canonically conjugate

Just watched Cloonie’s “Midnight Sky”

December 26, 2020 Incoherent ramblings 4 comments , , , , , ,

I just watched George Clooney’s “Midnight Sky” on netflix.

The movie is visually striking, set on a space ship and on an apocalyptic Earth in +30 years.  Some sort of unspecified radioactive disaster has pretty much wiped out all livable space on Earth.  The movie focuses on the attempt of a sick astronomer to communicate with a space ship that has been off exploring a newly found habitable moon of Jupiter.   They have been out of communication with Earth for a couple years.

I really didn’t understand the foundational premise of the movie.  We have been able to receive communications from satellites that we’ve sent to Jupiter, and a quick google says it’s only ~22 light minutes between Jupiter and Earth.  If that distance is the closest, let’s suppose that it’s a few times that at maximum separation — that’s still only a couple hours separation (guestimating).  Why would the ship have gone completely out of communication with Earth for years while they were on their mission?

There were lots of other holes in the movie, and I wonder if some of those missing pieces were detailed in the book?

Incidentally, the astronomy facility looked really cosy and comfortable for a something located in Antarctica!  There was mention of the poles late in the movie, but early on there was the famous picture of the explorer Scott with his four companions on the wall, which I assumed was meant to give away the location (I recognized that picture from Brian Keating’s book, “Loosing the Nobel Prize”.)

2D SHO xy perturbation

December 7, 2015 phy1520 No comments , , , , ,

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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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