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

Click For Summary

Homework Help Overview

The discussion revolves around a two-particle system characterized by their spins, focusing on calculating the probability of the total spin when particles are taken from different sources. The participants explore the implications of polarization on the problem and the mathematical framework involving Clebsch-Gordan coefficients.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the construction of wavefunctions and the role of polarization in determining probabilities. Questions arise about the calculation of Clebsch-Gordan coefficients without specific spin values and the nature of unpolarized states as linear combinations of spin states.

Discussion Status

The conversation is ongoing, with participants providing insights into the mathematical aspects of the problem. Some guidance has been offered regarding the relationship between the states and the coefficients, but explicit calculations or examples have not been fully developed.

Contextual Notes

There is mention of constraints related to the lack of specified spin values for the particles and the challenge of finding relevant information in existing resources. The nature of unpolarized states and their representation in the context of the problem is also under discussion.

kreil
Science Advisor
Insights Author
Messages
665
Reaction score
68

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?
 
Last edited:
Physics news on Phys.org
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
 

Similar threads

Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
Replies
16
Views
3K