A Rotated Stern-Gerlach Apparatus*

Imagine a Stern-Gerlach apparatus that first separates an \bgroup\color{black}$\ell=1$\egroup atomic beam with a strong B-field gradient in the z-direction. Let's assume the beam has atoms moving in the y-direction. The apparatus blocks two separated beams, leaving only the eigenstate of \bgroup\color{black}$L_z$\egroup with eigenvalue \bgroup\color{black}$+\hbar$\egroup. We follow this with an apparatus which separates in the u-direction, which is at an angle \bgroup\color{black}$\theta$\egroup from the z-direction, but still perpendicular to the direction of travel of the beam, y. What fraction of the (remaining) beam will go into each of the three beams which are split in the u-direction?

We could represent this problem with the following diagram.

\begin{displaymath}\bgroup\color{black}\mathrm{Oven}\rightarrow \left\{\matrix{+...
\left\{\matrix{+ D_+\cr 0 D_0\cr - D_-}\right\}_u\egroup\end{displaymath}

We put a detector in each of the beams split in \bgroup\color{black}$u$\egroup to determine the intensity.

To solve this with the rotation matrices, we first determine the state after the first apparatus. It is just \bgroup\color{black}$\psi^{(z)}_+=\left(\matrix{1\cr 0\cr 0}\right)$\egroup with the usual basis. Now we rotate to a new (primed) set of basis states with the \bgroup\color{black}$z'$\egroup along the \bgroup\color{black}$u$\egroup direction. This means a rotation through an angle \bgroup\color{black}$\theta$\egroup about the y direction. The problem didn't clearly define whether it is \bgroup\color{black}$+\theta$\egroup or \bgroup\color{black}$-\theta$\egroup, but, if we only need to know the intensities, it doesn't matter. So the state coming out of the second apparatus is

{1\over 2}(1+\cos(\th...
...\sqrt{2}}\sin(\theta_y)\cr {1\over 2}(1-\cos(\theta_y)) }\right)

The 3 amplitudes in this vector just need to be (absolute) squared to get the 3 intensities.

\begin{displaymath}\bgroup\color{black}I_+={1\over 4}(1+\cos(\theta_y))^2\qquad\...
I_-={1\over 4}(1-\cos(\theta_y))^2\egroup\end{displaymath}

These add up to 1.

An alternate solution would be to use the \bgroup\color{black}$L_u=\hat{u}\cdot\vec{L}=\cos\theta L_z+\sin\theta L_x$\egroup operator. Find the eigenvectors of this operator, like \bgroup\color{black}$\psi^{(u)}_+$\egroup. The intensity in the + beam is then \bgroup\color{black}$I_+=\vert\langle\psi^{(u)}_+\vert\psi^{(z)}_+\rangle\vert^2$\egroup.

Jim Branson 2013-04-22