A 90 degree rotation about the z axis.

If we rotate our coordinate system by 90 degrees about the z axis, the old x axis becomes the new -y axis. So we would expect that the state with angular momentum $+\hbar$ in the x direction, $\psi^{(x)}_+$, will rotate into $\psi^{(y)}_-$ within a phase factor. Lets do the rotation.

$$\bgroup\color{black}R_z(\theta_z)=\left(\matrix{e^{i\theta_z}&0&0\cr 0&1&0\cr 0&0&e^{-i\theta_z}}\right).\egroup$$

$$\bgroup\color{black}R_z(\theta_z=90)=\left(\matrix{i&0&0\cr 0&1&0\cr 0&0&-i}\right).\egroup$$

Before rotation the state is

$$\bgroup\color{black}\psi^{(x)}_{+\hbar}=\left(\matrix{{1\over 2}\cr {1\over\sqrt{2}}\cr {1\over 2} }\right)\egroup$$

The rotated state is.

$$\bgroup\color{black}\psi'=\left(\matrix{i&0&0\cr 0&1&0\cr 0&0... ...x{{i\over 2}\cr {1\over\sqrt{2}}\cr {-i\over 2} }\right)\egroup$$

Now, what remains is to check whether this state is the one we expect. What is $\psi^{(y)}_-$? We find that state by solving the eigenvalue problem.

$$\bgroup\color{black}L_y\psi^{(y)}_-=-\hbar\psi^{(y)}_-\egroup$$

$$\bgroup\color{black}{\hbar\over\sqrt{2}i}\left(\matrix{0&1&0\... ... b\cr c}\right) =-\hbar\left(\matrix{a\cr b\cr c}\right)\egroup$$

$$\bgroup\color{black} \left(\matrix{{ib\over\sqrt{2}}\cr {i(c-... ...over\sqrt{2}}}\right) =\left(\matrix{a\cr b\cr c}\right)\egroup$$

Setting $b=1$, we get the unnormalized result.

$$\bgroup\color{black}\psi^{(y)}_-=C\left(\matrix{{i\over\sqrt{2}}\cr 1\cr {-i\over\sqrt{2}}}\right)\egroup$$

Normalizing, we get.

$$\bgroup\color{black}\psi^{(y)}_-=\left(\matrix{{i\over 2}\cr {1\over\sqrt{2}}\cr {-i\over 2}}\right)\egroup$$

This is exactly the same as the rotated state. A 90 degree rotation about the z axis changes $\psi^{(x)}_+$ into $\psi^{(y)}_-$.