Two spin-1 objects: Addition of Angular Momenta

  • #1
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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?

mbid09.png



Homework Equations





The Attempt at a Solution



[tex]|j,j> = |2,2> = |(1,1)_A>|(1,1)_B>[/tex]

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

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

Is this right?
 

Answers and Replies

  • #2
CAF123
Gold Member
2,914
88
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 mi 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.
 
Last edited:

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