Stern-Gerlach Revisited

In this section we revisit the discussion of the Stern-Gerlach experiment and show that the observed behaviour of that system can be perfectly described using the mathematical framework developed so far.

In the previous discussion of the Stern-Gerlach experiment we saw that what was being measured was the component of spin angular momentum along a particular direction in space and that there were only ever two possible outcomes to such a measurement. Thus, it would appear that spin states of the electron live in a 2-dimensional state space, they are qubits, and it will be useful to employ the \(\ket{\mathbf{n},\pm}\) notation for states labeled by a particular direction in space. With respect to some (arbitrarily chosen) coordinate axes a Stern-Gerlach apparatus may be arranged to measure the component of spin in the \(z\)-direction and the corresponding spin states could be denoted \(\ket{z;\pm}\). We would be inclined to posit that these are eigenstates of an observable \(S_z\) corresponding to eigenvalues \(\hbar/2\) and \(-\hbar/2\) respectively. That is,
\begin{equation}
S_z\ket{z;\pm}=\pm\frac{\hbar}{2}\ket{z;\pm}
\end{equation}
and so the matrix representation of \(S_z\) in this basis is
\begin{equation}
\mathbf{S}_z=\frac{\hbar}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\frac{\hbar}{2}\boldsymbol{\sigma}_z
\end{equation}
where we have recalled the definition of the Pauli matrix \(\sigma_z\).

Indeed, the discussion in Qubit Mechanics I suggests that we should define spin observables for a general orientation, \(\mathbf{n}\), in space according to \begin{equation}
\mathbf{S}_\mathbf{n}=\frac{\hbar}{2}\mathbf{n}\cdot\boldsymbol{\sigma}
\end{equation}
with corresponding eigenstates \(\ket{\mathbf{n},\pm}\). So in particular we would have
\begin{equation}
\mathbf{S}_x=\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\frac{\hbar}{2}\boldsymbol{\sigma}_x
\end{equation}
and
\begin{equation}
\mathbf{S}_y=\frac{\hbar}{2}\begin{pmatrix}0&-i\\i&0\end{pmatrix}=\frac{\hbar}{2}\boldsymbol{\sigma}_y
\end{equation}
with respective orthonormal eigenstates,
\begin{eqnarray}
\ket{x;+}&=&\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}&=&\frac{1}{\sqrt{2}}\left(\ket{z;+}+\ket{z;-}\right)\\
\ket{x;-}&=&\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}&=&\frac{1}{\sqrt{2}}\left(\ket{z;+}-\ket{z;-}\right)
\end{eqnarray}
and
\begin{eqnarray}
\ket{y;+}&=&\frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}&=&\frac{1}{\sqrt{2}}\left(\ket{z;+}+i\ket{z;-}\right)\\
\ket{y;-}&=&\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix}&=&\frac{1}{\sqrt{2}}\left(\ket{z;+}-i\ket{z;-}\right).
\end{eqnarray}

Stern-Gerlach Explained

We now have all the quantum mechanical machinery in place to understand the Stern-Gerlach experiments. Recall the basic setup, which we referred to as SG1. We assume that the spin state of an atom entering SG1 is in some arbitrary state \(\ket{\psi}\) in a 2-dimensional state space. The measuring device in SG1 corresponds to the observable \(S_z\) whose spectral decomposition is
\begin{equation}
S_z=\frac{\hbar}{2}\ket{z;+}\bra{z;+}-\frac{\hbar}{2}\ket{z;-}\bra{z;-}
\end{equation}
and therefore the probability of measuring a particle to be spin up is \(p(z;+)\) given by
\begin{equation}
p(z;+)=\braket{\psi\ket{z;+}\bra{z;+}\psi}
\end{equation}
in other words it is the squared modulus of the probability amplitude \(\braket{z;+|\psi}\) for finding \(\ket{\psi}\) in the state \(\ket{z;+}\). Likewise, the probability amplitude for finding \(\ket{\psi}\) in the state \(\ket{z;-}\) is \(\braket{z;-|\psi}\) corresponding to the probability \(p(z;-)=|\braket{z;-|\psi}|^2\).

Now let us consider SG2. In this case, we retain only the atoms emerging with spin state \(\ket{z;+}\) from the initial \(S_z\) measuring device then subject these atoms to a second \(S_z\) device. In this case the amplitudes for measuring the \(z\)-component of spin up and down are respectively \(\braket{z;+|z;+}=1\) and \(\braket{z;-|z;+}=0\) so we are sure to confirm that the atom has spin state up.

If instead of passing the atoms retained from the first \(S_z\) device in SG2 into a second \(S_z\) device we pass them instead into an \(S_x\) device, as in SG3, then the relevant amplitudes are
\begin{equation}
\braket{x;+|z;+}=\frac{1}{\sqrt{2}}\braket{z;+|z;+}+\frac{1}{\sqrt{2}}\braket{z;-|z;+}=\frac{1}{\sqrt{2}}
\end{equation}
for finding the \(x\)-component of the spin to be up and
\begin{equation}
\braket{x;-|z;+}=\frac{1}{\sqrt{2}}\braket{z;+|z;+}-\frac{1}{\sqrt{2}}\braket{z;-|z;+}=-\frac{1}{\sqrt{2}}
\end{equation}
for finding the \(x\)-component of the spin to be down. That is, we find that there is an equal probability of \(1/2\) for the \(x\)-component of the spin to be up or down.

In SG4 we retain the atoms measured to be spin down by a \(S_x\) device which took as input atoms measured to be spin up by a \(S_z\) device. These atoms are then passed to another \(S_z\) measuring device. The relevant amplitudes are now
\begin{equation}
\braket{z;+|x;-}=\frac{1}{\sqrt{2}}\braket{z;+|z;+}-\frac{1}{\sqrt{2}}\braket{z;+|z;-}=\frac{1}{\sqrt{2}}
\end{equation}
for finding the \(z\)-component of the spin to be up and
\begin{equation}
\braket{z;-|x;-}=\frac{1}{\sqrt{2}}\braket{z;-|z;+}-\frac{1}{\sqrt{2}}\braket{z;-|z;-}=-\frac{1}{\sqrt{2}}
\end{equation}
for finding the \(z\)-component of the spin to be down. That is, we find that there is an equal probability of \(1/2\) for the \(z\)-component of the spin to be up or down. The quantum mechanical formalism makes it clear that there is no ‘memory’ that the atoms had previously, before the \(S_x\) measurement, been found with probability 1 to have \(z\)-component of their spin up!