Is there record of a bosonic Stern-Gerlach measurement?

I cannot seem to find any peer-reviewed (or other) reference to an integer-spin Stern-Gerlach experiment. It shouldn't be too hard to do: just find you friendly neighbourhood Deuterium ion and shoot it through a Stern-Gerlach magnet.

Can one devise a photonic Stern-Gerlach experiment, i.e. spatial seperation of polarization states? One should also see only two states in this case, because the spin-0 photon state is "reserved" for EM-interactions (this might be too simple a statement, but this is how I understand it currently).

EDIT it seems some of you are misunderstanding the question: I am inquiring about a Stern-Gerlach-like experiment, where spin states have been split, and by extension the perpendicular nature of non-commuting measurements. So only the concept of the S-G experiment as extensively described in introductory QM textbooks such as Sakurai.

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Weeeeeird, I was thinking of this exact question only a few hours ago. For what it's worth, this discussion on Physics Forums draws an interesting parallel between Stern-Gerlach and birefringence. –  Richard Terrett Dec 4 '12 at 10:01
–  Ben Crowell Aug 28 '13 at 23:10

Photons are gauge bosons, they do not have spins or magnetic moments!

For electrons, bose/fermi atoms in a magnetic field, we have the energy $$E({\bf r})=\boldsymbol{\mu}\cdot {\bf B}({\bf r})$$ where $\boldsymbol{\mu}$ is the magnetic moment.

Hence we have force due to the gradient of magnetic field, $${\bf F}=-\nabla E({\bf r})=-\mu\nabla{B}({\bf r})$$ which produces Stern-Gerlach effect. You cannot write down a "photonic version" of this.

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Photons do have spin, but they don't have magnetic moments, because photons don't couple directly to photons. –  user1504 Dec 4 '12 at 14:28
In the literature, it is called orbital angular momentum or polarization but not spin. –  ChenChao Dec 4 '12 at 14:59
Not in the literature I read (e.g., most particle physics books). Anyways, as long as it's clear that photons are not spin 0 objects, we don't need to argue about terminology. –  user1504 Dec 4 '12 at 15:15
Yes, you are right. Gauge bosons are spin-1 particles. But I mean in the context of atomic physics, photon only transfer orbital angular momenta to electrons. It is an important notion. –  ChenChao Dec 4 '12 at 15:41
Of course Stern-Gerlach doesn't work on photons. But it does on deuterium, and the link by @Richard links to birefringence in crystals, which spatially separates photons per spin state. –  rubenvb Dec 4 '12 at 18:15

You can absolutely do a corresponding experiment with light. In fact, it's the easiest way by far. Instead of a magnetic field, you would use a polarizing beam splitter to separate the two states, which as the name suggests is a cube that reflects light of one polarization and passes light of the other polarization. To do a Stern-Gerlach like experiment all one needs is a polarized photon source, a few of these cubes and a few half waveplates to change photon polarization, and then some photon detector looking at outputs.

This wouldn't normally be called a Stern-Gerlach experiment, which is specific to using a magnetic field to separate particles with magnetic moments, but the mathematics describing it is the same, as is the basic lesson that angular momentum is quantized and measurements in different directions don't commute.

As for atoms, a quick search found a Stern-Gerlach like experiment with not just single atoms, but a BEC: http://www.uibk.ac.at/exphys/ultracold/?http://www.uibk.ac.at/exphys/ultracold/projects/rubidium/rb87bec/ I can't immediately find a single-atom experiment with Rubidium, but I bet it's out there if you look around.

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If you're not doing it with a magnetic field, then it's not analogous to the Stern-Gerlach experiment. –  Ben Crowell Aug 29 '13 at 1:42
Well, I don't know what to say except that I disagree completely. To me this is like saying it isn't like the Stern-Gerlach experiment unless it uses silver ions. The reason we study S-G is because it showed angular momentum quantization and non-commutativity of measurements; the fact that this was originally done with a magnetic field is incidental. –  Rococo Aug 29 '13 at 3:31