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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.