# Homework Help: Addition of 4 spins

1. Jan 25, 2010

### G01

Hi Guys, this is from my grad quantum class. I'm pretty stuck and need some help:

1. The problem statement, all variables and given/known data

Given four spin-1/2 particles, derive an expression for the total spin state |S,m⟩ ≡ |1,0⟩ in terms of the the four bases |+⟩i , |−⟩i ; i = 1,2,3,4

2. Relevant equations

Clebsch Gordon Coefficients

Raising and lowering operators, etc.

3. The attempt at a solution

OK. So I know the solution has to be of this form:

$$|1,0> = a|+++->+b|++-+>+c|+-++>+d|-+++>$$

Now, here is my plan of attack:

First, the state $$|2,2>=|++++>$$

I applied the lowering operator to this state repeatedly to find the $|2,0>$ state.

Then I use the condition that:

$$<1,0|2,0>=0$$ to get: a+b+c+d=0

Also, there is the normalization condition:

$$a^2+b^2+c^2+d^2=1$$

So, I have two equations in 4 unknowns. This is my problem.

I can find a third equation by considering <0,0|1,0>=0 however, i don't know the form of the singlet configuration for 4 spins. Any hints on how I can find that?

Still that leaves me still with 3 equations in 4 unknowns. Where do I get the last equation?

Any hints at all would be appreciated. Thanks alot.

2. Jan 26, 2010

### kuruman

Last edited: Jan 26, 2010
3. Jan 26, 2010

### kuruman

I don't have any suggestions (yet) on how to proceed - I thought I did, but I was wrong and that is why I withdrew my previous posting. However, before you get too deep in your method consider this: State |1,0> is an eigenstate of S2 with eigenvalue S(S+1)=1*(1+1) = 2 and an eigenstate of Sz with eigenvalue zero. The latter eigenvalue means that you must have two pluses and two minuses in each of the four-spinor terms. Now the number of permutations of two pluses and two minuses is six not four.

In fact if you add four spins 1/2 the total number of states is 2*2*2*2 = 16. The possible total angular momenta are (a) One S = 2 (five states), (b) three S =1 (9 states) and (c) two S = 0 (2 states). Note that there are six states with |S,0> as expected.

4. Jan 26, 2010

### vela

Staff Emeritus
Perhaps you should try go from the |2,1> state to the |1,1> states since you know they have to be orthogonal. Then use the lowering operator to find the |1,0> states.

5. Jan 26, 2010

### G01

You beat us to it. After some office hours and group effort, we got the problem. This was the method suggested to us by our professor. Thanks!

6. Jan 26, 2010

### kuruman

OK, I got it. When you add two spins S = 1 to get spin S12=S1+S2, the state |S12,0> is in terms of the Clebsch-Gordan coefficients

|1,0>=Σ(<1,m1,1,-m1|1,0>)|1,m1>|1,-m1>

where the summation extends over all the m1 that are appropriate and the constant in parentheses is the Clebsch-Gordan coefficient.

For this problem, you need to substitute two-spinor states in the right side of the above expression. For example, you can pair spins 1 and 2 to get |1,1> = |++> and pair spins 3 and 4 to get |1,-1> = | - - >. So in the summation over all the m1 a two-spin term like |1,1>|1,-1> becomes the four-spinor term |+ + - - >.

You get additional total spin S = 1 states by pairing spins 1 & 3 and 2 & 4 and then pairing 1 & 4 and 2 & 3. This exhausts all the possible pairings and you end up with the three different S = 1 states that I mentioned earlier.

Of course, once you write state |1,0> as I indicated, you need to verify that it is an eigenstate of S2 with the correct eigenvalue S(S+1) = 2. It is a bit tedious, but straightforward if you are systematic about it.