A series of Stern-Gerlachs

Now that we have the shorthand notation for a Stern-Gerlach apparatus, we can put some together and think about what happens. The following is a simple example in which three successive apparati separate the atomic beam using a field gradient along the z direction.

\begin{displaymath}\bgroup\color{black} \mathrm{Oven}(I_0)\rightarrow \left\{\ma...
...ix{+ \vert \cr 0 \cr - \vert }\right\}_z(I_3)\rightarrow\egroup\end{displaymath}

If the intensity coming out of the oven is $I_0$, what are the intensities at positions 1, 2, and 3?We assume an unpolarized beam coming out of the oven so that 1/3 of the atoms will go into each initial beam in apparatus 1. This is essentially a classical calculation since we don't know the exact state of the particles coming from the oven. Now apparatus 1 removes the \bgroup\color{black}$m=1$\egroup component of the beam, leaving a state with a mixture of \bgroup\color{black}$m=0$\egroup and \bgroup\color{black}$m=-1$\egroup.

\begin{displaymath}\bgroup\color{black}I_1={2\over 3} I_0\egroup\end{displaymath}

We still don't know the relative phase of those two components and, in fact, different atoms in the beam will have different phases.

The beam will split into only two parts in the second apparatus since there is no \bgroup\color{black}$m=1$\egroup component left. Apparatus 2 blocks the \bgroup\color{black}$m=0$\egroup part, now leaving us with a state that we can write.

\begin{displaymath}\bgroup\color{black}I_2={1\over 3} I_0\egroup\end{displaymath}

All the particles in the beam are in the same state.


The beam in apparatus 3 all goes along the same path, the lower one. Apparatus 3 blocks that path.


The following is a more complex example using a field gradients in the z and x directions (assuming the beam is moving in y).

\begin{displaymath}\bgroup\color{black} \mathrm{Oven}(I_0)\rightarrow \left\{\ma...
...\matrix{+ \vert \cr 0 \cr - }\right\}_z(I_3)\rightarrow \egroup\end{displaymath}

If the intensity coming out of the oven is $I_0$, what are the intensities at positions 1, 2, and 3?

Now we have a Quantum Mechanics problem. After the first apparatus, we have an intensity as before

\begin{displaymath}\bgroup\color{black}I_1={1\over 3} I_0\egroup\end{displaymath}

and all the particles are in the state

\begin{displaymath}\bgroup\color{black}\psi^{(z)}_+=\left(\matrix{1 \cr 0 \cr 0}\right).\egroup\end{displaymath}

The second apparatus is oriented to separate the beam in the x direction. The beam separates into 3 parts. We can compute the intensity of each but lets concentrate on the bottom one because we block the other two.

\begin{displaymath}\bgroup\color{black}I_2=\left\vert\langle\psi^{(x)}_-\vert\psi^{(z)}_+\rangle\right\vert^2 I_1\egroup\end{displaymath}

We have written the probability that one particle, initially in the the state \bgroup\color{black}$\psi^{(z)}_+$\egroup, goes into the state \bgroup\color{black}$\psi^{(x)}_-$\egroup when measured in the x direction (times the intensity coming into the apparatus). Lets compute that probability.

\left(\matrix{1\cr 0 \cr 0}\right)=-{1\over 2}\egroup\end{displaymath}

So the probability is \bgroup\color{black}${1 \over 4}$\egroup.

\begin{displaymath}\bgroup\color{black}I_2={1\over 4} I_1={1\over 12} I_0\egroup\end{displaymath}

The third apparatus goes back to a separation in z and blocks the \bgroup\color{black}$m=1$\egroup component. The incoming state is

\begin{displaymath}\bgroup\color{black}\psi^{(x)}_-=\left(\matrix{-{1\over 2} \cr {1\over\sqrt{2}} \cr -{1\over 2}}\right)\egroup\end{displaymath}

Remember that the components of this vector are just the amplitudes to be in the different \bgroup\color{black}$m$\egroup states (using the z axis). The probability to get through this apparatus is just the probability to be in the \bgroup\color{black}$m=0$\egroup beam plus the probability to be in the \bgroup\color{black}$m=-1$\egroup beam.

\begin{displaymath}\bgroup\color{black}P=\left\vert-{1\over\sqrt{2}}\right\vert^2+\left\vert{1\over 2}\right\vert^2={3\over 4}\egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black}I_3={3\over 4}I_2={1\over 16}I_0\egroup\end{displaymath}

Now lets see what happens if we remove the blocking in apparatus 2.

\begin{displaymath}\bgroup\color{black} \mathrm{Oven}(I_0)\rightarrow \left\{\ma...
...\matrix{+ \vert \cr 0 \cr - }\right\}_z(I_3)\rightarrow \egroup\end{displaymath}

Assuming there are no bright lights in apparatus 2, the beam splits into 3 parts then recombines yielding the same state as was coming in, \bgroup\color{black}$\psi^{(z)}_+$\egroup. The intensity coming out of apparatus 2 is \bgroup\color{black}$I_2=I_1$\egroup. Now with the pure state \bgroup\color{black}$\psi^{(z)}_+$\egroup going into apparatus 3 and the top beam being blocked there, no particles come out of apparatus 3.


By removing the blocking in apparatus 2, the intensity dropped from \bgroup\color{black}${1\over 16}I_0$\egroup to zero. How could this happen?

What would happen if there were bright lights in apparatus 2?

Jim Branson 2013-04-22