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User:

Please generate the matrix of spin operators \( S_z \), \( S_+ \), \( S_- \) for spin=\(\frac{3}{2}\).

Assistant:

The spin operators for spin-\(\frac{3}{2}\) can be expressed in the basis states \( | \frac{3}{2}, \frac{3}{2} \rangle \), \( | \frac{3}{2}, \frac{1}{2} \rangle \), \( | \frac{3}{2}, -\frac{1}{2} \rangle \), and \( | \frac{3}{2}, -\frac{3}{2} \rangle \). Here are the matrices:

1. \( S_z \) Matrix

\[ S_z = \hbar \begin{pmatrix} \frac{3}{2} & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{3}{2} \end{pmatrix} \]

2. \( S_+ \) Matrix

\[ S_+ = \hbar \begin{pmatrix} 0 & 0 & 0 & 0 \\ \sqrt{3} & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix} \]

3. \( S_- \) Matrix

\[ S_- = \hbar \begin{pmatrix} 0 & \sqrt{3} & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & \sqrt{3} \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

User:

Write the Clebsch-Gordon relations for \( S_+ \) and \( S_- \).

Assistant:

The Clebsch-Gordon coefficients relate the states of combined angular momentum. For the raising and lowering operators, the relations are as follows:

\[ S_+ |S, m \rangle = \hbar \sqrt{(S - m)(S + m + 1)} |S, m + 1 \rangle \] \[ S_- |S, m \rangle = \hbar \sqrt{(S + m)(S - m + 1)} |S, m - 1 \rangle \]

User:

Write the equations of \( S_+ \) and \( S_- \) for general spin \( S \).

Assistant:

The equations for the raising and lowering operators for a general spin \( S \) are:

\[ S_+ |S, m \rangle = \hbar \sqrt{(S - m)(S + m + 1)} |S, m + 1 \rangle \] \[ S_- |S, m \rangle = \hbar \sqrt{(S + m)(S - m + 1)} |S, m - 1 \rangle \]

User:

Write the equations for \( S_+ \) and \( S_- \) on \( | \frac{3}{2}, m \rangle \) according to the equations above.

Assistant:

For a spin-\(\frac{3}{2}\) system, the states are \( | \frac{3}{2}, \frac{3}{2} \rangle \), \( | \frac{3}{2}, \frac{1}{2} \rangle \), \( | \frac{3}{2}, -\frac{1}{2} \rangle \), and \( | \frac{3}{2}, -\frac{3}{2} \rangle \). The actions of \( S_+ \) and \( S_- \) on these states are as follows:

Raising Operator \( S_+ \)

1. For \( | \frac{3}{2}, \frac{3}{2} \rangle \): \[ S_+ | \frac{3}{2}, \frac{3}{2} \rangle = 0 \]

2. For \( | \frac{3}{2}, \frac{1}{2} \rangle \): \[ S_+ | \frac{3}{2}, \frac{1}{2} \rangle = \sqrt{3} \hbar | \frac{3}{2}, \frac{3}{2} \rangle \]

3. For \( | \frac{3}{2}, -\frac{1}{2} \rangle \): \[ S_+ | \frac{3}{2}, -\frac{1}{2} \rangle = 2 \hbar | \frac{3}{2}, \frac{1}{2} \rangle \]

4. For \( | \frac{3}{2}, -\frac{3}{2} \rangle \): \[ S_+ | \frac{3}{2}, -\frac{3}{2} \rangle = \sqrt{3} \hbar | \frac{3}{2}, -\frac{1}{2} \rangle \]

Lowering Operator \( S_- \)

1. For \( | \frac{3}{2}, \frac{3}{2} \rangle \): \[ S_- | \frac{3}{2}, \frac{3}{2} \rangle = \sqrt{3} \hbar | \frac{3}{2}, \frac{1}{2} \rangle \]

2. For \( | \frac{3}{2}, \frac{1}{2} \rangle \): \[ S_- | \frac{3}{2}, \frac{1}{2} \rangle = 2 \hbar | \frac{3}{2}, -\frac{1}{2} \rangle \]

3. For \( | \frac{3}{2}, -\frac{1}{2} \rangle \): \[ S_- | \frac{3}{2}, -\frac{1}{2} \rangle = \sqrt{3} \hbar | \frac{3}{2}, -\frac{3}{2} \rangle \]

4. For \( | \frac{3}{2}, -\frac{3}{2} \rangle \): \[ S_- | \frac{3}{2}, -\frac{3}{2} \rangle = 0 \]

User:

Please generate the matrix of \( S_+ \) and \( S_- \) according to the above relations.

Assistant:

The matrices for the raising operator \( S_+ \) and the lowering operator \( S_- \) for a spin-\(\frac{3}{2}\) system are as follows:

Raising Operator \( S_+ \) Matrix

\[ S_+ = \hbar \begin{pmatrix} 0 & 0 & 0 & 0 \\ \sqrt{3} & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix} \]

Lowering Operator \( S_- \) Matrix

\[ S_- = \hbar \begin{pmatrix} 0 & \sqrt{3} & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & \sqrt{3} \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

User:

Now please generate \( S_x \) and \( S_y \) from \( S_+ \) and \( S_- \).

Assistant:

To construct the \( S_x \) and \( S_y \) operators from the raising \( S_+ \) and lowering \( S_- \) operators for a spin-\(\frac{3}{2}\) system, we can use the following relationships:

Definitions

1. Raising Operator: \[ S_+ = S_x + i S_y \]

2. Lowering Operator: \[ S_- = S_x - i S_y \]

From these definitions, we can isolate \( S_x \) and \( S_y \):

Formulas

To derive \( S_x \) and \( S_y \): 1. For \( S_x \): \[ S_x = \frac{1}{2} (S_+ + S_-) \]

2. For \( S_y \): \[ S_y = \frac{1}{2i} (S_+ - S_-) \]

User:

Now please calculate the matrices for \( S_x \) and \( S_y \).

Assistant:

Using the previously defined matrices for \( S_+ \) and \( S_- \):

Calculating \( S_x \)

\[ S_x = \frac{1}{2} (S_+ + S_-) = \frac{\hbar}{2} \begin{pmatrix} 0 & 0 & 0 & 0 \\ \sqrt{3} & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix} + \frac{\hbar}{2} \begin{pmatrix} 0 & \sqrt{3} & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & \sqrt{3} \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

\[ = \frac{\hbar}{2} \begin{pmatrix} 0 & \sqrt{3} & 0 & 0 \\ \sqrt{3} & 0 & 2 & 0 \\ 0 & 2 & 0 & \sqrt{3} \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix} \]

Calculating \( S_y \)

\[ S_y = \frac{1}{2i} (S_+ - S_-) = \frac{\hbar}{2i} \begin{pmatrix} 0 & 0 & 0 & 0 \\ \sqrt{3} & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix} - \frac{\hbar}{2i} \begin{pmatrix} 0 & \sqrt{3} & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & \sqrt{3} \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

\[ = \frac{\hbar}{2i} \begin{pmatrix} 0 & -\sqrt{3} & 0 & 0 \\ \sqrt{3} & 0 & -2 & 0 \\ 0 & 2 & 0 & -\sqrt{3} \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix} \]

Final Matrices

The final matrices for \( S_x \) and \( S_y \) are:

Matrix for \( S_x \)

\[ S_x = \frac{\hbar}{2} \begin{pmatrix} 0 & \sqrt{3} & 0 & 0 \\ \sqrt{3} & 0 & 2 & 0 \\ 0 & 2 & 0 & \sqrt{3} \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix} \]

Matrix for \( S_y \)

\[ S_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -\sqrt{3}i & 0 & 0 \\ \sqrt{3}i & 0 & -2i & 0 \\ 0 & 2i & 0 & -\sqrt{3}i \\ 0 & 0 & \sqrt{3}i & 0 \end{pmatrix} \]