The spin operator is $$\hat S_x=\frac\hbar2\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$ Which statement is correct?
Its positive and negative eigenstates are respectively $$\begin{pmatrix}1\\0\end{pmatrix},\quad\begin{pmatrix}0\\1\end{pmatrix}.$$
Its positive and negative eigenstates are respectively $$\frac1{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix},\quad\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}.$$
For state $$\frac12\begin{pmatrix}1\\\sqrt3\end{pmatrix},$$ the probability of the positive $S_x$ outcome is $1/4$.
For state $$\frac12\begin{pmatrix}1\\\sqrt3\end{pmatrix},$$ the probability of the positive $S_x$ outcome is $(2+\sqrt3)/4$.
Correct answer: (D) For state $$\frac12\begin{pmatrix}1\\\sqrt3\end{pmatrix},$$ the probability of the positive $S_x$ outcome is $(2+\sqrt3)/4$.
Explanation
The normalized positive $S_x$ eigenket is $(1,1)^T/\sqrt2$ and the negative one is $(1,-1)^T/\sqrt2$, so A and the reversed ordering in B are incorrect. The given state is normalized. Its positive-outcome amplitude is $(1+\sqrt3)/(2\sqrt2)$, giving probability $(1+\sqrt3)^2/8=(2+\sqrt3)/4$. Thus D.
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