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I have came across the following question:

Two particles, each of spin 1, are at rest. It is known that the $z$ component of the spin of each is zero. Show that the probability that the total angular momentum of the system is zero is 1/3. What is the probability that the total angular momentum is $J= 1$?

I am unsure how to go about this problem as I don't know how to calculate these probabilities. Is there a trick with using $J = |l-s| , |l-s|+1, . . .,|l+s|$?

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You have to compute $$|1,0 \rangle \otimes |1,0 \rangle = \sum_{J,m} \langle J,m|1,0;1,0 \rangle |J, m\rangle$$ where $\langle J,m|1,0;1,0 \rangle$ are Clebsch-Gordan coefficients. Indeed there are selection rules, and your comment shows that $J = 0,1,2$. Likewise you can easily show that only $m=0$ can appear. So you need to look up 3 possible CG coefficients (if you're really smart, you can argue that $J=1$ does not appear using an additional selection rule, so you only need to look up 2).

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  • $\begingroup$ Thank you. So adding and squaring the CG coefficients gives the probability? $\endgroup$ – Tyr Nov 23 '19 at 16:41
  • $\begingroup$ Indeed - that's the way to go. $\endgroup$ – Hans Moleman Nov 23 '19 at 16:47
  • $\begingroup$ I knew there was something simple and obvious that I was missing $\endgroup$ – Tyr Nov 23 '19 at 16:50
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Correction: Clearly M=0. There are three such states: $$ S=2: 2|0,0\rangle+|-1,1\rangle+|1,-1\rangle; \\ S=1: |1,-1\rangle -|-1,1\rangle; \\ S=0: |-1,1\rangle+|1,-1\rangle- |0,0\rangle; $$. The probability that S=1 is zero.

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  • $\begingroup$ This is plain wrong. For instance, the $J=1$ probability is 0. $\endgroup$ – Hans Moleman Nov 23 '19 at 19:32
  • $\begingroup$ From the CG coefficient's, $C_{1,-1} = \sqrt{\frac{1}{3}, C_{0,0} = -\sqrt{\frac{1}{3}, C_{-1,1} = \sqrt{\frac{1}{3}$. Adding and squaring gives the desired 1/3 $\endgroup$ – Tyr Nov 23 '19 at 19:39
  • $\begingroup$ @HansMoleman You are correct. I changed my answer. $\endgroup$ – my2cts Nov 23 '19 at 21:52

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