An electron whith spin down is in the state [tex] \psi _{510} [/tex] 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?(adsbygoogle = window.adsbygoogle || []).push({});

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

If so,accroding to Clebsch-Gordan table,[tex]

|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 [/tex],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.

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# Homework Help: Problem on spin of electron

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