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Next: The hydrogen molecule ion Up: lecture_13 Previous: Overview of molecular quantum

The Born-Oppenheimer approximation

In this section, we will discuss one of the most important and fundamental approximations in molecular quantum mechanics. This approximation was developed by Max Born and J. Robert Oppenheimer in 1927. We will consider a very general molecule with $N$ nuclei and $M$ electrons. The coordinates of the nuclei are ${\bf R}_1,...,{\bf R}_N$. The coordinates of the electrons are ${\bf r}_1,...,{\bf r}_M$, and their spin variables are $S_{z,1},...,S_{z,M}$. For shorthand, we will denote the complete set of nuclear coordinates as $R$ and the set of electrons coordinates as $r$ and $x$ the complete set of electron coordinates and spin variables. The total molecular wave function $\Psi(x,R)$ depends on $3N+4M$ variables, which makes it a very cumbersome object to deal with. The Born-Oppenheimer approximation leads to a very important simplifaction of the wave function.



If we could neglect the electron-nuclear interaction, then the wave function would be a simple product $\Psi(x,R) = \psi_{\rm elec}(x)\psi_{\rm nucl}(R)$. However, we cannot neglect this term, but it might still be possible to write the wave function as a product. We note, first, that most nuclei are 3-4 orders of magnitude heaver than an electron. For the lightest nucleus, the proton,

\begin{displaymath}
m_p \approx 2000 m_e
\end{displaymath}

This mass difference is large enough to have important physical consequences. Let us think classically about this mass difference first. If two particles interact in some way, and one is much heavier than the other, the light particle will move essentially as a ``slave'' of the heavy particle. That is, it will simply follow the heavy particle wherever it goes, and, it will move rapidly in reponse to the heavy particle motion. As an illustration of this phenomenon, consider the simple mechanical system pictured below:

\includegraphics[scale=0.5]{Adiabatic_springs.eps}
Considering this as a classical system, we expect that the motion will be dominated by the large heavy particle, which is attached to a fixed wall by a spring. The small, light particle, which is attached to the heavy particle by a spring will simply follow the heavy particle and execute rapid oscillations around it. The figure below (bottom panel) illustrates this:

\includegraphics[scale=0.5]{Adiabatic_motion.eps}
In this illustration, when the heavy particle moves even a tiny amount, the light particle executes many oscillations around the heavy particle. Thus, we see that the light particle moves quite a bit, and we can think of the light particle as generating ``on the fly'' an effective potential $V_{\rm eff}$ for the heavy particle, at least at the present position of the heavy particle. What we can conclude, therefore, is that we can approximately fix the position of the heavy particle and just allow the light particle to move around before we advance the heavy particle to its next position.



How do the different mass scales of electrons and nuclei manifest themselves in the time-independent Schrödinger equation? Light particles, such as electrons, tend to have very diffuse, delocalized wave functions, while heavy particles, such as the nuclei, tend to have wave functions that are very localized about the classical positions. This is illustrated in the figure below:

Figure 1: Crude picture of matter. Nuclear wave functions are shown above the axis, electronic wave functions are shown below.
\includegraphics[scale=0.5]{charge.eps}

Thus, taking the idea of solving the electronic problem for fixed nuclear positions seriously, we can write the molecular wave function as

\begin{displaymath}
\Psi^{\rm (mol)}(x,R) = \psi^{\rm (elec)}(x,R)\psi^{\rm (nuc)}(R)
\end{displaymath}

which suggests that the electronic wave function $\psi^{\rm (elec)}(x,R)$ is solved using the nuclear positions $R$ simply as parameters that characterize the wave function in the same way that the electron mass and electron charge do. This is not an exact wave function for the molecule but an approximate one. Hence, we call this the Born-Oppenheimer approximation to the wave function. The total classical energy of the molecule is

\begin{displaymath}
K_e + K_n + V_{ee}(r) + V_{en}(r,R) + V_{nn}(R) = E
\end{displaymath}

where $K_e$ and $K_n$ are the electronic and nuclear kinetic energies. Therefore, the total Hamiltonian of the molecule is

\begin{displaymath}
\hat{H} = \hat{K}_e + \hat{K}_n + V_{ee}(r) + V_{en}(r,R) + V_{nn}(R)
\end{displaymath}

where $\hat{K}_e$ and $\hat{K}_n$ are the kinetic energy operators that result from substituting momenta for derivatives. Thus, $\hat{K}_e$ and $\hat{K}_n$ contain second derivatives with respect to electronic and nuclear coordinates, respectively. The details of these operators are not that important for this discussion.



We now write the total molecular Hamiltonian as

\begin{displaymath}
\hat{H} = \hat{H}_{\rm elec} + \hat{H}_{\rm nucl}
\end{displaymath}

where we assign
$\displaystyle \hat{H}_{\rm elec}$ $\textstyle =$ $\displaystyle \hat{K}_e + V_{ee}(r) + V_{en}(r,R)$  
       
$\displaystyle \hat{H}_{\rm nucl}$ $\textstyle =$ $\displaystyle \hat{K}_n + V_{nn}(R)$  

If we now substitute this into the Schrödinger equation, we obtain

\begin{displaymath}
\left[\hat{K}_e + \hat{K}_n + V_{ee}(r) + V_{nn}(R) + V_{en}...
...i^{\rm (nuc)}(R) = E\psi^{\rm (elec)}(x,R)\psi^{\rm
(nuc)}(R)
\end{displaymath}

Let us consider how the electronic and nuclear kinetic energy operators act on the wave function. $\hat{K}_e$ contains derivatives with respect only to electronic coordinates. Thus, it has no effect on the nuclear wave function, and we can write

\begin{displaymath}
\hat{K}_e\psi^{\rm (elec)}(x,R)\psi^{\rm (nuc)}(R) =
\psi^{\rm (nuc)}(R)\hat{K}_e\psi^{\rm (elec)}(x,R)
\end{displaymath}

The operator $\hat{K}_n$ contains derivatives with respect to nuclear coordinates, thus it affects both parts of the wave function:

\begin{displaymath}
\hat{K}_n \psi^{\rm (elec)}(x,R)\psi^{\rm (nuc)}(R) =
\psi^...
...)\hat{K}_n \psi^{\rm (nuc)}(R) + {\rm mixed\ derivative\ term}
\end{displaymath}

The last term arises from an application of the product rule using the fact that $\hat{K}_n \sim d^2/dR^2$. What is important here is that if the nuclear wave function is highly localized, then its curvature is high in the vicnity of the nucleus, and hence, $\hat{K}_n \psi^{\rm (nuc)}$ is very large compared to $\hat{K}_n \psi^{\rm (elec)}$ since $\psi^{\rm (elec)}$ changes much less dramatically spatially due to the delocalized nature of the electronic wave function. Thus, we need only keep the term in the above equation that contains $\hat{K}_n \psi^{\rm (nuc)}$:

\begin{displaymath}
\hat{K}_n \psi^{\rm (elec)}(x,R)\psi^{\rm (nuc)}(R) \approx
\psi^{\rm (elec)}(x,R)\hat{K}_n \psi^{\rm (nuc)}(R)
\end{displaymath}

Using this fact in the Schrödinger equation, we can write the equation as

\begin{displaymath}
\psi^{\rm (nuc)}(R)
\left[\hat{K}_e + V_{ee}(r) + V_{en}(r,R...
...^{\rm (nuc)}(R)
= E \psi^{\rm (elec)}(x,R)\psi^{\rm (nuc)}(R)
\end{displaymath}

or

\begin{displaymath}
\psi^{\rm (nuc)}(R)
\hat{H}_{\rm elec}(r,R)\psi^{\rm (elec)}...
...^{\rm (nuc)}(R)
= E \psi^{\rm (elec)}(x,R)\psi^{\rm (nuc)}(R)
\end{displaymath}

We now divide both sides by $\psi^{\rm (elec)}(x,R)\psi^{\rm (nuc)}(R)$, which gives

\begin{displaymath}
{\hat{H}_{\rm elec}(r,R)\psi^{\rm (elec)}(x,R) \over \psi^{\...
...hat{H}_{\rm nuc}\psi^{\rm (nuc)}(R) \over \psi^{\rm (nuc)}(R)}
\end{displaymath}

Note that the right side depends only on the nuclear coordinates $R$, which means that we can write compactly as some function $\varepsilon(R)$. Hence, we obtain

\begin{displaymath}
{\hat{H}_{\rm elec}(r,R)\psi^{\rm (elec)}(x,R) \over \psi^{\rm (elec)}(x,R)}
= \varepsilon(R)
\end{displaymath}

or

\begin{displaymath}
\hat{H}_{\rm elec}(r,R)\psi^{\rm (elec)}(x,R) = \varepsilon(R)\psi^{\rm
(elec)}(x,R)
\end{displaymath}

which is called the electronic Schrödinger equation. it yields a set of wave functions $\psi^{\rm (elec)}_{\alpha}(x,R)$ and energies $\varepsilon_{\alpha}(R)$ characterized by a set of quantum numbers $\alpha$, which is the full set of quantum numbers needed to characterize the wave function, e.g. $n,l,m$ for hydrogen. Thus, the electronic Schrödinger equation can be written as

\begin{displaymath}
\hat{H}_{\rm elec}(r,R)\psi_{\alpha}^{\rm (elec)}(x,R) = \varepsilon_{\alpha}(R)
\psi_{\alpha}^{\rm (elec)}(x,R)
\end{displaymath}

Note that the potential energy $V_{en}(r,R)$ depends on the electronic coordinates and the fixed nuclear coordinates $R$. Thus, the nuclear coordinates can be thought of as additional parameters in the potential energy. Alternatively, we can think of the nuclear as providing a fixed background potential energy in addition to the internal electron-electron repulsion $V_{ee}(r)$ that governs the shape of the electronic wave function.



Now the nuclear part comes from setting the right side of the full Schrödinger equation equal to $\varepsilon_{\alpha}(R)$, which yields:

\begin{displaymath}
E - {\hat{H}_{\rm nuc}\psi^{\rm (nuc)}(R) \over \psi^{\rm (nuc)}(R)}
= \varepsilon_{\alpha}(R)
\end{displaymath}

and if we multiply by $\psi^{\rm (nuc)}(R)$, we obtain the nuclear Schrödinger equation
$\displaystyle \left[\hat{H}_{\rm nucl} + \varepsilon_{\alpha}(R)\right]
\psi^{\rm (nuc)}(R)$ $\textstyle =$ $\displaystyle E\psi^{\rm (nuc)}(R)$  
       
$\displaystyle \left[\hat{K}_n + V_{nn}(R) + \varepsilon_{\alpha}(R)\right]
\psi^{\rm (nuc)}(R)$ $\textstyle =$ $\displaystyle E\psi^{\rm (nuc)}(R)$  

The Schrödinger equations for the electronic and nuclear part of the wave function are not exact but approximate based on the separation of electronic and nuclear degrees of freedom we assume can be made based on the mass differential.



Two interesting facts about the approximate Schrödinger equations should be noted. First, the electronic Schrödinger equation yields a set of energy levels $\varepsilon_{\alpha}(R)$ that depend on the nuclear configuration. Thus, at each nuclear configuration we choose, we get a different set of energy levels. These energy levels change continuously with the nuclear configuration, in fact. At the same time, the nuclear Schrödinger equation involves a potential energy with two terms $V_{nn}(R) + \varepsilon_{\alpha}(R)$, i.e. the nuclear-nuclear repulsion plus the electronic energy level $\varepsilon_{\alpha}$ as a function of $R$. Therefore, there is a different nuclear Schrödinger equation for each electronic energy level. The potential energies $V_{nn}(R) + \varepsilon_{\alpha}(R)$ are called Born-Oppenheimer potential energies or Born-Oppenheimer surfaces. They are surfaces (or hypersurfaces) because they must be viewed as continuous functions of $R$ (see figure below):

\includegraphics[scale=5.0]{BO_surfaces.eps}
The potential energy $V_{nn}(R) + \varepsilon_{\alpha}(R)$ determines the geometry of the molecule by minimization, and the nuclear probability distribution by solution of the nuclear Schrödinger equation. It also determines the nuclear dynamics if we solve the nuclear time-dependent Schrödinger equation. Thus, $\varepsilon_{\alpha}(R)$, the electronic energy levels, are of utmost importance in molecular quantum mechanics. For this reason, we need to understand the distributions of electrons in molecules.



Incidentally, we could now introduce a further approximation in which we treat the nuclei as classical particles with a potential energy $V_{nn}(R) + \varepsilon_{\alpha}(R)$. The nuclei would then move as classical point particles via Newton's Second Law, which would now take the form

\begin{displaymath}
M{d^2 R \over dt^2} = -{d \over dR}
\left[V_{nn}(R) + \varepsilon_{\alpha}(R)\right]
\end{displaymath}


next up previous
Next: The hydrogen molecule ion Up: lecture_13 Previous: Overview of molecular quantum
Mark E. Tuckerman 2011-11-10