Total Angular Momentum in Nuclear Shell Model

[mackabra]

Homework Statement

Calculate the allowed total angular momentum quantum numbers J for 2 protons in a nuclear shell model state j = 3/2.

Homework Equations

J = j1 + j2 where j are the total angular momenta of each proton. Protons are spin 1/2, orbital angular momentum L is not given.

The Attempt at a Solution

Protons are identical, and therefore cannot occupy the same state. I need some clarification on the the addition of the angular momenta.
I know that for a completely filled state, J = 0. However does this come from the fact that all the mj's cancel out (mj = -j, -j+1, ... j) , and how is this related to J = j1+j2+j3+etc...?

Any help would be appreciated
Thank you

 

[mark726]

for your question. In the case of two identical particles, such as two protons, the allowed total angular momentum quantum numbers J will depend on the total spin of the system. Since protons have a spin of 1/2, the possible values for the total spin of the system can be 0 or 1.

If the total spin is 0, then the allowed values for J will be given by J = j1 + j2, where j1 and j2 are the individual angular momentum quantum numbers for each proton. In this case, since j = 3/2, the possible values for J will be 0, 1, 2, and 3.

If the total spin is 1, then the allowed values for J will be given by J = j1 + j2 + 1, where j1 and j2 are the individual angular momentum quantum numbers for each proton. In this case, since j = 3/2, the possible values for J will be 1, 2, 3, and 4.

In general, for N identical particles with spin j, the allowed values for J will be given by J = Nj, (N-1)j + 1, (N-2)j + 2, ... , j + (N-1), and j. Therefore, in the case of two protons with j = 3/2, the allowed values for J will be 0, 1, 2, 3, 4, and 5.

I hope this helps clarify your understanding of how total angular momentum is related to individual angular momenta in a system of identical particles. Please let me know if you have any further questions.