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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.