3 Spin-1/2 Ang.mom. Eigenstates?(just As Triplets)

[Gerenuk]

Hi,

I've learned about singlet and triplet states for 2 spin-1/2 particles. Now I wondered about 3 electrons. With some help I found the eigenstates of S^2=(S1+S2+S3)^2 in the 3 spin basis:

(u: spin up; d: spin down)
S^2=3/2(3/2+1):
uuu (Sz=+3/2)
uud+udu+duu (Sz=+1/2)
udd+dud+ddu (Sz=-1/2)
ddd (Sz=-3/2)

S^2=1/2(1/2+1):
uud-duu (Sz=+1/2)
uud-udu (Sz=+1/2)
udd-ddu (Sz=-1/2)
udd-dud (Sz=-1/2)

Apparently S^2 and Sz are not sufficient to remove all degeneracies. Even worse: if you antisymmetrize or symmetrize the last 4 spin function you get zero. Does that make sense?

Which other quantum operator should be used to distinguish between the degenerate states?

Anton

This post has been edited by Gerenuk on Sep 5 2007, 06:15 PM

[AlphaNumeric]

You should get 4 J=3/2 states and then 2 J=1/2 states. How are you generating these algebras, are you familiar with the whole J+, J-, J_3 system?

You generally don't have to mention the other quantum label but for 3 particles it'd be something like colour for quarks, you aren't going to be getting 3 electrons in such a setup I don't think.

The J=1/2 states are found by finding a state with linear combinations two ups and a down which are orthogonal to the |3/2 1/2> state you've found. Then you find the |1/2 -1/2> state from that and that's all the states you have. You should find that when you go from |J M> to |J-1 M>, the number of states in that 'multiplet' decreases by 2. Can you see why this is?

[Gerenuk]

What I know about angular momentum is the usual undergrad stuff (matrix representation, ladder operators, spin triplet/singlet,...). I don't know advanced schemes so I went back to basics:
I wrote down S^2 in the uuu,uud,udu,udd,duu,dud,ddu,ddd basis and diagonalized it.
The states I get are distinct orthogonal eigenstates of (S1+S2+S3)^2 and (S1z+S2z+S3z).

For more electrons even more degeneracies will appear (4 electrons: basis 16 states, momenta 1+3+5=9 eigenstates)

The J=1/2 state for my 3 spin system is on the basis of 3 spins. Obviously it is the tensor product of one spin in a definite state and the other two in a singlet state.

The states seem to be correct so the questions remain

Meanwhile I managed to actually orthogonalize the totalspin=1/2 states and made them eigenvectors of the cycling operator (123->231):
For example:
uud+e^(i pi /3) udu+e^(-i pi /3) duu
and so on...

[AlphaNumeric]

You're not working out the diagonalisation of S^2 when you compute the various possible spin multiplets.

The book by Georgi - Lie Algebras in Particle Physics is an excellent guide to how to do this stuff, both for 'simple' (it's a relative term!) SU(2) algebras like this, SU(3) for quark flavour and the often eye watering GUT stuff like SU(5) or SO(10).

Suppose you're considering 2 particles, both spin 1/2. Individually they have states written as |j m> and |j' m'>. j and j' are the 'total spin' for the particle and m and m' are the spin aligned in a given direction, typically the z axis or S_3 direction. Together they form a total state labelled |J M>,

|J M> = |j m> (x) |j m'>

The (x) is 'tensor product' I'm assuming you're familiar with.

The max value of M, m and m' is J, j and j' respectively. We should always have that M = m+m' and J = j+j'. As such, the 'top state' is going to be (for j=j'=1/2)

|1 1> = |1/2 1/2> (x) |1/2 1/2>

From here we can work out the next two states in the multiplet. This comes from applying a lowering operator, typically written as J-. There's various J-, depending on which multiplet you're acting on but it works as so :

J-(|J M>) = J-(|j m>) (x) |j m'> + |j m'> (x) J-(|j m>)

Using J-(|j m>) = sqrt( (j+m)(j-m+1) ) |j m-1> you can then work out what this is in terms of |J M-1> = .....

In the case of |1 1> = |1/2 1/2> (x) |1/2 1/2> you get

sqrt(2)|1 0> = |1/2 -1/2> (x) |1/2 1/2> + |1/2 1/2> (x) |1/2 -1/2>

Then applying J- again gets

2|1 -1> = |1/2 -1/2> (x) |1/2 -1/2> + |1/2 -1/2> (x) |1/2 -1/2>
so
|1 -1> = |1/2 -1/2> (x) |1/2 -1/2>

which is what you'd expect by symmetry.

Can you see how to use this now to work out what the first 5 multiplet of |3/2 3/2> = |1/2 1/2> (x) |1/2 1/2> (x) |1/2 1/2> is?

You then have the issue of finding the orthogonal state. Let's go back to my example of |1 1>. You know there'll be a state with J=0 and since M=< |J|, M=0 too (so there's only one of them). But how do I find |0 0> = ...? I want a state orthogonal to |1 0>, so a state orthogonal to

|1/2 -1/2> (x) |1/2 1/2> + |1/2 1/2> (x) |1/2 -1/2>

That's easy to do, just put a minus sign in the middle (it's more complex for things involving 3 or more particles but you want that <A|B> = 0, where |A> = |J J-1> and |B> = |J-1 J-1>).

Thus, |0,0> = (1/sqrt(2) (|1/2 -1/2> (x) |1/2 1/2> - |1/2 1/2> (x) |1/2 -1/2>)

And thus I've found all the states. Can you see how to use this to find all the states for your problem? Your first state is

|J M> = |j m>|j' m'>|j'' m''>
|3/2 3/2> = |1/2 1/2>|1/2 1/2>|1/2 1/2>

For ease of notation and the fact you're only working with spin 1/2 particles on the right hand side (though for higher spin particles you can't do this), you can use your usual notation of |1/2 1/2> = u and |1/2 -1/2> = d. It's good to practice doing the whole lot though since you'll often deal with higher spin particles.

[Gerenuk]

I'd like to stick to my method and see if that is OK. It uses the fundamental matrix representation of operators and evidently the states I get are eigenvectors.

But I think I understand the method you suggest. I'll surely get the same J=3/2 states and I'm fine with them. Now for J=1/2 M=+/- 1/2 you propose to search for orthogonal states. You will find 4 states two of them having the same M. All these states together form an orthogonal complete basis, but the J=1/2 states are degenerate!

uud+e^(i pi /3) udu+e^(-i pi /3) duu (M=+1/2)
uud+e^(-i pi /3) udu+e^(i pi /3) duu (M=+1/2)
udd+e^(i pi /3) ddu+e^(-i pi /3) dud (M=-1/2)
udd+e^(-i pi /3) ddu+e^(i pi /3) dud (M=-1/2)

These states are not fully antisymmetric (nor symmetric). Is that OK?
I'm personally fine with the cyclic operator (particle 1->2->3->1) having eigenvalues e(i pi/3) and e(-i pi/3) in this case unlike the eigenvalue 1 for usual symmetric or antisymmetric states.

(together with the J=3/2 states
uuu (M=+3/2)
uud+udu+duu (M=+1/2)
udd+dud+ddu (M=-1/2)
ddd (M=-3/2)
)