## 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.