I know that the elementary fermions are spin 1/2 and the elementary bosons are spin 1 (except the potential graviton), but are what is the highest known spin of any elementary particle (including their composites)?
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6What do you count as a particle? A hadron? An atom? A molecule? A drop of liquid? A crystal domain? – Lucas Baldo Aug 03 '21 at 18:30
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@LucasBaldo I'm not really sure I just meant generally, is there a maximum known spin? If that is too vague, then perhaps what about for Hadrons? – milkcookie Aug 03 '21 at 18:36
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3Spin for elementary particles (and their composites) is not the same thing as rotational spin for macroscopic objects like spherical cows. – StephenG - Help Ukraine Aug 03 '21 at 19:02
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@StephenG Thanks, in that case, I am definitely referring to elementary particles and their composites! :) – milkcookie Aug 03 '21 at 19:27
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Theoretically there's no maximum value. Observationally it will probably depend on exactly where you draw the line between a "composite particle" and a "macroscopic object". There's a reasonable argument that we should think of black holes as being a kind of elementary particle, in which case spins orders and orders of magnitude in excess of $\hbar$ have been observed. – Andrew Aug 04 '21 at 00:48
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I would suggest to rewrite the question to focus on hadrons. – Mauricio Aug 05 '21 at 11:19
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Possible duplicate in case of elementary particles: Why do we not have spin greater than 2? – Qmechanic Aug 05 '21 at 13:07
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The nuclear isomer tantalum-180m has spin $9\hbar$; its high spin effectively prevents its decay. – rob Aug 05 '21 at 16:14
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A partial answer. Let's talk about hadrons. In experiment, how does one discover new hadrons? They examine the invariant mass of a certain combination of decay products and search for peaks in this distribution, employing angular distributions to measure the spin. In order to claim reliably a new state in a dataset of moderate size, one should ideally see a reasonably narrow mass peak, which is not overlapping with other states (i.e. no large interference effects).
We know that for conventional hadrons (mesons and baryons, not talking about nuclei here), the higher its spin, the higher is its mass (figure taken from here):
The higher the mass, the more decay channels is available for a hadron. To a 0th order approximation, high-mass resonances are less stable, thus, broader than the low-mass resonances. (This is not always true due to conservation laws which might prohibit certain decays depending on quantum numbers of a given state). Also, the higher the mass, the lower is the cross-section to produce a given state in hadronic collisions or fixed-target experiments (once again, I am somewhat simplifying here).
Effectively, most states of a high spin have a high width, but low production cross-section, which makes the task of their separation from background rather difficult. And, philosophically, do we call something 'a hadron' if its lifetime is shorter than the characteristic timescale of a strong interaction?
For the hadrons made exclusively of the light quarks, there is a lot of data from hundreds of experiments, and states up to spin-6 (for mesons) such as $f_6(2510)$ and spin-15/2 (for baryons) such as $\Delta(2950)$, are established. There are some rumors about spin-7 mesons.
An important point is that in the quark model, it is often possible to interpret these states both as conventional meson/baryon states, or as exotic hadrons (tetraquarks, pentaquarks, glueballs etc).
Once we add heavier quarks, the experimental knowledge is less developed. For strange hadrons, the PDG tables know of a spin-5 $K^*_5(2380)$ meson and spin-9/2 $\Lambda(2350)$ baryon. For the charmonium, the recently discovered $\psi_3(3842)$ (spin 3) is the highest-spin known state, however, being exactly precise, its spin was not measured but rather 'guessed' from its properties.
Martino
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"An important point is that in the quark model, it is often possible to interpret these states both as conventional meson/baryon states, or as exotic hadrons (tetraquarks, pentaquarks, glueballs etc)." <- Could you elaborate on this? Do you mean that we can't tell experimentally which one it is, or that those QCD bound states are literally just different words for the same thing? – jacob1729 Aug 05 '21 at 11:12
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It depends. If the measured quantum numbers of a certain state are forbidden in a simple quark-antiquark model, then it is manifestly exotic. For the other cases, exotic states can only be deduced indirectly – from surprising decay modes, production mechanisms, or if their mass/width is vastly off the theoretical predictions. This is easier for states with heavy quarks (less soft QCD), where predicted spectra are well understood – that's why one hears so often of exotic particles with charm quarks. But we cannot just go and count the quarks. – Martino Aug 05 '21 at 11:32
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A fully magnetised iron particle has spin $S \approx N_A \frac{5}{2} h$. Such large single Fe crystals can be are commercially available. Note that an Fe crystal can be seen as a composite of elementary particles.
Surely you were not after this answer though.
my2cts
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2By $N_A$ do you mean Avogadro's number, so you are referring to a single-domain chunk of ferromagnetic iron with mass roughly 50 grams? That's not really a thing (iron is polycrystalline at that scale) and it doesn't seem to be what the asker had in mind. – rob Aug 05 '21 at 16:03