Eigenstates of 3 spin 1/2 particles

[tomdodd4598]

Hi,

I have learned about how to find the 4 spin states of 2 spin 1/2 particles, and how to find them by using the lowering operator twice on |1/2, 1/2> to find the triplet, then simply finding the orthogonal singlet state, |0, 0>.

I started to attempt finding the states of 3 spin 1/2 particles, and realize that there are 6:
|3/2, 3/2>, |3/2, 1/2>, |3/2, -1/2>, |3/2, -3/2>, |1/2, 1/2> and |1/2, -1/2>

I have found the first 4, by using the lowering operator three times on |3/2, 3/2>, to obtain the following:
|3/2, 3/2> = |uuu>
|3/2, 1/2> = 1/√3 (|uud> + |udu> + |duu>)
|3/2, -1/2> = 1/√3 (|udd> + |dud> + |ddu>)
|3/2, -3/2> = |ddd>

My problem is that I can't find the other two - there seem to be many possible orthogonal states, such as:
|1/2, 1/2> = 1/√2 (|uud> - |duu>) or 1/√2 (|uud> - |udu>) or 1/2√2 (2|uud> - |udu> - |duu>)
|1/2, -1/2> = 1/√2 (|udd> - |ddu>) or 1/√2 (|udd> - |dud>) or 1/2√2 (2|udd> - |dud> - |ddu>)

The question is really just the following - if they exist, what are the correct states for |1/2, 1/2> and |1/2, -1/2>?

Thanks in advance for any help,

Tom

 

[Orodruin]

There are not six but eight possible spin states (there must be! 2x2x2=8). To find them, you can first construct the possible states for two of the spins and then consider what happens when you add the third to either the triplet or singlet representation. Naturally, adding a spin to the singlet will result in a doublet and adding a spin to the triplet will result in one doublet and one quadruplet. Thus, in total you will have one quadruplet (spin 3/2) and two doublets (spin 1/2).

 

[tomdodd4598]

Ok, so there are 8 states, but there are two of |3/2, 1/2> and |3/2, -1/2>... so I guess the spin of the first two particles on their own needs to be taken into account to differentiate between the two of each. What I don't understand now is how you would do this - is another number required?

 

[Orodruin]

You can find the states by the procedure I described.

 

[tomdodd4598]

Ok, I have used that method, and I think I have found them - one of the spin 1/2 doublets has states antisymmetric with respect to swapping two of the particles, while the other has symmetric states with respect to this swap. This is what I got:

|1/2, 1/2>₁ = 1/√2 (|udu> − |duu>)
|1/2, -1/2>₁ = 1/√2 (|udd> − |dud>)

|1/2, 1/2>₂ = 1/√6 (2 |uud> − |udu> − |duu>)
|1/2, -1/2>₂ = 1/√6 (|udd> + |dud> − 2 |ddu>)

 

[Orodruin]

That looks fine. The things to look for is that the states are all orthogonal. Naturally, it does not matter which two states you start with. Since the states you ended up with transform in the same way, you could just as well take any orthogonal linear combinations of those.

 

[tomdodd4598]

Great - thanks for the help ;) Would the following states be viable then:

|1/2, 1/2>₁ = 1/√2 (|uud> − |udu>)
instead of
|1/2, 1/2>₁ = 1/√2 (|udu> − |duu>)
and
|1/2, -1/2>₂ = 1/√6 (2 |udd> - |dud> - |ddu>)
instead of
|1/2, -1/2>₂ = 1/√6 (|udd> + |dud> − 2 |ddu>) ?

Or do both states of each doublet need to be (anti)symmetric with respect to the same two particles being swapped?

 

[Orodruin]

What you need to make sure is that the states are orthogonal and that the step operators act the way you expect on them. If they are, then you are fine.

 

[tomdodd4598]

Thanks :) So the states I suggested secondarily would not work.
One question I do have though is that although I understood what you meant by adding one spin 1/2 particle to two others, I still don't see mathematically how or why adding a spin 1/2 particle to another two spin 1/2 particles would result in a quadruplet of spin 3/2 and two doublets of spin 1/2... in fact, I don't really understand why adding two spin 1/2s together results in a triplet and a singlet - is there somewhere I can learn about this, or is it quite simple?

 

[Orodruin]

The thing you will want to study to figure out exactly how it works is groups and representation theory.

 

[tomdodd4598]

I found a nice, concise set of notes on the addition of spins - from what I read, if I had 4 spin-1/2 particles together, I'd have a spin-2 quintuplet, three spin-1 triplets and two spin-0 singlets.

 

[Orodruin]

Yes, this is correct. Whenever you add a spin, you will get two irreps out of every irrep you had before adding the spin, one with one dimension lower and one with one dimension higher. Thus:

1 spin: 1 doublet
2 spin: 1 triplet, 1 singlet
3 spins: triplet -> (1 quadruplet + 1 doublet), singlet -> doublet. This gives in total 1 quadruplet + 2 doublets
4 spins: quadruplet -> 1 quintuplet + 1 triplet, doublet -> 1 triplet + 1 singlet. This give in total 1 quintuplet, 3 triplets, 2 singlets.

 

[tomdodd4598]

Awesome :)
Are the spin matrices for the 3 spin-1/2 particles then:
σ ⊗ I ⊗ I,
I ⊗ σ ⊗ I and
I ⊗ I ⊗ σ
for particles 1, 2 and 3 respectively?