# Two questions on Clebsch-Gordan coefficients

1. Feb 28, 2013

### Yoran91

Hello everyone,

I'm reading a bit about Clebsch-Gordan coefficients and I found two things in their general description I didn't quite understand. Can anyone help me with these questions?

First, I read that the Clebsch-Gordan coefficients are zero unless the total angular momentum satisfies $|j_1-j_2|\leq j \leq j_1+j_2$. How would you prove this?

Second, and this is a bit more difficult, I know the CG coefficients are elements of a change of basis matrix, which relates the two bases
1. $|j_1 m_1 \rangle \otimes |j_2 m_2 \rangle$
2. $|j_1j_2jm \rangle$.

I also know that with respect to the basis 1., the total angular momentum operator $J_i$ is represented by the matrix $J_{1i} \otimes \mathbb{1} + \mathbb{1} \otimes J_{2i}$. Now I wish to know how this matrix looks with respect to the other basis, perhaps using the change of basis matrix.

I know this is the 'tensor product decomposition' rule, but I don't fully understand what's going on. How does that matrix look in the other basis?

2. Feb 28, 2013

### fzero

From the formula you quote below, we have $J_z = J_{1z} \otimes \mathbb{1} + \mathbb{1} \otimes J_{2z}$. If we act on the state $|j_1 m_1 \rangle \otimes |j_2 m_2 \rangle$, we find the eigenvalue $m_1+m_2$. Therefore the allowed linear combinations of spins must have $m=m_1+m_2$. Now, recall that the allowed values of $m_1 = -j_1,\ldots, j_1$ and $m_2 = -j_2,\ldots, j_2$. These imply that $m$ can only take values $-(j_1+j_2), \ldots, j_1 - j_2, \ldots, j_1+j_2$.

We could proceed from here by making a list of states and then assigning them to particular $j$ multiplets. I won't do this explicitly, but it might help you to go over the procedure explicitly for some specific example. Recall that a state with $m=j$, which is $|j,j\rangle$, is called the highest weight state. In terms of the ladder operator basis, the raising operator $J_+ |j,j\rangle=0$, while acting with lowering operator $J_-$ gives the state $|j,j-1\rangle$. Obviously the state with the largest value of $m$, which is $j_1+j_2$, is a highest weight state of the multiplet with $j=j_1+j_2$. We can use the lowering operator on this state to determine the rest of the states in the multiplet. Now, we pull these out of the list and find the next largest value of $m$ in the states remaining. This is the highest weight state for the multiplet with $j=j_1+j_2-1$. We can continue the process until we've exhausted all of the possibilities after enumerating the $j = |j_1-j_2|$ states.

There are various ways to give the same argument, but this one seems rather straightforward, especially when worked out in a specific example.

You could indeed use the CG coefficients to perform a change of basis. But the total angular momentum operator $J_i$ has the standard action on the $|j,m\rangle$ states:

$$J_z |j,m\rangle = m |j,m\rangle,$$
$$J_\pm |j,m\rangle = \sqrt{ j(j+1) - m(m\pm1)} |j,m\pm 1\rangle.$$

In fact, you will always use these expressions any time that you want to work out explicit expressions for the CG coefficients.

3. Mar 1, 2013

### DrDu

Maybe again to your second question: The matrix of the CG coefficients is a unitary matrix U carrying the transformation from the basis labeled by m1,m2 to the one labeled by j,m_j.
So your J_i transforms like $UJ_i U^+$.
The transformed matrix can be written down directly using the formulas fzero provided.

4. Mar 1, 2013

### Yoran91

Ah of course, thanks! I completely forgot about the equations for $J_i$ in terms of the total angular momentum states.

One last question: I now see that we would have, for a given allowed $j$:
$J_{x}^{(j)} | j_1 j_2 j m \rangle = \frac{\hbar}{2} \sqrt{j(j+1)-m(m+1)} |j_1 j_2 j (m+1) \rangle + \frac{\hbar}{2} \sqrt{j(j+1)-m(m-1)}|j_1 j_2 j (m-1) \rangle$

and a similar equation for $J_{y}^{(j)}$.

What, then, is the matrix representing $J_i$? I think it is the direct sum, i.e.
$J_i= J_{i}^{j_1+j_2} \oplus J_{i}^{j_1+j_2-1}\oplus ... \oplus J_{i}^{|j_1-j_2|}$,

but I'm not really sure how you would prove this

EDIT: nevermind, I understand now. Thank you !

Last edited: Mar 1, 2013