I have a question about the sum of angular momenta of two $3/2$ spin particles (considering no orbital angular momentum). Let's suppose that I can with a magnetic field collide two $3/2$ spin particles, but the first particle has always $m > 0$ $(m = 1/2 \ \ or \ \ m = 3/2)$, $m$ being the projection of the spin on the $z$-axis. And the second particle has always $m < 0$ $(m = -1/2 \ \ or \ \ m = -3/2)$.
So my question is, will the final state angular momentum $\left|j, m\right>$, considering $j = 0, 1, 2, 3$ be equally probable?
Best Answer
The incoming state is of the form $|m_1, m_2\rangle$ (with $m_1 = \frac{1}{2}$ or $\frac{3}{2}$ and $m_2 = -\frac{1}{2}$ or $-\frac{3}{2}$). If we subsequently measure $j$ and $m$, the probability of measuring a particular final state $|j,m\rangle$ will be the square of the Clebsch-Gordan coefficient $\langle m_1, m_2 | j, m\rangle$. For this probability to be non-zero, of course, we must have $m = m_1 + m_2$. Wikipedia has a table of Clebsch-Gordan coefficients, including those for $j_1 = j_2 = \frac{3}{2}$; so it's just a matter of looking up the appropriate coefficients, taking their absolute values, squaring them, and summing them.
For example, let's look at the probability for $m_1 = \frac{1}{2}$ and $m_2 = - \frac{1}{2}$ to be measured as having $j = 1$ and $m = 0$. From the table, we see that $\langle \frac{1}{2}, -\frac{1}{2} | 1, 0\rangle = -\sqrt{\frac{1}{20}}$, so the probability of measuring $j = 1$ in this case will be $\frac{1}{20}$. It is evident from this result that the probabilities for different $j$ values cannot all be equally likely.