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Particles with Zero Spin?

  1. Feb 3, 2015 #1


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    Are there any elementary particles with zero intrinsic spin?

    Thanks in advance.
  2. jcsd
  3. Feb 3, 2015 #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.
    Last edited: Feb 3, 2015
  4. Feb 4, 2015 #3
    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.
    Last edited by a moderator: Feb 4, 2015
  5. Feb 4, 2015 #4

    Vanadium 50

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    Dirac62, that wasn't the question asked. A helium atom is not elementary.
  6. Feb 4, 2015 #5
    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.
  7. Feb 4, 2015 #6
    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.
  8. Feb 4, 2015 #7


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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...
  9. Feb 5, 2015 #8
    Would zero spin violate Heisenberg uncertainty principle....like absolute zero ?
  10. Feb 5, 2015 #9


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    What do you mean with "absolute zero" - temperature? That is possible and does not violate the uncertainty principle.
  11. Feb 5, 2015 #10


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    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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