Find the energy levels of a spin s = 3/2 particle

[pondzo]

Homework Statement

Find the energy levels of a spin $s=\frac{3}{2}$ particle whose Hamiltonian is given by:

$\hat{H}=\frac{a_1}{\hbar^2}(\hat{S}^2-\hat{S}_x^2-\hat{S}_y^2)-\frac{a_2}{\hbar}\hat{S}_z$ where $a_1$ and $a_2$ are constants.

Homework Equations

In the $\hat{S}_z$ basis $\hat{S}_z$ and $\hat{S}^2$ have the following matrix representations:

$\hat{S}_z=\frac{1}{2}\hbar\begin{bmatrix}1&&0\\0&&-1\end{bmatrix}$

$\hat{S}^2=\frac{3}{4}\hbar^2\begin{bmatrix}1&&0\\0&&1\end{bmatrix}$

The Attempt at a Solution

We can rewrite the Hamiltonian as follows:

$\hat{H}=\frac{a_1}{\hbar^2}(2\hat{S}_z^2-\hat{S}^2)-\frac{a_2}{\hbar}\hat{S}_z$

Subbing the matrices in the "relevant equations" I get the following matrix representation for the Hamiltonian:

$\hat{H}=\frac{-1}{4}\begin{bmatrix}a_1+2a_2&&0\\0&&a_1-2a_2\end{bmatrix}$

I'm not sure where to go from here... since $s=\frac{3}{2}$ then $m_s=-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}$. Is the next step to work out the energy for each of these $m_s$ values?
So find: $\langle\frac{3}{2},-\frac{3}{2}|\hat{H}|\frac{3}{2},-\frac{3}{2}\rangle,\langle\frac{3}{2},-\frac{1}{2}|\hat{H}|\frac{3}{2},-\frac{1}{2}\rangle,\langle\frac{3}{2},\frac{1}{2}|\hat{H}|\frac{3}{2},\frac{1}{2}\rangle,\langle\frac{3}{2},\frac{3}{2}|\hat{H}|\frac{3}{2},\frac{3}{2}\rangle,$? If so, I am a little confused as to the vector forms of some of those bras and kets.. Help is appreciated!

(Oh and by the way, my Lecturer made a mistake when he said a particle with spin s = 3/2, but he said to do the question regardless)

 

[blue_leaf77]

Homework Equations

In the $\hat{S}_z$ basis $\hat{S}_z$ and $\hat{S}^2$ have the following matrix representations:

$\hat{S}_z=\frac{1}{2}\hbar\begin{bmatrix}1&&0\\0&&-1\end{bmatrix}$

$\hat{S}^2=\frac{3}{4}\hbar^2\begin{bmatrix}1&&0\\0&&1\end{bmatrix}$

Those relations hold only for spin one-half particles. You don't need matrix representation actually, just use the fact that $S^2 = S_x^2 + S_y^2 + S_z^2$ to modify the first three terms contained in the bracket in the Hamiltonian.

 

[pondzo]

Hi Blue leaf, I realized this after I made the post and I think I have the correct answer now.

I just have one question: Is there a convention as to which order you evaluate the different $m_j$ values when computing the matrix elements? Is it ascending or descending?

 

[DrClaude]

I just have one question: Is there a convention as to which order you evaluate the different $m_j$ values when computing the matrix elements? Is it ascending or descending?

It is usually descending, just like for a spin-1/2 particle:
$$\hat{S}_z=\frac{1}{2}\hbar\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$

spin-3/2:
$$\hat{S}_z=\frac{1}{2}\hbar\begin{bmatrix}3&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-3 \end{bmatrix}$$
But it is always better to specify which convention is used.

 

[pondzo]

Ok thanks for the help!