The problem asks for the probability of measuring a spin $\frac{1}{2}$ particle in a spin-up state along a specific direction $\hat{x}'$, given that it is initially in a spin-up state along the $x$-axis.

Spin State Definition

The initial state is given as spin up along the $x$-axis, denoted by $|\frac{1}{2}, \frac{1}{2}\rangle_x$. This state is an eigenstate of the spin operator $S_x$ with eigenvalue $+\frac{\hbar}{2}$. In the standard basis of spin-up $|\uparrow_z\rangle$ and spin-down $|\downarrow_z\rangle$ states along the $z$-axis, this state is expressed as:

$ |\frac{1}{2}, \frac{1}{2}\rangle_x = \frac{1}{\sqrt{2}} (|\uparrow_z\rangle + |\downarrow_z\rangle) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} $

Measurement State Direction

The measurement is performed along the direction $\hat{x}'$, which lies in the $xy$-plane and forms an angle $\theta$ with the positive $x$-axis. The direction vector is $\hat{n} = (\cos\theta, \sin\theta, 0)$. The spin-up state along this direction, $|\uparrow_{\hat{x}'}\rangle$, is:

$ |\uparrow_{\hat{x}'}\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ e^{i\theta} \end{pmatrix} $

Probability Amplitude Calculation

The probability $P$ of measuring the spin-up state along $\hat{x}'$ is the squared magnitude of the probability amplitude (the inner product of the states):

$ P = |\langle \uparrow_{\hat{x}'} | \frac{1}{2}, \frac{1}{2}\rangle_x|^2 $

First, calculate the probability amplitude:

$ \langle \uparrow_{\hat{x}'} | \frac{1}{2}, \frac{1}{2}\rangle_x = \left( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & e^{-i\theta} \end{pmatrix} \right) \left( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right) $

$ = \frac{1}{2} (1 \cdot 1 + e^{-i\theta} \cdot 1) = \frac{1}{2}(1 + e^{-i\theta}) $

Next, calculate the probability by squaring the magnitude of this amplitude:

$ P = \left|\frac{1}{2}(1 + e^{-i\theta})\right|^2 = \frac{1}{4} |1 + e^{-i\theta}|^2 $

Using $|z|^2 = z z^*$ and the identity $e^{i\phi} + e^{-i\phi} = 2\cos\phi$:

$ |1 + e^{-i\theta}|^2 = (1 + e^{-i\theta})(1 + e^{i\theta}) = 1 + e^{i\theta} + e^{-i\theta} + 1 = 2 + 2\cos\theta $

Substituting this result back into the probability expression:

$ P = \frac{1}{4} (2 + 2\cos\theta) = \frac{1}{2}(1 + \cos\theta) $

Finally, applying the half-angle trigonometric identity $1 + \cos\theta = 2\cos^2(\frac{\theta}{2})$:

$ P = \frac{1}{2} \left( 2\cos^2\left(\frac{\theta}{2}\right) \right) = \cos^2\left(\frac{\theta}{2}\right) $

The probability of finding the particle in a spin-up state along $\hat{x}'$ is $\cos^2(\frac{\theta}{2})$.