# How does one experimentally determine chirality, helicity, spin and angular momentum?

If I've got an instance of a fundamental particle, how can I separate out the measurements of these three concepts?

(I think) I understand the theory behind them, and why the particles in the standard model are predicted to have the values they do. However, the process of validation of these numbers eludes me. How have they been historically measured?

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For spin measurements the original experiment was the Stern-Gerlach experiment in which you will see that a prior unpolarized beam will split up in two (Spin up and down) orientations.

For helicity, a very ingenious and fascinating experiment is the famours Goldhaber experiment that uses a very peculiar set of elements and also the Mößbauer effect to measure the helicity of neutrinos. The helicity is the projection of the Spin onto the momentum direction, and thus if you measure spin and momentum, you can compute helicity.

Link to original paper here: http://www.bnl.gov/nh50/

People often confuse helicity and chirality they are only the same in the case of massless particles. This is also the reason why we know that (at least interacting) neutrinos are always left-handed (the famous $$SU(2)_L$$ ), at least if we assume that neutrinos are massless (which is almost the case, but not strictly).

In contrast to helicity for massive particles (where you can always boost into a frame where you change the sign of the momentum direction and thus change helicity), chirality is a Lorentz-invariant property of the particle.

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This answer is fine +1, but the $SU(2)_L$ is before Higgs mechanism, where the neutrino is perfectly massless as well as the electron. –  Ron Maimon Apr 26 '12 at 4:54
one should also add that for decaying particles angular distributions of the decay products in the center of mass system can determine spin. –  anna v Apr 14 '13 at 10:50
in fact, at least in the Standard Model to talk about chirality makes only sense for the fermions (or the sfermions in the MSSM which have spin 0), because chirality is defined by how the corresponding field/particle transforms under $SU(2)_\mathrm{L}$ –  Andre Holzner Aug 20 at 8:52

By no means not a complete answer, more a criticism of @luksen’s one. It is posted here because the text is too long to fit in the comment field.

First of all, the spin is not a well-defined concept for composite particles. More precisely, whether the spin of a particle is defined depends on how the “particle” is defined. Look at an atom: it has the nucleus and electron shells. In many cases both have non-zero angular momentum, that implies they are coupled. There are several slimly different energy levels, and if you look at each of them as at a particle species, then yes, each has its spin.

Do such things as the Stern–Gerlach experiment capture such “particle species”? They don’t, because magnetic moment of electrons is strong (induced both by spin and orbital angular momentum), whereas the one of the nucleus is weak. Electrons vigorously interact with the external magnetic field, while nuclei reluctantly interact with either one. The species which are captured by a Stern–Gerlach-type device are superposed states of these slimly different energy levels, not spin states of a particle with a well-defined spin. By the way, for any silver atom of stable 107Ag and 109Ag isotopes a trivial calculation shows that it must be a boson as a whole and can’t be a fermion; this thing is downplayed in most QM texts.

What did the Stern–Gerlach experiment actually demonstrate, indeed? Electrons’ spin and their orbital angular momentum (more precisely, of the nucleus–electrons system) play all the drama. The nuclear spin was almost invisible because of much weaker magnetic moment. The experiment exposed electron shells and concealed nuclear spins – that’s why an impression of a fermion. I stress it: not an actual fermion. Only such apparition. No particle the spin is measured of.

So, why the apparition of spin ½ when the silver possesses 47 electrons? In fact, because of electronic configuration. Only the 5s electron contributed since the remainder forms completely filled subshells. What is the 5s electron? Since it is an s (ℓ = 0) orbital, only the electron’s spin matters. But it is only the silver. What happens to incomplete subshells in other cases? For d1 (it is the case for scandium, yttrium, lanthanum, lutetium) we have contributions of both electron’s spin and orbital angular momentum (ℓ = 2) to the magnetic moment; and there are several metals with more than one incomplete subshell. What will happen to other types of composite particles, such as numerous flavours of exotic atoms? I won’t predict. Finally, what remained of the fairy tale that spin and angular momentum are the same thing? Not so much.

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actually, the original poster asked for fundamental particles which usually means that they are not composite. –  Andre Holzner Aug 20 at 8:54
@Andre Holzner oops… now I see ¾ of my eloquence is not about the original question, only rebuffing the Stern–Gerlach stuff. Is some thread on the site more topical to place it? –  Incnis Mrsi Aug 20 at 11:00
You should consider editing out the irrelevant parts of your answer –  Danu Aug 20 at 18:15
@Danu: help me with meta.physics.stackexchange.com/questions/6071/… and I’ll remove it. –  Incnis Mrsi Aug 20 at 18:18
@IncnisMrsi I saw that post, but do not have any moderation power to help you. Save your text somewhere else in the meanwhile? As it stands, this answer is just (mostly) irrelevant. –  Danu Aug 20 at 18:24