Particles with Zero Spin?

  • #1
referframe
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Are there any elementary particles with zero intrinsic spin?

Thanks in advance.
 

Answers and Replies

  • #2
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.
 
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  • #3
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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.
regards,
 
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  • #4
Vanadium 50
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Dirac62, that wasn't the question asked. A helium atom is not elementary.
 
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  • #5
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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.
 
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  • #6
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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.
 
  • #7
ChrisVer
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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...
 
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  • #8
yaakov
Would zero spin violate Heisenberg uncertainty principle....like absolute zero ?
 
  • #9
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No.
What do you mean with "absolute zero" - temperature? That is possible and does not violate the uncertainty principle.
 
  • #10
ChrisVer
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Would zero spin violate Heisenberg uncertainty principle....like absolute zero ?
No.
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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