What Are the Possible Spin States After Particle Decay in Quantum Mechanics?

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The discussion addresses the spin states of a particle with spin S that decays into two electrons. The total spin of the two electrons must equal the original particle's spin, leading to possible values of S = 1, S_z = 0 or S = 0, S_z = 0. The electrons' spin configurations show a 50% probability for each arrangement, confirming these values. If the particle is equally likely to be in any of the 2S+1 eigenstates of S_z, the outcome remains S = 0, S_z = 0. This highlights the relationship between particle spin and decay products in quantum mechanics.
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Hi,
I'm having problems with this quantum mechanics problem. This is from a practice final I found online somewhere.
A particle of spin S has either spin S=0 or spin S=1. It decays into two electrons, and the spin of the two electrons is that of the original particle. The electrons come out along the x-axis. 50% of the time, electron 1 has s_z = +1/2 and electron 2 has s_z= -1/2, while the other 50% of the time electron 2 has s_z = +1/2 and electron 1 has s_z= -1/2



The question is what are the possible values of S and Sz consistent with this result? The answer is:
S = 1, S_z = 0
or
S = 0, S_z = 0



If, instead, the particle is equally likely to be in any of the 2S+1 eigenstates of S_z, how does the answer change?
S = 0, S_z = 0


Can someone explain this, at least the first part?

Thanks

Thanks.
 
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The first part of the answer is based on the fact that the total spin of the two electrons must be equal to the spin of the original particle. Since the electrons have spin 1/2, then it is impossible for their combined spin to be larger than 1. So if the total spin of both electrons is 1, then the spin of the original particle must also be 1. The possible values of Sz are then 0 (for a spin 1 particle) or -1, 0, 1 (for a spin 0 particle). For the second part, since the particle is equally likely to be in any of the 2S+1 eigenstates of S_z, then the possible values of S and Sz can be any combination of S=0, 1, and S_z = -1, 0, 1 (for a spin 0 particle) or S=1 and S_z = 0 (for a spin 1 particle).
 
The spin correlation problem is a common problem in quantum mechanics that involves understanding the relationship between the spin of a particle and its decay products. In this case, we are given a particle with spin S that decays into two electrons with the same spin as the original particle. The electrons are emitted along the x-axis and we are told that 50% of the time, electron 1 has s_z = +1/2 and electron 2 has s_z = -1/2, while the other 50% of the time, electron 2 has s_z = +1/2 and electron 1 has s_z = -1/2.

The question is asking for the possible values of S and S_z that are consistent with this result. In quantum mechanics, the spin of a particle can only take on certain discrete values, which are given by the eigenstates of the spin operator. In this case, since the electrons have the same spin as the original particle, the spin operator for the two electrons will be the same as the spin operator for the original particle.

The possible values of S and S_z can be determined by considering the possible combinations of spin states for the two electrons. Since the spin of each electron can take on either +1/2 or -1/2, the total spin of the system can either be 1 or 0. And since the spin of each electron is opposite to the other, the total spin projection along the z-axis (S_z) must be 0.

Therefore, the possible values of S and S_z consistent with the given result are S = 1, S_z = 0 or S = 0, S_z = 0.

If we change the scenario and consider that the particle is equally likely to be in any of the 2S+1 eigenstates of S_z, then the answer remains the same, S = 0, S_z = 0. This is because the probability of the particle being in any of these eigenstates is still 50%, and the result of the decay will still be the same, with one electron having s_z = +1/2 and the other having s_z = -1/2.

I hope this explanation helps to clarify the first part of the problem. If you have any further questions, please let me know. Best of luck with your practice final!
 
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