Discussion
Consider a spin-1 particle in the state \(|S_{z}, +1\rangle\). Let \(P_{1}^{x}, P_{0}^{x}\) and \(P_{-1}^{x}\) denote the probabilities that the measurement of \(S_{x}\) gives the values \(\hbar\), 0 and \(-\hbar\), respectively. Then :
Answer: C
The probabilities are \(P_{1}^{x} = 1/4, P_{0}^{x} = 1/2, P_{-1}^{x} = 1/4\).
For spin 1, the eigenvectors of \(S_{x}\) written in the \(S_{z}\) basis \((|{+1}\rangle, |0\rangle, |{-1}\rangle)\) are
\[ \begin{aligned} |{+1}\rangle_{x} &= \tfrac{1}{2}(1, \sqrt{2}, 1) \\ |0\rangle_{x} &= \tfrac{1}{\sqrt{2}}(1, 0, -1) \\ |{-1}\rangle_{x} &= \tfrac{1}{2}(1, -\sqrt{2}, 1) \end{aligned} \]
The probability of each outcome is the square of the first component (the overlap with \(|S_{z}, +1\rangle\)): \((1/2)^{2} = 1/4\), \((1/\sqrt{2})^{2} = 1/2\) and \(1/4\). They add to 1, and the symmetry between \(+\hbar\) and \(-\hbar\) is expected because \(|S_{z}, +1\rangle\) has \(\langle S_{x} \rangle = 0\).
The equal split 1/3 each and the asymmetric splits fail that symmetry check.
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