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### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] ch. 5 content.

### Another approach (for last time?)

Imagine we perturb a potential, say a harmonic oscillator with an electric field

\label{eqn:qmLecture22:20}
V_0(x) = \inv{2} k x^2

\label{eqn:qmLecture22:40}
V(x) = \mathcal{E} e x

After minimizing the energy, using $$\PDi{x}{V} = 0$$, we get

\label{eqn:qmLecture22:60}
\inv{2} k x^2 + \mathcal{E} e x \rightarrow k x^\conj = – e \mathcal{E}

\label{eqn:qmLecture22:80}
p^\conj = -e x^\conj = – \frac{e^2 \mathcal{E}}{k}

For such a system the polarizability is

\label{eqn:qmLecture22:100}
\alpha = \frac{e^2 }{k}

\label{eqn:qmLecture22:120}
\begin{aligned}
\inv{2} k \lr{ -\frac{ e \mathcal{E}}{k} }^2 + \mathcal{E} e \lr{ – \frac{e \mathcal{E}}{k} }
&= – \inv{2} \lr{ \frac{e^2}{k} } \mathcal{E}^2 \\
&= – \inv{2} \alpha \mathcal{E}^2
\end{aligned}

## Van der Wall potential

\label{eqn:qmLecture22:140}
H_0 =
H_{0 1} + H_{0 2},

where

\label{eqn:qmLecture22:160}
H_{0 \alpha} = \frac{p_\alpha^2}{2m} – \frac{e^2}{4 \pi \epsilon_0 \Abs{ \Br_\alpha – \BR_\alpha} }, \qquad \alpha = 1,2

The full interaction potential is

\label{eqn:qmLecture22:180}
V =
\frac{e^2}{4 \pi \epsilon_0} \lr{
\inv{\Abs{\BR_1 – \BR_2}}
+
\inv{\Abs{\Br_1 – \Br_2}}

\inv{\Abs{\Br_1 – \BR_2}}

\inv{\Abs{\Br_2 – \BR_1}}
}

Let

\label{eqn:qmLecture22:200}
\Bx_\alpha = \Br_\alpha – \BR_\alpha,

\label{eqn:qmLecture22:220}
\BR = \BR_1 – \BR_2,

as sketched in fig. 1.

fig. 1. Two atom interaction.

\label{eqn:qmLecture22:240}
H_{0 \alpha}
=
\frac{\Bp^2}{2m}
-\frac{e^2}{4 \pi \epsilon_0 \Abs{\Bx_\alpha}}

which allows the total interaction potential to be written
\label{eqn:qmLecture22:260}
V =
\frac{e^2}{4 \pi \epsilon_0 R}
\lr{
1
+
\frac{R}{\Abs{\Bx_1 – \Bx_2 + \BR}}

\frac{R}{\Abs{\Bx_1 + \BR}}

\frac{R}{\Abs{-\Bx_2 + \BR}}
}

For $$R \gg x_1, x_2$$, this interaction potential, after a multipole expansion, is approximately

\label{eqn:qmLecture22:280}
V =
\frac{e^2}{4 \pi \epsilon_0} \lr{
\frac{\Bx_1 \cdot \Bx_2}{\Abs{\BR}^3}
-3 \frac{
(\Bx_1 \cdot \BR)
(\Bx_2 \cdot \BR)
}{\Abs{\BR}^5}
}

### 1. $$O(\lambda)$$

.

With

\label{eqn:qmLecture22:300}
\psi_0 = \ket{ 1s, 1s }

\label{eqn:qmLecture22:320}
\Delta E^{(1)} = \bra{\psi_0} V \ket{\psi_0}

The two particle wave functions are of the form

\label{eqn:qmLecture22:340}
\braket{ \Bx_1, \Bx_2 }{\psi_0} =
\psi_{1s}(\Bx_1)
\psi_{1s}(\Bx_2),

so braket integrals must be evaluated over a six-fold space. Recall that

\label{eqn:qmLecture22:740}
\psi_{1s} = \inv{\sqrt{\pi} a_0^{3/2} } e^{-r/a_0},

so

\label{eqn:qmLecture22:760}
\bra{\psi_{1s}} x_i \ket{\psi_{1s}}
\propto
\int_0^\pi \sin\theta d\theta \int_0^{2\pi} d\phi x_i

where
\label{eqn:qmLecture22:780}
x_i \in \setlr{ r \sin\theta \cos\phi, r \sin\theta \sin\phi, r \cos\theta }.

The $$x, y$$ integrals are zero because of the $$\phi$$ integral, and the $$z$$ integral is proportional to $$\int_0^\pi \sin(2 \theta) d\theta$$, which is also zero. This leads to zero averages

\label{eqn:qmLecture22:360}
\expectation{\Bx_1} = 0 = \expectation{\Bx_2}

so

\label{eqn:qmLecture22:380}
\Delta E^{(1)} = 0.

### 2. $$O(\lambda^2)$$

.

\label{eqn:qmLecture22:400}
\begin{aligned}
\Delta E^{(2)}
&= \sum_{n \ne 0} \frac{ \Abs{ \bra{\psi_n } V \ket{\psi_0} }^2 }{E_0 – E_n} \\
&= \sum_{n \ne 0} \frac{ \bra{\psi_0 } V \ket{\psi_n} \bra{\psi_n } V \ket{\psi_0} }{E_0 – E_n}.
\end{aligned}

This is a sum over all excited states. We expect that this will be of the form

\label{eqn:qmLecture22:420}
\Delta E^{(2)} = – \lr{ \frac{e^2}{4 \pi \epsilon_0} }^2 \frac{C_6}{R^6}

$$\Bx_1$$ and $$\Bx_2$$ are dipole operators. The first time this has a non-zero expectation is when we go from the 1s to the 2p states (both 1s and 2s states are spherically symmetric).

Noting that $$E_n = -e^2/2 n^2 a_0$$, we can compute a minimum bound for the energy denominator

\label{eqn:qmLecture22:440}
\begin{aligned}
\lr{E_n – E_0}^{\mathrm{min}}
&= 2 \lr{ E_{2p} – E_{1s} } \\
&= 2 E_{1s} \lr{ \inv{4} – 1 } \\
&= 2 \frac{3}{4} \Abs{E_{1s}} \\
&= \frac{3}{2} \Abs{E_{1s}}.
\end{aligned}

Note that the factor of two above comes from summing over the energies for both electrons. This gives us

\label{eqn:qmLecture22:460}
C_6
=
\frac{3}{2} \Abs{E_{1s}}
\bra{\psi_0 } \tilde{V} \ket{\psi_0},

where

\label{eqn:qmLecture22:480}
\tilde{V} =
\lr{
\Bx_1 \cdot \Bx_2
-3
(\Bx_1 \cdot \Rcap)
(\Bx_2 \cdot \Rcap)
}

### What about degeneracy?

\label{eqn:qmLecture22:500}
\Delta E^{(2)}_n
= \sum_{m \ne n} \frac{ \Abs{ \bra{\psi_n } V \ket{\psi_0} }^2 }{E_0 – E_n}

If $$\bra{\psi_n} V \ket{\psi_m} \propto \delta_{n m}$$ then it’s okay.
In general the we can’t expect the matrix element will be anything but fully populated, say

\label{eqn:qmLecture22:520}
V =
\begin{bmatrix}
V_{11} & V_{12} & V_{13} & V_{14} \\
V_{21} & V_{22} & V_{23} & V_{24} \\
V_{31} & V_{32} & V_{33} & V_{34} \\
V_{41} & V_{42} & V_{43} & V_{44} \\
\end{bmatrix},

If we choose a basis so that

\label{eqn:qmLecture22:540}
V =
\begin{bmatrix}
V_{11} & & & \\
& V_{22} & & \\
& & V_{33} & \\
& & & V_{44} \\
\end{bmatrix}.

When this is the case, we have no mixing of elements in the sum of \ref{eqn:qmLecture22:500}

### Degeneracy in the Stark effect

\label{eqn:qmLecture22:560}
H = H_0 + e \mathcal{E} z,

where

\label{eqn:qmLecture22:580}
H_0 = \frac{\Bp^2}{2m} – \frac{e}{4 \pi \epsilon_0} \inv{\Abs{\Bx}}

Consider the states $$2s, 2 p_x, 2p_y, 2p_z$$, for which $$E_n^{(0)} \equiv E_{2 s}$$, as sketched in fig. 2.

fig. 2. 2s 2p degeneracy.

Because of spherical symmetry

\label{eqn:qmLecture22:600}
\begin{aligned}
\bra{2 s} e \mathcal{E} z \ket{ 2 s} &= 0 \\
\bra{2 p_x} e \mathcal{E} z \ket{ 2 p_x} &= 0 \\
\bra{2 p_y} e \mathcal{E} z \ket{ 2 p_y} &= 0 \\
\bra{2 p_z} e \mathcal{E} z \ket{ 2 p_z} &= 0 \\
\end{aligned}

Looking at odd and even properties, it turns out that the only off-diagonal matrix element is

\label{eqn:qmLecture22:620}
\bra{2 s} e \mathcal{E} z \ket{ 2 p_z } = V_1 = -3 e \mathcal{E} a_0.

With a $$\setlr{ 2s, 2p_x, 2p_y, 2p_z }$$ basis the potential matrix is

\label{eqn:qmLecture22:640}
\begin{bmatrix}
0 & 0 & 0 & V_1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
V_1^\conj & 0 & 0 & 0 \\
\end{bmatrix}

\label{eqn:qmLecture22:660}
\begin{bmatrix}
0 & -\Abs{V_1} \\
-\Abs{V_1} & 0 \\
\end{bmatrix}

implies that the energy splitting goes as

\label{eqn:qmLecture22:680}
E_{2s} \rightarrow
E_{2s} \pm \Abs{V_1},

as sketched in fig. 3.

fig. 3. Stark effect energy level splitting.

The diagonalizing states corresponding to eigenvalues $$\pm 3 a_0 \mathcal{E}$$, are $$(\ket{2s} \mp \ket{2p_z})/\sqrt{2}$$.

The matrix element above is calculated explicitly in lecture22Integrals.nb.

The degeneracy that is left unsplit here, and has to be accounted for should we attempt higher order perturbation calculations.

### Appendix. Multipole expansion

Noting that

\label{eqn:qmLecture22:700}
\begin{aligned}
\lr{1 + \epsilon}^{-1/2}
&=
1 -\inv{2} \epsilon -\inv{2}\lr{\frac{-3}{2}}\inv{2!} \epsilon^2 \\
&=
1 -\inv{2} \epsilon + \frac{3}{8} \epsilon^2,
\end{aligned}

we have

\label{eqn:qmLecture22:720}
\begin{aligned}
\frac{R}{\Abs{\Bepsilon + \BR}}
&=
\frac{1}{\Abs{\frac{\Bepsilon}{R} + \Rcap}} \\
&=
\lr{ 1 + 2 \frac{\Bepsilon}{R} \cdot \Rcap + \lr{\frac{\Bepsilon}{R}}^2 }^{-1/2} \\
&=
1 – \frac{\Bepsilon}{R} \cdot \Rcap -\inv{2} \lr{\frac{\Bepsilon}{R}}^2
+ \frac{3}{8}
\lr{ 2 \frac{\Bepsilon}{R} \cdot \Rcap + \lr{\frac{\Bepsilon}{R}}^2 }^2 \\
&=
1 – \frac{\Bepsilon}{R} \cdot \Rcap -\inv{2} \lr{\frac{\Bepsilon}{R}}^2
+ \frac{3}{8}
\lr{ 4 \lr{ \frac{\Bepsilon}{R} \cdot \Rcap}^2 + \lr{\frac{\Bepsilon}{R}}^4
+ 4 \frac{\Bepsilon}{R} \cdot \Rcap \lr{\frac{\Bepsilon}{R}}^2
} \\
&\approx
1 – \frac{\Bepsilon}{R} \cdot \Rcap -\inv{2} \lr{\frac{\Bepsilon}{R}}^2
+ \frac{3}{2}
\lr{ \frac{\Bepsilon}{R} \cdot \Rcap}^2 .
\end{aligned}

Inserting the values from the brackets of \ref{eqn:qmLecture22:260} we have

\label{eqn:qmLecture22:800}
\begin{aligned}
1
+
\frac{R}{\Abs{\Bx_1 – \Bx_2 + \BR}}
&-
\frac{R}{\Abs{\Bx_1 + \BR}}

\frac{R}{\Abs{-\Bx_2 + \BR}} \\
&=
– \frac{\lr{ \Bx_1 – \Bx_2 }}{R} \cdot \Rcap -\inv{2} \lr{\frac{\lr{ \Bx_1 – \Bx_2 }}{R}}^2
+ \frac{3}{2}
\lr{ \frac{\lr{ \Bx_1 – \Bx_2 }}{R} \cdot \Rcap}^2 \\
&\quad + \frac{\Bx_1}{R} \cdot \Rcap +\inv{2} \lr{\frac{\Bx_1}{R}}^2
– \frac{3}{2}
\lr{ \frac{\Bx_1}{R} \cdot \Rcap}^2 \\
&\quad – \frac{\Bx_2}{R} \cdot \Rcap +\inv{2} \lr{\frac{\Bx_2}{R}}^2
– \frac{3}{2}
\lr{ \frac{\Bx_2}{R} \cdot \Rcap}^2 \\
&=
\frac{\Bx_1}{R} \cdot \frac{\Bx_2 }{R}
+ \frac{3}{2}
\lr{ \frac{\lr{ \Bx_1 – \Bx_2 }}{R} \cdot \Rcap}^2 \\
– \frac{3}{2}
\lr{ \frac{\Bx_1}{R} \cdot \Rcap}^2 \\