Discussion
A spin \(\frac{1}{2}\) particle is in a spin up state along the \(x\)-axis (with unit vector \(\hat{x}\)) and is denoted as \(\left|\frac{1}{2}, \frac{1}{2}\right\rangle_{x}\). What is the probability of finding the particle to be in a spin up state along the direction \(\hat{x}'\), which lies in the \(xy\)-plane and makes an angle \(\theta\) with respect to the positive \(x\)-axis, if such a measurement is made?
Answer: D
The probability is \(\cos^{2}\frac{\theta}{2}\).
The spin-up state along a direction in the \(xy\)-plane at angle \(\phi\) from \(x\) is \(\frac{1}{\sqrt{2}}(|\uparrow\rangle + e^{i\phi}|\downarrow\rangle)\) in the \(z\) basis. For \(\hat{x}\) (\(\phi = 0\)) and \(\hat{x}'\) (\(\phi = \theta\)):
\[ \begin{aligned} \langle x'+|x+\rangle &= \tfrac{1}{2}(1 + e^{-i\theta}) \\ |\langle x'+|x+\rangle|^{2} &= \tfrac{1}{2}(1 + \cos\theta) = \cos^{2}\frac{\theta}{2} \end{aligned} \]
This is the general spin-1/2 rule: the overlap between spin-up states along two directions an angle \(\theta\) apart is \(\cos^{2}(\theta/2)\). Check: \(\theta = 0\) gives certainty and \(\theta = \pi\) gives zero. The options with \(\theta/4\) or an extra factor 1/2 fail these checks.
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