A.33 Explanation of the London forces

To fully understand the details of the London forces, it helps to first understand the popular explanation of them, and why it is all wrong. To keep things simple, the example will be the London attraction between two neutral hydrogen atoms that are well apart. (This will also correct a small error that the earlier discussion of the hydrogen molecule made; that discussion implied incorrectly that there is no attraction between two neutral hydrogen atoms that are far apart. The truth is that there really is some Van der Waals attraction. It was ignored because it is small compared to the chemical bond that forms when the atoms are closer together and would distract from the real story.)

Figure A.23: Possible polarizations of a pair of hydrogen atoms.
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\end{picture}
\end{figure}

The popular explanation for the London force goes something like this: “Sure, there would not be any attraction between two distant hydrogen atoms if they were perfectly spherically symmetric. But according to quantum mechanics, nature is uncertain. So sometimes the electron clouds of the two atoms are somewhat to the left of the nuclei, like in figure A.23 (b). This polarization [dipole creation] of the atoms turns out to produce some electrostatic attraction between the atoms. At other times, the electron clouds are somewhat to the right of the nuclei like in figure A.23 (c); it is really the same thing seen in the mirror. In cases like figure A.23 (a), where the electron clouds move towards each other, and (b), where they move away from each other, there is some repulsion between the atoms; however, the wave functions become correlated so that (b) and (c) are more likely than (a) and (d). Hence a net attraction results.”

Before examining what is wrong with this explanation, first consider what is right. It is perfectly right that figure A.23 (b) and (c) produce some net attraction between the atoms, and that (a) and (d) produce some repulsion. This follows from the net Coulomb potential energy between the atoms for given positions of the electrons:

\begin{displaymath}
V_{\rm {lr}} = \frac{e^2}{4\pi\epsilon_0}
\left(
\frac...
...
- \frac{1}{r_{\rm {r}}} + \frac{1}{r_{\rm {lr}}}
\right)
\end{displaymath}

where $e$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.6 10$\POW9,{-19}$ C is the magnitude of the charges of the protons and electrons, $\epsilon_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8.85 10$\POW9,{-12}$ C$\POW9,{2}$/J m is the permittivity of space, $d$ is the distance between the nuclei, $r_{\rm {l}}$ is the distance between the left electron and the right nucleus, $r_{\rm {r}}$ the one between the right electron and the left nucleus, and $r_{\rm {lr}}$ is the distance between the two electrons. If the electrons charges are distributed over space according to densities $n_{\rm {l}}({\skew0\vec r}_{\rm {l}})$ and $n_{\rm {r}}({\skew0\vec r}_{\rm {r}})$, the classical potential energy is

\begin{displaymath}
V_{\rm {lr}} =
\frac{e^2}{4\pi\epsilon_0}
\int_{{\rm a...
...d}^3{\skew0\vec r}_{\rm {l}}{\rm d}^3{\skew0\vec r}_{\rm {r}}
\end{displaymath}

(Since the first, 1/$d$, term represents the repulsion between the nuclei, it may seem strange to integrate it against the electron charge distributions, but the charge distributions integrate to one, so they disappear. Similarly in the second and third term, the charge distribution of the uninvolved electron integrates away.)

Since it is assumed that the atoms are well apart, the integrand above can be simplified using Taylor series expansions to give:

\begin{displaymath}
V_{\rm {lr}} =
\frac{e^2}{4\pi\epsilon_0}
\int_{{\rm a...
...3{\skew0\vec r}_{\rm {l}}
{\rm d}^3{\skew0\vec r}_{\rm {r}}
\end{displaymath}

where the positions of the electrons are measured from their respective nuclei. Also, the two $z$ axes are both taken horizontal and positive towards the left. For charge distributions as shown in figure A.23, the $x_{\rm {l}}x_{\rm {r}}$ and $y_{\rm {l}}y_{\rm {r}}$ terms integrate to zero because of odd symmetry. However, for a distribution like in figure A.23 (c), $n_{\rm {l}}$ and $n_{\rm {r}}$ are larger at positive $z_{\rm {l}}$, respectively $z_{\rm {r}}$, than at negative one, so the integral will integrate to a negative number. That means that the potential is lowered, there is attraction between the atoms. In a similar way, distribution (b) produces attraction, while (a) and (d) produce repulsion.

So there is nothing wrong with the claim that (b) and (c) produce attraction, while (a) and (d) produce repulsion. It is also perfectly right that the combined quantum wave function gives a higher probability to (b) and (c) than to (a) and (d).

So what is wrong? There are two major problems with the story.

1.
Energy eigenstates are stationary. If the wave function oscillated in time like the story suggests, it would require uncertainty in energy, which would act to kill off the lowering of energy. True, states with the electrons at the same side of their nuclei are more likely to show up when you measure them, but to reap the benefits of this increased probability, you must not do such a measurement and just let the electron wave function sit there unchanging in time.
2.
The numbers are all wrong. Suppose the wave functions in figures (b) and (c) shift (polarize) by a typical small amount $\varepsilon$. Then the attractive potential is of order $\varepsilon^2$$\raisebox{.5pt}{$/$}$$d^3$. Since the distance $d$ between the atoms is assumed large, the energy gained is a small amount times $\varepsilon^2$. But to shift atom energy eigenfunctions by an amount $\varepsilon$ away from their ground state takes an amount of energy $C\varepsilon^2$ where $C$ is some constant that is not small. So it would take more energy to shift the electron clouds than the dipole attraction could recover. In the ground state, the electron clouds should therefore stick to their original centered positions.

On to the correct quantum explanation. First the wave function is needed. If there were no Coulomb potentials linking the atoms, the combined ground-state electron wave function would simply take the form

\begin{displaymath}
\psi({\skew0\vec r}_{\rm {l}},{\skew0\vec r}_{\rm {r}})
...
...{\skew0\vec r}_{\rm {l}})\psi_{100}({\skew0\vec r}_{\rm {r}})
\end{displaymath}

where $\psi_{100}$ is the ground state wave function of a single hydrogen atom. To get a suitable correlated polarization of the atoms, throw in a bit of the $\psi_{210}$ 2p$_z$ states, as follows:

\begin{displaymath}
\psi({\skew0\vec r}_{\rm {l}},{\skew0\vec r}_{\rm {r}})=
...
...\skew0\vec r}_{\rm {l}})\psi_{210}({\skew0\vec r}_{\rm {r}}).
\end{displaymath}

For $\varepsilon$ $\raisebox{.3pt}{$>$}$ 0, it produces the desired correlation between the wave functions: $\psi_{100}$ is always positive, and $\psi_{210}$ is positive if the electron is at the positive-$z$ side of its nucleus and negative otherwise. So if both electrons are at the same side of their nucleus, the product $\psi_{210}({\skew0\vec r}_{\rm {l}})\psi_{210}({\skew0\vec r}_{\rm {r}})$ is positive, and the wave function is increased, giving increased probability of such states. Conversely, if the electrons are at opposite sides of their nucleus, $\psi_{210}({\skew0\vec r}_{\rm {l}})\psi_{210}({\skew0\vec r}_{\rm {r}})$ is negative, and the wave function is reduced.

Now write the expectation value of the energy:

\begin{displaymath}
\big\langle E\big\rangle =
\langle
\sqrt{1-\varepsilon...
...{100}\psi_{100}
+ \varepsilon\psi_{210}\psi_{210}
\rangle
\end{displaymath}

where $H_{\rm {l}}$ and $H_{\rm {r}}$ are the Hamiltonians of the individual electrons and

\begin{displaymath}
V_{\rm {lr}} = \frac{e^2}{4\pi\epsilon_0}
\frac{x_{\rm {...
...\rm {r}}+y_{\rm {l}}y_{\rm {r}}-2z_{\rm {l}}z_{\rm {r}}}{d^3}
\end{displaymath}

is again the potential between atoms. Working out the inner product, noting that the $\psi_{100}$ and $\psi_{210}$ are orthonormal eigenfunctions of the atom Hamiltonians $H_{\rm {l}}$ and $H_{\rm {r}}$ with eigenvalues $E_1$ and $E_2$, and that most $V_{\rm {lr}}$ integrals are zero on account of odd symmetry, you get

\begin{displaymath}
\big\langle E\big\rangle = 2 E_1 +
2 \varepsilon^2 (E_2-...
...\vert z_{\rm {l}}z_{\rm {r}}\vert\psi_{210}\psi_{210}\rangle.
\end{displaymath}

The final term is the savior for deriving the London force. For small values of $\varepsilon$, for which the square root can be approximated as one, this energy-lowering term dominates the energy $2\varepsilon^2(E_2-E_1)$ needed to distort the atom wave functions. The best approximation to the true ground state is then obtained when the quadratic in $\varepsilon$ is minimal. That happens when the energy has been lowered by an amount

\begin{displaymath}
\frac{2}{E_2-E_1}
\left(
\frac{e^2}{4\pi\epsilon_0}\la...
...}\vert z\vert\psi_{210}\rangle^2
\right)^2
\frac{1}{d^6}.
\end{displaymath}

Since the assumed eigenfunction is not exact, this variational approximation will underestimate the actual London force. For example, it can be seen that the energy can also be lowered similar amounts by adding some of the 2p$_x$ and 2p$_y$ states; these cause the atom wave functions to move in opposite directions normal to the line between the nuclei.

So what is the physical meaning of the savior term? Consider the inner product that it represents:

\begin{displaymath}
\langle\psi_{100}\psi_{100}\vert V_{\rm {lr}}\vert\psi_{210}\psi_{210}\rangle.
\end{displaymath}

That is the energy if both electrons are in the spherically symmetric $\psi_{100}$ ground state if both electrons are in the antisymmetric 2p$_z$ state. The savior term is a twilight term, like the ones discussed earlier in chapter 5.3 for chemical bonds. It reflects nature’s habit of doing business in terms of an unobservable wave function instead of observable probabilities.