# Clebsch-Gordan coefficient

1. Sep 11, 2012

### semc

Hi, for the deutron triplet state, I am trying to find the coefficient using the lowering operator but I kept getting 0 hope someone can help me.

Using
J-|j,m>=[(j+m)(j-m+1)]1/2|j,m-1>
I can get from |++> to 1/$\sqrt{2}$(|-+> + |+->) but from here I am stuck since j+m=0 for |+->.

What is wrong with my working? Thanks

2. Sep 11, 2012

### Oxvillian

Perhaps you are not remembering that J- has two pieces?

$$J_- = J_{1-} + J_{2-}$$

3. Sep 12, 2012

### semc

$$J_- |+-> + |-+> = J_{1-} + J_{2-}|+-> + |-+>$$ I am not sure about this part but i split the operators into
$$J_- = J_{1-}|+-> + |-+> + J_{2-}|+-> + |-+>$$
so $$J_{1-}|+-> + |-+>$$ is where I am stuck since $$m_1=1/2 , m_2=-1/2$$ so everything will be reduced to 0.

4. Sep 12, 2012

### Oxvillian

Don't forget the brackets!

$$\left( J_{1-} + J_{2-} \right) \left(\lvert +- \rangle + \lvert -+ \rangle \right)$$

There are 4 terms here, 2 of which survive the lowering operators.

5. Sep 12, 2012

### semc

Yeah I am aware there are 4 terms but is this correct?
$$J_- = J_{1-}|+-> + |-+> + J_{2-}|+-> + |-+>$$
If its correct then I should be able to evaluate this $$J_{1-}|+-> + |-+>$$
I am aware that when the lowering operator acts on the lowest eigenvalue it will disappear but I cant continue since the formula has (j+m) term which is 0 for different spins. By the way I am doing revision for this so it would be better if anyone can just tell me what I did wrong here. This is not homework. Thanks

6. Sep 12, 2012

### Oxvillian

You're not using the distributive law correctly: (A+B)(x+y) = Ax + Ay + Bx + By. Every state in the sum should have a lowering operator acting on it.

Two of the terms vanish but two survive because you're acting on a + with the lowering operator.

7. Sep 12, 2012

### semc

I really still don't see what I did wrong there. I simply use (A+B)(x+y)=A(x+y)+B(x+y). I am having problem here because I am using the lowering operator on + but I am getting 0 which is clearly wrong but I just don't know why I got it wrong. I will show my working

Using the lowering operator $$J_-|jm>=((j+m)(j-m+1))^{1/2}|jm-1>$$
on |-+> I got this $$J_-|-1/2, 1/2>=((-1/2+1/2)(-1/2-1/2+1))^{1/2}|-1/2,-1/2>$$
$$(-1/2+1/2)(-1/2-1/2+1)=(0)(0)=0$$ and for |+-> the operator is also acting on the second entry which is - so it goes to 0 as well.
Clearly I am making a stupid mistake but I really don't see it myself. Hopefully someone can tell me whats wrong

8. Sep 12, 2012

### Oxvillian

eg.

$$S_{1-} \lvert +- \rangle = \lvert -- \rangle$$

9. Sep 12, 2012

### Oxvillian

Sorry, we cross-posted...

$$\sqrt{(j+m)(j-m+1)} = \sqrt{(1/2 + 1/2)(1/2 - 1/2 + 1)} = 1$$

10. Sep 12, 2012

### semc

Yes but I believe this is for the case of |++> since j=m=1/2?

11. Sep 12, 2012

### Oxvillian

It's also the case for |+-> because J1- only acts on the +. It leaves the - alone.

Maybe that's the point you're missing? J1- lowers the first index and leaves the second alone; J2- acts on the second index and leaves the first alone.

12. Sep 12, 2012

### semc

Yes I am aware of that but in the working I wrote a few post ago, the coefficient is 0 since we have (j+m) which is 0.

13. Sep 12, 2012

### Oxvillian

but j=1/2 and m=1/2.

14. Sep 12, 2012

### semc

Think this is the part that I got confuse. For |+-> , does it mean that j=1/2 , m=-1/2?

15. Sep 12, 2012

### Oxvillian

Aha! I think we've nailed the problem. What you have here is the tensor product of two spin-1/2 states. So j1 = j2= 1/2. The + and - refer to m1 and m2 respectively.

For instance |+-> would have j1 = j2= 1/2, m1 = 1/2 and m2 = -1/2.

16. Sep 12, 2012

### semc

Haha I get it now. Thank you so much for explaining this to me!

17. Sep 12, 2012

No problem