D.51 Born-Oppenheimer nuclear motion

This note gives a derivation of the Born-Oppenheimer Hamiltonian eigenvalue problems (9.13) 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

\begin{displaymath}
\psi = \sum_{\underline n}c_{\underline n}\psi^{\rm E}_{\underline n}
\end{displaymath}

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}_...
...^{\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

\begin{displaymath}
\fbox{$\displaystyle
\psi=\sum\psi^{\rm N}_{\underline n}\psi^{\rm E}_{\underline n}
$}
\end{displaymath} (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 (9.13) 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:

\begin{displaymath}
\left[{\widehat T}^{\rm N}+ {\widehat T}^{\rm E}+ V^{\rm N...
...line n}\psi^{\rm N}_{\underline n}\psi^{\rm E}_{\underline n}
\end{displaymath}

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 (9.13).

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

\begin{displaymath}
\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}
\end{displaymath}

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 (9.13) that you need.

That leaves only the nuclear kinetic term, and that one is a bit tricky. Recalling the definition (9.4) of the kinetic energy operator ${\widehat T}^{\rm N}$ in terms of the nuclear coordinate Laplacians, you have

\begin{displaymath}
-\sum_{j=1}^J\sum_{\alpha=1}^3\sum_{\underline n}\frac{\hb...
...}^2}
\psi^{\rm N}_{\underline n}\psi^{\rm E}_{\underline n}
\end{displaymath}

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

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

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,

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

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


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

The first thing to note is the final sum in (D.32). Unless you can talk away this sum as negligible, (9.13) 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 (9.13) 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 (9.13) becomes invalid.

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

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

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}{\partia...
...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,

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

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.