Question

A spin $\dfrac{1}{2}$ particle is in a spin up state along the $x$-axis (with unit vector $\hat{x}$) and is denoted as $\left|\dfrac{1}{2}, \dfrac{1}{2}\right\rangle_x$. 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?

• A. $\dfrac{1}{2} \cos^2{\dfrac{\theta}{4}}$
• B. $\cos^2{\dfrac{\theta}{4}}$
• C. $\dfrac{1}{2} \cos^2{\dfrac{\theta}{2}}$
• D. $\cos^2{\dfrac{\theta}{2}}$

Hint:

For spin-1/2 particles, the probability amplitude of finding the particle in a spin state along a new direction is given by the square of the cosine of half the angle between the two directions.

Answer 1

The Correct Option is D

For a spin $\dfrac{1}{2}$ particle, the probability amplitude for finding the particle in the state $\left| \dfrac{1}{2}, \dfrac{1}{2} \right\rangle$ along an axis making an angle $\theta$ with respect to the $x$-axis is given by: \[ \left| \langle \dfrac{1}{2}, \dfrac{1}{2} \mid \hat{x'} \rangle \right|^2 = \cos^2{\left(\dfrac{\theta}{2}\right)}. \] Thus, the probability of finding the particle in the spin up state along the direction $\hat{x'}$ is $\cos^2{\left(\dfrac{\theta}{2}\right)}$, which corresponds to option (D).