Are there particles with zero spin?

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Discussion Overview

The discussion centers on the existence of elementary particles with zero intrinsic spin, exploring theoretical implications and examples from particle physics. Participants examine specific particles, their classifications, and the relationship between spin and fundamental principles like the Heisenberg uncertainty principle.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • Some participants propose that the Higgs boson is an example of a particle with zero intrinsic spin, classifying it as a scalar boson.
  • Others mention that while the Higgs is confirmed to have zero spin, there are predicted particles with zero spin that have not yet been observed.
  • A participant argues that the helium atom behaves like a boson and has a total spin of zero, although this is contested as helium is not an elementary particle.
  • There is mention of various theoretical particles, such as axions and dilatons, which are predicted to have zero spin in different models beyond the Standard Model.
  • Some participants question whether zero spin would violate the Heisenberg uncertainty principle, with responses clarifying that it would not.
  • Discussion includes the idea that measuring a particle's spin with absolute certainty does not contradict the uncertainty principle, as it pertains to other observables.

Areas of Agreement / Disagreement

Participants generally agree that the Higgs boson is a confirmed example of a zero spin particle, but there are multiple competing views regarding the classification of other particles and the implications of zero spin on fundamental principles. The discussion remains unresolved on the broader implications and existence of other zero spin particles.

Contextual Notes

Some claims depend on definitions of elementary particles and the context of spin measurements. The discussion includes speculative elements regarding unobserved particles and theoretical models.

LarryS
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Are there any elementary particles with zero intrinsic spin?

Thanks in advance.
 
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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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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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Dirac62, that wasn't the question asked. A helium atom is not elementary.
 
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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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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...
 
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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.
 
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yaakov said:
Would zero spin violate Heisenberg uncertainty principle...like absolute zero ?
mfb said:
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 (\Delta p= 0) ... 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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