Understanding Decay Angle of Spin Angular Momentum

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The discussion centers on the decay process of Lambda particles into protons and pions, focusing on the implications of spin and orbital angular momentum on angular distribution. It is noted that while spin angular momentum is conserved in the decay, the presence of non-zero orbital angular momentum leads to a cosine dependence in the angular distribution of decay products. The conversation highlights that if parity is conserved, the angular distribution should be flat, but if parity is not conserved, as in weak decays, angular dependence arises. The participants explore how the interference between different angular momentum states contributes to the observed angular distribution. The discussion concludes with a request for further reading on the relationship between angular dependence and spin in particle decays.
Manojg
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Hi,

I have a question about decay angle. For example,
Let us consider the decay

\Lambda \rightarrow p + \pi^{-}

Here, \Lambda and p are spin 1/2 and \pi^{-} is spin 0. So, spin angular momentum is conserved. So, should not the cosine of angular distribution of p ( or \pi^{-} ) in center of mass frame of \Lambda be flat?

If the spin angular momentum is not conserved ( like in \rho \rightarrow \pi^{+} \pi^{-} ) then total angular momentum will be conserved because pions system has orbital angular momentum. So, the cosine of angular distribution of one of the pions in center of mass frame of \rho is not flat.

So, if the spin angular momentum is conserved and there is no orbital angular momentum then should not the cosine of angular distribution of one of the decay product ( two particle decay) in mother reference frame be flat?

Thanks.
 
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Because parity is not conserved in hyperon decay, the final state is a mixture of orbital angular momentum L=1 and L=0. In each case L+S=1/2 so total angular momentum is conserved. The cos\theta dependence of the angular distribution is evidence that parity is not conserved in this decay.
 
Thanks clem.

So, angular dependence comes from non-zero orbital momentum of the final particles, and given by spherical harmonics? In case of zero orbital momentum, angular distribution should be flat? Is the distribution integrated over the z-components of the momentum (m for given J)?
 
The angular distribution is summed over m_J. m_L and m_s are summed over, weighted by
Clesch-Gordan coefficients (to make an eigenstate of J). It turns out that pure L=1 or pure L=0 for the orbital angle momentum gives a flat distribution. It is the interference between the two that gives a cos\theta dependence.
 
Now, I am confused.

If parity is conserved (gives L=0) then angular distribution should be flat. If parity is not conserved and L=1, then cos(theta) term will come in the spherical harmonics and as a result in the amplitude. So, even for pure L=1 state, should not be there angular dependence?

What about weak decay (parity does not conserve)?

Thanks.
 
It turns out that the pure state J=1/2, L=1 still has no angular dependence.
The decay Lambda--> p + pi is a weak decay. Only weak interaction violate parity conservation. All beta decays are also weak and violate parity conservation.
 
Sorry for the offtopic, could you please point me to the source of information I can read about the angle dependence on the spin, angular momentum etc.

Thanks
 

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