What is the Probability of Total Spin in a Two Particle System?

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The discussion focuses on calculating the probability of total spin in a two-particle system, where one particle has spin s1 and the other has spin s2. Participants explore how to construct wavefunctions for different spin states and the implications of using unpolarized sources. The importance of Clebsch-Gordan coefficients in determining the overlap between total spin states and individual particle states is highlighted. It is noted that explicit numerical probabilities may not be achievable, emphasizing the need for proper linear combinations of states. Overall, the conversation aims to clarify the relationship between polarization and the calculation of total spin probabilities.
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Homework Statement


Consider a two particle system of which one particle has spin s1 and the other s2.

1. If one particle is taken from each of two sources characterized by the state vectors |s1,m1> and |s2,m2> respectively, what is the probability that the resultant two particle system will have total spin S?

2. If the particles are taken from unpolarized sources, what is the probability that the two particle system will have total spin S?

Homework Equations


S^2|s_n,m_n \rangle = \hbar^2 s_n(s_n+1)|s_n,m_n \rangle

The Attempt at a Solution



I need help getting started, and also I don't understand how polarization affects the nature of the problem in part 2. To calculate the probabilities don't I just construct a wavefunction for the particle in that spin state and square it? If so what goes in the bra?

| \langle \psi | s_1,s_2,m_1,m_2 \rangle |^2

Edit: I think this is a good start:

\langle \hat S \rangle = \langle \psi|\hat S|\psi \rangle = \Sigma \langle \psi | \hat S |s_1, s_2,m_1,m_2\rangle \langle s_1, s_2,m_1,m_2|\psi \rangle

= \hbar^2 \left [ s_1(s_1+1)+s_2(s_2+1) \right ] \Sigma |\langle s_1, s_2,m_1,m_2|\psi \rangle|^2

But how do I quantify the squared wavefunction on the RHS? Should I simple solve for it and say that is the probability?
 
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For part 1, the states of total spin S are the states |S,m_S\rangle. Note that m_S is not specified. The overlap with the states | s_1,s_2,m_1,m_2 \rangle are the Clebsch-Gordan coefficients.

For part 2, the unpolarized states are linear combinations of |s,m_s\rangle.
 
Is it possible to calculate the CB coefficients without specifying a spin value for each particle?
 
kreil said:
Is it possible to calculate the CB coefficients without specifying a spin value for each particle?

No, you're probably not going to get explicit numbers in this problem. It's mainly about finding the proper linear combinations.
 
fzero said:
For part 2, the unpolarized states are linear combinations of |s,m_s\rangle.

I'm having trouble with part 2, and I can't find much information about the situation in my notes or book (Ballentine). Can you expand a little bit more on this? Perhaps an example from a similar problem?
 
instead of \left| s_1 , m_1 \right\rangle \otimes \left| s_2 , m_2 \right\rangle forming up the composite system

you have

( \alpha \left| s_1 , m_1 \right\rangle + \alpha' \left| s_2 , m_2 \right\rangle) \otimes ( \alpha \left| s_1 , m_1 \right\rangle + \alpha' \left| s_2 , m_2 \right\rangle)

so you get a mixture of s_1+s_2, 2s_1, 2s_2 states
 

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