# Problem on spin of electron

1. Jan 4, 2011

### kakarotyjn

An electron whith spin down is in the state $$\psi _{510}$$ of the hydrogen atom.If you could measure the total angular momentum squared of the electron alone(not including the proton spin),what values might you get,and what is the probability of each?

About this problem,I want to ask what does $$\psi _{510}$$ mean?Does it mean the proton in the hydrogen atom is in state |1 0>?

If so,accroding to Clebsch-Gordan table,$$|1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 0\rangle |\frac{1}{2}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - \frac{1}{2}\rangle = \sqrt {\frac{2}{3}{\kern 1pt} } {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} |\frac{3}{2}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - \frac{1}{2}\rangle + \sqrt {\frac{1}{3}} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} |\frac{1}{2}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - \frac{1}{2}\rangle$$,then I will get 15/4 and 3/4,the probability is 2/3 and 1/3 in respect.

Do I consider it right?Thank you.

2. Jan 4, 2011

### vela

Staff Emeritus
For the hydrogen atom, the spatial eigenfunctions are labeled by the quantum numbers n, l, and m, so yes, l=1 and m=0 as you inferred.

Your results for the measurements are missing a factor of $\hbar^2$, but otherwise your work looks fine.

3. Jan 4, 2011

### kakarotyjn

Oh,thank you!