Four spin 1/2 particles at the Vertices of tetrahedron

[Diracobama2181]

For a tetrahedron with four spin (1/2) particles, I know there are three separate energy levels at $$l=2,l=1,and l=0$$. My question is how I would go about finding the degeneracy of each level. I know that the number of states must be $$2^4$$. Any clues on where to start would be appreciated. Thank you.

 

[Orodruin]

Are you familiar with the representation theory of SU(2)?

 

[Diracobama2181]

Unfortunately not. It still be interested in your explanation though.

 

[Orodruin]

Are you familiar with how two spin-1/2 couple to form a triplet (spin 1) and a singlet (spin 0) state?

 

[Diracobama2181]

Yes I am. I know you can essentially form separate pairs, and hence two groups of spin 1 and spin 0 states. Would I just take a tensor product of these two states?

 

[Orodruin]

Your full space is the tensor product of four spin 1/2 states. You can split them into different irreps product by product. So first do two 1/2, then do another 1/2 with the results from the first two, then do another 1/2 with the results from the first three. This will give you the decomposition into different irreps.

 

[Diracobama2181]

So $$ (1/2)\bigotimes(1/2)\bigotimes(1/2)\bigotimes(1/2)$$? When I work it out, I get $$2\bigoplus1\bigoplus1\bigoplus1\bigoplus0$$ which would give me 5 states for l=2, 9 for l=1, and 2 for l=0. Thanks!

 

[Orodruin]

Almost. You are missing a singlet state (5+9+1=15, which is not 16). Edit: I see now that you said two l=0, so I assume this was just an error in the writing of the direct sum.

Note that, in general, different copies of irreps may have different energies even if they correspond to the same irrep. For example, the singlet states could a priori have different energies without violating the tetrahedral symmetry.

It is also the case that the tetrahedral symmetry is just a subgroup of rotations. In general you should check that this restricted symmetry does not split the irreps further.

Also note that the standard notation is to use the dimensionality of the representation, not the spin, ie, $2\otimes 2\otimes2\otimes 2 = 5\oplus 3\oplus 3 \oplus 3 \oplus 1 \oplus 1$.