# How to measure the spin of a neutral particle?

If a charged particle with charge $q$ and mass $m$ has spin $s \neq 0$ we can measure an intrinsic magnetic moment $\mu = g \frac{q}{2m}\hbar \sqrt{s(s+1)}$. This is how spin was discovered in the first place in the Stern-Gerlach Experiment.

But for a neutral particle $\mu = 0$, so we cannot measure the spin of the particle in the same manner. But it is said, that e.g. the Neutron or the Neutrino both have a spin $s=1/2$. How was or can this be measured?

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First of all, the Stern Gerlach experiment used a beam of neutral particles. Second, as Fabian pointed out neutral particle does not mean $\mu=0$. – Approximist Apr 12 '11 at 18:32
I think this is a great lead in for a high energy experimentalist to explain some of their techniques. How did they measure the Z boson to have spin 1? This isn't my field and since we can't even see the Z boson track directly, I really don't know. It would be neat to see more than the easy case of the neutron with a magnetic moment. – Edward Apr 13 '11 at 8:14

Conservation of angular momentum is invoked for the neutrinos because beams of neutrinos cannot be collimated for an experimental measurement. Neutron spin can be measured in a Stern Gerlach setup.

The interactions and decays were carefully examined in various experiments and the only consistent spin values are the ones assigned.

Edit: I see that the question should be formulated as : why the neutron has a Dirac magnetic moment, although it is neutral, which is the formula that is displayed above, and does the neutrino have a Dirac magnetic moment?

The neutron, and other baryons, has a magnetic moment because the quarks that compose it have a Dirac magnetic moment. See for example Perkins, Introduction to High Energy Physics, section: baryon magnetic moments for the derivation.

Whether the neutrino has a magnetic moment due to higher order loop diagrams is a research question.

So, though spin in charged point like particles is connected to magnetic moment with the formula above, analogous to classical charges circulating in a loop having a magnetic moment, , charge is not necessary for spin to appear. There is intrinsic spin which for the neutrino comes from the angular momentum balance in the interactions where it appears. The neutrino is a spinor in the Dirac formalism.

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It's not the spin that is measured, but the magnetic moment of the neutron. There is a relation between the magnetic moment and the spin which works for charged particles. What I'm searching for is a relation between the magnetic moment and the spin that is also valid for a neutral particle. – asmaier Apr 12 '11 at 19:00
@asmair Any magnet has a magnetic moment (dipole) that interacts with magnetic fields, see en.wikipedia.org/wiki/Magnetic_moment . The electric field is a hindrance not a help in the stern gerlach experiment. You need to think a bit about what you are reading. – anna v Apr 13 '11 at 3:42
In the stern gerlach experiment the magnetic moment is used to deflect the particles with an external magnetic field, and show that they either are polarized up, or down. i.e their spin is quantized. – anna v Apr 13 '11 at 4:13
And charge is also not necessary for a magnetic moment to appear? – asmaier Apr 13 '11 at 9:32
Yes, the charge is necessary to have a magnetic moment. The wiki article en.wikipedia.org/wiki/Spin_%28physics%29 describes the state of art on this, in "magnetic moments". They arise because of the charge and the intrinsic spin. The neutrino gets a tiny one by the charges of particles in the higher order feynman diagram loops . – anna v Apr 13 '11 at 12:04

Also the neutron has a magnetic moment. Check this out. The reason is that the neutron is not an elementary particle but built up from quarks which have charge...

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But what is then the relation between the magnetic moment of a neutron and it's spin? What about the neutrino? – asmaier Apr 12 '11 at 18:25
@asmaier $\mathbf{\mu}= g \frac{q}{2mc}\mathbf{S}$ – Approximist Apr 12 '11 at 18:36
@Approximist But q = 0 for a neutron, so your formula cannot work, if the neutron has $\mu \neq 0$ and $S \neq 0$. – asmaier Apr 12 '11 at 18:40
@asmaier: Did you read the link? – Fabian Apr 12 '11 at 18:43
Sure, but there is no explanation of the relation between the neutrons magnetic moment and it's assumed spin. Spin is not even mentioned in the article. – asmaier Apr 12 '11 at 18:55