Are there particles with zero spin?

[LarryS]

Are there any elementary particles with zero intrinsic spin?

Thanks in advance.

 

[danielakkerma]

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.

 

[Dirac62]

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,

 

[Vanadium 50]

Dirac62, that wasn't the question asked. A helium atom is not elementary.

 

[Vagn]

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.

 

[Dirac62]

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.

 

[ChrisVer]

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...

 

[yaakov]

Would zero spin violate Heisenberg uncertainty principle...like absolute zero ?

 

[mfb]

No.
What do you mean with "absolute zero" - temperature? That is possible and does not violate the uncertainty principle.

 

[ChrisVer]

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).