Splitting the Eigenstates with Stern-Gerlach

A beam of atoms can be split into the eigenstates of \bgroup\color{black}$L_z$\egroup with a Stern-Gerlach apparatus. A magnetic moment is associated with angular momentum.

\begin{displaymath}\bgroup\color{black} \vec{\mu}={-e\over 2mc}\vec{L}=\mu_B{\vec{L}\over\hbar} \egroup\end{displaymath}

This magnetic moment interacts with an external field, adding a term to the Hamiltonian.

\begin{displaymath}\bgroup\color{black} H=-\vec{\mu}\cdot\vec{B} \egroup\end{displaymath}

If the magnetic field has a gradient in the z direction, there is a force exerted (classically).

\begin{displaymath}\bgroup\color{black} F=-{\partial U\over\partial z}=\mu_z{\partial B\over\partial z} \egroup\end{displaymath}

A magnet with a strong gradient to the field is shown below.

Lets assume the field gradient is in the z direction.

In the Stern-Gerlach experiment, a beam of atoms (assume \bgroup\color{black}$\ell=1$\egroup) is sent into a magnet with a strong field gradient. The atoms come from an oven through some collimator to form a beam. The beam is said to be unpolarized since the three m states are equally likely no particular state has been prepared. An unpolarized, \bgroup\color{black}$\ell=1$\egroup beam of atoms will be split into the three beams (of equal intensity) corresponding to the different eigenvalues of \bgroup\color{black}$L_z$\egroup.

The atoms in the top beam are in the \bgroup\color{black}$m=1$\egroup state. If we put them through another Stern-Gerlach apparatus, they will all go into the top beam again. Similarly for the middle beam in the \bgroup\color{black}$m=0$\egroup state and the lower beam in the \bgroup\color{black}$m=-1$\egroup state.

We can make a fancy Stern-Gerlach apparatus which puts the beam back together as shown below.

\epsfig{file=figs/FancySternGerlach.eps,height=1.2in} \bgroup\color{black}$\left\{\matrix{+ \cr 0 \cr - }\right\}_z\rightarrow $\egroup
We can represent the apparatus by the symbol to the right.

We can use this apparatus to prepare an eigenstate. The apparatus below picks out the \bgroup\color{black}$m=1$\egroup state

\epsfig{file=figs/FancySternGerlachP1.eps,height=1.5in} \bgroup\color{black}$\left\{\matrix{+ \cr 0 \vert \cr - \vert}\right\}_z\rightarrow $\egroup
again representing the apparatus by the symbol at the right. We could also represent our apparatus plus blocking by an operator

\begin{displaymath}\bgroup\color{black} O=\vert+\rangle\;\langle +\vert \egroup\end{displaymath}

where we are writing the states according to the \bgroup\color{black}$m$\egroup value, either +, -, or 0. This is a projection operator onto the + state.

An apparatus which blocks both the + and - beams

\epsfig{file=figs/FancySternGerlachP0.eps,height=1.5in} \bgroup\color{black}$\left\{\matrix{+ \vert \cr 0 \cr - \vert}\right\}_z\rightarrow $\egroup
would be represented by the projection operator

\begin{displaymath}\bgroup\color{black} O=\vert\rangle\;\langle 0\vert \egroup\end{displaymath}

Similarly, an apparatus with no blocking could be written as the sum of the three projection operators.

\begin{displaymath}\bgroup\color{black} \left\{\matrix{+ \cr 0 \cr - }\right\}_z...
...imits_{m=-1}^{+1} \vert z_m\rangle\;\langle z_m\vert =1 \egroup\end{displaymath}

If we block only the \bgroup\color{black}$m=1$\egroup beam, the apparatus would be represented by

\begin{displaymath}\bgroup\color{black} \left\{\matrix{+ \vert \cr 0 \cr - }\rig...
...\;\langle 0\vert  +  \vert-\rangle\;\langle -\vert .\egroup\end{displaymath}

* Example: A series of Stern-Gerlachs.*

Jim Branson 2013-04-22