How would the eigenstates of a particle with spin 3/2 look like?

Question

I learnt in an introductory course about quantum mecanics how to work with spin 1/2 particles. I saw how the algebra is almost the same as for angular momentum, but no one ever told me about particles having a spin different from 1/2. I know there are no known particles of spin 3/2, but I am wondering how the eigenstates of the spin operator in z direction would look like, to get a better understanding of what spin really is.

Answer 1

It should follow most of the same ideas of spin-1/2 and spin-1, in terms of satisfying the general algebra of angular momentum. We can construct the operators with a little bit of work. Firstly in general,

$$[J_i,J_j] = i\hbar\epsilon_{ijk}J_k$$

Following that a particle with spin $s$ will have $2s+1$ a spin-3/2 particle will have four states, with z-component of angular momentum $\frac{3}{2}\hbar$, $\frac{1}{2}\hbar$, $-\frac{1}{2}\hbar$, $-\frac{3}{2}\hbar$. We can take these states as a basis, and immediately write the $\hat{S}_z$ operator in this basis,

$$\hat{S}_z = \frac{1}{2}\hbar\left(\begin{matrix} 3 & 0 & 0 & 0\\0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -3 \end{matrix}\right)$$

Now it comes down to finding two matrices $\hat{S}_x$ and $\hat{S}_y$ that satisfy the algebraic equation above given this matrix representing $\hat{S}_z$. Two such examples taken from wikipedia (https://en.wikipedia.org/wiki/3D_rotation_group#A_note_on_Lie_algebra),

$$\hat{S}_x = \frac{1}{2}\hbar\left(\begin{matrix} 0 & \sqrt{3} & 0 & 0\\\sqrt{3} & 0 & 2 & 0\\ 0 & 2 & 0 & \sqrt{3}\\ 0 & 0 & \sqrt{3} & 0 \end{matrix}\right)$$ and,

$$\hat{S}_y = \frac{1}{2}\hbar\left(\begin{matrix} 0 & -i\sqrt{3} & 0 & 0\\ i\sqrt{3} & 0 & -2i & 0\\ 0 & 2i & 0 & -i\sqrt{3}\\ 0 & 0 & i\sqrt{3} & 0 \end{matrix}\right)$$

You could follow this same logic to represent the higher spins as well.

Edit: Instead of pulling these matrices out of nowhere I will show a way to find them. Following directly from the angular momentum algebra, you can construct raising and lowering operators. These take the form $J_{+} = J_x + iJ_y$ and $J_- = J_x - iJ_y$. They take a state with a given z-projection and return a state one higher (or lower) multiplied by a number. Specifically, $$J_{\pm}|J,m\rangle = \hbar\sqrt{j(j+1)-m(m\pm 1)}|J,m\pm 1\rangle$$

Given our basis and representation of $\hat{S}_z$ we can calculate the coefficients for the matrices $S_+$ and $S_-$. For example this would give for $S_+$ matrix element $\langle \frac{3}{2}, -\frac{1}{2}|\hat{S}_+|\frac{3}{2},-\frac{3}{2}\rangle = \hbar\sqrt{\frac{3}{2}(\frac{3}{2}+1)+\frac{3}{2}(-\frac{3}{2}+1)}=\frac{\hbar}{2}\sqrt{15-3} = \sqrt{3}\hbar$

$$\hat{S}_+ = \hbar\left(\begin{matrix} 0 & \sqrt{3} & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & \sqrt{3}\\ 0 & 0 & 0 & 0 \end{matrix}\right)$$

and,

$$\hat{S}_- = \hbar\left(\begin{matrix} 0 & 0 & 0 & 0\\ \sqrt{3} & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & \sqrt{3} & 0 \end{matrix}\right)$$

The above versions of $\hat{S}_x$ and $\hat{S}_y$ can be calculated by plugging these into the definitions of the raising and lowering operators. I find this method of construction clarifies the structure of the matrices ($\hat{S}_x$ and $\hat{S}_y$ only having off diagonal components, and the pattern of minus signs) at least in the basis where you have diagonalized $\hat{S}_z$

Answer 2

For spin $\frac32$, there are four eigenstates of $S_z$ with eigenvalues $-\frac32\hbar$, $-\frac12\hbar$, $+\frac12\hbar$, and $+\frac32\hbar$. They are typically written as $\left|\frac32,-\frac32\right\rangle$, $\left|\frac32,-\frac12\right\rangle$, $\left|\frac32,\frac12\right\rangle$, and $\left|\frac32,\frac32\right\rangle$ to show the quantum numbers $s$ and $m_s$ for each eigenstate.

The generalization to arbitrary spin $s$ should be obvious. There are $2s+1$ eigenstates of $S_z$. Their $m_s$ quantum numbers (specifying the $S_z$ eigenvalue as $m_s\hbar$) range from $-s$ to $+s$ in steps of $1$.