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### Q: [1] pr. 5.4

Given a 2D SHO with Hamiltonian

\begin{equation}\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 },

\end{equation}

- (a)

What are the energies and degeneracies of the three lowest states?
- (b)

With perturbation
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:40}

V = m \omega^2 x y,

\end{equation}

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

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:60}

H \ket{n_1, n_2} =

\Hbar\omega

\lr{

n_1 + n_2 + 1

}

\ket{n_1, n_2},

\end{equation}

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

\begin{equation}\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}

\end{equation}

\begin{equation}\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}

\end{equation}

\begin{equation}\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}

\end{equation}

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

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:260}

x y

\sim

\inv{2} \Hbar \omega

\begin{bmatrix}

0 & 0 & 0 \\

0 & 0 & 1 \\

0 & 1 & 0 \\

\end{bmatrix}.

\end{equation}

This diagonalizes with

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:300}

U

=

\begin{bmatrix}

1 & 0 \\

0 & \tilde{U}

\end{bmatrix}

\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:320}

\tilde{U}

=

\inv{\sqrt{2}}

\begin{bmatrix}

1 & 1 \\

1 & -1 \\

\end{bmatrix}

\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:340}

D =

\inv{2} \Hbar \omega

\begin{bmatrix}

0 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & -1 \\

\end{bmatrix}

\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:360}

x y = U D U^\dagger = U D U.

\end{equation}

The unperturbed Hamiltonian in the original basis is

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:380}

H_0

=

\Hbar \omega

\begin{bmatrix}

1 & 0 \\

0 & 2 I

\end{bmatrix},

\end{equation}

So the transformation to the diagonal \( x y \) basis leaves the initial Hamiltonian unaltered

\begin{equation}\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}

\end{equation}

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

\begin{equation}\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}

\end{equation}

The new energies are

\begin{equation}\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}

\end{equation}

### 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

\begin{equation}\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}

\end{equation}

Similarly, for the momentum operators, let

\begin{equation}\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}

\end{equation}

For the commutators of the new operators we have

\begin{equation}\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}

\end{equation}

\begin{equation}\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}

\end{equation}

\begin{equation}\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}

\end{equation}

\begin{equation}\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}

\end{equation}

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

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:460}

\begin{aligned}

x &= \inv{\sqrt{2}} \lr{ u + v } \\

y &= \inv{\sqrt{2}} \lr{ u – v }.

\end{aligned}

\end{equation}

We have

\begin{equation}\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}

\end{equation}

\begin{equation}\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}

\end{equation}

and

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:520}

\begin{aligned}

x y

&=

\inv{2} \lr{ (u+v)(u-v) } \\

&=

\inv{2} \lr{ u^2 – v^2 }.

\end{aligned}

\end{equation}

The perturbed Hamiltonian is

\begin{equation}\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}

\end{equation}

In this coordinate system, the corresponding eigensystem is

\begin{equation}\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}.

\end{equation}

For small \( \delta \)

\begin{equation}\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,

\end{equation}

so

\begin{equation}\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}.

\end{equation}

The lowest order perturbed energy levels are

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:620}

\ket{0,0} \rightarrow \Hbar \omega

\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:640}

\ket{1,0} \rightarrow \Hbar \omega \lr{ 2 + \inv{2} \delta }

\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:660}

\ket{0,1} \rightarrow \Hbar \omega \lr{ 2 – \inv{2} \delta }

\end{equation}

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.

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