# Clebsch-Gordan coefficients calculation

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1. Aug 30, 2016

### abcs22

1. The problem statement, all variables and given/known data
So i have to calculate the Clebsch-gordan coefficients for the state j1=3/2 and j2=1/2

2. Relevant equations
Recursiom formula, lowering and uppering operator

3. The attempt at a solution
I have tried to calculate the first set of the coefficients stating that:
L- l2,2> = L- l3/2,1/2>l1/2,1/2>

For the left side I got that it is equal to 2hl2,1>. Is that correct? I can't get the right coefficients on the right side. In general, I have trouble understanding which combinations of j and m to use tobget what I have to do. Is there some procedure to get all of them in order? Right now I am trying to guess what to use and see if I get the right combination.

2. Aug 30, 2016

### blue_leaf77

Yes.

You should start from the states in the two representations which are proportional to each other. One of such states are $|3/2,3/2\rangle |1/2,1/2\rangle$ and $|2,2\rangle$. Therefore, $|2,2\rangle = c|3/2,3/2\rangle |1/2,1/2\rangle$. The constant of proportionality $c$ is in general a unimodular complex number, but conventionally it is chosen to be $1$. Thus $|2,2\rangle = |3/2,3/2\rangle |1/2,1/2\rangle$. Then apply the lowering operator on both sides like you have done and in the RHS, use $L_- = L_{1-}+L_{2-}$.

Last edited: Aug 30, 2016
3. Aug 30, 2016

### Staff: Mentor

Yes, assuming that h means $\hbar$.

Given that you are adding together $j_1$ and $j_2$, then the allowed values of $J$ are
$$J = j_1 + j_2, j_1+j_2 - 1, \ldots, \left| j_1 - j_2 \right|$$
Second, the triangle rule has to be followed: $M = m_1 + m_2$. So in the case you have, the result will have to be 0 since $M = 2$ while $m_1+m_2 = 1$.

4. Aug 30, 2016

### abcs22

Thank you! I got the first ones:
I2,1> =√3/2 I1/2,1/2> + 1/2 I3/2,-1/2>

Is that ok? I tried applying lowering again on this expression but didn't get the solution. How do I know where to use the operator next?

5. Aug 30, 2016

### blue_leaf77

If that square root applies only on the 3 in the numerator, then it's fine.
Why won't it work? To get |2,0> you should indeed apply the lowering operator once more on both sides. Maybe you just made a mistake during calculation.

6. Aug 30, 2016

### Staff: Mentor

That's not correct, check the values of $j_!$ and $j_2$.

7. Aug 30, 2016

### blue_leaf77

Actually he is using $|m_1,m_2\rangle$ notation on the RHS while the values of $j_1$ and $j_2$ are only implied.

8. Aug 30, 2016

### Staff: Mentor

Right Nevermind....