Are there any elementary particles with zero intrinsic spin?
Thanks in advance.
I believe the Higgs boson is just the very example of such a particle. It is the reason for its classification as a "scalar" boson.
Hello dear Sir
Helium atom is likely to have spin 0, then acts like a boson. It contains quarks and electrons which are fermions, but as a hole, helium can be boson.
Dirac62, that wasn't the question asked. A helium atom is not elementary.
The only confirmed particle with zero spin is the Higgs, there are others predicted by various models, however none have yet been observed.
A (probably) non-exhaustive list of such bosons is given in wikipedia's list of particles.
Elementary bosons may have spins (0,1,2,....), but 0 and 2, for example, were just anticipated. By 2013, it has been supposed that the Higgs boson with spin zero exist. Besides, spin 2 for Graviton is predicted.
spin-0 = all the scalar fields you can find in any theory...
The Higgs that was discovered is such a particle.
The axion that is predicted as a resolution to the Strong CP-problem and can account for CDM, is such a particle.
Then other theories beyond the SM can add a vast number of such particles ( dilatons, supersymmetry which gives a scalar partner field to each fermion etc)
Would you consider the mesons elementary particles? then the pion is such... and many more...
Would zero spin violate Heisenberg uncertainty principle....like absolute zero ?
What do you mean with "absolute zero" - temperature? That is possible and does not violate the uncertainty principle.
I think by zero he meant that its spin can be determined 100% to be zero , no uncertainty.
But it's still a no. Because that is not what the uncertainty principle tells you.
You can as well measure a particle's momentum with infinite accuracy ([itex]\Delta p= 0 [/itex]) ... the only thing stopping you could be your device... that however means that you will get some uncertainty to some other observable that [its operator] does not commute with momentum's one (doesn't share the same eigenstates / are not mutually diagonalizible in some basis).
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