The decay#

Particle A (spin $s_A = 3/2$) decays at rest into:

• Particle B: spin $s_B = 1/2$
• Particle C: spin $s_C = 0$

Working in the rest frame of A, there is no orbital angular momentum initially. Conservation of total angular momentum gives $j_f = j_i = 3/2$.

Part (a): What orbital angular momenta are allowed?#

The final state angular momentum comes from the orbital motion of B and C about their common centre of mass, plus the spin of B. The total spin of the final state is $s_f = s_B + s_C = \tfrac12$, and $\vec{l}_f$ and $\vec{s}_f$ must couple to $j_f = \tfrac32$. By the triangle rule $|l_f - s_f| \le j_f \le l_f + s_f$:

$j_f = l_f \pm \frac{1}{2} = \frac{3}{2} \quad\Longrightarrow\quad l_f = \frac{3}{2} \mp \frac{1}{2} = \{1, 2\}$

So the relative orbital angular momentum can be $l = 1$ (spin coupled “parallel” to $\vec{l}$, $j = l + \tfrac12$) or $l = 2$ (spin coupled “antiparallel”, $j = l - \tfrac12$). This is a statement about how the magnitudes couple, not about whether B is spin-up or spin-down along some axis. As Part (d) shows, each channel contains both $m_s = +\tfrac12$ and $m_s = -\tfrac12$ components.

Part (b): All possible states $|l, m_l; \tfrac{1}{2}, m_s\rangle$#

For $l = 1$ (three values of $m_l$, two of $m_s$, giving 6 states):

$\left|1,1;\tfrac{1}{2},\tfrac{1}{2}\right\rangle, \quad \left|1,1;\tfrac{1}{2},-\tfrac{1}{2}\right\rangle$

$\left|1,0;\tfrac{1}{2},\tfrac{1}{2}\right\rangle, \quad \left|1,0;\tfrac{1}{2},-\tfrac{1}{2}\right\rangle$

$\left|1,-1;\tfrac{1}{2},\tfrac{1}{2}\right\rangle, \quad \left|1,-1;\tfrac{1}{2},-\tfrac{1}{2}\right\rangle$

For $l = 2$ (five values of $m_l$, two of $m_s$, giving 10 states):

$\left|2,2;\tfrac{1}{2},\tfrac{1}{2}\right\rangle, \quad \left|2,2;\tfrac{1}{2},-\tfrac{1}{2}\right\rangle$

$\left|2,1;\tfrac{1}{2},\tfrac{1}{2}\right\rangle, \quad \left|2,1;\tfrac{1}{2},-\tfrac{1}{2}\right\rangle$

$\left|2,0;\tfrac{1}{2},\tfrac{1}{2}\right\rangle, \quad \left|2,0;\tfrac{1}{2},-\tfrac{1}{2}\right\rangle$

$\left|2,-1;\tfrac{1}{2},\tfrac{1}{2}\right\rangle, \quad \left|2,-1;\tfrac{1}{2},-\tfrac{1}{2}\right\rangle$

$\left|2,-2;\tfrac{1}{2},\tfrac{1}{2}\right\rangle, \quad \left|2,-2;\tfrac{1}{2},-\tfrac{1}{2}\right\rangle$

(Of these uncoupled states, only the combinations with total $j = \tfrac32$ are actually reached in the decay. In particular, $|2,\pm2;\tfrac12,\pm\tfrac12\rangle$ have $|m_j| = \tfrac52$ and can’t appear.)

Part (c): Parity#

The parity of a state with orbital angular momentum $l$ is $(-1)^l$:

• $l = 1$: parity $= (-1)^1 = -1$ (odd)
• $l = 2$: parity $= (-1)^2 = +1$ (even)

So the two allowed decay channels have opposite parity. The full parity of the final state also includes the intrinsic parities, $P_f = P_B P_C (-1)^l$. If parity is conserved in the decay (which it is for strong and electromagnetic decays, but not weak), only one of the two channels is allowed, depending on whether $P_A = P_BP_C(-1)^1$ or $P_BP_C(-1)^2$.

Part (d): Spin measurement probability via Clebsch-Gordan coefficients#

Suppose particle A is prepared in the state $\left|s_A = \tfrac{3}{2},\, m_A = \tfrac{1}{2}\right\rangle$. What is the probability of measuring particle B in the spin-up state $\left|s_B = \tfrac{1}{2}, m_B = \tfrac{1}{2}\right\rangle$?

We work in the $l = 1$ sector and expand the total state in the uncoupled basis using Clebsch-Gordan coefficients for $1 \otimes \tfrac{1}{2} \to \tfrac{3}{2}$:

$\left|\tfrac{3}{2}, \tfrac{1}{2}\right\rangle = \sqrt{\frac{1}{3}}\left|1,1;\tfrac{1}{2},-\tfrac{1}{2}\right\rangle + \sqrt{\frac{2}{3}}\left|1,0;\tfrac{1}{2},\tfrac{1}{2}\right\rangle$

The state $|s_B = \tfrac{1}{2}, m_B = \tfrac{1}{2}\rangle$ appears only in the second term. The probability is the squared coefficient:

$\boxed{P\!\left(m_B = +\tfrac{1}{2}\right) = \left(\sqrt{\frac{2}{3}}\right)^2 = \frac{2}{3}}$

The complementary probability $P(m_B = -\tfrac{1}{2}) = 1/3$ comes from the first term.

Why this is the right way to think about it#

Notice what we did: we never solved a differential equation or computed an integral. We used:

1. Angular momentum conservation to constrain $l_f$
2. The Clebsch-Gordan table to decompose the initial state in the measurement basis
3. Born rule — probability = squared coefficient

This is the standard workflow for any spin/angular momentum measurement problem in nuclear or particle physics. The CG coefficients encode all the geometry of how angular momenta add, and once you have them, probabilities follow immediately.