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## Homework Statement

The electron in a hydrogen atom is described by the following superposition of two states:

[itex]|\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>) [/itex]

(b) Let J = L + S be the total angular momentum. Express state [itex]\psi[/itex] in basis [itex]|n, l, J, J_z>[/itex]

Hint: use the Clebsch -Gordan coffecients from the table on pg. 188 of Griffiths.

## Homework Equations

[itex] |s_1 m_1>|s_2 m_2> = \sum C^{s_1s_2s}_{m_1m_2m}|s m> [/itex]

## The Attempt at a Solution

For the first state:

[itex]|n=2,l=1,m=0,s_z= +1/2> ====> l=1, s = 1/2 [/itex]

[itex]|1,0>|1/2,1/2> = \sqrt{\frac{2}{3}}|3/2,1/2> - \sqrt{\frac{1}{3}}|1/2,1/2> [/itex]

For the second state:

[itex]|n=2,l=0,m=0,s_z= +1/2> [/itex] ====> 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.