9.2 The Born-Oppenheimer Approximation

Exact solutions in quantum mechanics are hard to come by. In almost all cases, approximation is needed. The Born-Oppenheimer approximation in particular is a key part of real-life quantum analysis of atoms and molecules and the like. The basic idea is that the uncertainty in the nuclear positions is too small to worry about when you are trying to find the wave function for the electrons. That was already assumed in the earlier approximate solutions for the hydrogen molecule and molecular ion. This section discusses the approximation, and how it can be used, in more depth.

9.2.1 The Hamiltonian

The general problem to be discussed in this section is that of a number of electrons around a number of nuclei. You first need to know what is the true problem to be solved, and for that you need the Hamiltonian.

This discussion will be restricted to the strictly nonrelativistic case. Corrections for relativistic effects on energy, including those involving spin, can in principle be added later, though that is well beyond the scope of this book. The physical problem to be addressed is that there are a finite number $I$ of electrons around a finite number $J$ of nuclei in otherwise empty space. That describes basic systems of atoms and molecules, but modifications would have to be made for ambient electric and magnetic fields and electromagnetic waves, or for the infinite systems of electrons and nuclei used to describe solids.

The electrons will be numbered using an index $i$, and whenever there is a second electron involved, its index will be called ${\underline i}$. Similarly, the nuclei will be numbered with an index $j$, or ${\underline j}$ where needed. The nuclear charge of nucleus number $j$, i.e. the number of protons in that nucleus, will be indicated by $Z_j$, and the mass of the nucleus by $m^{\rm n}_j$. Roughly speaking, the mass $m^{\rm n}_j$ will be the sum of the masses of the protons and neutrons in the nucleus; however, internal nuclear energies are big enough that there are noticeable relativistic deviations in total nuclear rest mass from what you would think. All the electrons have the same mass $m_{\rm e}$ since relativistic mass changes due to motion are ignored.

Under the stated assumptions, the Hamiltonian of the system consists of a number of contributions that will be looked at one by one. First there is the kinetic energy of the electrons, the sum of the kinetic energy operators of the individual electrons:

{\widehat T}^{\rm E}
= - \sum_{i=1}^I \frac{\hbar^2}{2m_...
...}}^2} +
\frac{\partial^2}{\partial {r_{3i}}^2}
\right). %
\end{displaymath} (9.3)

where ${\skew0\vec r}_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(r_{1i},r_{2i},r_{3i})$ is the position of electron number $i$. Note the use of $(r_1,r_2,r_3)$ as the notation for the components of position, rather than $(x,y,z)$. For more elaborate mathematics, the index notation $(r_1,r_2,r_3)$ is often more convenient, since you can indicate any generic component by the single expression $r_\alpha$, (with the understanding that $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, or 3,) instead of writing them out all three separately.

Similarly, there is the kinetic energy of the nuclei,

{\widehat T}^{\rm N}
= - \sum_{j=1}^J \frac{\hbar^2}{2m^...
\frac{\partial^2}{\partial {r^{\rm n}_{3j}}^2}
\right). %
\end{displaymath} (9.4)

where ${\skew0\vec r}^{\,\rm {n}}_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(r^{\rm n}_{1j},r^{\rm n}_{2j},r^{\rm n}_{3j})$ is the position of nucleus number $j$.

Next there is the potential energy due to the attraction of the $I$ electrons by the $J$ nuclei. That potential energy is, summing over all electrons and over all nuclei:

V^{\rm NE}=
- \sum_{i=1}^I \sum_{j=1}^J
\frac{Z_j e^2}{4\pi\epsilon_0} \frac{1}{r_{ij}} %
\end{displaymath} (9.5)

where $r_{ij}\equiv\vert{\skew0\vec r}_i-{\skew0\vec r}^{\,\rm {n}}_j\vert$ is the distance between electron number $i$ and nucleus number $j$, and $\epsilon_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8.85 10$\POW9,{-12}$ C$\POW9,{2}$/J m is the permittivity of space.

Next there is the potential energy due to the electron-electron repulsions:

V^{\rm EE}=
{\textstyle\frac{1}{2}} \sum_{i=1}^I
\frac{e^2}{4\pi\epsilon_0} \frac{1}{r_{i{\underline i}}} %
\end{displaymath} (9.6)

where $r_{i{\underline i}}\equiv\vert{\skew0\vec r}_i-{\skew0\vec r}_{\underline i}\vert$ is the distance between electron number $i$ and electron number ${\underline i}$. Half of this repulsion energy will be attributed to electron $i$ and half to electron ${\underline i}$, accounting for the factor $\frac12$.

Finally, there is the potential energy due to the nucleus-nucleus repulsions,

V^{\rm NN}= {\textstyle\frac{1}{2}}
\sum_{j=1}^J \sum_{\...
...rline j}e^2}{4\pi\epsilon_0} \frac{1}{r_{j{\underline j}}}, %
\end{displaymath} (9.7)

where $r_{j{\underline j}}\equiv\vert{\skew0\vec r}^{\,\rm n}_j-{\skew0\vec r}^{\,\rm {n}}_{\underline j}\vert$ is the distance between nucleus number $j$ and nucleus number ${\underline j}$.

Solving the full quantum problem for this system of electrons and nuclei exactly would involve finding the eigenfunctions $\psi$ to the Hamiltonian eigenvalue problem

\left[{\widehat T}^{\rm E}+ {\widehat T}^{\rm N}+ V^{\rm NE}+ V^{\rm EE}+ V^{\rm NN}\right] \psi
= E \psi %
\end{displaymath} (9.8)

Here $\psi$ is a function of the position and spin coordinates of all the electrons and all the nuclei, in other words:
{\skew0\vec r}_1,S_{z1}, {\skew0\vec r}_2,S_{...
..._{z2}, \ldots,
{\skew0\vec r}^{\,\rm n}_J,S^{\rm n}_{zJ}) %
\end{displaymath} (9.9)

You might guess solving this problem is a tall order, and you would be perfectly right. It can only be done analytically for the very simplest case of one electron and one nucleus. That is the hydrogen atom solution, using an effective electron mass to include the nuclear motion. For any decent size system, an accurate numerical solution is a formidable task too.

9.2.2 The basic Born-Oppenheimer approximation

The general idea of the Born-Oppenheimer approximation is simple. First note that the nuclei are thousands of times heavier than the electrons. A proton is almost two thousand times heavier than an electron, and that does not even count any neutrons in the nuclei.

So, if you take a look at the kinetic energy operators of the two,

&& {\widehat T}^{\rm E}
= - \sum_{i=1}^I \frac{\hbar^2}{2m...
...2} +
\frac{\partial^2}{\partial {r^{\rm n}_{3j}}^2}

then what would seem more reasonable than to ignore the kinetic energy ${\widehat T}^{\rm N}$ of the nuclei? It has those heavy masses in the bottom.

An alternative, and better, way of phrasing the assumption that ${\widehat T}^{\rm N}$ can be ignored is to say that you ignore the uncertainty in the positions of the nuclei. For example, visualize the hydrogen molecule, figure 5.2. The two protons, the nuclei, have pretty well defined positions in the molecule, while the electron wave function extends over the entire region like a big blob of possible measurable positions. So how important could the uncertainty in position of the nuclei really be?

Assuming that the nuclei do not suffer from quantum uncertainty in position is really equivalent to putting $\hbar$ to zero in their kinetic energy operator above, making the operator disappear, because $\hbar$ is nature’s measure of uncertainty. And without a kinetic energy term for the nuclei, there is nothing left in the mathematics to force them to have uncertain positions. Indeed, you can now just guess numerical values for the positions of the nuclei, and solve the approximated eigenvalue problem $H\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$ for those assumed values.

That thought is the Born-Oppenheimer approximation in a nutshell. Just do the electrons, assuming suitable positions for the nuclei a priori. The solutions that you get doing so will be called $\psi^{\rm E}$ to distinguish them from the true solutions $\psi$ that do not use the Born-Oppenheimer approximation. Mathematically $\psi^{\rm E}$ will still be a function of the electron and nuclear positions:

\psi^{\rm E}=\psi^{\rm E}(
{\skew0\vec r}_1,S_{z1}, {\sk...
...{z2}, \ldots,
{\skew0\vec r}^{\,\rm n}_J,S^{\rm n}_{zJ}). %
\end{displaymath} (9.10)

But physically it will be a quite different thing: it describes the probability of finding the electrons, given the positions of the nuclei. That is why there is a semi-colon between the electron positions and the nuclear positions. The nuclear positions are here assumed positions, while the electron positions are potential positions, for which the square magnitude of the wave function $\psi^{\rm E}$ gives the probability. This is an electron wave function only.

In application, it is usually most convenient to write the Hamiltonian eigenvalue problem for the electron wave function as

\left[{\widehat T}^{\rm E}+ V^{\rm NE}+ V^{\rm EE}+ V^{\rm...
...right] \psi^{\rm E}
= (E^{\rm E}+ V^{\rm NN}) \psi^{\rm E},

which just means that the eigenvalue is called $E^{\rm E}+V^{\rm NN}$ instead of simply $E^{\rm E}$. The reason is that you can then get rid of $V^{\rm NN}$, and obtain the electron wave function eigenvalue problem in the more concise form
\left[{\widehat T}^{\rm E}+ V^{\rm ...
... V^{\rm EE}\right] \psi^{\rm E}= E^{\rm E}\psi^{\rm E}
$} %
\end{displaymath} (9.11)

After all, for given nuclear coordinates, $V^{\rm NN}$ is just a bothersome constant in the solution of the electron wave function that you may just as well get rid of.

Of course, after you compute your electron eigenfunctions, you want to get something out of the results. Maybe you are looking for the ground state of a molecule, like was done earlier for the hydrogen molecule and molecular ion. In that case, the simplest approach is to try out various nuclear positions and for each likely set of nuclear positions compute the electronic ground state energy $E^{\rm E}_{\rm {gs}}$, the lowest eigenvalue of the electronic problem (9.11) above.

For different assumed nuclear positions, you will get different values for the electronic ground state energy, and the nuclear positions corresponding to the actual ground state of the molecule will be the ones for which the total energy is least:

\mbox{nominal ground state condition: } E^{\rm E}_{\rm gs}+V^{\rm NN}
\mbox{ is minimal}
$} %
\end{displaymath} (9.12)

This is what was used to solve the hydrogen molecule cases discussed in earlier chapters; a computer program was written to print out the energy $E^{\rm E}_{\rm {gs}}+V^{\rm NN}$ for a lot of different spacings between the nuclei, allowing the spacing that had the lowest total energy to be found by skimming down the print-out. That identified the ground state. The biggest error in those cases was not in using the Born-Oppenheimer approximation or the nominal ground state condition above, but in the crude way in which the electron wave function for given nuclear positions was approximated.

For more accurate work, the nominal ground state condition (9.12) above does have big limitations, so the next subsection discusses a more advanced approach.

9.2.3 Going one better

Solving the wave function for electrons only, given positions of the nuclei is definitely a big simplification. But identifying the ground state as the position of the nuclei for which the electron energy plus nuclear repulsion energy is minimal is much less than ideal.

Such a procedure ignores the motion of the nuclei, so it is no use for figuring out any molecular dynamics beyond the ground state. And even for the ground state, it is really wrong to say that the nuclei are at the position of minimum energy, because the uncertainty principle does not allow precise positions for the nuclei.

Instead, the nuclei behave much like the particle in a harmonic oscillator. They are stuck in an electron blob that wants to push them to their nominal positions. But uncertainty does not allow that, and the wave function of the nuclei spreads out a bit around the nominal positions, adding both kinetic and potential energy to the molecule. One example effect of this “zero point energy” is to lower the required dissociation energy a bit from what you would expect otherwise.

It is not a big effect, maybe on the order of tenths of electron volts, compared to typical electron energies described in terms of multiple electron volts (and much more for the inner electrons in all but the lightest atoms.) But it is not as small as might be guessed based on the fact that the nuclei are at least thousands of times heavier than the electrons.

Moreover, though relatively small in energy, the motion of the nuclei may actually be the one that is physically the important one. One reason is that the electrons tend to get stuck in single energy states. That may be because the differences between electron energy levels tend to be so large compared to a typical unit $\frac12kT$ of thermal energy, about one hundredth of an electron volt, or otherwise because they tend to get stuck in states for which the next higher energy levels are already filled with other electrons. The interesting physical effects then become due to the seemingly minor nuclear motion.

For example, the heat capacity of typical diatomic gases, like the hydrogen molecule or air under normal conditions, is not in any direct sense due to the electrons; it is kinetic energy of translation of the molecules plus a comparable energy due to angular momentum of the molecule; read, angular motion of the nuclei around their mutual center of gravity. The heat capacity of solids too is largely due to nuclear motion, as is the heat conduction of nonmetals.

For all those reasons, you would really, really, like to actually compute the motion of the nuclei, rather than just claim they are at fixed points. Does that mean that you need to go back and solve the combined wave function for the complete system of electrons plus nuclei anyway? Throw away the Born-Oppenheimer approximation results?

Fortunately, the answer is mostly no. It turns out that nature is quite cooperative here, for a change. After you have done the electronic structure computations for all relevant positions of the nuclei, you can proceed with computing the motion of nuclei as a separate problem. For example, if you are interested in the ground state nuclear motion, it is governed by the Hamiltonian eigenvalue problem

\left[{\widehat T}^{\rm N}+ V^{\rm NN}+ E^{\rm E}_1\right] \psi^{\rm N}_1 = E \psi^{\rm N}_1

where $\psi^{\rm N}_1$ is a wave function involving the nuclear coordinates only, not any electronic ones. The trick is in the potential energy to use in such a computation; it is not just the potential energy of nucleus to nucleus repulsions, but you must include an additional energy $E^{\rm E}_1$.

So, what is this $E^{\rm E}_1$? Easy, it is the electronic ground state energy $E^{\rm E}_{\rm {gs}}$ that you computed for assumed positions of the nuclei. So it will depend on where the nuclei are, but it does not depend on where the electrons are. You can just computed $E^{\rm E}_1$ for a sufficient number of relevant nuclear positions, tabulate the results somehow, and interpolate them as needed. $E^{\rm E}_1$ is then a known function function of the nuclear positions and so is $V^{\rm NN}$. Proceed to solve for the wave function for the nuclei $\psi^{\rm N}_1$ as a problem not directly involving any electrons.

And it does not necessarily have to be just to compute the ground state. You might want to study thermal motion or whatever. As long as the electrons are not kicked strongly enough to raise them to the next energy level, you can assume that they are in their ground state, even if the nuclei are not. The usual way to explain this is to say something like that the electrons “move so fast compared to the slow nuclei that they have all the time in the world to adjust themselves to whatever the electronic ground state is for the current nuclear positions.“

You might even decide to use classical molecular dynamics based on the potential $V^{\rm NN}+E^{\rm E}_1$ instead of quantum mechanics. It would be much faster and easier, and the results are often good enough.

So what if you are interested in what your molecule is doing when the electrons are at an elevated energy level, instead of in their ground state? Can you still do it? Sure. If the electrons are in an elevated energy level $E^{\rm E}_n$, (for simplicity, it will be assumed that the electron energy levels are numbered with a single index $n$,) just solve

\left[{\widehat T}^{\rm N}+ V^{\rm NN}+ E^{\rm E}_n\right] \psi^{\rm N}_n = E \psi^{\rm N}_n
$} %
\end{displaymath} (9.13)

or equivalent.

Note that for a different value of $n$, this is truly a different motion problem for the nuclei, since the potential energy will be different. If you are a visual sort of person, you might vaguely visualize the potential energy for a given value of $n$ plotted as a surface in some high-di­men­sion­al space, and the state of the nuclei moving like a roller-coaster along that potential energy surface, speeding up when the surface goes down, slowing down if it goes up. There is one such surface for each value of $n$. Anyway. The bottom line is that people refer to these different potential energies as “potential energy surfaces.” They are also called “adiabatic surfaces” because adiabatic normally means processes sufficiently fast that heat transfer can be ignored. So, some quantum physicists figured that it would be a good idea to use the same term for quantum processes that are so slow that quasi-equilibrium conditions persist throughout, and that have nothing to do with heat transfer.

Of course, any approximation can fail. It is possible to get into trouble solving your problem for the nuclei as explained above. The difficulties arise if two electron energy levels, call them $E^{\rm E}_n$ and $E^{\rm E}_{\underline n}$, become almost equal, and in particular when they cross. In simple terms, the difficulty is that if energy levels are equal, the energy eigenfunctions are not unique, and the slightest thing can throw you from one eigenfunction to the completely different one.

You might now get alarmed, because for example the hydrogen molecular ion does have two different ground state solutions with the same energy. Its single electron can be in either the spin-up state or the spin down state, and it does not make any difference for the energy because the assumed Hamiltonian does not involve spin. In fact, all systems with an odd number of electrons will have a second solution with all spins reversed and the same energy {D.50}. There is no need to worry, though; these reversed-spin solutions go their own way and do not affect the validity of (9.13). It is spatial, rather than spin nonuniqueness that is a concern.

There is a derivation of the nuclear eigenvalue problem (9.13) in derivation {D.51}, showing what the ignored terms are and why they can usually be ignored.