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!

Recall that a current circulating in a closed loop induces a magnetic dipole moment (a quantity that determines the torque which the loop will experience in an external magnetic field). This is a vector quantity, \(\boldsymbol{\mu}\), given by, \(\boldsymbol{\mu}=I\mathbf{A}\), where \(I\) is the current in the loop and \(\mathbf{A}\) is the oriented area enclosed by the loop, the orientation given by the right hand rule to be consistent with the direction of the current. The torque \(\boldsymbol{\tau}\) experienced by such a current loop in a magnetic field \(\mathbf{B}\) is then given by \(\boldsymbol{\tau}=\boldsymbol{\mu}\times\mathbf{B}\). This turning force works to align the magnetic moment with the magnetic field.

More generally a rotating charge distribution results in a magnetic moment and if the distribution has mass then an angular momentum which must be related to the magnetic moment. Indeed, consider a ring of charge with radius \(r\) that has a uniform charge distribution and total charge \(Q\). We assume the ring is rotating about an axis perpendicular to the plane of the ring and going through its centre. If the tangential velocity is \(v\), then the current at the loop is given by \(I=\lambda v\) where \(\lambda\) is the charge density, that is, \(\lambda=Q/2\pi r\). Thus we have that the magnitude \(\mu\) of \(\boldsymbol{\mu}\) is given by,
\begin{equation}
\mu=IA=\frac{Q}{2\pi r}v\pi r^2=\frac{Q}{2}rv.
\end{equation}
Now if the mass of the ring is \(M\), then recalling that the angular momentum is given by \(\mathbf{L}=M\mathbf{r}\times\mathbf{v}\), then
\begin{equation}
\boldsymbol{\mu}=\frac{Q}{2M}\mathbf{L}
\end{equation}
In particular, for a single electron with charge \(-e\) and mass \(m_e\),
\begin{equation}
\boldsymbol{\mu}=\frac{-e}{2m_e}\mathbf{L}.
\end{equation}
The ratio of the magnetic moment to the angular momentum is called the gyromagnetic ratio and denoted \(\gamma\). It depends only on the total charge and total mass. Generally we have,
\begin{equation}
\gamma=\frac{\mu}{L}=\frac{Q}{2M},
\end{equation}
and in the case of a single electron,
\begin{equation}
\gamma=\frac{-e}{2m_e}.
\end{equation}

It might be thought that the motion of electrons inside an atom and the motion of protons within a nucleus would account for, respectively, observed atomic and nuclear magnetism. However this is not found to be the case. Rather, such particles possess a wholly intrinsic angular momentum, quite distinct from the usual spatial, or orbital angular momentum, called spin and it is only when this extra contribution is incorporated that agreement with experiment is achieved.

The gyromagnetic ratio associated with spin is different to that associated with spatial or orbital angular momentum. For example, for the electron, this ratio, denoted \(\gamma_e\), is given by,
\begin{equation}
\gamma_e=-\frac{e}{m_e},
\end{equation}
and we have a relationship between the magnetic moment and spin angular momentum, \(\mathbf{S}\), given by,
\begin{equation}
\boldsymbol{\mu}=-g\mu_B\frac{\mathbf{S}}{\hbar},
\end{equation}
where we have introduced the so called “g-factor”, which for an electron is 2, and the Bohr magneton,
\begin{equation}
\mu_B=\frac{e\hbar}{2m_e}.
\end{equation}
Note that despite the motivation in terms of moments induced by current loops, the spin induced magnetic moment has nothing to do with electric charge. Indeed, a neutron carries no charge yet possesses a magnetic moment with gyromagnetic ratio given by,
\begin{equation}
\gamma_n=-3.83\frac{q_p}{2m_p},
\end{equation}
where \(q_p\) and \(m_p\) are respectively the charge and mass of a proton.

The Stern-Gerlach experiment depicted below probes the mysterious intrinsic angular momentum of an electron. Silver atoms have forty-seven electrons. Forty-six of them fill completely the \(n=1,2,3\) and \(4\) energy levels leaving a solitary \(n=5\) electron with zero orbital angular momentum. In the Stern-Gerlach apparatus silver is vaporised in an oven then collimated to create a beam of such atoms which are directed through a magnetic field behind which is a detector screen.
The potential energy, \(U\), of a magnetic moment \(\boldsymbol{\mu}\) in a magnetic field \(\mathbf{B}\) is given by \(U=-\boldsymbol{\mu}\cdot\mathbf{B}\) and the corresponding force is thus,
\begin{equation}
\mathbf{F}=-\nabla U=\nabla(\boldsymbol{\mu}\cdot\mathbf{B}).
\end{equation}
Thus the force points in the direction for which \(\boldsymbol{\mu}\cdot\mathbf{B}\) increases fastest. In the Stern-Gerlach setup the magnetic field is highly inhomogeneous, and to a good approximation,
\begin{equation}
\mathbf{F}=\mu_z\frac{\partial B_z}{\partial z}\mathbf{e}_z.
\end{equation}
Note that in the arrangement depicted, \(\partial B_z/\partial z\) is negative. Now, thanks to the high temperature of the oven generating the beam of silver atoms we would expect, reasoning classically, that the distribution of the magnetic moments of the sliver atoms passing through the apparatus would be isotropic. In particular, the component of the magnetic moment in the \(z\)-direction would be expected to be \(\mu_z=|\boldsymbol{\mu}|\cos\theta\) with no preferred angle \(\theta\) between the moment and the \(z\)-axis. Thus we would expect a spread of deflections with the upper and lower bounds corresponding respectively to \(-|\boldsymbol{\mu}|\) and \(|\boldsymbol{\mu}|\) and therefore a distribution detected by the screen looking something like,In fact what is observed is a distribution of the form,The atoms are deflected either up or down with nothing in between. It is as if all the atoms have either a fixed positive \(\mu_z\), corresponding to the lower screen distribution, or fixed negative \(\mu_z\), corresponding to the upper screen distribution. We therefore conclude that the dipole moment, and therefore the spin angular momentum, of an electron is quantized. The two values of \(\mu_z\) can be calculated and leads to a determination of the two possible values of \(S_z\),
\begin{equation}
S_z=\pm\frac{\hbar}{2}.
\end{equation}
The Stern-Gerlach experiment is effectively measuring the component of spin angular momentum along a particular direction in space of electrons in a beam and finds that they can take just two discrete values which we call up and down. Startling though this is the mystery certainly doesn’t stop here.

We’ll now consider a series of thought experiments, involving two or more Stern-Gerlach experiments in series. A single such experiment, which we’ll subsequently refer to as SG1, will be represented schematically as

The label \(\mathbf{e}_z\) on this machine indicates that the beam of electrons entering from the left will be subjected to a measurement of the electron spin in the \(z\)-direction. From such a machine two beams may emerge, in this case corresponding respectively to the \(z\)-component of spin ‘up’, \(S_z=\hbar/2\), and ‘down’, \(S_z=-\hbar/2\).

Let us now consider the following experiment, SG2

We begin by sending a beam of electrons of undetermined spin (i.e. silver atoms produced in an oven) to an \(\mathbf{e}_z\)-machine. Of the two beams emerging from this machine we discard the spin down beam passing only the spin up beam into another \(\mathbf{e}_z\)-machine. From this machine only one beam emerges corresponding to the spin up atoms with \(S_z=\hbar/2\). So if electrons are already in a \(S_z=\hbar/2\) state then another measurement of the \(z\)-component of the spin of the electrons is certain to find that \(S_z=\hbar/2\). There will be no electrons found with \(S_z=-\hbar/2\). In some sense these two states, spin up and spin down are ‘orthogonal’, an electron in a spin up state has no ‘component’ of spin in the spin down state of the same direction.

Now consider replacing the second machine above with an \(\mathbf{e}_x\)-machine whose two outputs correspond respectively to \(S_x=\hbar/2\) and \(S_x=-\hbar/2\), SG3.

Intuitively we think of the \(x\) and \(z\) directions as being orthogonal and indeed if we were dealing here with measurements of the orbital angular momentum of some object then of course there could be no component in the \(x\)-direction of \(z\)-oriented angular momentum. The result of this spin measurement however is that we find about half the electrons entering the second apparatus emerge from the \(S_x=\hbar/2\) output and half from the \(S_x=-\hbar/2\) output. Thus in the quantum world, we conclude that if we measure the \(x\)-component of spin, \(S_x\), of a particle known to have \(z\)-component of spin, \(S_z=\hbar/2\), then we will measure \(S_x\) to be either \(S_x=\hbar/2\) or \(S_x=-\hbar/2\) with equal probability.

Finally, let us consider taking the apparatus of the previous experiment and directing the \(S_x=-\hbar/2\) beam through a \(\mathbf{e}_z\)-machine, SG4.

One might perhaps think that having, in our first machine, selected only atoms carrying spin \(S_z=\hbar/2\) to enter the second machine that only such atoms would emerge through the final \(\mathbf{e}_z\)-machine. However, we find that about half emerge from the \(S_z=\hbar/2\) and half through the \(S_x=-\hbar/2\) output. It’s as if the intervening \(\mathbf{e}_x\)-machine has scrambled any memory of the output of the first machine.