Question 4.40: (a) Aparticle of spin 1 and a particle of spin 2 are at rest...

(a) Aparticle of spin 1 and a particle of spin 2 are at rest in a configuration such that the total spin is 3, and its z component is \hbar   . If you measured the z-component of the angular momentum of the spin-2 particle, what values might you get, and what is the probability of each one? Comment: Using Clebsch–Gordan tables is like driving a stick-shift—scary and frustrating when you start out, but easy once you get the hang of it.

(b) An electron with 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?

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(a) From the 2 × 1 Clebsch-Gordan table we get

|31\rangle=\sqrt{\frac{1}{15}}|22\rangle|1-1\rangle+\sqrt{\frac{8}{15}}|21\rangle|10\rangle+\sqrt{\frac{6}{15}}|20\rangle|11\rangle .

so you might get 2 \hbar (probability 1/15), \hbar (probability 8/15), or 0 (probability 6/15).

(b) From the 1 \times \frac{1}{2} \text { table: }|10\rangle\left|\frac{1}{2}-\frac{1}{2}\right\rangle=\sqrt{\frac{2}{3}}\left|\frac{3}{2}-\frac{1}{2}\right\rangle+\sqrt{\frac{1}{3}}\left|\frac{1}{2}-\frac{1}{2}\right\rangle .

So the total is 3/2 or 1/2, with l(l+1) \hbar^{2}= 15 / 4 \hbar^{2} \text { and } 3 / 4 \hbar^{2} , respectively. Thus you get \frac{15}{4} \hbar^{2} (probability 2/3), or \frac{3}{4} \hbar^{2} (probability 1/3).

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