Van der Waals Bonding in Detail*
(a) Here we will do a much more precise calculation of the van der Waals force between two hydrogen atoms. First, let the positions of the two nuclei be separated by a vector $\mathbf{R}$, and let the vector from nucleus 1 to electron 1 be $\mathbf{r}_{1}$ and let the vector from nucleus 2 to electron 2 be $\mathbf{r}_{2}$ as shown in the following figure.
Let us now write the Hamiltonian for both atoms (assuming fixed positions of nuclei, i.e., using BornOppenheimer approximation) as
$$
\begin{aligned}
H &=H_{0}+H_{1} \\
H_{0} &=\frac{\mathbf{p}_{1}^{2}}{2 m}+\frac{\mathbf{p}_{2}^{2}}{2 m}-\frac{e^{2}}{4 \pi \epsilon_{0}\left|\mathbf{r}_{1}\right|}-\frac{e^{2}}{4 \pi \epsilon_{0}\left|\mathbf{r}_{2}\right|} \\
H_{1} &=\frac{e^{2}}{4 \pi \epsilon_{0}|\mathbf{R}|}+\frac{e^{2}}{4 \pi \epsilon_{0}\left|\mathbf{R}-\mathbf{r}_{1}+\mathbf{r}_{2}\right|} \\
&-\frac{e^{2}}{4 \pi \epsilon_{0}\left|\mathbf{R}-\mathbf{r}_{1}\right|}-\frac{e^{2}}{4 \pi \epsilon_{0}\left|\mathbf{R}+\mathbf{r}_{2}\right|}
\end{aligned}
$$
Here $H_{0}$ is the Hamiltonian for two non-interacting hydrogen atoms, and $H_{1}$ is the interaction between the atoms.
Without loss of generality, let us assume that $\mathbf{R}$ is in the $x$ direction. Show that for large $\mathbf{R}$ and small $\mathrm{r}_{i}$, the interaction Hamiltonian can be written as
$$
H_{1}=\frac{e^{2}}{4 \pi \epsilon_{0}|\mathbf{R}|^{3}}\left(z_{1} z_{2}+y_{1} y_{2}-2 x_{1} x_{2}\right)+\mathcal{O}\left(1 / R^{4}\right)
$$
where $x_{i}, y_{i}, z_{i}$ are the components of $\mathrm{r}_{1}$. Show that this is just the interaction between two dipoles.
(b) Perturbation Theory:
The eigenvalues of $H_{0}$ can be given as the eigenvalues of the two atoms separately. Recall that the eigenstates of hydrogen are written in the usual notation as $|n, l, m\rangle$ and have energies $E_{n}=$ $-\mathrm{Ry} / n^{2}$ with $\mathrm{Ry}=m e^{4} /\left(32 \pi^{2} \epsilon_{0}^{2} \hbar^{2}\right)=e^{2} /\left(8 \pi \epsilon_{0} \alpha_{0}\right)$ the Rydberg (here $l \geqslant 0,|m| \leqslant l$ and $n \geqslant$
$l+1)$. Thus the eigenstates of $H_{0}$ are written as $\left.\mid n_{1}, l_{l}, m_{1} ; n_{2}, l_{2}, m_{2}\right)$ with energies $E_{n_{1}, n_{2}}=$ $-\mathrm{Ry}\left(1 / n_{1}^{2}+1 / n_{2}^{2}\right)$. The ground state of $H_{0}$ is $|1,0,0 ; 1,0,0\rangle$.
D Perturbing $H_{0}$ with the interaction $H_{1}$, show that to first order in $H_{1}$ there is no change in the ground-state energy. Thus conclude that the leading correction to the grotund-state energy is proportional to $1 / R^{6}$ (and hence the force is proportional to $1 / R^{7}$ ).
D Recalling second-order perturbation theory show that we have a correction to the total energy given by
$$
\delta E=
$$
$\sum_{n_{1}, n_{2}} \begin{aligned}&|<1,0,0 ; 1,0,0| H_{1} \mid n_{1}, l_{l}, m_{1} ; n_{2}, l_{2} \\&l_{1}, l_{2} \atop m_{1}, m_{2}\end{aligned}$ $$ \begin{array}{l}\text { E0,0 }-E_{n_{1}, n_{2}} \\ \text { Show that the force must be attractive. }\end{array} $$
(c)*Bounding the binding eneryy:
First, show that the numerator in this expression is zero if either $n_{1}=1$ or $n_{2}=1$. Thus the smallest $E_{n_{1}, n_{2}}$ that appears in the denominator is $E_{2,2}$. If we replace $E_{n_{1}, n_{2}}$ in the denominator with $E_{2,2}$ then the $|\delta E|$ we calculate will be greater than than the $|\delta E|$ in the exact calculation. On the other hand, if we replace $E_{n_{1}, n_{2}}$ by 0 , then the $|\delta E|$ will always be less than the $\delta E$ of the exact calculation.
D Make these replacements, and perform the remaining sum by identifying a complete set. Derive the bound
at the force must be attractive.
$g$ the binding energy:
that the numerator in this expression
her $n_{1}=1$ or $n_{2}=1$. Thus the small-
that appears in the denominator is $E_{2,2}$.
the exact calculation. On the other
replace $E_{n_{1}, n_{2}}$ by 0, then the $|\delta E|$ will
ws than the $\delta E$ of the exact calculation.
ese replacements, and perform the re-
by identifying a complete set. Derive