# Homework Help: Calculation of Clebsch-Gordan coefficients

1. Feb 9, 2012

### PineApple2

1. The problem statement, all variables and given/known data
I have two spin-1/2 particles and I need to calculate their Clebsch-Gordan coefficients.

2. Relevant equations

3. The attempt at a solution

I followed the procedure of applying $J_-$ to $|{j,m}\rangle$
and $J_{1-}$ and $J_{2-}$ to $|{m_1,m_2}\rangle$ and comparing them. I got correctly
$\langle{1,1}|{1/2,1/2}\rangle =1$,
$\langle{1,0}|{1/2,-1/2}\rangle=1/\sqrt{2}$,
$\langle{1,0}|{-1/2,1/2}\rangle =1/\sqrt{2}$,
$\langle{1,-1}|{-1/2,-1/2}\rangle=1$.
Now I want to find $\langle{0,0}|{1/2,-1/2}\rangle$
and $\langle{0,0}|{-1/2,1/2}\rangle$.
Therefore I denoted $|{0,0}\rangle = \alpha|{1/2,-1/2}\rangle + \beta|{-1/2,1/2}\rangle$
and used the normalization condition $|\alpha|^2 + |\beta|^2 = 1$ and orthogonality to the $|{1,0}\rangle$ state. I got the equation
$|\alpha|^2 = 1/2$ from which there are 2 options:
$\alpha = 1/\sqrt{2}$ and $\alpha = -1/\sqrt{2}$ (only real coefficients by convention). How do I know which is the right option out of the two?

Thanks!

Last edited: Feb 9, 2012
2. Feb 9, 2012

### vela

Staff Emeritus
The overall sign of $\alpha$ doesn't really matter. What's important is the relative phase between $\alpha$ and $\beta$. What combination of $\alpha$ and $\beta$ will yield a state orthogonal to $|1, 0\rangle$?

3. Feb 9, 2012

### PineApple2

Ok. I agree. But the standard table has a certain convention. How do I know how to pick the sign to fit this convention?

4. Feb 9, 2012

### vela

Staff Emeritus