(a) A particle of spin1and a particle of spin 2 are at rest in a configuration such that the total spin is 3, and its z component is $\mathbf{\hslash }$ . If you measured the z component of the angular momentum of the spin-2particle, what values might you get, and what is the probability of each one?

(b) An electron with spin down is in the state${\mathit{\psi }}_{\mathbf{510}}$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?

The probability of an event occurring. The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

Expand the composite spin $\mid 3,1>$from the individual spins. For spin 2, the expected states are as follows,

$|2,2>,|2,1>,|2,0>,|2,-1>$, and $\mid 2,-2>$.

Write the possible states for spin 1.

localid="1658127583657" $|1,1>,|1,0>,$and$\mid 1,-1>$.

The combinations that have a z projection equal to one are needed, so the expansion can be written as follows,

$|3,1⟩=\alpha |2,2>|1,-1>+\beta |2,1>|1,0>+\gamma |2,0>|1,1⟩$

Determine the three expansion coefficients a ,$\beta$ and $\gamma$ in the Clebsch-Gordon tables. Then the probabilities are $|\alpha {|}^2,|\beta {|}^2$ and ${\left|\gamma \right|}^2$.

Return to the Clebsch-Gorden table and using the equation.

$|sm⟩=c_{\begin{array}{c}{m}_1{m}_{2}m\\ m_1+m_2=m\end{array}}^{s_1{s}_{2}s}|s_1{m}_1>|s_2{m}_2>$

Write the outcomes using the above information.

Thus, $2\hslash$ is obtained with a probability equal to 1/15 , or$\hslash$with a probability of 8/15 or with a probability of 6/15 .

Look the table $1\times 1/2$ and write the outcome.

$|10⟩|\frac{1}{2}-\frac{1}{2}⟩=\sqrt{\frac{2}{3}}\overline{)\frac{3}{4}}-\frac{1}{2}⟩+\sqrt{\frac{1}{3}}\overline{)\frac{1}{2}}-\frac{1}{2}⟩$

So the total is 3/2 or 1/2 with $l(l+1){\hslash }^2=\frac{15}{4}{\hslash }^2$and $\frac{3}{4}{\hslash }^2$respectively.

Thus, for $\frac{15}{4}{\hslash }^2$the probability is 2/3 , and for $\frac{3}{4}{\hslash }^2$it is 1/3 .