Two spin-1 objects: Addition of Angular Momenta

[unscientific]

Homework Statement

Suppose in a container there are two spin-1 objects, A and B. It is found that $J=j_1+j_22$ and $M=m_1+m_2=1$. What is the probability when $J_z$ is measured for A, values of m=0, m=-1 and m=+1 will be obtained?

Homework Equations

The Attempt at a Solution

|j,j> = |2,2> = |(1,1)_A>|(1,1)_B>

Starting:
J_{-}|2,2> = \left(J_{-}^A + J_{-}^B\right)|(1,1)_A>|(1,1)_B>
\sqrt{2(2+1)-2(2-1)}|2,1> = \sqrt{1(1+1)-1(1-1)}\left(|(1,0)_A>|(1,1)_B> + |(1,1)_A>|(1,0)_B>\right)
|2,1> = \frac{1}{\sqrt{2}}\left(|(1,0)_A>|(1,1)_B> + |(1,1)_A>|(1,0)_B>\right)

For m=0, probability = 1/2
For m=1, probability = 1/2
For m=-1, probability = 0

Is this right?

 

[CAF123]

I would say yes, that is correct. So essentially you are writing the coupled state |2,1> in the uncoupled basis. Since j₁+j₂=1+1 = 2 = J and m₁+m₂=1, you know that the only possible values for the m_i are 0 and 1, since any other combination does not give a sum of 1.

So the coupled state |J,M> decomposed in the uncoupled basis |j₁j₂m₁m₂> → |m₁m₂> would usually consist of 9 terms for fixed j₁ and j₂ both equal to 1. But by the above restriction, the only possibilities are |0,1> and |1,0>, as you found.

So the decomposition must be |2,1> = c₁|0,1> + c₂|1,0>. For normalized states, |c₁|²+ |c₂|² = 1 => c₁ = c₂ = 1/√2.

Thought I would give a more descriptive solution, which helps in understanding what you are actually doing.