# Homework Help: Addition of Angular Momenta, problem with notation

1. Dec 12, 2012

### joelcponte

1. The problem statement, all variables and given/known data
The electron in a hydrogen atom is described by the following superposition of two states:

$|\psi> = \frac{1}{\sqrt{2}}(|n=2,l=1,m=0,s_z= +1/2> + |n=2,l=0,m=0,s_z= +1/2>)$

(b) Let J = L + S be the total angular momentum. Express state $\psi$ in basis $|n, l, J, J_z>$
Hint: use the Clebsch -Gordan coffecients from the table on pg. 188 of Griffiths.

2. Relevant equations

$|s_1 m_1>|s_2 m_2> = \sum C^{s_1s_2s}_{m_1m_2m}|s m>$

3. The attempt at a solution

For the first state:

$|n=2,l=1,m=0,s_z= +1/2> ====> l=1, s = 1/2$
$|1,0>|1/2,1/2> = \sqrt{\frac{2}{3}}|3/2,1/2> - \sqrt{\frac{1}{3}}|1/2,1/2>$

For the second state:

$|n=2,l=0,m=0,s_z= +1/2>$ ====> there's no orbital angular momentum

My question is: I don't know if what I've done is correct and, if it is, I don't know how to transform to the notation he asks. For me, the "l" in the new notation doesn't make sense or have to be 0.

2. Dec 16, 2012

### vela

Staff Emeritus
That's right.

So what are j and jz equal to in this case?

You have one basis with states of the form $\lvert \ l\ m_l \rangle \lvert\ s\ m_s \rangle$. These states are eigenstates of L2, Lz, S2, and Sz. The other basis has states of the form $\lvert \ l\ s\ j\ j_z\rangle$, which are eigenstates of L2, S2, J2, and Jz. These two bases are related through the Clebsch-Gordan coefficients.

When you wrote
$$\lvert 1,0\rangle \lvert 1/2,1/2\rangle = \sqrt{\frac{2}{3}} \lvert 3/2,1/2\rangle - \sqrt{\frac{1}{3}} \lvert1/2,1/2\rangle,$$ what it really means is
$$\lvert 1,0\rangle \lvert 1/2,1/2\rangle = \sqrt{\frac{2}{3}} \lvert l=1, s=1/2, j=3/2, j_z=1/2\rangle - \sqrt{\frac{1}{3}} \lvert l=1, s=1/2, j=1/2, j_z=1/2\rangle,$$