Day: December 15, 2015

Another aggregation of notes for phy1520, Graduate Quantum Mechanics.

December 15, 2015 phy1520, Uncategorized No comments

I’ve posted a fourth (pre-exam) update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains the remainder of my lecture notes, more problem set solutions (not posted separately), and additional worked practice problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

Time reversal behavior of solutions to crystal spin Hamiltonian

December 15, 2015 phy1520 No comments , , , ,

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Q: [1] pr 4.12

Solve the spin 1 Hamiltonian
\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:20}
H = A S_z^2 + B(S_x^2 – S_y^2).
\end{equation}

Is this Hamiltonian invariant under time reversal?

How do the eigenkets change under time reversal?

A:

In spinMatrices.nb the matrix representation of the Hamiltonian is found to be
\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:40}
H =
\Hbar^2
\begin{bmatrix}
A+\frac{B}{2} & 0 & \frac{B}{2} \\
-\frac{i B}{\sqrt{2}} & B & -\frac{i B}{\sqrt{2}} \\
\frac{B}{2} & 0 & A+\frac{B}{2} \\
\end{bmatrix}.
\end{equation}

The eigenvalues are
\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:60}
\setlr{ \Hbar^2 A, \Hbar^2 B, \Hbar^2(A + B)},
\end{equation}

and the respective eigenvalues (unnormalized) are

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:80}
\setlr{
\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
1 \\
-\frac{i \sqrt{2} B}{A} \\
1 \\
\end{bmatrix}
}.
\end{equation}

Under time reversal, the Hamiltonian is

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:100}
H \rightarrow A (-S_z)^2 + B ( (-S_x)^2 – (-S_y)^2 ) = H,
\end{equation}

so we expect the eigenkets for this Hamiltonian to vary by at most a phase factor. To check this, first recall that the time reversal action on a spin one state is

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:120}
\Theta \ket{1, m} = (-1)^m \ket{1, -m},
\end{equation}

or

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:140}
\begin{aligned}
\Theta \ket{1} &= -\ket{-1} \\
\Theta \ket{0} &= \ket{0} \\
\Theta \ket{-1} &= -\ket{1}.
\end{aligned}
\end{equation}

Let’s write the eigenkets respectively as

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:160}
\begin{aligned}
\ket{A} &= -\ket{1} + \ket{-1} \\
\ket{B} &= \ket{0} \\
\ket{A+B} &= \ket{1} + \ket{-1} – \frac{i \sqrt{2} B}{A} \ket{0}.
\end{aligned}
\end{equation}

Noting that the time reversal operator maps complex numbers onto their conjugates, the time reversed eigenkets are

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:180}
\begin{aligned}
\ket{A} &\rightarrow \ket{-1} – \ket{-1} = -\ket{A} \\
\ket{B} &\rightarrow \ket{0} = \ket{B} \\
\ket{A+B} &\rightarrow -\ket{1} – \ket{-1} + \frac{i \sqrt{2} B}{A} \ket{0} = -\ket{A+B}.
\end{aligned}
\end{equation}

Up to a sign, the time reversed states match the unreversed states.

References

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

Spin three halves spin interaction

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Q: [1] pr 3.33

A spin \( 3/2 \) nucleus subjected to an external electric field has an interaction Hamiltonian of the form

\begin{equation}\label{eqn:spinThreeHalvesNucleus:20}
H = \frac{e Q}{2 s(s-1) \Hbar^2} \lr{
\lr{\PDSq{x}{\phi}}_0 S_x^2
+\lr{\PDSq{y}{\phi}}_0 S_y^2
+\lr{\PDSq{z}{\phi}}_0 S_z^2
}.
\end{equation}

Show that the interaction energy can be written as

\begin{equation}\label{eqn:spinThreeHalvesNucleus:40}
A(3 S_z^2 – \BS^2) + B(S_{+}^2 + S_{-}^2).
\end{equation}

Find the energy eigenvalues for such a Hamiltonian.

A:

Reordering
\begin{equation}\label{eqn:spinThreeHalvesNucleus:60}
\begin{aligned}
S_{+} &= S_x + i S_y \\
S_{-} &= S_x – i S_y,
\end{aligned}
\end{equation}

gives
\begin{equation}\label{eqn:spinThreeHalvesNucleus:80}
\begin{aligned}
S_x &= \inv{2} \lr{ S_{+} + S_{-} } \\
S_y &= \inv{2i} \lr{ S_{+} – S_{-} }.
\end{aligned}
\end{equation}

The squared spin operators are
\begin{equation}\label{eqn:spinThreeHalvesNucleus:100}
\begin{aligned}
S_x^2
&=
\inv{4} \lr{ S_{+}^2 + S_{-}^2 + S_{+} S_{-} + S_{-} S_{+} } \\
&=
\inv{4} \lr{ S_{+}^2 + S_{-}^2 + 2( S_x^2 + S_y^2 ) } \\
&=
\inv{4} \lr{ S_{+}^2 + S_{-}^2 + 2( \BS^2 – S_z^2 ) },
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:spinThreeHalvesNucleus:120}
\begin{aligned}
S_y^2
&=
-\inv{4} \lr{ S_{+}^2 + S_{-}^2 – S_{+} S_{-} – S_{-} S_{+} } \\
&=
-\inv{4} \lr{ S_{+}^2 + S_{-}^2 – 2( S_x^2 + S_y^2 ) } \\
&=
-\inv{4} \lr{ S_{+}^2 + S_{-}^2 – 2( \BS^2 – S_z^2 ) }.
\end{aligned}
\end{equation}

This gives
\begin{equation}\label{eqn:spinThreeHalvesNucleus:140}
\begin{aligned}
H &= \frac{e Q}{2 s(s-1) \Hbar^2} \biglr{ \inv{4} \lr{\PDSq{x}{\phi}}_0 \lr{ S_{+}^2 + S_{-}^2 + 2( \BS^2 – S_z^2 ) }
-\lr{\PDSq{y}{\phi}}_0 \lr{ S_{+}^2 + S_{-}^2 – 2( \BS^2 – S_z^2 ) }
+\lr{\PDSq{z}{\phi}}_0 S_z^2 } \\
&= \frac{e Q}{2 s(s-1) \Hbar^2} \biglr{ \inv{4} \lr{ \lr{\PDSq{x}{\phi}}_0 -\lr{\PDSq{y}{\phi}}_0 } \lr{ S_{+}^2 + S_{-}^2 }
+ \inv{2} \lr{ \lr{\PDSq{x}{\phi}}_0 + \lr{\PDSq{y}{\phi}}_0 } \BS^2
+ \lr{ \lr{\PDSq{z}{\phi}}_0 – \inv{2} \lr{\PDSq{x}{\phi}}_0 – \inv{2} \lr{\PDSq{y}{\phi}}_0 } S_z^2
}.
\end{aligned}
\end{equation}

For a static electric field we have

\begin{equation}\label{eqn:spinThreeHalvesNucleus:160}
\spacegrad^2 \phi = -\frac{\rho}{\epsilon_0},
\end{equation}

but are evaluating it at a point away from the generating charge distribution, so \( \spacegrad^2 \phi = 0 \) at that point. This gives

\begin{equation}\label{eqn:spinThreeHalvesNucleus:180}
H
=
\frac{e Q}{4 s(s-1) \Hbar^2}
\biglr{
\inv{2} \lr{ \lr{\PDSq{x}{\phi}}_0 -\lr{\PDSq{y}{\phi}}_0
} \lr{ S_{+}^2 + S_{-}^2 }
+
\lr{
\lr{\PDSq{x}{\phi}}_0 + \lr{\PDSq{y}{\phi}}_0
} (\BS^2 – 3 S_z^2)
},
\end{equation}

so
\begin{equation}\label{eqn:spinThreeHalvesNucleus:200}
A =
-\frac{e Q}{4 s(s-1) \Hbar^2} \lr{
\lr{\PDSq{x}{\phi}}_0 + \lr{\PDSq{y}{\phi}}_0
}
\end{equation}
\begin{equation}\label{eqn:spinThreeHalvesNucleus:220}
B =
\frac{e Q}{8 s(s-1) \Hbar^2}
\lr{ \lr{\PDSq{x}{\phi}}_0 – \lr{\PDSq{y}{\phi}}_0 }.
\end{equation}

A: energy eigenvalues

Using sakuraiProblem3.33.nb, matrix representations for the spin three halves operators and the Hamiltonian were constructed with respect to the basis \( \setlr{ \ket{3/2}, \ket{1/2}, \ket{-1/2}, \ket{-3/2} } \)

\begin{equation}\label{eqn:spinThreeHalvesNucleus:240}
\begin{aligned}
S_{+} &=
\Hbar
\begin{bmatrix}
0 & \sqrt{3} & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & \sqrt{3} \\
0 & 0 & 0 & 0 \\
\end{bmatrix} \\
S_{-} &=
\Hbar
\begin{bmatrix}
0 & 0 & 0 & 0 \\
\sqrt{3} & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & \sqrt{3} & 0 \\
\end{bmatrix} \\
S_x &=
\Hbar
\begin{bmatrix}
0 & \sqrt{3}/2 & 0 & 0 \\
\sqrt{3}/2 & 0 & 1 & 0 \\
0 & 1 & 0 & \sqrt{3}/2 \\
0 & 0 & \sqrt{3}/2 & 0 \\
\end{bmatrix} \\
S_y &=
i \Hbar
\begin{bmatrix}
0 & -\ifrac{\sqrt{3}}{2} & 0 & 0 \\
\ifrac{\sqrt{3}}{2} & 0 & -1 & 0 \\
0 & 1 & 0 & -\ifrac{\sqrt{3}}{2} \\
0 & 0 & \ifrac{\sqrt{3}}{2} & 0 \\
\end{bmatrix} \\
S_z &=
\frac{\Hbar}{2}
\begin{bmatrix}
3 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -3 \\
\end{bmatrix} \\
H &=
\begin{bmatrix}
3 A & 0 & 2 \sqrt{3} B & 0 \\
0 & -3 A & 0 & 2 \sqrt{3} B \\
2 \sqrt{3} B & 0 & -3 A & 0 \\
0 & 2 \sqrt{3} B & 0 & 3 A \\
\end{bmatrix}.
\end{aligned}
\end{equation}

The energy eigenvalues are found to be

\begin{equation}\label{eqn:spinThreeHalvesNucleus:260}
E = \pm \Hbar^2 \sqrt{9 A^2 + 12 B^2 },
\end{equation}

with two fold degeneracies for each eigenvalue.

References

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