## PHY1520H Graduate Quantum Mechanics. Lecture 22: Van der Wall potential and Stark effect. Taught by Prof. Arun Paramekanti

December 10, 2015 phy1520 , , ,

### 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)
}

\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 \\
– \frac{3}{2}
\lr{ \frac{\Bx_2}{R} \cdot \Rcap}^2 \\
&=
\frac{\Bx_1}{R} \cdot \frac{\Bx_2 }{R}
– 3 \frac{\Bx_1}{R} \cdot \Rcap \frac{\Bx_2}{R} \cdot \Rcap.
\end{aligned}

This proves \ref{eqn:qmLecture22:280}.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Simplest perturbation two by two Hamiltonian

### Q: two state Hamiltonian.

Given a two-state system

\label{eqn:simplestTwoByTwoPerturbation:20}
H = H_0 + \lambda V
=
\begin{bmatrix}
E_1 & \lambda \Delta \\
\lambda \Delta & E_2
\end{bmatrix}

• (a) Solve the system exactly.
• (b) Find the first order perturbed states and second order energy shifts, and compare to the exact solution.
• (c) Solve the degenerate case for $$E_1 = E_2$$, and compare to the exact solution.

### A: part (a)

The energy eigenvalues $$\epsilon$$ are given by

\label{eqn:simplestTwoByTwoPerturbation:40}
0
=
\lr{ E_1 – \epsilon }
\lr{ E_2 – \epsilon }
– (\lambda \Delta)^2,

or

\label{eqn:simplestTwoByTwoPerturbation:60}
\epsilon^2 – \epsilon\lr{ E_1 + E_2 } + E_1 E_2 = (\lambda \Delta)^2.

After rearranging this is
\label{eqn:simplestTwoByTwoPerturbation:80}
\epsilon = \frac{ E_1 + E_2 }{2} \pm \sqrt{ \lr{ \frac{ E_1 – E_2 }{2} }^2 + (\lambda \Delta)^2 }.

Notice that for $$E_2 = E_1$$ we have

\label{eqn:simplestTwoByTwoPerturbation:100}
\epsilon = E_1 \pm \lambda \Delta.

Since a change of basis can always put the problem in a form so that $$E_1 > E_2$$, let’s assume that to make an approximation of the energy eigenvalues for $$\Abs{\lambda \Delta} \ll \ifrac{ (E_1 – E_2) }{2}$$

\label{eqn:simplestTwoByTwoPerturbation:120}
\begin{aligned}
\epsilon
&=
\frac{ E_1 + E_2 }{2} \pm \frac{ E_1 – E_2 }{2} \sqrt{ 1 + \frac{(2 \lambda \Delta)^2}{(E_1 – E_2)^2} } \\
&\approx
\frac{ E_1 + E_2 }{2} \pm \frac{ E_1 – E_2 }{2} \lr{ 1 + 2 \frac{(\lambda
\Delta)^2}{(E_1 – E_2)^2} } \\
&=
\frac{ E_1 + E_2 }{2} \pm \frac{ E_1 – E_2 }{2}
\pm
\frac{(\lambda \Delta)^2}{E_1 – E_2} \\
&=
E_1 + \frac{(\lambda \Delta)^2}{E_1 – E_2}, E_2 + \frac{(\lambda \Delta)^2}{E_2 – E_1}.
\end{aligned}

For the perturbed states, starting with the plus case, if

\label{eqn:simplestTwoByTwoPerturbation:140}
\ket{+} \propto
\begin{bmatrix}
a \\
b
\end{bmatrix},

we must have
\label{eqn:simplestTwoByTwoPerturbation:160}
\begin{aligned}
0
&=
\biglr{ E_1 – \lr{ E_1 + \frac{(\lambda \Delta)^2}{E_1 – E_2} } } a + \lambda
\Delta b \\
&=
\biglr{ – \frac{(\lambda \Delta)^2}{E_1 – E_2} } a + \lambda \Delta b,
\end{aligned}

so

\label{eqn:simplestTwoByTwoPerturbation:180}
\ket{+} \rightarrow
\begin{bmatrix}
1 \\
\frac{\lambda \Delta}{E_1 – E_2}
\end{bmatrix}
= \ket{+} + \frac{\lambda \Delta}{E_1 – E_2} \ket{-}.

Similarly for the minus case we must have

\label{eqn:simplestTwoByTwoPerturbation:200}
\begin{aligned}
0
&=
\lambda \Delta a + \biglr{ E_2 – \lr{ E_2 + \frac{(\lambda \Delta)^2}{E_2 – E_1} } } b \\
&=
\lambda \Delta b + \biglr{ – \frac{(\lambda \Delta)^2}{E_2 – E_1} } b,
\end{aligned}

for
\label{eqn:simplestTwoByTwoPerturbation:220}
\ket{-} \rightarrow
\ket{-} + \frac{\lambda \Delta}{E_2 – E_1} \ket{+}.

### A: part (b)

For the perturbation the first energy shift for perturbation of the $$\ket{+}$$ state is

\label{eqn:simplestTwoByTwoPerturbation:240}
\begin{aligned}
E_{+}^{(1)}
&= \ket{+} V \ket{+} \\
&=
\lambda \Delta
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\lambda \Delta
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 \\
1
\end{bmatrix} \\
&=
0.
\end{aligned}

The first order energy shift for the perturbation of the $$\ket{-}$$ state is also zero. The perturbed $$\ket{+}$$ state is

\label{eqn:simplestTwoByTwoPerturbation:260}
\begin{aligned}
\ket{+}^{(1)}
&= \frac{\overline{{P}}_{+}}{E_1 – H_0} V \ket{+} \\
&= \frac{\ket{-}\bra{-}}{E_1 – E_2} V \ket{+}
\end{aligned}

The numerator matrix element is

\label{eqn:simplestTwoByTwoPerturbation:280}
\begin{aligned}
\bra{-} V \ket{+}
&=
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
0 & \Delta \\
\Delta & 0
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
0 \\
\Delta
\end{bmatrix} \\
&=
\Delta,
\end{aligned}

so

\label{eqn:simplestTwoByTwoPerturbation:300}
\ket{+} \rightarrow \ket{+} + \ket{-} \frac{\Delta}{E_1 – E_2}.

Observe that this matches the first order series expansion of the exact value above.

For the perturbation of $$\ket{-}$$ we need the matrix element

\label{eqn:simplestTwoByTwoPerturbation:320}
\begin{aligned}
\bra{+} V \ket{-}
&=
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 & \Delta \\
\Delta & 0
\end{bmatrix}
\begin{bmatrix}
0 \\
1
\end{bmatrix} \\
&=
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
\Delta \\
0 \\
\end{bmatrix} \\
&=
\Delta,
\end{aligned}

so it’s clear that the perturbed ket is

\label{eqn:simplestTwoByTwoPerturbation:340}
\ket{-} \rightarrow \ket{-} + \ket{+} \frac{\Delta}{E_2 – E_1},

also matching the approximation found from the exact computation. The second order energy shifts can now be calculated

\label{eqn:simplestTwoByTwoPerturbation:360}
\begin{aligned}
\bra{+} V \ket{+}’
&=
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 & \Delta \\
\Delta & 0
\end{bmatrix}
\begin{bmatrix}
1 \\
\frac{\Delta}{E_1 – E_2}
\end{bmatrix} \\
&=
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
\frac{\Delta^2}{E_1 – E_2} \\
\Delta
\end{bmatrix} \\
&=
\frac{\Delta^2}{E_1 – E_2},
\end{aligned}

and

\label{eqn:simplestTwoByTwoPerturbation:380}
\begin{aligned}
\bra{-} V \ket{-}’
&=
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
0 & \Delta \\
\Delta & 0
\end{bmatrix}
\begin{bmatrix}
\frac{\Delta}{E_2 – E_1} \\
1 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
\Delta \\
\frac{\Delta^2}{E_2 – E_1} \\
\end{bmatrix} \\
&=
\frac{\Delta^2}{E_2 – E_1},
\end{aligned}

The energy perturbations are therefore
\label{eqn:simplestTwoByTwoPerturbation:400}
\begin{aligned}
E_1 &\rightarrow E_1 + \frac{(\lambda \Delta)^2}{E_1 – E_2} \\
E_2 &\rightarrow E_2 + \frac{(\lambda \Delta)^2}{E_2 – E_1}.
\end{aligned}

This is what we had by approximating the exact case.

### A: part (c)

For the $$E_2 = E_1$$ case, we’ll have to diagonalize the perturbation potential. That is

\label{eqn:simplestTwoByTwoPerturbation:420}
\begin{aligned}
V &= U \bigwedge U^\dagger \\
\bigwedge &=
\begin{bmatrix}
\Delta & 0 \\
0 & -\Delta
\end{bmatrix} \\
U &= U^\dagger = \inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.
\end{aligned}

A change of basis for the Hamiltonian is

\label{eqn:simplestTwoByTwoPerturbation:440}
\begin{aligned}
H’
&=
U^\dagger H U \\
&=
U^\dagger H_0 U + \lambda U^\dagger V U \\
&=
E_1 U^\dagger + \lambda U^\dagger V U \\
&=
H_0 + \lambda \bigwedge.
\end{aligned}

We can now calculate the perturbation energy with respect to the new basis, say $$\setlr{ \ket{1}, \ket{2} }$$. Those energy shifts are

\label{eqn:simplestTwoByTwoPerturbation:460}
\begin{aligned}
\Delta^{(1)} &= \bra{1} V \ket{1} = \Delta \\
\Delta^{(2)} &= \bra{2} V \ket{2} = -\Delta.
\end{aligned}

The perturbed energies are therefore

\label{eqn:simplestTwoByTwoPerturbation:480}
\begin{aligned}
E_1 &\rightarrow E_1 + \lambda \Delta \\
E_2 &\rightarrow E_2 – \lambda \Delta,
\end{aligned}

which matches \ref{eqn:simplestTwoByTwoPerturbation:100}, the exact result.

# References

## PHY1520H Graduate Quantum Mechanics. Lecture 20: Perturbation theory. Taught by Prof. Arun Paramekanti

### 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.

### Perturbation theory

Given a $$2 \times 2$$ Hamiltonian $$H = H_0 + V$$, where

\label{eqn:qmLecture20:20}
H =
\begin{bmatrix}
a & c \\
c^\conj & b
\end{bmatrix}

which has eigenvalues

\label{eqn:qmLecture20:40}
\lambda_\pm = \frac{a + b}{2} \pm \sqrt{ \lr{ \frac{a – b}{2}}^2 + \Abs{c}^2 }.

If $$c = 0$$,

\label{eqn:qmLecture20:60}
H_0 =
\begin{bmatrix}
a & 0 \\
0 & b
\end{bmatrix},

so

\label{eqn:qmLecture20:80}
V =
\begin{bmatrix}
0 & c \\
c^\conj & 0
\end{bmatrix}.

Suppose that $$\Abs{c} \ll \Abs{a – b}$$, then

\label{eqn:qmLecture20:100}
\lambda_\pm \approx \frac{a + b}{2} \pm \Abs{ \frac{a – b}{2} } \lr{ 1 + 2 \frac{\Abs{c}^2}{\Abs{a – b}^2} }.

If $$a > b$$, then

\label{eqn:qmLecture20:120}
\lambda_\pm \approx \frac{a + b}{2} \pm \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} }.

\label{eqn:qmLecture20:140}
\begin{aligned}
\lambda_{+}
&= \frac{a + b}{2} + \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&= a + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= a + \frac{\Abs{c}^2}{a – b},
\end{aligned}

and
\label{eqn:qmLecture20:680}
\begin{aligned}
\lambda_{-}
&= \frac{a + b}{2} – \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&=
b + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= b + \frac{\Abs{c}^2}{a – b}.
\end{aligned}

This adiabatic evolution displays a “level repulsion”, quadradic in $$\Abs{c}$$ as sketched in fig. 1, and is described as a non-degenerate perbutation.

If $$\Abs{c} \gg \Abs{a -b}$$, then

\label{eqn:qmLecture20:160}
\begin{aligned}
\lambda_\pm
&= \frac{a + b}{2} \pm \Abs{c} \sqrt{ 1 + \inv{\Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&\approx \frac{a + b}{2} \pm \Abs{c} \lr{ 1 + \inv{2 \Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&= \frac{a + b}{2} \pm \Abs{c} \pm \frac{\lr{a – b}^2}{8 \Abs{c}}.
\end{aligned}

Here we loose the adiabaticity, and have “level repulsion” that is linear in $$\Abs{c}$$, as sketched in fig. 2. We no longer have the sign of $$a – b$$ in the expansion. This is described as a degenerate perbutation.

fig. 2. Degenerate perbutation

### General non-degenerate perturbation

Given an unperturbed system with solutions of the form

\label{eqn:qmLecture20:180}
H_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}},

we want to solve the perturbed Hamiltonian equation

\label{eqn:qmLecture20:200}
\lr{ H_0 + \lambda V } \ket{ n } = \lr{ E_n^{(0)} + \Delta n } \ket{n}.

Here $$\Delta n$$ is an energy shift as that goes to zero as $$\lambda \rightarrow 0$$. We can write this as

\label{eqn:qmLecture20:220}
\lr{ E_n^{(0)} – H_0 } \ket{ n } = \lr{ \lambda V – \Delta_n } \ket{n}.

We are hoping to iterate with application of the inverse to an initial estimate of $$\ket{n}$$

\label{eqn:qmLecture20:240}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n}.

This gets us into trouble if $$\lambda \rightarrow 0$$, which can be fixed by using

\label{eqn:qmLecture20:260}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },

which can be seen to be a solution to \ref{eqn:qmLecture20:220}. We want to ask if

\label{eqn:qmLecture20:280}
\lr{ \lambda V – \Delta_n } \ket{n} ,

contains a bit of $$\ket{ n^{(0)} }$$? To determine this act with $$\bra{n^{(0)}}$$ on the left

\label{eqn:qmLecture20:300}
\begin{aligned}
\bra{ n^{(0)} } \lr{ \lambda V – \Delta_n } \ket{n}
&=
\bra{ n^{(0)} } \lr{ E_n^{(0)} – H_0 } \ket{n} \\
&=
\lr{ E_n^{(0)} – E_n^{(0)} } \braket{n^{(0)}}{n} \\
&=
0.
\end{aligned}

This shows that $$\ket{n}$$ is entirely orthogonal to $$\ket{n^{(0)}}$$.

Define a projection operator

\label{eqn:qmLecture20:320}
P_n = \ket{n^{(0)}}\bra{n^{(0)}},

which has the idempotent property $$P_n^2 = P_n$$ that we expect of a projection operator.

Define a rejection operator
\label{eqn:qmLecture20:340}
\overline{{P}}_n
= 1 –
\ket{n^{(0)}}\bra{n^{(0)}}
= \sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}.

Because $$\ket{n}$$ has no component in the direction $$\ket{n^{(0)}}$$, the rejection operator can be inserted much like we normally do with the identity operator, yielding

\label{eqn:qmLecture20:360}
\ket{n}’ = \lr{ E_n^{(0)} – H_0 }^{-1} \overline{{P}}_n \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },

valid for any initial $$\ket{n}$$.

### Power series perturbation expansion

Instead of iterating, suppose that the unknown state and unknown energy difference operator can be expanded in a $$\lambda$$ power series, say

\label{eqn:qmLecture20:380}
\ket{n}
=
\ket{n_0}
+ \lambda \ket{n_1}
+ \lambda^2 \ket{n_2}
+ \lambda^3 \ket{n_3} + \cdots

and

\label{eqn:qmLecture20:400}
\Delta_{n} = \Delta_{n_0}
+ \lambda \Delta_{n_1}
+ \lambda^2 \Delta_{n_2}
+ \lambda^3 \Delta_{n_3} + \cdots

We usually interpret functions of operators in terms of power series expansions. In the case of $$\lr{ E_n^{(0)} – H_0 }^{-1}$$, we have a concrete interpretation when acting on one of the unpertubed eigenstates

\label{eqn:qmLecture20:420}
\inv{ E_n^{(0)} – H_0 } \ket{m^{(0)}} =
\inv{ E_n^{(0)} – E_m^0 } \ket{m^{(0)}}.

This gives

\label{eqn:qmLecture20:440}
\ket{n}
=
\inv{ E_n^{(0)} – H_0 }
\sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}
\lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },

or

\label{eqn:qmLecture20:460}
\boxed{
\ket{n}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_n } \ket{n}.
}

From \ref{eqn:qmLecture20:220}, note that

\label{eqn:qmLecture20:500}
\Delta_n =
\frac{\bra{n^{(0)}} \lambda V \ket{n}}{\braket{n^0}{n}},

however, we will normalize by setting $$\braket{n^0}{n} = 1$$, so

\label{eqn:qmLecture20:521}
\boxed{
\Delta_n =
\bra{n^{(0)}} \lambda V \ket{n}.
}

### to $$O(\lambda^0)$$

If all $$\lambda^n, n > 0$$ are zero, then we have

\label{eqn:qmLecture20:780}
\label{eqn:qmLecture20:740}
\ket{n_0}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ – \Delta_{n_0} } \ket{n_0}

\label{eqn:qmLecture20:800}
\Delta_{n_0} \braket{n^{(0)}}{n_0} = 0

so

\label{eqn:qmLecture20:540}
\begin{aligned}
\ket{n_0} &= \ket{n^{(0)}} \\
\Delta_{n_0} &= 0.
\end{aligned}

### to $$O(\lambda^1)$$

Requiring identity for all $$\lambda^1$$ terms means

\label{eqn:qmLecture20:760}
\ket{n_1} \lambda
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_{n_1} \lambda } \ket{n_0},

so

\label{eqn:qmLecture20:560}
\ket{n_1}
=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}} \bra{ m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} } \ket{n_0}.

With the assumption that $$\ket{n^{(0)}}$$ is normalized, and with the shorthand

\label{eqn:qmLecture20:600}
V_{m n} = \bra{ m^{(0)}} V \ket{n^{(0)}},

that is

\label{eqn:qmLecture20:580}
\begin{aligned}
\ket{n_1}
&=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}
\\
\Delta_{n_1} &= \bra{ n^{(0)} } V \ket{ n^0} = V_{nn}.
\end{aligned}

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

The second order perturbation states are found by selecting only the $$\lambda^2$$ contributions to

\label{eqn:qmLecture20:820}
\lambda^2 \ket{n_2}
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – (\lambda \Delta_{n_1} + \lambda^2 \Delta_{n_2}) }
\lr{
\ket{n_0}
+ \lambda \ket{n_1}
}.

Because $$\ket{n_0} = \ket{n^{(0)}}$$, the $$\lambda^2 \Delta_{n_2}$$ is killed, leaving

\label{eqn:qmLecture20:840}
\begin{aligned}
\ket{n_2}
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\ket{n_1} \\
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\sum_{l \ne n}
\frac{
\ket{l^{(0)}}
}
{
E_n^{(0)} – E_l^{(0)}
}
V_{l n},
\end{aligned}

which can be written as

\label{eqn:qmLecture20:620}
\ket{n_2}
=
\sum_{l,m \ne n}
\ket{m^{(0)}}
\frac{V_{m l} V_{l n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}

\sum_{m \ne n}
\ket{m^{(0)}}
\frac{V_{n n} V_{m n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.

For the second energy perturbation we have

\label{eqn:qmLecture20:860}
\lambda^2 \Delta_{n_2} =
\bra{n^{(0)}} \lambda V \lr{ \lambda \ket{n_1} },

or

\label{eqn:qmLecture20:880}
\begin{aligned}
\Delta_{n_2}
&=
\bra{n^{(0)}} V \ket{n_1} \\
&=
\bra{n^{(0)}} V
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}.
\end{aligned}

That is

\label{eqn:qmLecture20:900}
\Delta_{n_2}
=
\sum_{m \ne n} \frac{V_{n m} V_{m n} }{E_n^{(0)} – E_m^{(0)}}.

### to $$O(\lambda^3)$$

Similarily, it can be shown that

\label{eqn:qmLecture20:640}
\Delta_{n_3} =
\sum_{l, m \ne n} \frac{V_{n m} V_{m l} V_{l n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}

\sum_{ m \ne n} \frac{V_{n m} V_{n n} V_{m n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.

In general, the energy perturbation is given by

\label{eqn:qmLecture20:660}
\Delta_n^{(l)} = \bra{n^{(0)}} V \ket{n^{(l-1)}}.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.