Derive Expression for Total Spin State |1,0> w/ 4 Spin-1/2 Particles

[G01]

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

Homework Statement

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

Homework Equations

Clebsch Gordon Coefficients

Raising and lowering operators, etc.

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.

 

[kuruman]

*** Reply withdrawn. I need to think about this some more. ***

 

[kuruman]

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

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 S_z 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.

 

[vela]

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.

 

[G01]

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.

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!

 

[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.