vec n},0}$ and the cor­re­spond­ing eigen­val­ues by $E_{{\vec n},0}$. Note the use of the generic ${\vec n}$ to in­di­cate the quan­tum num­bers of the eigen­func­tions. If the ba­sic sys­tem is an hy­dro­gen atom, as is of­ten the case in text­book ex­am­ples, and spin is unim­por­tant, ${\vec n}$ would likely stand for the set of quan­tum num­bers $n$, $l$, and $m$. But for a three-di­men­sion­al har­monic os­cil­la­tor, ${\vec n}$ might stand for the quan­tum num­bers $n_x$, $n_y$, and $n_z$. In a three-di­men­sion­al prob­lem with one spin­less par­ti­cle, it takes three quan­tum num­bers to de­scribe an en­ergy eigen­func­tion. How­ever, which three de­pends on the prob­lem and your ap­proach to it. The ad­di­tional sub­script 0 in $\psi_{{\vec n},0}$ and $E_{{\vec n},0}$ in­di­cates that they ig­nore the per­tur­ba­tion Hamil­ton­ian $H_1$. They are called the un­per­turbed wave func­tions and en­er­gies.

The key to per­tur­ba­tion the­ory are the “Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients” de­fined as

$$\fbox{$\displaystyle H_{\underline{\vec n}{\vec n},1} \equ... ...\underline{\vec n},0} \vert H_1 \psi_{{\vec n},0}\rangle $} %$$ (A.239)

If you can eval­u­ate these for every pair of en­ergy eigen­func­tions, you should be OK. Note that eval­u­at­ing in­ner prod­ucts is just sum­ma­tion or in­te­gra­tion; it is gen­er­ally a lot sim­pler than try­ing to solve the eigen­value prob­lem $\left(H_0+H_1\right)\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$.

In the ap­pli­ca­tion of per­tur­ba­tion the­ory, the idea is to pick one un­per­turbed eigen­func­tion $\psi_{{\vec n},0}$ of $H_0$ of in­ter­est and then cor­rect it to ac­count for $H_1$, and es­pe­cially cor­rect its en­ergy $E_{{\vec n},0}$. Cau­tion! If the en­ergy $E_{{\vec n},0}$ is de­gen­er­ate, i.e. there is more than one un­per­turbed eigen­func­tion $\psi_{{\vec n},0}$ of $H_0$ with that en­ergy, you must use a good eigen­func­tion to cor­rect the en­ergy. How to do that will be dis­cussed in sub­sec­tion A.38.3.

For now just as­sume that the en­ergy is not de­gen­er­ate or that you picked a good eigen­func­tion $\psi_{{\vec n},0}$. Then a first cor­rec­tion to the en­ergy $E_{{\vec n},0}$ to ac­count for the per­tur­ba­tion $H_1$ is very sim­ple, {D.79}; just add the cor­re­spond­ing Hamil­ton­ian per­tur­ba­tion co­ef­fi­cient:

$$\fbox{$\displaystyle E_{\vec n}= E_{{\vec n},0} + H_{{\vec n}{\vec n},1} + \ldots $} %$$ (A.240)

This is a quite user-friendly re­sult, be­cause it only in­volves the se­lected en­ergy eigen­func­tion $\psi_{{\vec n},0}$. The other en­ergy eigen­func­tions are not in­volved. In a nu­mer­i­cal so­lu­tion, you might only have com­puted one state, say the ground state of $H_0$. Then you can use this re­sult to cor­rect the ground state en­ergy for a per­tur­ba­tion even if you do not have data about any other en­ergy states of $H_0$.

Un­for­tu­nately, it does hap­pen quite a lot that the above cor­rec­tion $H_{{\vec n}{\vec n},1}$ is zero be­cause of some sym­me­try or the other. Or it may sim­ply not be ac­cu­rate enough. In that case, to find the en­ergy change you have to use what is called “sec­ond or­der per­tur­ba­tion the­ory:”

$$\fbox{$\displaystyle E_{\vec n}= E_{{\vec n},0} + H_{{\vec... ...rt^2}{E_{\underline{\vec n},0}-E_{{\vec n},0}} + \ldots $} %$$ (A.241)

Now all eigen­func­tions of $H_0$ will be needed, which makes sec­ond or­der the­ory a lot nas­tier. Then again, even if the “first or­der” cor­rec­tion $H_{{\vec n}{\vec n},1}$ to the en­ergy is nonzero, the sec­ond or­der for­mula will likely give a much more ac­cu­rate re­sult.

Some­times you may also be in­ter­ested in what hap­pens to the en­ergy eigen­func­tions, not just the en­ergy eigen­val­ues. The cor­re­spond­ing for­mula is

$$\fbox{$\displaystyle \psi_{\vec n}= \psi_{{\vec n},0} - \... ...underline{\vec n}} \psi_{\underline{\vec n},0} + \ldots $} %$$ (A.242)

That is the first or­der re­sult. The sec­ond sum is zero if the prob­lem is not de­gen­er­ate. Oth­er­wise its co­ef­fi­cients $c_{\underline{\vec n}}$ are de­ter­mined by con­sid­er­a­tions found in de­riva­tion {D.79}.

In some cases, in­stead of us­ing sec­ond or­der the­ory as above, it may be sim­pler to com­pute the first or­der wave func­tion per­tur­ba­tion and the sec­ond or­der en­ergy change from

$$\fbox{$\displaystyle (H_0-E_{{\vec n},0})\psi_{{\vec n},1}... ...c n},0}\vert(H_1-E_{{\vec n},1})\psi_{{\vec n},1}\rangle $} %$$ (A.243)

Eigen­func­tion $\psi_{{\vec n},0}$ must be good. The good news is that this does not re­quire all the un­per­turbed eigen­func­tions. The bad news is that it re­quires so­lu­tion of a non­triv­ial equa­tion in­volv­ing the un­per­turbed Hamil­ton­ian in­stead of just in­te­gra­tion. It may be the best way to pro­ceed for a per­tur­ba­tion of a nu­mer­i­cal so­lu­tion.

One ap­pli­ca­tion of per­tur­ba­tion the­ory is the “Hell­mann-Feyn­man the­o­rem.” Here the per­tur­ba­tion Hamil­ton­ian is an in­fin­i­tes­i­mal change $\partial{H}$ in the un­per­turbed Hamil­ton­ian caused by an in­fin­i­tes­i­mal change in some pa­ra­me­ter that it de­pends on. If the pa­ra­me­ter is called $\lambda$, per­tur­ba­tion the­ory says that the first or­der en­ergy change is

$$\fbox{$\displaystyle \frac{\partial E_{\vec n}}{\partial\l... ...rtial H}{\partial\lambda} \psi_{{\vec n},0} \Bigg\rangle $} %$$ (A.244)

when di­vided by the change in pa­ra­me­ter $\partial\lambda$. If you can fig­ure out the in­ner prod­uct, you can fig­ure out the change in en­ergy. But more im­por­tant is the re­verse: if you can find the de­riv­a­tive of the en­ergy with re­spect to the pa­ra­me­ter, you have the in­ner prod­uct. For ex­am­ple, the Hell­mann-Feyn­man the­o­rem is help­ful for find­ing the ex­pec­ta­tion value of 1$\raisebox{.5pt}{$/$}$​$r^2$ for the hy­dro­gen atom, a nasty prob­lem, {D.83}. Of course, al­ways make sure the eigen­func­tion $\psi_{{\vec n},0}$ is a good one for the de­riv­a­tive of the Hamil­ton­ian.

A.38.2 Ion­iza­tion en­ergy of he­lium

One promi­nent de­fi­ciency in the ap­prox­i­mate analy­sis of the heav­ier atoms in chap­ter 5.9 was the poor ion­iza­tion en­ergy that it gave for he­lium. The pur­pose of this ex­am­ple is to de­rive a much more rea­son­able value us­ing per­tur­ba­tion the­ory.

Ex­actly speak­ing, the ion­iza­tion en­ergy is the dif­fer­ence be­tween the en­ergy of the he­lium atom with both its elec­trons in the ground state and the he­lium ion with its sec­ond elec­tron re­moved. Now the en­ergy of the he­lium ion with elec­tron 2 re­moved is easy; the Hamil­ton­ian for the re­main­ing elec­tron 1 is

$$H_{\rm He ion} = \mbox{} - \frac{\hbar^2}{2m_{\rm e}} \nabla_1^2 - 2 \frac{e^2}{4\pi\epsilon_0} \frac{1}{r_1}$$

where the first term rep­re­sents the ki­netic en­ergy of the elec­tron and the sec­ond its at­trac­tion to the two-pro­ton nu­cleus. The he­lium nu­cleus nor­mally also con­tains two neu­trons, but they do not at­tract the elec­tron.

This Hamil­ton­ian is ex­actly the same as the one for the hy­dro­gen atom in chap­ter 4.3, ex­cept that it has $2e^2$ where the hy­dro­gen one, with just one pro­ton in its nu­cleus, has $e^2$. So the so­lu­tion for the he­lium ion is sim­ple: just take the hy­dro­gen so­lu­tion, and every­where where there is an $e^2$ in that so­lu­tion, re­place it by $2e^2$. In par­tic­u­lar, the Bohr ra­dius $a$ for the he­lium ion is half the Bohr ra­dius $a_0$ for hy­dro­gen,

$$a=\frac{4\pi\epsilon_0\hbar^2}{m_{\rm e}2e^2} = {\textstyle\frac{1}{2}} a_0$$

and so its en­ergy and wave func­tion be­come

$$E_{\rm gs,ion} = - \frac{\hbar^2}{2 m_{\rm e}a^2} = 4 E_1 ... ...rm gs,ion}({\skew0\vec r}) = \frac{1}{\sqrt{\pi a^3}} e^{-r/a}$$

where $E_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$13.6 eV is the en­ergy of the hy­dro­gen atom.

It is in­ter­est­ing to see that the he­lium ion has four times the en­ergy of the hy­dro­gen atom. The rea­sons for this much higher en­ergy are both that the nu­cleus is twice as strong, and that the elec­tron is twice as close to it: the Bohr ra­dius is half the size. More gen­er­ally, in heavy atoms the elec­trons that are poorly shielded from the nu­cleus, which means the in­ner elec­trons, have en­er­gies that scale with the square of the nu­clear strength. For such elec­trons, rel­a­tivis­tic ef­fects are much more im­por­tant than they are for the elec­tron in a hy­dro­gen atom.

The neu­tral he­lium atom is not by far as easy to an­a­lyze as the ion. Its Hamil­ton­ian is, from (5.34):

$$H_{\rm He} = \mbox{} - \frac{\hbar^2}{2m_{\rm e}} \nabla_... ...psilon_0} \frac1{\vert{\skew0\vec r}_2 -{\skew0\vec r}_1\vert}$$

The first two terms are the ki­netic en­ergy and nu­clear at­trac­tion of elec­tron 1, and the next two the same for elec­tron 2. The fi­nal term is the elec­tron to elec­tron re­pul­sion, the curse of quan­tum me­chan­ics. This fi­nal term is the rea­son that the ground state of he­lium can­not be found an­a­lyt­i­cally.

Note how­ever that the re­pul­sion term is qual­i­ta­tively sim­i­lar to the nu­clear at­trac­tion terms, ex­cept that there are four of these nu­clear at­trac­tion terms ver­sus a sin­gle re­pul­sion term. So maybe then, it may work to treat the re­pul­sion term as a small per­tur­ba­tion, call it $H_1$, to the Hamil­ton­ian $H_0$ given by the first four terms? Of course, if you ask math­e­mati­cians whether 25% is a small amount, they are go­ing to ve­he­mently deny it; but then, so they would for any amount if there is no limit process in­volved, so just don’t ask them, OK?

The so­lu­tion of the eigen­value prob­lem $H_0\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$ is sim­ple: since the elec­trons do not in­ter­act with this Hamil­ton­ian, the ground state wave func­tion is the prod­uct of the ground state wave func­tions for the in­di­vid­ual elec­trons, and the en­ergy is the sum of their en­er­gies. And the wave func­tions and en­er­gies for the sep­a­rate elec­trons are given by the so­lu­tion for the ion above, so

$$\psi_{\rm gs,0} = \frac{1}{\pi a^3} e^{-(r_1+r_2)/a} \qquad E_{\rm gs,0} = 8 E_1$$

Ac­cord­ing to this re­sult, the en­ergy of the atom is $8E_1$ while the ion had $4E_1$, so the ion­iza­tion en­ergy would be $4\vert E_1\vert$, or 54.4 eV. Since the ex­per­i­men­tal value is 24.6 eV, this is no bet­ter than the 13.6 eV chap­ter 5.9 came up with.

To get a bet­ter ion­iza­tion en­ergy, try per­tur­ba­tion the­ory. Ac­cord­ing to first or­der per­tur­ba­tion the­ory, a bet­ter value for the en­ergy of the hy­dro­gen atom should be

$$E_{\rm gs} = E_{\rm gs,0} + \langle\psi_{\rm gs,0}\vert H_1\psi_{\rm gs,0}\rangle$$

or sub­sti­tut­ing in from above,

$$E_{\rm gs} = 8 E_1 + \frac{e^2}{4\pi\epsilon_0} \bigg\lang... ...0\vec r}_1\vert}\frac{1}{\pi a^3} e^{-(r_1+r_2)/a}\bigg\rangle$$

The in­ner prod­uct of the fi­nal term can be writ­ten out as

$$\frac{e^2}{4\pi\epsilon_0} \frac{1}{\pi^2a^6} \int_{{\rm a... ..._1\vert} { \rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}_2$$

This in­te­gral can be done an­a­lyt­i­cally. Try it, if you are so in­clined; in­te­grate ${\rm d}^3{\skew0\vec r}_1$ first, us­ing spher­i­cal co­or­di­nates with ${\skew0\vec r}_2$ as their axis and do­ing the az­imuthal and po­lar an­gles first. Be care­ful, $\sqrt{(r_1-r_2)^2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert r_1-r_2\vert$, not $r_1-r_2$, so you will have to in­te­grate $r_1$ $\raisebox{.3pt}{$<$}$ $r_2$ and $r_1$ $\raisebox{.3pt}{$>$}$ $r_2$ sep­a­rately in the fi­nal in­te­gra­tion over ${\rm d}{r}_1$. Then in­te­grate ${\rm d}^3{\skew0\vec r}_2$.

The re­sult of the in­te­gra­tion is

$$\frac{e^2}{4\pi\epsilon_0} \bigg\langle\frac{1}{\pi a^3}e^... ...^2}{4\pi\epsilon_0} \frac{5}{8a} = \frac{5}{2} \vert E_1\vert$$

There­fore, the he­lium atom en­ergy in­creases by $2.5\vert E_1\vert$ due to the elec­tron re­pul­sion, and with it, the ion­iza­tion en­ergy de­creases to $1.5\vert E_1\vert$, or 20.4 eV. It is not 24.6 eV, but it is clearly much more rea­son­able than 54 or 13.6 eV were.

The sec­ond or­der per­tur­ba­tion re­sult should give a much more ac­cu­rate re­sult still. How­ever, if you did the in­te­gral above, you may feel lit­tle in­cli­na­tion to try the ones in­volv­ing all pos­si­ble prod­ucts of hy­dro­gen en­ergy eigen­func­tions.

In­stead, the re­sult can be im­proved us­ing a vari­a­tional ap­proach, like the ones that were used ear­lier for the hy­dro­gen mol­e­cule and mol­e­c­u­lar ion, and this re­quires al­most no ad­di­tional work. The idea is to ac­cept the hint from per­tur­ba­tion the­ory that the wave func­tion of he­lium can be ap­prox­i­mated as $\psi_a({\skew0\vec r}_1)\psi_a({\skew0\vec r}_2)$ where $\psi_a$ is the hy­dro­gen ground state wave func­tion us­ing a mod­i­fied Bohr ra­dius $a$ in­stead of $a_0$:

$$\psi_{\rm gs} = \psi_a({\skew0\vec r}_1)\psi_a({\skew0\vec ... ...psi_a({\skew0\vec r}) \equiv \frac{1}{\sqrt{\pi a^3}} e^{-r/a}$$

How­ever, in­stead of ac­cept­ing the per­tur­ba­tion the­ory re­sult that $a$ should be half the nor­mal Bohr ra­dius $a_0$, let $a$ be op­ti­mized to make the ex­pec­ta­tion en­ergy for the ground state

$$E_{\rm gs} = \langle\psi_{\rm gs}\vert H_{\rm He}\psi_{\rm gs}\rangle$$

as small as pos­si­ble. This will pro­duce the most ac­cu­rate ground state en­ergy pos­si­ble for a ground state wave func­tion of this form, guar­an­teed no worse than as­sum­ing that $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12a_0$, and prob­a­bly bet­ter.

No new in­te­grals need to be done to eval­u­ate the in­ner prod­uct above. In­stead, not­ing that for the hy­dro­gen atom ac­cord­ing to the vir­ial the­o­rem of chap­ter 7.2 the ex­pec­ta­tion ki­netic en­ergy equals $-E_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2$$\raisebox{.5pt}{$/$}$​$2{m_{\rm e}}a_0^2$ and the po­ten­tial en­ergy equals $2E_1$, two of the needed in­te­grals can be in­ferred from the hy­dro­gen so­lu­tion: chap­ter 4.3,

$$\bigg\langle\psi_a\bigg\vert{-}\frac{\hbar^2}{2m_{\rm e}}\nabla^2\psi_a\bigg\rangle = \frac{\hbar^2}{2m_{\rm e}a^2}$$

$$- \frac{e^2}{4\pi\epsilon_0} \bigg\langle\psi_a\bigg\vert\... ...si_a\bigg\rangle = - \frac{\hbar^2}{m_{\rm e}a_0} \frac{1}{a}$$

and this sub­sec­tion added

$$\bigg\langle\psi_a\psi_a\bigg\vert\frac{1}{\vert{\skew0\vec... ...{\skew0\vec r}_1\vert}\psi_a\psi_a\bigg\rangle = \frac{5}{8a}$$

Us­ing these re­sults with the he­lium Hamil­ton­ian, the ex­pec­ta­tion en­ergy of the he­lium atom can be writ­ten out to be

$$\langle\psi_a\psi_a\vert H_{\rm He}\psi_a\psi_a\rangle = \... ...^2}{m_{\rm e}a^2} - \frac{27}{8} \frac{\hbar^2}{m_{\rm e}a_0a}$$

Set­ting the de­riv­a­tive with re­spect to $a$ to zero lo­cates the min­i­mum at $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac{16}{27}a_0$, rather than $\frac12a_0$. Then the cor­re­spond­ing ex­pec­ta­tion en­ergy is $\vphantom{0}\raisebox{1.5pt}{$-$}$$3^6\hbar^2$$\raisebox{.5pt}{$/$}$​$2^8{m_{\rm e}}a_0^2$, or $3^6E_1$$\raisebox{.5pt}{$/$}$​$2^7$. Putting in the num­bers, the ion­iza­tion en­ergy is now found as 23.1 eV, in quite good agree­ment with the ex­per­i­men­tal 24.6 eV.

A.38.3 De­gen­er­ate per­tur­ba­tion the­ory

En­ergy eigen­val­ues are de­gen­er­ate if there is more than one in­de­pen­dent eigen­func­tion with that en­ergy. Now, if you try to use per­tur­ba­tion the­ory to cor­rect a de­gen­er­ate eigen­value of a Hamil­ton­ian $H_0$ for a per­tur­ba­tion $H_1$, there may be a prob­lem. As­sume that there are $d$ $\raisebox{.3pt}{$>$}$ 1 in­de­pen­dent eigen­func­tions with en­ergy $E_{{\vec n},0}$ and that they are num­bered as

$$\psi_{{\vec n}_1,0},\psi_{{\vec n}_2,0},\ldots,\psi_{{\vec n}_d,0}$$

Then as far as $H_0$ is con­cerned, any com­bi­na­tion

$$\psi_{{\vec n},0} = c_1 \psi_{{\vec n}_1,0} + c_2\psi_{{\vec n}_2,0} + \ldots + c_d\psi_{{\vec n}_d,0}$$

with ar­bi­trary co­ef­fi­cients $c_1,c_2,\ldots,c_d$, (not all zero, of course), is just as good an eigen­func­tion with en­ergy $E_{{\vec n},0}$ as any other.

Un­for­tu­nately, the full Hamil­ton­ian $H_0+H_1$ is not likely to agree with $H_0$ about that. As far as the full Hamil­ton­ian is con­cerned, nor­mally only very spe­cific com­bi­na­tions are ac­cept­able, the good eigen­func­tions. It is said that the per­tur­ba­tion $H_1$ breaks the de­gen­er­acy of the en­ergy eigen­value. The sin­gle en­ergy eigen­value splits into sev­eral eigen­val­ues of dif­fer­ent en­ergy. Only good com­bi­na­tions will show up these changed en­er­gies; the bad ones will pick up un­cer­tainty in en­ergy that hides the ef­fect of the per­tur­ba­tion.

The var­i­ous ways of en­sur­ing good eigen­func­tions are il­lus­trated in the fol­low­ing sub­sec­tions for ex­am­ple per­tur­ba­tions of the en­ergy lev­els of the hy­dro­gen atom. Re­call that the un­per­turbed en­ergy eigen­func­tions of the hy­dro­gen atom elec­tron, as de­rived in chap­ter 4.3, and also in­clud­ing spin, are given as $\psi_{nlm}{\uparrow}$ and $\psi_{nlm}{\downarrow}$. They are highly de­gen­er­ate: all the eigen­func­tions with the same value of $n$ have the same en­ergy $E_n$, re­gard­less of what is the value of the az­imuthal quan­tum num­ber 0 $\raisebox{-.3pt}{$\leqslant$}$ $l$ $\raisebox{-.3pt}{$\leqslant$}$ $n-1$ cor­re­spond­ing to the square or­bital an­gu­lar mo­men­tum $L^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l(l+1)\hbar^2$; re­gard­less of what is the mag­netic quan­tum num­ber $\vert m\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $l$ cor­re­spond­ing to the or­bital an­gu­lar mo­men­tum $L_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m\hbar$ in the $z$-​di­rec­tion; and re­gard­less of what is the spin quan­tum num­ber $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm\frac12$ cor­re­spond­ing to the spin an­gu­lar mo­men­tum $m_s\hbar$ in the $z$-​di­rec­tion. In par­tic­u­lar, the ground state en­ergy level $E_1$ is two-fold de­gen­er­ate, it is the same for both $\psi_{100}{\uparrow}$, i.e. $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$ and $\psi_{100}{\downarrow}$, $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\frac12$. The next en­ergy level $E_2$ is eight-fold de­gen­er­ate, it is the same for $\psi_{200}{\updownarrow}$, $\psi_{211}{\updownarrow}$, $\psi_{210}{\updownarrow}$, and $\psi_{21{-1}}{\updownarrow}$, and so on for higher val­ues of $n$.

There are two im­por­tant rules to iden­tify the good eigen­func­tions, {D.79}:

1.

Look for good quan­tum num­bers. The quan­tum num­bers that make the en­ergy eigen­func­tions of the un­per­turbed Hamil­ton­ian $H_0$ unique cor­re­spond to the eigen­val­ues of ad­di­tional op­er­a­tors be­sides the Hamil­ton­ian. If the per­tur­ba­tion Hamil­ton­ian $H_1$ com­mutes with one of these ad­di­tional op­er­a­tors, the cor­re­spond­ing quan­tum num­ber is good. You do not need to com­bine eigen­func­tions with dif­fer­ent val­ues of that quan­tum num­ber.

In par­tic­u­lar, if the per­tur­ba­tion Hamil­ton­ian com­mutes with all ad­di­tional op­er­a­tors that make the eigen­func­tions of $H_0$ unique, stop wor­ry­ing: every eigen­func­tion is good al­ready.

For ex­am­ple, for the usual hy­dro­gen en­ergy eigen­func­tions $\psi_{nlm}{\updownarrow}$, the quan­tum num­bers $l$, $m$, and $m_s$ make the eigen­func­tions at a given un­per­turbed en­ergy level $n$ unique. They cor­re­spond to the op­er­a­tors $\L ^2$, $\L _z$, and ${\widehat S}_z$. If the per­tur­ba­tion Hamil­ton­ian $H_1$ com­mutes with any one of these op­er­a­tors, the cor­re­spond­ing quan­tum num­ber is good. If the per­tur­ba­tion com­mutes with all three, all eigen­func­tions are good al­ready.

2.

Even if some quan­tum num­bers are bad be­cause the per­tur­ba­tion does not com­mute with that op­er­a­tor, eigen­func­tions are still good if there are no other eigen­func­tions with the same un­per­turbed en­ergy and the same good quan­tum num­bers.

Oth­er­wise lin­ear al­ge­bra is re­quired. For each set of en­ergy eigen­func­tions

$$\psi_{{\vec n}_1,0},\psi_{{\vec n}_2,0},\ldots$$

with the same un­per­turbed en­ergy and the same good quan­tum num­bers, but dif­fer­ent bad ones, form the ma­trix of Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients

$$\left( \begin{array}{ccc} \langle\psi_{{\vec n}_1,0}\vert... ... & \cdots \\ \vdots & \vdots & \ddots \end{array} \right)$$

The eigen­val­ues of this ma­trix are the first or­der en­ergy cor­rec­tions. Also, the co­ef­fi­cients $c_1,c_2,\ldots$ of each good eigen­func­tion

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

must be an eigen­vec­tor of the ma­trix.

Un­for­tu­nately, if the eigen­val­ues of this ma­trix are not all dif­fer­ent, the eigen­vec­tors are not unique, so you re­main un­sure about what are the good eigen­func­tions. In that case, if the sec­ond or­der en­ergy cor­rec­tions are needed, the de­tailed analy­sis of de­riva­tion {D.79} will need to be fol­lowed.

If you are not fa­mil­iar with lin­ear al­ge­bra at all, in all cases men­tioned here the ma­tri­ces are just two by two, and you can find that so­lu­tion spelled out in the no­ta­tions un­der eigen­vec­tor.

The fol­low­ing, re­lated, prac­ti­cal ob­ser­va­tion can also be made:

Hamil­ton­ian per­tur­ba­tion co­ef­fi­cients can only be nonzero if all the good quan­tum num­bers are the same.

A.38.4 The Zee­man ef­fect

If you put an hy­dro­gen atom in an ex­ter­nal mag­netic field $\skew2\vec{\cal B}_{\rm {ext}}$, the en­ergy lev­els of the elec­tron change. That is called the Zee­man ef­fect.

If for sim­plic­ity a co­or­di­nate sys­tem is used with its $z$-​axis aligned with the mag­netic field, then ac­cord­ing to chap­ter 13.4, the Hamil­ton­ian of the hy­dro­gen atom ac­quires an ad­di­tional term

$$H_1 = \frac{e}{2m_{\rm e}}{\cal B}_{\rm ext}\left(\L _z + 2 {\widehat S}_z\right)$$ (A.245)

be­yond the ba­sic hy­dro­gen atom Hamil­ton­ian $H_0$ of chap­ter 4.3.1. Qual­i­ta­tively, it ex­presses that a spin­ning charged par­ti­cle is equiv­a­lent to a tiny elec­tro­mag­net, and a mag­net wants to align it­self with a mag­netic field, just like a com­pass nee­dle aligns it­self with the mag­netic field of earth.

For this per­tur­ba­tion, the $\psi_{nml}{\updownarrow}$ en­ergy eigen­func­tions are al­ready good ones, be­cause $H_1$ com­mutes with all of $\L ^2$, $\L _z$ and ${\widehat S}_z$. So, ac­cord­ing to per­tur­ba­tion the­ory, the en­ergy eigen­val­ues of an hy­dro­gen atom in a mag­netic field are ap­prox­i­mately

$$E_n + \langle \psi_{nml}{\updownarrow}\vert H_1\vert\psi_{n... ... E_n + \frac{e}{2m_{\rm e}}{\cal B}_{\rm ext}(m + 2 m_s)\hbar$$

Ac­tu­ally, this is not ap­prox­i­mate at all; it is the ex­act eigen­value of $H_0+H_1$ cor­re­spond­ing to the ex­act eigen­func­tion $\psi_{nml}{\updownarrow}$.

The Zee­man ef­fect can be seen in an ex­per­i­men­tal spec­trum. Con­sider first the ground state. If there is no elec­tro­mag­netic field, the two ground states $\psi_{100}{\uparrow}$ and $\psi_{100}{\downarrow}$ would have ex­actly the same en­ergy. There­fore, in an ex­per­i­men­tal spec­trum, they would show up as a sin­gle line. But with the mag­netic field, the two en­ergy lev­els are dif­fer­ent,

$$E_{100{\downarrow}} = E_1 - \frac{e\hbar}{2m_{\rm e}}{\cal ... ...}{2m_{\rm e}}{\cal B}_{\rm ext} \qquad E_1 = - 13.6\;{\rm eV}$$

so the sin­gle line splits into two! Do note that the en­ergy change due to even an ex­tremely strong mag­netic field of 100 Tesla is only 0.006 eV or so, chap­ter 13.4, so it is not like the spec­trum would be­come un­rec­og­niz­able. The sin­gle spec­tral line of the eight $\psi_{2lm}{\updownarrow}$ L shell states will sim­i­larly split in five closely spaced but sep­a­rate lines, cor­re­spond­ing to the five pos­si­ble val­ues $\vphantom{0}\raisebox{1.5pt}{$-$}$2, $\vphantom{0}\raisebox{1.5pt}{$-$}$1, 0, 1 and 2 for the fac­tor $m+2m_s$ above.

Some dis­claimers should be given here. First of all, the 2 in $m+2m_s$ is only equal to 2 up to about 0.1% ac­cu­racy. More im­por­tantly, even in the ab­sence of a mag­netic field, the en­ergy lev­els at a given value of $n$ do not re­ally form a sin­gle line in the spec­trum if you look closely enough. There are small er­rors in the so­lu­tion of chap­ter 4.3 due to rel­a­tivis­tic ef­fects, and so the the­o­ret­i­cal lines are al­ready split. That is dis­cussed in ad­den­dum {A.39}. The de­scrip­tion given above is a good one for the strong Zee­man ef­fect, in which the mag­netic field is strong enough to swamp the rel­a­tivis­tic er­rors.

A.38.5 The Stark ef­fect

If an hy­dro­gen atom is placed in an ex­ter­nal elec­tric field $\skew3\vec{\cal E}_{\rm {ext}}$ in­stead of the mag­netic one of the pre­vi­ous sub­sec­tion, its en­ergy lev­els will change too. That is called the Stark ef­fect. Of course a Zee­man, Dutch for sea-man, would be most in­ter­ested in mag­netic fields. A Stark, maybe in a spark? (Apolo­gies.)

If the $z$-​axis is taken in the di­rec­tion of the elec­tric field, the con­tri­bu­tion of the elec­tric field to the Hamil­ton­ian is given by:

$$H_1 = e{\cal E}_{\rm ext}z$$ (A.246)

It is much like the po­ten­tial en­ergy $mgh$ of grav­ity, with the elec­tron charge $e$ tak­ing the place of the mass $m$, ${\cal E}_{\rm {ext}}$ that of the grav­ity strength $g$, and $z$ that of the height $h$.

Since the typ­i­cal mag­ni­tude of $z$ is of the or­der of a Bohr ra­dius $a_0$, you would ex­pect that the en­ergy lev­els will change due to the elec­tric field by an amount of rough size $e{\cal E}_{\rm {ext}}a_0$. A strong lab­o­ra­tory elec­tric field might have $e{\cal E}_{\rm {ext}}a_0$ of the or­der of 0.000 5 eV, [25, p. 339]. That is re­ally small com­pared to the typ­i­cal elec­tron en­ergy lev­els.

And ad­di­tion­ally, it turns out that for many eigen­func­tions, in­clud­ing the ground state, the first or­der cor­rec­tion to the en­ergy is zero. To get the en­ergy change in that case, you need to com­pute the sec­ond or­der term, which is a pain. And that term will be much smaller still than even $e{\cal E}_{\rm {ext}}a_0$ for rea­son­able field strengths.

Now first sup­pose that you ig­nore the warn­ings on good eigen­func­tions, and just com­pute the en­ergy changes us­ing the in­ner prod­uct $\langle\psi_{nlm}{\updownarrow}\vert H_1\psi_{nlm}{\updownarrow}\rangle$. You will then find that this in­ner prod­uct is zero for what­ever en­ergy eigen­func­tion you take:

$$\langle\psi_{nlm}{\updownarrow}\vert e{\cal E}_{\rm ext}z\p... ...pdownarrow}\rangle=0 \mbox{ for all $n$, $l$, $m$, and $m_s$}$$

The rea­son is that neg­a­tive $z$ val­ues in­te­grate away against pos­i­tive ones. (The in­ner prod­ucts are in­te­grals of $z$ times $\vert\psi_{nlm}\vert^2$, and $\vert\psi_{nlm}\vert^2$ is the same at op­po­site sides of the nu­cleus while $z$ changes sign, so the con­tri­bu­tions of op­po­site sides to the in­ner prod­uct pair­wise can­cel.)

So, since all first or­der en­ergy changes that you com­pute are zero, you would nat­u­rally con­clude that to first or­der ap­prox­i­ma­tion none of the en­ergy lev­els of a hy­dro­gen atom changes due to the elec­tric field. But that con­clu­sion is wrong for any­thing but the ground state en­ergy. And the rea­son it is wrong is be­cause the good eigen­func­tions have not been used.

Con­sider the op­er­a­tors $\L ^2$, $\L _z$, and $S_z$ that make the en­ergy eigen­func­tions $\psi_{nlm}{\updownarrow}$ unique. If $H_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e{\cal E}_{\rm {ext}}z$ com­muted with them all, the $\psi_{nlm}{\updownarrow}$ would be good eigen­func­tions. Un­for­tu­nately, while $z$ com­mutes with $\L _z$ and $S_z$, it does not com­mute with $\L ^2$, see chap­ter 4.5.4. The quan­tum num­ber $l$ is bad.

Still, the two states $\psi_{100}{\updownarrow}$ with the ground state en­ergy are good states, be­cause there are no states with the same en­ergy and a dif­fer­ent value of the bad quan­tum num­ber $l$. Re­ally, spin has noth­ing to do with the Stark prob­lem. If you want, you can find the purely spa­tial en­ergy eigen­func­tions first, then for every spa­tial eigen­func­tion, there will be one like that with spin up and one with spin down. In any case, since the two eigen­func­tions $\psi_{100}{\updownarrow}$ are both good, the ground state en­ergy does in­deed not change to first or­der.

But now con­sider the eight-fold de­gen­er­ate $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 en­ergy level. Each of the four eigen­func­tions $\psi_{211}{\updownarrow}$ and $\psi_{21{-1}}{\updownarrow}$ is a good one be­cause for each of them, there is no other $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 eigen­func­tion with a dif­fer­ent value of the bad quan­tum num­ber $l$. The en­er­gies cor­re­spond­ing to these good eigen­func­tions too do in­deed not change to first or­der.

How­ever, the re­main­ing two $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 spin-up states $\psi_{200}{\uparrow}$ and $\psi_{210}{\uparrow}$ have dif­fer­ent val­ues for the bad quan­tum num­ber $l$, and they have the same val­ues $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}$ for the good quan­tum num­bers of or­bital and spin $z$-​mo­men­tum. These eigen­func­tions are bad and will have to be com­bined to pro­duce good ones. And sim­i­larly the re­main­ing two spin-down states $\psi_{200}{\downarrow}$ and $\psi_{210}{\downarrow}$ are bad and will have to be com­bined.

It suf­fices to just an­a­lyze the spin up states, be­cause the spin down ones go ex­actly the same way. The co­ef­fi­cients $c_1$ and $c_2$ of the good com­bi­na­tions $c_1\psi_{200}{\uparrow}+c_2\psi_{210}{\uparrow}$ must be eigen­vec­tors of the ma­trix

$$\left( \begin{array}{cc} \langle\psi_{200}{\uparrow}\vert... ...w}\rangle \end{array} \right) \qquad H_1=e{\cal E}_{\rm ext}z$$

The di­ag­o­nal el­e­ments of this ma­trix (top left cor­ner and bot­tom right cor­ner) are zero be­cause of can­cel­la­tion of neg­a­tive and pos­i­tive $z$ val­ues as dis­cussed above. And the top right and bot­tom left el­e­ments are com­plex con­ju­gates, (2.16), so only one of them needs to be ac­tu­ally com­puted. And the spin part of the in­ner prod­uct pro­duces one and can there­fore be ig­nored. What is left is a mat­ter of find­ing the two spa­tial eigen­func­tions in­volved ac­cord­ing to (4.36), look­ing up the spher­i­cal har­mon­ics in ta­ble 4.2 and the ra­dial func­tions in ta­ble 4.4, and in­te­grat­ing it all against $e{\cal E}_{\rm {ext}}z$. The re­sult­ing ma­trix is

$$\left( \begin{array}{cc} 0 & - 3 e{\cal E}_{\rm ext}a_0 \\ - 3 e{\cal E}_{\rm ext}a_0 & 0 \end{array} \right)$$

The eigen­vec­tors of this ma­trix are sim­ple enough to guess; they have ei­ther equal or op­po­site co­ef­fi­cients $c_1$ and $c_2$:

$$\begin{eqnarray*} & \displaystyle \left( \begin{array}{cc} 0 & - 3 e{\cal E... ...n{array}{c} \sqrt{\frac12} \ -\sqrt{\frac12} \end{array}\right) \end{eqnarray*}$$

If you want to check these ex­pres­sions, note that the prod­uct of a ma­trix times a vec­tor is found by tak­ing dot prod­ucts be­tween the rows of the ma­trix and the vec­tor. It fol­lows that the good com­bi­na­tion $\sqrt{\frac12}\psi_{200}{\uparrow}+\sqrt{\frac12}\psi_{210}{\uparrow}$ has a first or­der en­ergy change $-3e{\cal E}_{\rm {ext}}a_0$, and the good com­bi­na­tion $\sqrt{\frac12}\psi_{200}{\uparrow}-\sqrt{\frac12}\psi_{210}{\uparrow}$ has $+3e{\cal E}_{\rm {ext}}a_0$. The same ap­plies for the spin down states. It fol­lows that to first or­der the $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 level splits into three, with en­er­gies $E_2-3e{\cal E}_{\rm {ext}}a_0$, $E_2$, and $E_2+3e{\cal E}_{\rm {ext}}a_0$, where the value $E_2$ ap­plies to the eigen­func­tions $\psi_{211}{\updownarrow}$ and $\psi_{21{-1}}{\updownarrow}$ that were al­ready good. The con­clu­sion, based on the wrong eigen­func­tions, that the en­ergy lev­els do not change was all wrong.

Re­mark­ably, the good com­bi­na­tions of $\psi_{200}$ and $\psi_{210}$ are the sp hy­brids of car­bon fame, as de­scribed in chap­ter 5.11.4. Note from fig­ure 5.13 in that sec­tion that these hy­brids do not have the same mag­ni­tude at op­po­site sides of the nu­cleus. They have an in­trin­sic “elec­tric di­pole mo­ment,” with the charge shifted to­wards one side of the atom, and the elec­tron then wants to align this di­pole mo­ment with the am­bi­ent elec­tric field. That is much like in Zee­man split­ting, where elec­tron wants to align its or­bital and spin mag­netic di­pole mo­ments with the am­bi­ent mag­netic field.

The cru­cial thing to take away from all this is: al­ways, al­ways, check whether the eigen­func­tion is good be­fore ap­ply­ing per­tur­ba­tion the­ory.

It is ob­vi­ously some­what dis­ap­point­ing that per­tur­ba­tion the­ory did not give any in­for­ma­tion about the en­ergy change of the ground state be­yond the fact that it is sec­ond or­der, i.e. very small com­pared to $e{\cal E}_{\rm {ext}}a_0$. You would like to know ap­prox­i­mately what it is, not just that it is very small. Of course, now that it is es­tab­lished that $\psi_{100}{\uparrow}$ is a good state with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\textstyle\frac{1}{2}}$, you could think about eval­u­at­ing the sec­ond or­der en­ergy change (A.241), by in­te­grat­ing $\langle\psi_{100}{\uparrow}\vert e{\cal E}_{\rm {ext}}z\psi_{nl0}{\uparrow}\rangle$ for all val­ues of $n$ and $l$. But af­ter re­fresh­ing your mem­ory about the an­a­lyt­i­cal ex­pres­sion (D.8) for the $\psi_{nlm}$, you might think again.

It is how­ever pos­si­ble to find the per­tur­ba­tion in the wave func­tion from the al­ter­nate ap­proach (A.243), {D.80}. In that way the sec­ond or­der ground state en­ergy is found to be

$$E_{100} = E_1 - \frac{3e{\cal E}_{\rm {ext}}a_0}{8\vert E_1\vert} 3e{\cal E}_{\rm {ext}}a_0 \qquad E_1 = - 13.6\;{\rm eV}$$

Note that the atom likes an elec­tric field: it low­ers its ground state en­ergy. Also note that the en­ergy change is in­deed sec­ond or­der; it is pro­por­tional to the square of the elec­tric field strength. You can think of the at­trac­tion of the atom to the elec­tric field as a two-stage process: first the elec­tric field po­lar­izes the atom by dis­tort­ing its ini­tially sym­met­ric charge dis­tri­b­u­tion. Then it in­ter­acts with this po­lar­ized atom in much the same way that it in­ter­acts with the sp hy­brids. But since the po­lar­iza­tion is now only pro­por­tional to the field strength, the net en­ergy drop is pro­por­tional to the square of the field strength.

Fi­nally, note that the typ­i­cal value of 0.000 5 eV or so for $e{\cal E}_{\rm {ext}}a_0$ quoted ear­lier is very small com­pared to the about 100 eV for $8\vert E_1\vert$, mak­ing the frac­tion in the ex­pres­sion above very small. So, in­deed the sec­ond or­der change in the ground state en­ergy $E_1$ is much smaller than the first or­der en­ergy changes $\pm3e{\cal E}_{\rm {ext}}a_0$ in the $E_2$ en­ergy level.

A weird pre­dic­tion of quan­tum me­chan­ics is that the elec­tron will even­tu­ally es­cape from the atom, leav­ing it ion­ized. The rea­son is that the po­ten­tial is lin­ear in $z$, so if the elec­tron goes out far enough in the $z$-​di­rec­tion, it will even­tu­ally en­counter po­ten­tial en­er­gies that are lower than the one it has in the atom. Of course, to get at such large val­ues of $z$, the elec­tron must pass po­si­tions where the re­quired en­ergy far ex­ceeds the $\vphantom{0}\raisebox{1.5pt}{$-$}$13.6 eV it has avail­able, and that is im­pos­si­ble for a clas­si­cal par­ti­cle. How­ever, in quan­tum me­chan­ics the po­si­tion of the elec­tron is un­cer­tain, and the elec­tron does have some minis­cule chance of tun­nel­ing out of the atom through the en­ergy bar­rier, chap­ter 7.12.2. Re­al­is­ti­cally, though, for even strong ex­per­i­men­tal fields like the one men­tioned above, the “life time” of the elec­tron in the atom be­fore it has a de­cent chance of be­ing found out­side it far ex­ceeds the age of the uni­verse.