# Clebsch-Gordan coefficients (l+1/2)

1. Jul 19, 2017

### Daviddc

1. The problem statement, all variables and given/known data
I was going through Sakurai's textbook and I tried to work out by myself the following expression for the Clebsch-Gordan coefficients (Addition of orbital angular momentum and spin 1/2)

\langle m-1/2,1/2|l+1/2,m \rangle = \sqrt{\frac{l+m+1/2}{2l+1}}

2. Relevant equations

J_{\pm} |jm \rangle=\sqrt{(j \mp m)(j \pm m +1)} \\
J_{i,\pm} |m_1 m_2 \rangle=\sqrt{(j_i \mp m_i)(j_i \pm m_i +1)}

3. The attempt at a solution

|l+1/2,m \rangle = \sum_{m'} \left[|m'-1/2,1/2\rangle \langle m'-1/2,1/2|l+1/2,m \rangle + |m'+1/2,-1/2\rangle \langle m'+1/2,-1/2|l+1/2,m \rangle \right]

Aplying J_+

\sqrt{(l+1/2-m)(l+1/2+m+1)}|l+1/2,m+1 \rangle = \sum_{m'} \left[\sqrt{(l-m'+1/2)(l+m'-1/2+1)}|m'+1/2,1/2\rangle \langle m'-1/2,1/2|l+1/2,m \rangle + \sqrt{(1/2+1/2)(1/2-1/2+1)} |m'+1/2,+1/2\rangle \langle m'+1/2,-1/2|l+1/2,m \rangle +\sqrt{(l-m'-1/2)(l+m'+1/2+1)}(|m'+3/2,-1/2\rangle \langle m'+1/2,-1/2|l+1/2,m \rangle \right]

Taking the inner product $$\langle m+1/2, 1/2 |$$

\sqrt{(l+1/2-m)(l+1/2+m+1)} \langle m+1/2,1/2 | l+1/2, m+1 \rangle = \sqrt{ (l-m+1/2)(l+m+1/2)} \langle m-1/2,1/2 | l+1/2, m \rangle

\langle m-1/2,1/2 | l+1/2, m \rangle=\sqrt{\frac{l+m+3/2}{l+m+1/2}} \langle m+1/2,1/2 | l+1/2, m+1 \rangle

Aplying the ladder operator until m reaches its maximum value (i.e. l+1/2),

\langle m-1/2,1/2 | l+1/2, m \rangle=\sqrt{\frac{2l+1}{l+m+1/2}}

As you can see, I obtain the answer upside down. It is obviously wrong since you get values larger than the unity. In Sakurai's textbook he starts with this:

\sqrt{(l+1/2-m)(l+1/2+m+1)} \langle m-1/2,1/2 | l+1/2, m \rangle = \sqrt{ (l-m+1/2)(l+m+1/2)} \langle m+1/2,1/2 | l+1/2, m+1 \rangle

But I don't understand why it is like that instead of the one I derived. I'm sure that my mistake is something stupid, but I can't find it and it is pissing me off since it's 1 am and I didn't have dinner yet because of this. If any of you can point out where I am failing I will be thankful.