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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?

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  • $\begingroup$ Do you mean that the first particle is always in the $S_z$ eigenstate $|m = \frac12\rangle$ or the eigenstate $|m = \frac32\rangle$? Or do you allow a superposition of the two, of the form $|\psi\rangle = \alpha|m = \frac12\rangle + \beta|m = \frac32\rangle$? I suspect you may get different answers depending on that. $\endgroup$
    – Michael Seifert
    Commented Nov 17, 2021 at 20:07
  • $\begingroup$ Hi @MichaelSeifert they are always in the eigenstate |m=1/2⟩ or the eigenstate |m=3/2⟩. $\endgroup$
    – Matthew D.
    Commented Nov 17, 2021 at 20:09

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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.

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  • $\begingroup$ Yeah, i think i got it. But, can I calculate the probability of obtaining a final state with j = 1 or j = 2 independent of the final m? (i am looking for the probability to measure a state with j=1 for example, independent of the m). I could just sum them clebsch coefficient for j=1 and every m, but the final probability wouldn't be one. $\endgroup$
    – Matthew D.
    Commented Nov 17, 2021 at 20:59
  • $\begingroup$ @MatthewD.: If you know $m_1$ and $m_2$ are, then there's only one possible choice for $m$: $m = m_1 + m_2$. So there's no need to sum over all possible $m$ (or alternately, the sum is trivial.) $\endgroup$
    – Michael Seifert
    Commented Nov 17, 2021 at 21:03
  • $\begingroup$ but i dont know m1 and m2, rigth? I mean, I know they can be m1=1/2,3/2 and m2=-1/2,-3/2, so they can assume four pairs (3/2,-3/2);(3/2,-1/2),(1/2,-3/2) and (1/2,-1/2). Given these four possible pairs, what is the probability that the final state has j=1. There are 3 final states in the table which are possible, |1,-1>, |1,0> and |1,-1>. What I do not understand, is that if I just know that m1 will assume values m1>0 and m2<0, without knowing exactly which value the pair assumes, will the probably of the final state |1,0> be the same as |1,-1> for example? $\endgroup$
    – Matthew D.
    Commented Nov 17, 2021 at 21:18
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    $\begingroup$ @MatthewD.: That was the point of my comment on your original question, which I interpreted to mean "both particles are in a definite, known eigenstate of $m_1$ and $m_2$." If you're just told that $m_1>0$ and $m_2 < 0$, that's not enough information to answer the question; you would need to know which superposition of the above four states you were dealing with. Or it could be that you're dealing with a mixed state, which would require a different technique to solve. In either event, you need more information about your incoming state. $\endgroup$
    – Michael Seifert
    Commented Nov 17, 2021 at 21:28
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    $\begingroup$ Oh okay, now I got it. Thank you for the discussion! $\endgroup$
    – Matthew D.
    Commented Nov 17, 2021 at 21:33

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