# Angular momentum from spin

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 $$J = |l-s| , |l-s|+1, . . .,|l+s|$$?

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).
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.
• This is plain wrong. For instance, the $J=1$ probability is 0. – Hans Moleman Nov 23 '19 at 19:32
• 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 – Tyr Nov 23 '19 at 19:39