# Clebcsh Gordan Coefficients - QM

1. May 14, 2014

### binbagsss

I'm looking at how a l j1,j2,jm> state can be operated on by the ladder operators, changing the value of m, and how this state can be expressed as a linear combination of l j1, j2, m1, m2> states.

Where the following relationships must be obeyed (x2):

m=m1+m2
l j1-j2 l ≤ j ≤ j1+j2

So I'm looking at an example where we start with state j1=1/2 and j2=1/2.

From the above relations I know :

m1=m2=$\pm$1/2
j=j1+j2=1
m=-1,0,0,1

I am in state l j1,j1,j,m> = l 1/2,1/2,1,1>.

I operate with L- to get:

l 1/2, 1/2, 1,0>

Now I want to express this in terms of l j1, j2, m1, m2> bases. So looking at the possible values of m1 and m2, I can see this can be produced by a linear combination of l j1, j2, m1, m2> = l 1/2,1/2,1/2,-1/2> , l 1/2,1/2,-1/2,1/2>

So I need to figure out the coefficients.

Now here is my question:

I have the relation: J$\pm$ l j1,j2,j,m> = (j(j+1)-m(m$\pm$1))$^{\frac{1}{2}}$ l j1,j2,j,m$\pm$1>, *

Which would give the coefficients as $\sqrt{\frac{3}{4}}$

So I dont understand where the coefficients come from : ($\sqrt{\frac{1}{2}}$ is the solution).

I can see (i think ) that this makes sense from a normalization point of view. But then what happens to the coefficients attained from * ?

Many thanks for any assistance, greatly appreciated !

2. May 15, 2014

### Fightfish

I'm afraid you are a little confused here. You are still in the $|j_{1},j_{2}, j ,m \rangle$ basis and not the $|j_{1},j_{2}, m_{1} ,m_{2} \rangle$ basis, so what you've got there is not the Clebsch-Gordan coefficient.

The correct way to derive the Clebsch-Gordan coefficients is to recognise that the state $|j_{1},j_{2}, j ,m \rangle = |\frac{1}{2},\frac{1}{2},1,1 \rangle$ is equivalent to $|j_{1},j_{2}, m_{1} ,m_{2} \rangle = |\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2} \rangle$. Now, you apply the lowering operator on this state. Give it a try, if you still need any help, I'll be happy to provide more details.

3. May 15, 2014

### binbagsss

Okay thanks.
So do you consider the lowering operator acting on m1 and m2 individually, so effectively defining two different ladder operators - one for the '1 space' (L$_{1}$$_{\pm}$)and (L$_{2}$$_{\pm}$) for the '2 space'

If we then operate on l 1/2, 1/2, 1/2, 1/2 > in turn with (L$_{1}$$_{\pm}$)and (L$_{2}$$_{\pm}$)

we get L$_{1}$$_{-}$ = l 1/2, 1/2, -1/2, 1/2 >

and L$_{2}$$_{-}$= l 1/2, 1/2, 1/2, -1/2 >

So this gives a coefficient of 1 for both
Would this be along the right lines?

Last edited: May 15, 2014
4. May 15, 2014

### Fightfish

Yup! The trick is to consider the total lowering operator, which is just the sum of the lowering operators on each subspace. So we have:
$$\ell^{-} \left|\frac{1}{2}, \frac{1}{2}, 1, 1 \right\rangle = \ell^{-}_{1} \left|\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right\rangle + \ell^{-}_{2} \left|\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right\rangle$$
where the LHS is in the $\left|j_{1}, j_{2}, j, m \right\rangle$ basis and the RHS in the $\left|j_{1}, j_{2}, m_{1}, m_{2} \right\rangle$ basis.

5. May 15, 2014

### binbagsss

Ok thanks. and the coefficients are 1, so does (1/2)^1/2 come from normalization?

6. May 16, 2014

### Fightfish

No, there is no need for an additional step. The factor that you are looking for comes out naturally from the expression that i wrote. If you compute it explicitly, it yields
$$\sqrt{2}\left|\frac{1}{2}, \frac{1}{2}, 1, 0 \right\rangle = \left|\frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, \frac{1}{2} \right\rangle + \left|\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2} \right\rangle$$
Move the factor of $\sqrt{2}$ to the RHS and it yields you exactly what you want.

7. May 17, 2014

ah. thanks.