Discussion
An electron is in a state \(\chi = \begin{pmatrix} \frac{1+i}{\sqrt{6}} \\ \sqrt{2}/\sqrt{3} \end{pmatrix}\) where the basis functions are \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\). The respective probabilities that a measurement of \(S_{z}\) yields values \(+\hbar/2\) and \(-\hbar/2\) are :
Answer: D
The probabilities are 1/3 and 2/3.
The probability of each outcome is the squared modulus of the corresponding component:
\[ \begin{aligned} P(+\hbar/2) &= \frac{|1+i|^{2}}{6} = \frac{2}{6} = \frac{1}{3} \\ P(-\hbar/2) &= \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^{2} = \frac{2}{3} \end{aligned} \]
They add to 1, so the state is normalised. The trap is \(|1+i|^{2}\): it is 2, not 1 (which would give 1/6), and not \((1+i)^{2} = 2i\). Swapping the two values gives 2/3 and 1/3.
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