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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.

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