 Solution

Here the dimensionless spin one-half operator is $S_x={\sigma }_x/2$. Since the Pauli matrix obeys ${\sigma }_x^2=I$, the even and odd terms of the matrix exponential giveThis is the exponential of a Pauli vector for an $x$-axis rotation. With a physical spin angular momentum $S_x^{\mathrm{phys}}=\hslash {\sigma }_x/2$, the exponent is $-i\theta S_x^{\mathrm{phys}}/\hslash$.