# Homework Help: Calculating Clebsch-Gordan coefficients

1. Feb 9, 2017

### Kara386

1. The problem statement, all variables and given/known data
The question asks me to calculate the non-zero Clebsch Gordan coefficients
$\langle j_1+j_2, j_s-2|j_1,m_1;j_2,m_2\rangle$
Where $j_s=j_1+j_2$.
2. Relevant equations

3. The attempt at a solution
The $j_s-2$ part is the $m$ of a $|j,m\rangle$ and I know that m has to equal $m_1+m_2$ or that equation would just be zero. And I also know that I should apply the lowering operators $J_-$ to one of these two equations:

$\langle j_s, j_s-1|j_1,j_1-1;j_2,j_2\rangle = \sqrt{\frac{j_1}{j_s}}$
Or
$\langle j_s, j_s-1|j_1,j_1;j_2,j_2-1\rangle = \sqrt{\frac{j_2}{j_s}}$

I don't know which of these two I should be applying the lowering operator to. And then I don't know how to apply it because there are two $j, m$ sets in the ket, so when calculating the coefficient using $\sqrt{j+m}\sqrt{j-m+1}$ which m and j exactly is this thing supposed to change? How do you apply the operator to an inner product? And you can't apply $J_-$ to the RHS so I'm not sure what to do about that either.

Essentially we're being taught via lots of examples, but it's hard to extract the rules from the examples without an explanation. Which is why my questions probably involve incredibly basic and fundamental concepts. Basically I have no idea what's going on, but it isn't for lack of reading or trying. I've definitely looked in lots of textbooks and on the internet, I just don't understand what they say! So any help is hugely appreciated, I'm trying to use this question to work out all the stuff I don't know. :)

2. Feb 10, 2017

### blue_leaf77

You can't, operators only act on kets.
To start, we have the convention
$$|j_1+j_2,j_1+j_2\rangle = |j_1,j_1;j_2,j_2\rangle$$
In order to get $|j_1+j_2,j_1+j_2-2\rangle$, you have to apply $(J_-)^2$ to the left side and hence also to the right side. This gives you
$$(J_-)^2 |j_1+j_2,j_1+j_2\rangle = (J_-)^2 |j_1,j_1;j_2,j_2\rangle$$
Then project both sides by $\langle j_1,m_2;j_2,m_2|$
$$\langle j_1,m_2;j_2,m_2|(J_-)^2 |j_1+j_2,j_1+j_2\rangle = \langle j_1,m_1;j_2,m_2|(J_-)^2 |j_1,j_1;j_2,j_2\rangle$$
In the end, it looks like that you should get a piecewise answer.

3. Feb 11, 2017

### Kara386

I don't know how to do the projecting thing. What I ended up with is:
$|j_s,j_s-2\rangle = \sqrt{\frac{2j_1(2j_1-1)}{j_s(4j_s-2)}} |j_1,j_1-2;j_2,j_2\rangle + \sqrt{\frac{2j_1j_2}{j_s(4j_s-2)}}|j_1,j_1-1;j_2,j_2-1\rangle + \sqrt{\frac{2j_2(2j_2-1)}{j_s(4j_s-2)}} |j_1,j_1;j_2,j_2-2\rangle + \sqrt{\frac{2j_1j_2}{j_s(4j_s-2)}}|j_1,j_1-1;j_2,j_2-1\rangle$

$=\sqrt{\frac{2j_1(2j_1-1)}{j_s(4j_s-2)}} |j_1,j_1-2;j_2,j_2\rangle + 2\sqrt{\frac{2j_1j_2}{j_s(4j_s-2)}}|j_1,j_1-1;j_2,j_2-1\rangle + \sqrt{\frac{2j_2(2j_2-1)}{j_s(4j_s-2)}} |j_1,j_1;j_2,j_2-2\rangle$

So when I do the inner product, what's the rule there? On the LHS it's the inner product of something with itself so do I get the square of the LHS? As for the RHS I know you can move the constants out from the middle so you get inner products there too, and in this basis all these vectors are orthogonal I think - except if they were the RHS would just be zero.

4. Feb 11, 2017

### blue_leaf77

No, you are not projecting something on itself. You should project $|j_s,j_s-2\rangle$ onto $\langle j_1,m_1;j_2m_2|$.
For example for the first term in RHS you get
$$\langle j_1,m_1;j_2m_2|j_1,j_1-2;j_2,j_2\rangle$$
What's the condition on $m_1$ and $m_2$ so that the above inner product does not vanish?

5. Feb 11, 2017

### Kara386

Oh, the inner product vanishes except for $m=m_1+m_2$. Thanks! :)

6. Feb 11, 2017

### Kara386

Ah, except which one is m? If $m = j_s -2$ then none of them vanish.

7. Feb 11, 2017

### blue_leaf77

I don't understand what you mean.

8. Feb 11, 2017

### Kara386

For an inner product to exist it has to have $m_1+m_2=m$. I can identify $m_1$ and $m_2$ in the kets, but is $m=j_s-2$? From the ket on the RHS? Or actually in this case $m_1,m_2$ are in the bra, but I still don't know which value is m. Do you get it from the RHS or from the ket which is part of the inner product?

9. Feb 11, 2017

### vela

Staff Emeritus
It seems a bit strange to claim that the inner product doesn't vanish only if $m_1+m_2=m$ when you have no idea what $m$ stands for. Anyway, you should rethink what the condition is for an inner product not to be 0.

10. Feb 11, 2017