## 2D SHO xy perturbation

### Q: [1] pr. 5.4

Given a 2D SHO with Hamiltonian

\label{eqn:2dHarmonicOscillatorXYPerturbation:20}
H_0 = \inv{2m} \lr{ p_x^2 + p_y^2 } + \frac{m \omega^2}{2} \lr{ x^2 + y^2 },

• (a)
What are the energies and degeneracies of the three lowest states?

• (b)
With perturbation

\label{eqn:2dHarmonicOscillatorXYPerturbation:40}
V = m \omega^2 x y,

calculate the first order energy perturbations and the zeroth order perturbed states.

• (c)
Solve the $$H_0 + \delta V$$ problem exactly, and compare.

### A: part (a)

Recall that we have

\label{eqn:2dHarmonicOscillatorXYPerturbation:60}
H \ket{n_1, n_2} =
\Hbar\omega
\lr{
n_1 + n_2 + 1
}
\ket{n_1, n_2},

So the three lowest energy states are $$\ket{0,0}, \ket{1,0}, \ket{0,1}$$ with energies $$\Hbar \omega, 2 \Hbar \omega, 2 \Hbar \omega$$ respectively (with a two fold degeneracy for the second two energy eigenkets).

### A: part (b)

Consider the action of $$x y$$ on the $$\beta = \setlr{ \ket{0,0}, \ket{1,0}, \ket{0,1} }$$ subspace. Those are

\label{eqn:2dHarmonicOscillatorXYPerturbation:200}
\begin{aligned}
x y \ket{0,0}
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \lr{ b + b^\dagger } \ket{0,0} \\
&=
\frac{x_0^2}{2} \lr{ b + b^\dagger } \ket{1,0} \\
&=
\frac{x_0^2}{2} \ket{1,1}.
\end{aligned}

\label{eqn:2dHarmonicOscillatorXYPerturbation:220}
\begin{aligned}
x y \ket{1, 0}
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \lr{ b + b^\dagger } \ket{1,0} \\
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \ket{1,1} \\
&=
\frac{x_0^2}{2} \lr{ \ket{0,1} + \sqrt{2} \ket{2,1} } .
\end{aligned}

\label{eqn:2dHarmonicOscillatorXYPerturbation:240}
\begin{aligned}
x y \ket{0, 1}
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \lr{ b + b^\dagger } \ket{0,1} \\
&=
\frac{x_0^2}{2} \lr{ b + b^\dagger } \ket{1,1} \\
&=
\frac{x_0^2}{2} \lr{ \ket{1,0} + \sqrt{2} \ket{1,2} }.
\end{aligned}

The matrix representation of $$m \omega^2 x y$$ with respect to the subspace spanned by basis $$\beta$$ above is

\label{eqn:2dHarmonicOscillatorXYPerturbation:260}
x y
\sim
\inv{2} \Hbar \omega
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{bmatrix}.

This diagonalizes with

\label{eqn:2dHarmonicOscillatorXYPerturbation:300}
U
=
\begin{bmatrix}
1 & 0 \\
0 & \tilde{U}
\end{bmatrix}

\label{eqn:2dHarmonicOscillatorXYPerturbation:320}
\tilde{U}
=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1 \\
\end{bmatrix}

\label{eqn:2dHarmonicOscillatorXYPerturbation:340}
D =
\inv{2} \Hbar \omega
\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1 \\
\end{bmatrix}

\label{eqn:2dHarmonicOscillatorXYPerturbation:360}
x y = U D U^\dagger = U D U.

The unperturbed Hamiltonian in the original basis is

\label{eqn:2dHarmonicOscillatorXYPerturbation:380}
H_0
=
\Hbar \omega
\begin{bmatrix}
1 & 0 \\
0 & 2 I
\end{bmatrix},

So the transformation to the diagonal $$x y$$ basis leaves the initial Hamiltonian unaltered

\label{eqn:2dHarmonicOscillatorXYPerturbation:400}
\begin{aligned}
H_0′
&= U^\dagger H_0 U \\
&=
\Hbar \omega
\begin{bmatrix}
1 & 0 \\
0 & \tilde{U} 2 I \tilde{U}
\end{bmatrix} \\
&=
\Hbar \omega
\begin{bmatrix}
1 & 0 \\
0 & 2 I
\end{bmatrix}.
\end{aligned}

Now we can compute the first order energy shifts almost by inspection. Writing the new basis as $$\beta’ = \setlr{ \ket{0}, \ket{1}, \ket{2} }$$ those energy shifts are just the diagonal elements from the $$x y$$ operators matrix representation

\label{eqn:2dHarmonicOscillatorXYPerturbation:420}
\begin{aligned}
E^{{(1)}}_0 &= \bra{0} V \ket{0} = 0 \\
E^{{(1)}}_1 &= \bra{1} V \ket{1} = \inv{2} \Hbar \omega \\
E^{{(1)}}_2 &= \bra{2} V \ket{2} = -\inv{2} \Hbar \omega.
\end{aligned}

The new energies are

\label{eqn:2dHarmonicOscillatorXYPerturbation:440}
\begin{aligned}
E_0 &\rightarrow \Hbar \omega \\
E_1 &\rightarrow \Hbar \omega \lr{ 2 + \delta/2 } \\
E_2 &\rightarrow \Hbar \omega \lr{ 2 – \delta/2 }.
\end{aligned}

### A: part (c)

For the exact solution, it’s possible to rotate the coordinate system in a way that kills the explicit $$x y$$ term of the perturbation. That we could do this for $$x, y$$ operators wasn’t obvious to me, but after doing so (and rotating the momentum operators the same way) the new operators still have the required commutators. Let

\label{eqn:2dHarmonicOscillatorXYPerturbation:80}
\begin{aligned}
\begin{bmatrix}
u \\
v
\end{bmatrix}
&=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} \\
&=
\begin{bmatrix}
x \cos\theta + y \sin\theta \\
-x \sin\theta + y \cos\theta
\end{bmatrix}.
\end{aligned}

Similarly, for the momentum operators, let
\label{eqn:2dHarmonicOscillatorXYPerturbation:100}
\begin{aligned}
\begin{bmatrix}
p_u \\
p_v
\end{bmatrix}
&=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
p_x \\
p_y
\end{bmatrix} \\
&=
\begin{bmatrix}
p_x \cos\theta + p_y \sin\theta \\
-p_x \sin\theta + p_y \cos\theta
\end{bmatrix}.
\end{aligned}

For the commutators of the new operators we have

\label{eqn:2dHarmonicOscillatorXYPerturbation:120}
\begin{aligned}
\antisymmetric{u}{p_u}
&=
\antisymmetric{x \cos\theta + y \sin\theta}{p_x \cos\theta + p_y \sin\theta} \\
&=
\antisymmetric{x}{p_x} \cos^2\theta + \antisymmetric{y}{p_y} \sin^2\theta \\
&=
i \Hbar \lr{ \cos^2\theta + \sin^2\theta } \\
&=
i\Hbar.
\end{aligned}

\label{eqn:2dHarmonicOscillatorXYPerturbation:140}
\begin{aligned}
\antisymmetric{v}{p_v}
&=
\antisymmetric{-x \sin\theta + y \cos\theta}{-p_x \sin\theta + p_y \cos\theta} \\
&=
\antisymmetric{x}{p_x} \sin^2\theta + \antisymmetric{y}{p_y} \cos^2\theta \\
&=
i \Hbar.
\end{aligned}

\label{eqn:2dHarmonicOscillatorXYPerturbation:160}
\begin{aligned}
\antisymmetric{u}{p_v}
&=
\antisymmetric{x \cos\theta + y \sin\theta}{-p_x \sin\theta + p_y \cos\theta} \\
&= \cos\theta \sin\theta \lr{ -\antisymmetric{x}{p_x} + \antisymmetric{y}{p_p} } \\
&=
0.
\end{aligned}

\label{eqn:2dHarmonicOscillatorXYPerturbation:180}
\begin{aligned}
\antisymmetric{v}{p_u}
&=
\antisymmetric{-x \sin\theta + y \cos\theta}{p_x \cos\theta + p_y \sin\theta} \\
&= \cos\theta \sin\theta \lr{ -\antisymmetric{x}{p_x} + \antisymmetric{y}{p_p} } \\
&=
0.
\end{aligned}

We see that the new operators are canonical conjugate as required. For this problem, we just want a 45 degree rotation, with

\label{eqn:2dHarmonicOscillatorXYPerturbation:460}
\begin{aligned}
x &= \inv{\sqrt{2}} \lr{ u + v } \\
y &= \inv{\sqrt{2}} \lr{ u – v }.
\end{aligned}

We have
\label{eqn:2dHarmonicOscillatorXYPerturbation:480}
\begin{aligned}
x^2 + y^2
&=
\inv{2} \lr{ (u+v)^2 + (u-v)^2 } \\
&=
\inv{2} \lr{ 2 u^2 + 2 v^2 + 2 u v – 2 u v } \\
&=
u^2 + v^2,
\end{aligned}

\label{eqn:2dHarmonicOscillatorXYPerturbation:500}
\begin{aligned}
p_x^2 + p_y^2
&=
\inv{2} \lr{ (p_u+p_v)^2 + (p_u-p_v)^2 } \\
&=
\inv{2} \lr{ 2 p_u^2 + 2 p_v^2 + 2 p_u p_v – 2 p_u p_v } \\
&=
p_u^2 + p_v^2,
\end{aligned}

and
\label{eqn:2dHarmonicOscillatorXYPerturbation:520}
\begin{aligned}
x y
&=
\inv{2} \lr{ (u+v)(u-v) } \\
&=
\inv{2} \lr{ u^2 – v^2 }.
\end{aligned}

The perturbed Hamiltonian is

\label{eqn:2dHarmonicOscillatorXYPerturbation:540}
\begin{aligned}
H_0 + \delta V
&=
\inv{2m} \lr{ p_u^2 + p_v^2 }
+ \inv{2} m \omega^2 \lr{ u^2 + v^2 + \delta u^2 – \delta v^2 } \\
&=
\inv{2m} \lr{ p_u^2 + p_v^2 }
+ \inv{2} m \omega^2 \lr{ u^2(1 + \delta) + v^2 (1 – \delta) }.
\end{aligned}

In this coordinate system, the corresponding eigensystem is

\label{eqn:2dHarmonicOscillatorXYPerturbation:560}
H \ket{n_1, n_2}
= \Hbar \omega \lr{ 1 + n_1 \sqrt{1 + \delta} + n_2 \sqrt{ 1 – \delta } } \ket{n_1, n_2}.

For small $$\delta$$

\label{eqn:2dHarmonicOscillatorXYPerturbation:580}
n_1 \sqrt{1 + \delta} + n_2 \sqrt{ 1 – \delta }
\approx
n_1 + n_2
+ \inv{2} n_1 \delta
– \inv{2} n_2 \delta,

so
\label{eqn:2dHarmonicOscillatorXYPerturbation:600}
H \ket{n_1, n_2}
\approx \Hbar \omega \lr{ 1 + n_1 + n_2 + \inv{2} n_1 \delta – \inv{2} n_2 \delta
} \ket{n_1, n_2}.

The lowest order perturbed energy levels are

\label{eqn:2dHarmonicOscillatorXYPerturbation:620}
\ket{0,0} \rightarrow \Hbar \omega

\label{eqn:2dHarmonicOscillatorXYPerturbation:640}
\ket{1,0} \rightarrow \Hbar \omega \lr{ 2 + \inv{2} \delta }

\label{eqn:2dHarmonicOscillatorXYPerturbation:660}
\ket{0,1} \rightarrow \Hbar \omega \lr{ 2 – \inv{2} \delta }

The degeneracy of the $$\ket{0,1}, \ket{1,0}$$ states has been split, and to first order match the zeroth order perturbation result.

# 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

## Harmonic oscillator with energy shift

December 5, 2015 phy1520 No comments , ,

### Q: [1] pr 5.1

Given a perturbed 1D SHO Hamiltonian

\label{eqn:harmonicOscillatorEnergyShiftPertubation:20}
H = \inv{2m} p^2 + \inv{2} m \omega^2 x^2 + \lambda b x,

calculate the first non-zero perturbation to the ground state energy. Then solve for that energy directly and compare.

### A:

The first order energy shift is seen to be zero

\label{eqn:harmonicOscillatorEnergyShiftPertubation:40}
\begin{aligned}
\Delta_0^{(0)}
&= V_{00} \\
&= \bra{0} b x \ket{0} \\
&= \frac{x_0}{\sqrt{2}} \bra{0} a + a^\dagger \ket{0} \\
&= \frac{x_0}{\sqrt{2}} \braket{0}{1} \\
&= 0.
\end{aligned}

The first order perturbation to the ground state is

\label{eqn:harmonicOscillatorEnergyShiftPertubation:60}
\begin{aligned}
\ket{0^{(1)}}
&= \sum_{m \ne 0} \frac{ \ket{m} \bra{m} b x \ket{0} }{ \Hbar \omega/2 – \Hbar
\omega (m – 1/2) } \\
&= -b \frac{x_0}{\sqrt{2} \Hbar \omega} \sum_{m \ne 0} \frac{ \ket{m}
\braket{m}{1} }{ m } \\
&= -b \frac{x_0}{\sqrt{2} \Hbar \omega} \ket{1}.
\end{aligned}

The second order ground state energy perturbation is

\label{eqn:harmonicOscillatorEnergyShiftPertubation:80}
\begin{aligned}
\Delta_0^{(2)}
&=
\bra{0} b x \ket{0^{(1)}} \\
&=
\frac{b x_0}{\sqrt{2}} \bra{0} a + a^\dagger \lr{ -b \frac{x_0}{\sqrt{2} \Hbar \omega} \ket{1} } \\
&=
\frac{b x_0}{\sqrt{2}} \lr{ -b \frac{x_0}{\sqrt{2} \Hbar \omega} } \\
&=
-\frac{b^2 x_0^2}{ 2 \Hbar \omega } \\
&=
-\frac{b^2 }{ 2 \Hbar \omega } \frac{\Hbar}{m \omega} \\
&=
-\frac{b^2 }{ 2 m \omega^2 },
\end{aligned}

so the total energy perturbation up to second order is

\label{eqn:harmonicOscillatorEnergyShiftPertubation:100}
\Delta_0 = -\lambda^2 \frac{b^2 }{ 2 m \omega^2 }.

To compare to the exact result, rewrite the Hamiltonian as

\label{eqn:harmonicOscillatorEnergyShiftPertubation:120}
\begin{aligned}
H
&= \inv{2m} p^2 + \inv{2} m \omega^2 \lr{ x^2 + \frac{2 \lambda b x}{m \omega^2} } \\
&= \inv{2m} p^2 + \inv{2} m \omega^2 \lr{ x + \frac{\lambda b }{m \omega^2} }^2 – \inv{2} m \omega^2 \lr{ \frac{\lambda b }{m \omega^2} }^2.
\end{aligned}

The Hamiltonian is subject to a constant energy shift

\label{eqn:harmonicOscillatorEnergyShiftPertubation:140}
\begin{aligned}
\Delta E
&=
– \inv{2} m \omega^2 \frac{\lambda^2 b^2 }{m^2 \omega^4} \\
&=
– \frac{\lambda^2 b^2 }{2 m \omega^2}.
\end{aligned}

This is an exact match with the second order perturbation result of \ref{eqn:harmonicOscillatorEnergyShiftPertubation:100}.

# References

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

## PHY1520H Graduate Quantum Mechanics. Lecture 21: Non-degenerate perturbation. 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 [2] chap. 5 content.

### Non-degenerate perturbation theory. Recap.

\label{eqn:qmLecture21:20}
\ket{n} = \ket{n_0}
+ \lambda \ket{n_1}
+ \lambda^2 \ket{n_2}
+ \lambda^3 \ket{n_3} + \cdots

and

\label{eqn:qmLecture21:40}
\Delta_{n} = \Delta_{n_0}
+ \lambda \Delta_{n_1}
+ \lambda^2 \Delta_{n_2}
+ \lambda^3 \Delta_{n_3} + \cdots

\label{eqn:qmLecture21:60}
\begin{aligned}
\Delta_{n_1} &= \bra{n^{(0)}} V \ket{n^{(0)}} \\
\ket{n_0} &= \ket{n^{(0)}}
\end{aligned}

\label{eqn:qmLecture21:80}
\begin{aligned}
\Delta_{n_2} &= \sum_{m \ne n} \frac{\Abs{\bra{n^{(0)}} V \ket{m^{(0)}}}^2}{E_n^{(0)} – E_m^{(0)}} \\
\ket{n_1} &= \sum_{m \ne n} \frac{ \ket{m^{(0)}} V_{mn} }{E_n^{(0)} – E_m^{(0)}}
\end{aligned}

### Example: Stark effect

\label{eqn:qmLecture21:100}
H = H_{\textrm{atom}} + e \mathcal{E} z,

where $$H_{\textrm{atom}}$$ is assumed to be Hydrogen-like with Hamiltonian

\label{eqn:qmLecture21:120}
H_{\textrm{atom}} = \frac{\BP^2}{2m} – \frac{e^2}{4 \pi \epsilon_0 r},

and wave functions

\label{eqn:qmLecture21:140}
\braket{\Br}{\psi_{n l m}} = R_{n l}(r) Y_{lm}( \theta, \phi )

For the first level correction to the energy

\label{eqn:qmLecture21:160}
\begin{aligned}
\Delta_1
&= \bra{\psi_{100}} e \mathcal{E} z \ket{ \psi_{100}} \\
&= e \mathcal{E} \int \frac{d\Omega}{4 \pi} \cos \theta \int dr r^2 R_{100}^2(r)
\end{aligned}

The cosine integral is obliterated, so we have $$\Delta_1 = 0$$.

How about the second order energy correction? That is

\label{eqn:qmLecture21:180}
\Delta_2 = \sum_{n l m \ne 100} \frac{
\Abs{ \bra{\psi_{100}} e \mathcal{E} z \ket{ n l m }}^2
}{
E_{100}^{(0)} – E_{n l m}
}

The matrix element in the numerator is the absolute square of

\label{eqn:qmLecture21:200}
V_{100,nlm}
=
e \mathcal{E} \int d\Omega \inv{\sqrt{ 4 \pi } }
\cos\theta Y_{l m}(\theta, \phi)
\int dr r^3 R_{100}(r) R_{n l}(r).

For all $$m \ne 0$$, $$Y_{lm}$$ includes a $$e^{i m \phi}$$ factor, so this cosine integral is zero. For $$m = 0$$, each of the $$Y_{lm}$$ functions appears to contain either even or odd powers of cosines. For example:

\label{eqn:qmLecture21:760}
\begin{aligned}
Y_{00} &= \frac{1}{2 \sqrt{\pi}} \\
Y_{10} &= \frac{1}{2} \sqrt{\frac{3}{\pi }} \cos(t) \\
Y_{20} &= \frac{1}{4} \sqrt{\frac{5}{\pi }} \lr{(3 \cos^2(t)-1} \\
Y_{30} &= \frac{1}{4} \sqrt{\frac{7}{\pi }} \lr{(5 \cos^3(t)-3 \cos(t)} \\
Y_{40} &= \frac{3 \lr{(35 \cos^4(t)-30 \cos^2(t)+3}}{16 \sqrt{\pi }} \\
Y_{50} &= \frac{1}{16} \sqrt{\frac{11}{\pi }} \lr{(63 \cos^5(t)-70 \cos^3(t)+15 \cos(t)} \\
Y_{60} &= \frac{1}{32} \sqrt{\frac{13}{\pi }} \lr{(231 \cos^6(t)-315 \cos^4(t)+105 \cos^2(t)-5} \\
Y_{70} &= \frac{1}{32} \sqrt{\frac{15}{\pi }} \lr{(429 \cos^7(t)-693 \cos^5(t)+315 \cos^3(t)-35 \cos(t)} \\
Y_{80} &= \frac{1}{256} \sqrt{\frac{17}{\pi }} \lr{(6435 \cos^8(t)-12012 \cos^6(t)+6930 \cos^4(t)-1260 \cos^2(t)+35 } \\
\end{aligned}

This shows that for even $$2k = l$$, the cosine integral is zero

\label{eqn:qmLecture21:780}
\int_0^\pi \sin\theta \cos\theta \sum_k a_k \cos^{2k}\theta d\theta
=
0,

since $$\cos^{2k}(\theta)$$ is even and $$\sin\theta \cos\theta$$ is odd over the same interval. We find zero for $$\int_0^\pi \sin\theta \cos\theta Y_{30}(\theta, \phi) d\theta$$, and Mathematica appears to show that the rest of these integrals for $$l > 1$$ are also zero.

FIXME: find the property of the spherical harmonics that can be used to prove that this is true in general for $$l > 1$$.

This leaves

\label{eqn:qmLecture21:220}
\begin{aligned}
\Delta_2
&= \sum_{n \ne 1} \frac{
\Abs{ \bra{\psi_{100}} e \mathcal{E} z \ket{ n 1 0 }}^2
}{
E_{100}^{(0)} – E_{n 1 0}
} \\
&=
-e^2 \mathcal{E}^2
\sum_{n \ne 1} \frac{
\Abs{ \bra{\psi_{100}} z \ket{ n 1 0 }}^2
}{
E_{n 1 0}
-E_{100}^{(0)}
}.
\end{aligned}

This is sometimes written in terms of a polarizability $$\alpha$$

\label{eqn:qmLecture21:260}
\Delta_2 = -\frac{\mathcal{E}^2}{2} \alpha,

where

\label{eqn:qmLecture21:280}
\alpha =
2 e^2
\sum_{n \ne 1} \frac{
\Abs{ \bra{\psi_{100}} z \ket{ n 1 0 }}^2
}{
E_{n 1 0}
-E_{100}^{(0)}
}.

With
\label{eqn:qmLecture21:840}
\BP = \alpha \boldsymbol{\mathcal{E}},

the energy change upon turning on the electric field from $$0 \rightarrow \mathcal{E}$$ is simply $$– \BP \cdot d\boldsymbol{\mathcal{E}}$$ integrated from $$0 \rightarrow \mathcal{E}$$. Putting $$\BP = \alpha \mathcal{E} \zcap$$, we have

\label{eqn:qmLecture21:400}
\begin{aligned}
– \int_0^\mathcal{E} p_z d\mathcal{E}
&=
– \int_0^\mathcal{E} \alpha \mathcal{E} d\mathcal{E} \\
&=
– \inv{2} \alpha \mathcal{E}^2
\end{aligned}

leading to an energy change $$– \alpha \mathcal{E}^2/2$$, so we can directly compute $$\expectation{\BP}$$ or we can compute change in energy, and both contain information about the polarization factor $$\alpha$$.

There is an exact answer to the sum \ref{eqn:qmLecture21:280}, but we aren’t going to try to get it here. Instead let’s look for bounds

\label{eqn:qmLecture21:240}
\Delta_2^{\mathrm{min}} < \Delta_2 < \Delta_2^{\mathrm{max}} $$\label{eqn:qmLecture21:320} \alpha^{\mathrm{min}} = 2 e^2 \frac{ \Abs{ \bra{\psi_{100}} z \ket{\psi_{210}} }^2 }{E_{210}^{(0)} - E_{100}^{(0)}}$$ For the hydrogen atom we have $$\label{eqn:qmLecture21:820} E_n = -\frac{ e^2}{ 2 n^2 a_0 },$$ allowing any difference of energy levels to be expressed as a fraction of the ground state energy, such as $$\label{eqn:qmLecture21:340} E_{210}^{(0)} = \inv{4} E_{100}^{(0)} = \inv{4} \frac{ -\Hbar^2 }{ 2 m a_0^2 }$$ So $$\label{eqn:qmLecture21:360} E_{210}^{(0)} - E_{100}^{(0)} = \frac{3}{4} \frac{ \Hbar^2 }{ 2 m a_0^2 }$$ In the numerator we have \label{eqn:qmLecture21:380} \begin{aligned} \bra{\psi_{100}} z \ket{\psi_{210}} &= \int r^2 d\Omega \lr{ \inv{\sqrt{\pi} a_0^{3/2}} e^{-r/a_0} } r \cos\theta \lr{ \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \cos\theta } \\ &= (2 \pi) \inv{\sqrt{\pi}} \inv{4 \sqrt{2 \pi} } a_0 \int_0^\pi d\theta \sin\theta \cos^2\theta \int_0^\infty \frac{dr}{a_0} \frac{r^4}{a_0^4} e^{-r/a_0 - r/2 a_0} \\ &= (2 \pi) \inv{\sqrt{\pi}} \inv{4 \sqrt{2 \pi} } a_0 \lr{ \evalrange{-\frac{u^3}{3}}{1}{-1} } \int_0^\infty s^4 ds e^{- 3 s/2 } \\ &= 2 \inv{4 \sqrt{2} } a_0 \lr{ \evalrange{-\frac{u^3}{3}}{1}{-1} } \int_0^\infty s^4 ds e^{- 3 s/2 } \\ &= \inv{2 \sqrt{2}} \frac{2}{3} a_0 \frac{256}{81} \\ &= \frac{1}{3 \sqrt{2} } \frac{ 256}{81} a_0 \approx 0.75 a_0. \end{aligned} This gives \label{eqn:qmLecture21:420} \begin{aligned} \alpha^{\mathrm{min}} &= \frac{ 2 e^2 (0.75)^2 a_0^2 }{ \frac{3}{4} \frac{\Hbar^2}{2 m a_0^2} } \\ &= \frac{6}{4} \frac{2 m e^2 a_0^4}{ \Hbar^2 } \\ &= 3 \frac{m e^2 a_0^4}{ \Hbar^2 } \\ &= 3 \frac{ 4 \pi \epsilon_0 }{a_0} a_0^4 \\ &\approx 4 \pi \epsilon_0 a_0^3 \times 3. \end{aligned} The factor $$4 \pi \epsilon_0 a_0^3$$ are the natural units for the polarizability. There is a neat trick that generalizes to many problems to find the upper bound. Recall that the general polarizability was $$\label{eqn:qmLecture21:440} \alpha = 2 e^2 \sum_{nlm \ne 100} \frac{ \Abs{ \bra{100} z \ket{ n l m }}^2 }{ E_{n l m} -E_{100}^{(0)} }.$$ If we are looking for the upper bound, and replace the denominator by the smallest energy difference that will be encountered, it can be brought out of the sum, for $$\label{eqn:qmLecture21:460} \alpha^{\mathrm{max}} = 2 e^2 \inv{E_{2 1 0} -E_{100}^{(0)} } \sum_{nlm \ne 100} \bra{100} z \ket{ n l m } \bra{nlm} z \ket{ 100 }$$ Because $$\bra{nlm} z \ket{100} = 0$$, the constraint in the sum can be removed, and the identity summation evaluated \label{eqn:qmLecture21:480} \begin{aligned} \alpha^{\mathrm{max}} &= 2 e^2 \inv{E_{2 1 0} -E_{100}^{(0)} } \sum_{nlm} \bra{100} z \ket{ n l m } \bra{nlm} z \ket{ 100 } \\ &= \frac{2 e^2 }{ \frac{3}{4} \frac{\Hbar^2}{ 2 m a_0^2} } \bra{100} z^2 \ket{ 100 } \\ &= \frac{16 e^2 m a_0^2 }{ 3 \Hbar^2 } \times a_0^2 \\ &= 4 \pi \epsilon_0 a_0^3 \times \frac{16}{3}. \end{aligned} The bounds are $$\label{eqn:qmLecture21:520} \boxed{ 3 \ge \frac{\alpha}{\alpha^{\mathrm{at}}} < \frac{16}{3}, }$$ where $$\label{eqn:qmLecture21:560} \alpha^{\mathrm{at}} = 4 \pi \epsilon_0 a_0^3.$$ The actual value is $$\label{eqn:qmLecture21:580} \frac{\alpha}{\alpha^{\mathrm{at}}} = \frac{9}{2}.$$

### Example: Computing the dipole moment

\label{eqn:qmLecture21:600}
\expectation{P_z}
= \alpha \mathcal{E}
= \bra{\psi_{100}} e z \ket{\psi_{100}}.

Without any perturbation this is zero. After perturbation, retaining only the terms that are first order in $$\delta \psi_{100}$$ we have

\label{eqn:qmLecture21:620}
\bra{\psi_{100} + \delta \psi_{100}} e z \ket{\psi_{100} + \delta \psi_{100}}
\approx
\bra{\psi_{100}} e z \ket{\delta \psi_{100}}
+
\bra{\delta \psi_{100}} e z \ket{\psi_{100}}.

### Next time: Van der Walls

We will look at two hyrdogenic atomic systems interacting where the pair of nuclei are supposed to be infinitely heavy and stationary. The wave functions each set of atoms are individually known, but we can consider the problem of the interactions of atom 1’s electrons with atom 2’s nucleus and atom 2’s electrons, and also the opposite interactions of atom 2’s electrons with atom 1’s nucleus and its electrons. This leads to a result that is linear in the electric field (unlike the above result, which is called the quadratic Stark effect).

### Appendix. Hydrogen wavefunctions

From [3], with the $$a_0$$ factors added in.

\label{eqn:qmLecture21:660}
\psi_{1 s} = \psi_{100} = \inv{\sqrt{\pi} a_0^{3/2}} e^{-r/a_0}

\label{eqn:qmLecture21:680}
\psi_{2 s} = \psi_{200} = \inv{4 \sqrt{2 \pi} a_0^{3/2}} \lr{ 2 – \frac{r}{a_0} } e^{-r/2a_0}

\label{eqn:qmLecture21:700}
\psi_{2 p_x} = \inv{\sqrt{2}} \lr{ \psi_{2,1,1} – \psi_{2,1,-1} }
= \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \sin\theta\cos\phi

\label{eqn:qmLecture21:720}
\psi_{2 p_y} = \frac{i}{\sqrt{2}} \lr{ \psi_{2,1,1} + \psi_{2,1,-1} }
= \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \sin\theta\sin\phi

\label{eqn:qmLecture21:740}
\psi_{2 p_z} = \psi_{210} = \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \cos\theta

I looked to [1] to see where to add in the $$a_0$$ factors.

# References

[1] Carl R. Nave. Hydrogen Wavefunctions, 2015. URL http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html. [Online; accessed 03-Dec-2015].

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

[3] Robert Field Troy Van Voorhis. Hydrogen Atom, 2013. URL http://ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2013/lecture-notes/MIT5_61F13_Lecture19-20.pdf. [Online; accessed 03-Dec-2015].

## PHY1520H Graduate Quantum Mechanics. Lecture 19: Variational method. Taught by Prof. Arun Paramekanti

November 27, 2015 phy1520 No comments , ,

### 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 \textchapref{{5}} [1] content.

### Variational method

Today we want to use the variational degree of freedom to try to solve some problems that we don’t have analytic solutions for.

### Anharmonic oscillator

\label{eqn:qmLecture19:20}
V(x) = \inv{2} m \omega^2 x^2 + \lambda x^4, \qquad \lambda \ge 0.

With the potential growing faster than the harmonic oscillator, which had a ground state solution

\label{eqn:qmLecture19:40}
\psi(x) = \inv{\pi^{1/4}} \inv{a_0^{1/2} } e^{- x^2/2 a_0^2},

where
\label{eqn:qmLecture19:60}
a_0 = \sqrt{\frac{\Hbar}{m \omega}}.

Let’s try allowing $$a_0 \rightarrow a$$, to be a variational degree of freedom

\label{eqn:qmLecture19:80}
\psi_a(x) = \inv{\pi^{1/4}} \inv{a^{1/2} } e^{- x^2/2 a^2},

\label{eqn:qmLecture19:100}
\bra{\psi_a} H \ket{\psi_a}
=
\bra{\psi_a} \frac{p^2}{2m} + \inv{2} m \omega^2 x^2 + \lambda x^4 \ket{\psi_a}

We can find
\label{eqn:qmLecture19:120}
\expectation{x^2} = \inv{2} a^2

\label{eqn:qmLecture19:140}
\expectation{x^4} = \frac{3}{4} a^4

Define

\label{eqn:qmLecture19:160}
\tilde{\omega} = \frac{\Hbar}{m a^2},

so that

\label{eqn:qmLecture19:180}
\overline{{E}}_a
=
\bra{\psi_a} \lr{ \frac{p^2}{2m} + \inv{2} m \tilde{\omega}^2 x^2 }
+ \lr{
\inv{2} m \lr{ \omega^2 – \tilde{\omega}^2 } x^2
+
\lambda x^4 }
\ket{\psi_a}
=
\inv{2} \Hbar \tilde{\omega} + \inv{2} m \lr{ \omega^2 – \tilde{\omega}^2 } \inv{2} a^2 + \frac{3}{4} \lambda a^4.

Write this as
\label{eqn:qmLecture19:200}
\overline{{E}}_{\tilde{\omega}}
=
\inv{2} \Hbar \tilde{\omega} + \inv{4} \frac{\Hbar}{\tilde{\omega}} \lr{ \omega^2 – \tilde{\omega}^2 } + \frac{3}{4} \lambda \frac{\Hbar^2}{m^2 \tilde{\omega}^2 }.

This might look something like fig. 1.

fig. 1: Energy after perturbation.

Demand that

\label{eqn:qmLecture19:220}
0
= \PD{\tilde{\omega}}{ \overline{{E}}_{\tilde{\omega}}}
=
\frac{\Hbar}{2} – \frac{\Hbar}{4} \frac{\omega^2}{\tilde{\omega}^2}
– \frac{\Hbar}{4}
+ \frac{3}{4} (-2) \frac{\lambda \Hbar^2}{m^2 \tilde{\omega}^3}
=
\frac{\Hbar}{4}
\lr{
1 – \frac{\omega^2}{\tilde{\omega}^2}
– 6 \frac{\lambda \Hbar}{m^2 \tilde{\omega}^3}
}

or
\label{eqn:qmLecture19:260}
\tilde{\omega}^3 – \omega^2 \tilde{\omega} – \frac{6 \lambda \Hbar}{m^2} = 0.

for $$\lambda a_0^4 \ll \Hbar \omega$$, we have something like $$\tilde{\omega} = \omega + \epsilon$$. Expanding \ref{eqn:qmLecture19:260} to first order in $$\epsilon$$, this gives

\label{eqn:qmLecture19:280}
\omega^3 + 3 \omega^2 \epsilon – \omega^2 \lr{ \omega + \epsilon } – \frac{6 \lambda \Hbar}{m^2} = 0,

so that

\label{eqn:qmLecture19:300}
2 \omega^2 \epsilon = \frac{6 \lambda \Hbar}{m^2},

and

\label{eqn:qmLecture19:320}
\Hbar \epsilon = \frac{ 3 \lambda \Hbar^2}{m^2 \omega^2 } = 3 \lambda a_0^4.

Plugging into

\label{eqn:qmLecture19:340}
\overline{{E}}_{\omega + \epsilon}
=
\inv{2} \Hbar \lr{ \omega + \epsilon }
+ \inv{4} \frac{\Hbar}{\omega} \lr{ -2 \omega \epsilon + \epsilon^2 } + \frac{3}{4} \lambda \frac{\Hbar^2}{m^2 \omega^2 }
\approx
\inv{2} \Hbar \lr{ \omega + \epsilon }
– \inv{2} \Hbar \epsilon
+ \frac{3}{4} \lambda \frac{\Hbar^2}{m^2 \omega^2 }
=
\inv{2} \Hbar \omega + \frac{3}{4} \lambda a_0^4.

With \ref{eqn:qmLecture19:320}, that is

\label{eqn:qmLecture19:540}
\overline{{E}}_{\tilde{\omega} = \omega + \epsilon} \approx \inv{2} \Hbar \lr{ \omega + \frac{\epsilon}{2} }.

The energy levels are shifted slightly for each shift in the Hamiltonian frequency.

What do we have in the extreme anharmonic limit, where $$\lambda a_0^4 \gg \Hbar \omega$$. Now we get

\label{eqn:qmLecture19:360}
\tilde{\omega}^\conj = \lr{ \frac{ 6 \Hbar \lambda }{m^2} }^{1/3},

and
\label{eqn:qmLecture19:380}
\overline{{E}}_{\tilde{\omega}^\conj} = \frac{\Hbar^{4/3} \lambda^{1/3}}{m^{2/3}} \frac{3}{8} 6^{1/3}.

(this last result is pulled from a web treatment somewhere of the anharmonic oscillator). Note that the first factor in this energy, with $$\Hbar^4 \lambda/m^2$$ traveling together could have been worked out on dimensional grounds.

This variational method tends to work quite well in these limits. For a system where $$m = \omega = \Hbar = 1$$, for this problem, we have

tab. 1: Comparing numeric and variational solutions.

### Example: (sketch) double well potential

fig. 2: Double well potential.

\label{eqn:qmLecture19:400}
V(x) = \frac{m \omega^2}{8 a^2} \lr{ x – a }^2\lr{ x + a}^2.

Note that this potential, and the Hamiltonian, both commute with parity.

We are interested in the regime where $$a_0^2 = \frac{\Hbar}{m \omega} \ll a^2$$.

Near $$x = \pm a$$, this will be approximately

\label{eqn:qmLecture19:420}
V(x) = \inv{2} m \omega^2 \lr{ x \pm a }^2.

Guessing a wave function that is an eigenstate of parity

\label{eqn:qmLecture19:440}
\Psi_{\pm} = g_{\pm} \lr{ \phi_{\textrm{R}}(x) \pm \phi_{\textrm{L}}(x) }.

perhaps looking like the even and odd functions sketched in fig. 3, and fig. 4.

fig. 3. Even double well function

fig. 4. Odd double well function

Using harmonic oscillator functions

\label{eqn:qmLecture19:460}
\begin{aligned}
\phi_{\textrm{L}}(x) &= \Psi_{{\textrm{H}}.{\textrm{O}}.}(x + a) \\
\phi_{\textrm{R}}(x) &= \Psi_{{\textrm{H}}.{\textrm{O}}.}(x – a)
\end{aligned}

After doing a lot of integral (i.e. in the problem set), we will see a splitting of the variational energy levels as sketched in fig. 5.

fig. 5. Splitting for double well potential.

This sort of level splitting was what was used in the very first mazers.

### Perturbation theory (outline)

Given

\label{eqn:qmLecture19:480}
H = H_0 + \lambda V,

where $$\lambda V$$ is “small”. We want to figure out the eigenvalues and eigenstates of this Hamiltonian

\label{eqn:qmLecture19:500}
H \ket{n} = E_n \ket{n}.

We don’t know what these are, but do know that

\label{eqn:qmLecture19:520}
H_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}}.

We are hoping that the level transitions have adiabatic transitions between the original and perturbed levels as sketched in fig. 6.

and not crossed level transitions as sketched in fig. 7.

fig. 7. Crossed level transitions.

If we have level crossings (which can in general occur), as opposed to adiabatic transitions, then we have no hope of using perturbation theory.

# References

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

## Final notes for ECE1254, Modelling of Multiphysics Systems

I’ve now finished my first grad course, Modelling of Multiphysics Systems, taught by Prof Piero Triverio.

I’ve posted notes for lectures and other material as I was taking the course, but now have an aggregated set of notes for the whole course posted.
This is now updated with all my notes from the lectures, solved problems, additional notes on auxillary topics I wanted to explore (like SVD), plus the notes from the Harmonic Balance report that Mike and I will be presenting in January.

This version of my notes also includes all the matlab figures regenerating using http://www.mathworks.com/matlabcentral/fileexchange/23629-export-fig, which allows a save-as pdf, which rescales much better than Matlab saveas() png’s when embedded in latex.  I’m not sure if that’s the best way to include Matlab figures in latex, but they are at least not fuzzy looking now.

All in all, I’m pretty pleased with my notes for this course.  They are a lot more readable than any of the ones I’ve done for the physics undergrad courses I was taking (http://peeterjoot.com/writing/).  While there was quite a lot covered in this course, the material really only requires an introductory circuits course and some basic math (linear algebra and intro calculus), so is pretty accessible.

This was a fun course.  I recall, back in ancient times when I was a first year student, being unsatisfied with all the ad-hoc strategies we used to solve circuits problems.  This finally answers the questions of how to tackle things more systematically.

Here’s the contents outline for these notes:

Preface
Lecture notes
1 nodal analysis
1.1 In slides
1.2 Mechanical structures example
1.3 Assembling system equations automatically. Node/branch method
1.4 Nodal Analysis
1.5 Modified nodal analysis (MNA)
2 solving large systems
2.1 Gaussian elimination
2.2 LU decomposition
2.3 Problems
3 numerical errors and conditioning
3.1 Strict diagonal dominance
3.2 Exploring uniqueness and existence
3.3 Perturbation and norms
3.4 Matrix norm
4 singular value decomposition, and conditioning number
4.1 Singular value decomposition
4.2 Conditioning number
5 sparse factorization
5.1 Fill ins
5.2 Markowitz product
5.3 Markowitz reordering
5.4 Graph representation
6.1 Summary of factorization costs
6.2 Iterative methods
6.4 Recap: Summary of Gradient method
6.6 Full Algorithm
6.7 Order analysis
6.9 Gershgorin circle theorem
6.10 Preconditioning
6.11 Symmetric preconditioning
6.13 Problems
7 solution of nonlinear systems
7.1 Nonlinear systems
7.2 Richardson and Linear Convergence
7.3 Newton’s method
7.4 Solution of N nonlinear equations in N unknowns
7.5 Multivariable Newton’s iteration
7.6 Automatic assembly of equations for nonlinear system
7.7 Damped Newton’s method
7.8 Continuation parameters
7.9 Singular Jacobians
7.10 Struts and Joints, Node branch formulation
7.11 Problems
8 time dependent systems
8.1 Assembling equations automatically for dynamical systems
8.2 Numerical solution of differential equations
8.3 Forward Euler method
8.4 Backward Euler method
8.5 Trapezoidal rule (TR)
8.6 Nonlinear differential equations
8.7 Analysis, accuracy and stability (Dt ! 0)
8.8 Residual for LMS methods
8.9 Global error estimate
8.10 Stability
8.11 Stability (continued)
8.12 Problems
9 model order reduction
9.1 Model order reduction
9.2 Moment matching
9.3 Model order reduction (cont).
9.4 Moment matching
9.5 Truncated Balanced Realization (1000 ft overview)
9.6 Problems
Final report
10 harmonic balance
10.1 Abstract
10.2 Introduction
10.2.1 Modifications to the netlist syntax
10.3 Background
10.3.1 Discrete Fourier Transform
10.3.2 Harmonic Balance equations
10.3.3 Frequency domain representation of MNA equations
10.3.4 Example. RC circuit with a diode.
10.3.5 Jacobian
10.3.6 Newton’s method solution
10.3.7 Alternative handling of the non-linear currents and Jacobians
10.4 Results
10.4.1 Low pass filter
10.4.2 Half wave rectifier
10.4.3 AC to DC conversion
10.4.4 Bridge rectifier
10.4.5 Cpu time and error vs N
10.4.6 Taylor series non-linearities
10.4.7 Stiff systems
10.5 Conclusion
10.6 Appendices
10.6.1 Discrete Fourier Transform inversion
Appendices
a singular value decomposition
b basic theorems and definitions
c norton equivalents
d stability of discretized linear differential equations
e laplace transform refresher
f discrete fourier transform
g harmonic balance, rough notes
g.1 Block matrix form, with physical parameter ordering
g.2 Block matrix form, with frequency ordering
g.3 Representing the linear sources
g.4 Representing non-linear sources
g.5 Newton’s method
g.6 A matrix formulation of Harmonic Balance non-linear currents
h matlab notebooks
i mathematica notebooks
Index
Bibliography