D.51 Born-Op­pen­heimer nu­clear mo­tion

This note gives a de­riva­tion of the Born-Op­pen­heimer Hamil­ton­ian eigen­value prob­lems (9.14) for the wave func­tions of the nu­clei.

First con­sider an ex­act eigen­func­tion $\psi$ of the com­plete sys­tem, in­clud­ing both the elec­trons and the nu­clei fully. Can it be re­lated some­how to the sim­pler elec­tron eigen­func­tions $\psi^{\rm E}_1,\psi^{\rm E}_2,\ldots$ that ig­nored nu­clear ki­netic en­ergy? Yes it can. For any given set of nu­clear co­or­di­nates, the elec­tron eigen­func­tions are com­plete; they are the eigen­func­tions of an Her­mit­ian elec­tron Hamil­ton­ian. And that means that you can for any given set of nu­clear co­or­di­nates write the ex­act wave func­tion as

$$\psi = \sum_{\underline n}c_{\underline n}\psi^{\rm E}_{\underline n}$$

You can do this for any set of nu­clear co­or­di­nates that you like, but the co­ef­fi­cients $c_{\underline n}$ will be dif­fer­ent for dif­fer­ent sets of nu­clear co­or­di­nates. That is just an­other way of say­ing that the $c_{\underline n}$ are func­tions of the nu­clear co­or­di­nates.

So, to be re­ally pre­cise, the wave func­tion of $I$ elec­trons and $J$ nu­clei can be writ­ten as:

$$\begin{eqnarray*} \lefteqn{\psi({\skew0\vec r}_1,S_{z1},\ldots,{\skew0\vec r}_I... ...S^{\rm n}_{z1},\ldots,{\skew0\vec r}^{ \rm n}_J,S^{\rm n}_{zJ}) \end{eqnarray*}$$

where su­per­scripts n in­di­cate nu­clear co­or­di­nates. (The nu­clear spins are re­ally ir­rel­e­vant, but it can­not hurt to keep them in.)

Con­sider what this means phys­i­cally. By con­struc­tion, the square elec­tron eigen­func­tions $\vert\psi^{\rm E}_{\underline n}\vert^2$ give the prob­a­bil­ity of find­ing the elec­trons as­sum­ing that they are in eigen­state ${\underline n}$ and that the nu­clei are at the po­si­tions listed in the fi­nal ar­gu­ments of the elec­tron eigen­func­tion. But then the prob­a­bil­ity that the nu­clei are ac­tu­ally at those po­si­tions, and that the elec­trons are ac­tu­ally in eigen­state $\psi^{\rm E}_{\underline n}$, will have to be $\vert c_{\underline n}\vert^2$. Af­ter all, the full wave func­tion $\psi$ must de­scribe the prob­a­bil­ity for the en­tire sys­tem to ac­tu­ally be in a spe­cific state. That means that $c_{\underline n}$ must be the nu­clear wave func­tion $\psi^{\rm N}_{\underline n}$ for when the elec­trons are in en­ergy eigen­state $\psi^{\rm E}_{\underline n}$. So from now on, just call it $\psi^{\rm N}_{\underline n}$ in­stead of $c_{\underline n}$. The full wave func­tion is then

$$\fbox{$\displaystyle \psi=\sum\psi^{\rm N}_{\underline n}\psi^{\rm E}_{\underline n} $}$$ (D.31)

In the un­steady case, the $c_{\underline n}$, hence the $\psi^{\rm N}_{\underline n}$, will also be func­tions of time. The $\psi^{\rm E}_{\underline n}$ will re­main time in­de­pen­dent as long as no ex­plic­itly time-de­pen­dent terms are added. The de­riva­tion then goes ex­actly the same way as the time-in­de­pen­dent Schrö­din­ger equa­tion (Hamil­ton­ian eigen­value prob­lem) de­rived be­low, with ${\rm i}\hbar\partial$$\raisebox{.5pt}{$/$}$​$\partial{t}$ re­plac­ing $E$.

So far, no ap­prox­i­ma­tions have been made; the only thing that has been done is to de­fine the nu­clear wave func­tions $\psi^{\rm N}_{\underline n}$. But the ob­jec­tive is still to de­rive the claimed equa­tion (9.14) for them. To do so plug the ex­pres­sion $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sum\psi^{\rm N}_{\underline n}\psi^{\rm E}_{\underline n}$ into the ex­act Hamil­ton­ian eigen­value prob­lem:

$$\left[{\widehat T}^{\rm N}+ {\widehat T}^{\rm E}+ V^{\rm NE... ...rline n}\psi^{\rm N}_{\underline n}\psi^{\rm E}_{\underline n}$$

Note first that the eigen­func­tions can be taken to be real since the Hamil­ton­ian is real. If the eigen­func­tions were com­plex, then their real and imag­i­nary parts sep­a­rately would be eigen­func­tions, and both of these are real. This ar­gu­ment ap­plies to both the elec­tron eigen­func­tions sep­a­rately as well as to the full eigen­func­tion. The trick is now to take an in­ner prod­uct of the equa­tion above with a cho­sen elec­tron eigen­func­tion $\psi^{\rm E}_n$. More pre­cisely, mul­ti­ply the en­tire equa­tion by $\psi^{\rm E}_n$, and in­te­grate/sum over the elec­tron co­or­di­nates and spins only, keep­ing the nu­clear po­si­tions and spins at fixed val­ues.

What do you get? Con­sider the terms in re­verse or­der, from right to left. In the right hand side, the elec­tron-co­or­di­nate in­ner prod­uct $\langle\psi^{\rm E}_n\vert\psi^{\rm E}_{\underline n}\rangle_e$ is zero un­less ${\underline n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$, and then it is one, since the elec­tron wave func­tions are or­tho­nor­mal for given nu­clear co­or­di­nates. So all we have left in the right-hand side is $E\psi^{\rm N}_n$, Check, $E\psi^{\rm N}_n$ is the cor­rect right hand side in the nu­clear-wave-func­tion Hamil­ton­ian eigen­value prob­lem (9.14).

Turn­ing to the lat­ter four terms in the left-hand side, re­mem­ber that by de­f­i­n­i­tion the elec­tron eigen­func­tions sat­isfy

$$\left[{\widehat T}^{\rm E}+ V^{\rm NE}+ V^{\rm EE}+ V^{\rm ... ...^{\rm E}_{\underline n}+V^{\rm NN})\psi^{\rm E}_{\underline n}$$

and if you then take an in­ner prod­uct of $\sum\psi^{\rm N}_{\underline n}(E^{\rm E}_{\underline n}+V^{\rm NN})\psi^{\rm E}_{\underline n}$ with $\psi^{\rm E}_n$, it is just like the ear­lier term, and you get $(E^{\rm E}_n+V^{\rm NN})\psi^{\rm N}_n$. Check, that are two of the terms in the left-hand side of (9.14) that you need.

That leaves only the nu­clear ki­netic term, and that one is a bit tricky. Re­call­ing the de­f­i­n­i­tion (9.5) of the ki­netic en­ergy op­er­a­tor ${\widehat T}^{\rm N}$ in terms of the nu­clear co­or­di­nate Lapla­cians, you have

$$-\sum_{j=1}^J\sum_{\alpha=1}^3\sum_{\underline n}\frac{\hba... ...pt}^2} \psi^{\rm N}_{\underline n}\psi^{\rm E}_{\underline n}$$

Re­mem­ber that not just the nu­clear wave func­tions, but also the elec­tron wave func­tions de­pend on the nu­clear co­or­di­nates. So, if you dif­fer­en­ti­ate out the prod­uct, you get

$$- \sum_{j=1}^J\sum_{\alpha=1}^3\sum_{\underline n} \left[ ... ...e n}}{\partial r^{\rm n}_{\alpha j}\rule{0pt}{8pt}^2} \right]$$

Now if you take the in­ner prod­uct with elec­tron eigen­func­tion $\psi^{\rm E}_n$, the first term in the brack­ets gives you what you need, the ex­pres­sion for the ki­netic en­ergy of the nu­clei. But you do not want the other two terms; these terms have the nu­clear ki­netic en­ergy dif­fer­en­ti­a­tions at least in part on the elec­tron wave func­tion in­stead of on the nu­clear wave func­tion.

Well, whether you like it or not, the ex­act equa­tion is, col­lect­ing all terms and re­ar­rang­ing,

$$\fbox{$\displaystyle \left[{\widehat T}^{\rm N}+ V^{\rm NN... ...erline n}a_{n{\underline n}} \psi^{\rm N}_{\underline n} $} %$$ (D.32)

where

$$\fbox{$\displaystyle {\widehat T}^{\rm N}= -\sum_{j=1}^J\... ...rtial^2}{\partial r^{\rm n}_{\alpha j}\rule{0pt}{8pt}^2} $} %$$ (D.33)

$$\fbox{$\displaystyle a_{n{\underline n}} = \sum_{j=1}^J\s... ...{\rm n}_{\alpha j}\rule{0pt}{8pt}^2}\Big\rangle \right) $} %$$ (D.34)

The first thing to note is the fi­nal sum in (D.32). Un­less you can talk away this sum as neg­li­gi­ble, (9.14) is not valid. The off-di­ag­o­nal co­ef­fi­cients, the $a_{n{\underline n}}$ for ${\underline n}$ $\raisebox{.2pt}{$\ne$}$ $n$, are par­tic­u­larly bad news, be­cause they pro­duce in­ter­ac­tions be­tween the dif­fer­ent po­ten­tial en­ergy sur­faces, shift­ing en­ergy from one value of $n$ to an­other. These off-di­ag­o­nal terms are called “vi­bronic cou­pling terms.” (The word is a con­trac­tion of vi­bra­tion” and “elec­tronic, if you are won­der­ing.)

Let’s have a closer look at (D.33) and (D.34) to see how big the var­i­ous terms re­ally are. At first ap­pear­ance it might seem that both the nu­clear ki­netic en­ergy ${\widehat T}^{\rm N}$ and the co­ef­fi­cients $a_{n{\underline n}}$ can be ig­nored, since both are in­versely pro­por­tional to the nu­clear masses, hence ap­par­ently thou­sands of times smaller than the elec­tronic ki­netic en­ergy in­cluded in $E^{\rm E}_n$. But do not go too quick here. First ball­park the typ­i­cal de­riv­a­tive, $\partial$$\raisebox{.5pt}{$/$}$​$\partial{r^{\rm n}_{{\alpha}j}}$ when ap­plied to the nu­clear wave func­tion. You can es­ti­mate such a de­riv­a­tive as 1/$\ell^{\rm N}$, where $\ell^{\rm N}$ is the typ­i­cal length over which there are sig­nif­i­cant changes in a nu­clear wave func­tion $\psi^{\rm N}_n$. Well, there are sig­nif­i­cant changes in nu­clear wave func­tions if you go from the mid­dle of a nu­cleus to its out­side, and that is a very small dis­tance com­pared to the typ­i­cal size of the elec­tron blob $\ell^{\rm E}$. It means that the dis­tance $\ell^{\rm N}$ is small. So the rel­a­tive im­por­tance of the nu­clear ki­netic en­ergy in­creases by a fac­tor $(\ell^{\rm E}/\ell^{\rm N})^2$ rel­a­tive to the elec­tron ki­netic en­ergy, com­pen­sat­ing quite a lot for the much higher nu­clear mass. So keep­ing the nu­clear ki­netic en­ergy is def­i­nitely a good idea.

How about the co­ef­fi­cients $a_{n{\underline n}}$? Well, nor­mally the elec­tron eigen­func­tions only change ap­pre­cia­ble when you vary the nu­clear po­si­tions over a length com­pa­ra­ble to the elec­tron blob scale $\ell^{\rm E}$. Think back of the ex­am­ple of the hy­dro­gen mol­e­cule. The ground state sep­a­ra­tion be­tween the nu­clei was found as 0.87Å. But you would not see a dra­matic change in elec­tron wave func­tions if you made it a few per­cent more or less. To see a dra­matic change, you would have to make the nu­clear dis­tance 1.5Å, for ex­am­ple. So the de­riv­a­tives $\partial$$\raisebox{.5pt}{$/$}$​$\partial{r^{\rm n}_{\alpha{j}}}$ ap­plied to the elec­tron wave func­tions are nor­mally not by far as large as those ap­plied to the nu­clear wave func­tions, hence the $a_{n{\underline n}}$ terms are rel­a­tively small com­pared to the nu­clear ki­netic en­ergy, and ig­nor­ing them is usu­ally jus­ti­fied. So the fi­nal con­clu­sion is that equa­tion (9.14) is usu­ally jus­ti­fied.

But there are ex­cep­tions. If dif­fer­ent en­ergy lev­els get close to­gether, the elec­tron wave func­tions be­come very sen­si­tive to small ef­fects, in­clud­ing small changes in the nu­clear po­si­tions. When the wave func­tions have be­come sen­si­tive enough that they vary sig­nif­i­cantly un­der nu­clear po­si­tion changes com­pa­ra­ble in size to the nu­clear wave func­tion blobs, you can no longer ig­nore the $a_{n{\underline n}}$ terms and (9.14) be­comes in­valid.

You can be a bit more pre­cise about that claim with a few tricks. Con­sider the fac­tors

$$\Big\langle\psi^{\rm E}_n\Big\vert \frac{\partial\psi^{\rm E}_{\underline n}}{\partial r^{\rm n}_{\alpha j}}\Big\rangle$$

ap­pear­ing in the $a_{n{\underline n}}$, (D.34). First of all, these fac­tors are zero when ${\underline n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$. The rea­son is that be­cause of or­tho­nor­mal­ity, $\langle\psi^{\rm E}_n\vert\psi^{\rm E}_n\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, and tak­ing the $\partial$$\raisebox{.5pt}{$/$}$​$\partial{r}^{\rm n}_{\alpha{j}}$ de­riv­a­tive of that, not­ing that the eigen­func­tions are real, you see that the fac­tor is zero.

For ${\underline n}$ $\raisebox{.2pt}{$\ne$}$ $n$, the fol­low­ing trick works:

$$\begin{eqnarray*} \Big\langle\psi^{\rm E}_n\Big\vert \frac{\partial}{\partial ... ...a i}}{r_{ij}^3} \Big\vert\psi^{\rm E}_{\underline n}\Big\rangle \end{eqnarray*}$$

The first equal­ity is just a mat­ter of the de­f­i­n­i­tion of the elec­tron eigen­func­tions and tak­ing the sec­ond $H^{\rm E}$ to the other side, which you can do since it is Her­mit­ian. The sec­ond equal­ity is a mat­ter of look­ing up the Hamil­ton­ian in chap­ter 9.2.1 and then work­ing out the com­mu­ta­tor in the left­most in­ner prod­uct. ($V^{\rm NN}$ does not com­mute with the de­riv­a­tive, but you can use or­thog­o­nal­ity on the cleaned up ex­pres­sion.) The bot­tom line is that the fi­nal in­ner prod­uct is fi­nite, with no rea­son for it to be­come zero when en­ergy lev­els ap­proach. So, look­ing at the sec­ond equal­ity, the first term in $a_{n{\underline n}}$, (D.34), blows up like 1$\raisebox{.5pt}{$/$}$​$(E^{\rm E}_{\underline n}-E^{\rm E}_n)$ when those en­ergy lev­els be­come equal.

As far as the fi­nal term in $a_{n{\underline n}}$ is con­cerned, like the sec­ond term, you would ex­pect it to be­come im­por­tant when the scale of non­triv­ial changes in elec­tron wave func­tions with nu­clear po­si­tions be­comes com­pa­ra­ble to the size of the nu­clear wave func­tions. You can be a lit­tle bit more pre­cise by tak­ing one more de­riv­a­tive of the in­ner prod­uct ex­pres­sion de­rived above,

$$\Big\langle \frac{\partial\psi^{\rm E}_n}{\partial r^{\rm ... ...a i}}{r_{ij}} \Big\vert\psi^{\rm E}_{\underline n}\Big\rangle$$

The first term should not be large: while the left hand side of the in­ner prod­uct has a large com­po­nent along $\psi^{\rm E}_{\underline n}$, the other side has zero com­po­nent and vice-versa. The fi­nal term should be of or­der 1/$(E^{\rm E}_{\underline n}-E^{\rm E}_n)^2$, as you can see if you first change the ori­gin of the in­te­gra­tion vari­able in the in­ner prod­uct to be at the nu­clear po­si­tion, to avoid hav­ing to dif­fer­en­ti­ate the po­ten­tial de­riv­a­tive. So you con­clude that the sec­ond term of co­ef­fi­cient $a_{n{\underline n}}$ is of or­der 1/$(E^{\rm E}_{\underline n}-E^{\rm E}_n)^2$. In view of the fact that this term has one less de­riv­a­tive on the nu­clear wave func­tion, that is just enough to al­low it to be­come sig­nif­i­cant at about the same time that the first term does.

The di­ag­o­nal part of ma­trix $a_{n{\underline n}}$, i.e. the $a_{nn}$ terms, is some­what in­ter­est­ing since it pro­duces a change in ef­fec­tive en­ergy with­out in­volv­ing in­ter­ac­tions with the other po­ten­tial en­ergy sur­faces, i.e. with­out in­ter­ac­tion with the $\psi^{\rm N}_{\underline n}$ for ${\underline n}$ $\raisebox{.2pt}{$\ne$}$ $n$. The di­ag­o­nal part is called the “Born-Op­pen­heimer di­ag­o­nal cor­rec­tion.” Since as noted above, the first term in the ex­pres­sion (D.34) for the $a_{n{\underline n}}$ does not have a di­ag­o­nal part, the di­ag­o­nal cor­rec­tion is given by the sec­ond term.

Note that in a tran­sient case that starts out as a sin­gle nu­clear wave func­tion $\psi^{\rm N}_n$, the di­ag­o­nal term $a_{nn}$ mul­ti­plies the pre­dom­i­nant nu­clear wave func­tion $\psi^{\rm N}_n$, while the off-di­ag­o­nal terms only mul­ti­ply the small other nu­clear wave func­tions. So de­spite not in­volv­ing any de­riv­a­tive of the nu­clear wave func­tion, the di­ag­o­nal term will ini­tially be the main cor­rec­tion to the Born-Op­pen­heimer ap­prox­i­ma­tion. It will re­main im­por­tant at later times.