Why is the Higgs boson spin 0? Detailed equation-form answers would be great, but if possible, some explanation of the original logic behind this feature of the Higgs mechanism (e.g., "to provide particles with mass without altering their spin states") would also be appreciated.
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asked Mar 31, 2012 at 15:51
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4$\begingroup$ It doesn't have to be spin-0 AFAIK, but the simplest prediction, the "standard model Higgs" just happens to be. Are you asking for an explanation of why the SM Higgs turns out to have zero spin, or were you thinking about more general requirements that would mean any Higgs would have had to be spinless? $\endgroup$– David ZMar 31, 2012 at 16:44
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$\begingroup$ My intent was the latter: Were there drivers other than simplicity and Occam's razor at work back when Higgs was first postulated to be spin 0? Since no known examples of point-like spin 0 bosons exist, that seemingly trivial assumption actually steps a wee bit outside the boundaries of known physics. I wondered if that point might have driven any early exploration of more complicated Higgs options. For example, pion-like compositions of two spin 1 bosons comes to mind as the next simplest option if spin 0 is avoided. Hmm. Maybe I just need to do a good literature search on my own question?... $\endgroup$– Terry BollingerMar 31, 2012 at 17:44
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1$\begingroup$ I think it came out of the calculation, but I don't remember the details - I'd have to look some things up to address that. As for Higgs models beyond the SM Higgs, someone else would have to address that. $\endgroup$– David ZMar 31, 2012 at 17:57
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$\begingroup$ David, thanks. The recent LHC concerns hopefully have re-activated multiple "other options?" lines of Higgs thinking, so that may make my literature scan easier. $\endgroup$– Terry BollingerMar 31, 2012 at 18:45
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4 Answers
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The Higgs boson is, by definition, the excitation of the field behind the Higgs mechanism. The Higgs mechanism is a spontaneous symmetry breaking. Spontaneous symmetry breaking means that the laws of physics, or the action $S$, is symmetric with respect to some symmetry $G$, i.e. $$\delta_G S = 0$$ however, the vacuum state of the quantum field theory isn't symmetric under the generators of this symmetry, $$ G_i|0\rangle \neq 0$$ If we want to satisfy these conditions at the level of classical field theory, there must exist a field $\phi$ such that the vacuum expectation value $$\langle \phi\rangle \equiv \langle 0 | \phi (x)| 0 \rangle$$ isn't symmetric under $G$, $$\delta_G \langle \phi\rangle \neq 0 $$ However, if the field $\phi$ with the nonzero vev had a nonzero spin, the vacuum expectation value would also fail to be symmetric under the Lorentz symmetry because particular components of a vector or a tensor would be nonzero and every nonzero vector or tensor, except for functions of $g_{\mu\nu}$ and $\epsilon_{\lambda\mu\nu\kappa}$, breaks the Lorentz symmetry.
Because one only wants to break the (global part of the) gauge symmetry but not the Lorentz symmetry, the field with the nonzero vev has to be Lorentz-invariant i.e. singlet i.e. spin-zero $j=0$ field, but it must transform in a nontrivial representation of the group that should be broken, e.g. $SU(2)\times U(1)$. The Standard Model Higgs is a doublet under this $SU(2)$ with some charge under the $U(1)$ so that the vev is still invariant under a different "diagonal" $U(1)$, the electromagnetic one. The Higgs component that has a vev is electrically neutral, thus keeping the electromagnetic group unbroken, photons massless, and electromagnetism being a long-range force.
Aside from the Higgs mechanism, there exist other, less well-established proposed mechanisms how to break the electroweak symmetry and make the W-bosons and Z-bosons massive. They go under the names "technicolor", "Higgs compositeness", and so on. The de facto discovery of the 125 GeV Higgs at the LHC has more or less excluded these theories for good. The Higgs boson seems to be comparably light to W and Z-bosons and weakly coupled, close to the Standard Model predictions, and the Higgs mechanism sketched above has to be the right low-energy description (up to energies well above the electroweak scale).
answered Mar 31, 2012 at 18:56
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Just to highlight one simple but important argument.
If Higgs is to be responsible for giving particles mass then it has to be a scalar (spin-0) particle because a particle's mass is reference frame independent, just like the values of the field $\psi$ of a spin-0 particle are reference frame independent.
Compare this for instance with charged particles at rest which gain potential energy in a static electromagnetic potential field $A^o$ (which is the same anywhere). As long as you stay in the rest frame then it looks like the particles have gained an extra mass $qA^o$. However, as soon as you change to another reference frame then the illusion of extra mass breaks down. The reason it breaks down is that $A^\mu$ is a spin-1 field which therefor transforms like a vector going from one reference frame to another.
Hans.
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3$\begingroup$ @LubošMotl, Hans, my thanks to both of you. These are both superb explanations, very much of the type for which I was hoping. I will attempt to Accept both of you answers, Luboš first for the level of detail enabling deeper study, then Hans for his deft highlighting of a nicely critical point. If I can only do one, Hans, my intent was there even if the software was uncooperative. [And I see I really am a newbie: Stack Exchange would not let me accept either answer yet...!] $\endgroup$ Mar 31, 2012 at 22:32
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I would like to add to these above answers another non-rigourous way, if one had forgotten, to remember that the Higgs boson has spin 0, through elimination.
In 2012, it was directly observed that the Higgs boson could decay to two photons (massless particles of spin 1). However, by the Landau–Yang theorem in quantum mechanics, a massive particle with spin 1 $\rightarrow$ $2\gamma$ is impossible. Hence, the experimental result eliminates the spin 1 option for the Higgs, leaving the other integer option of spin 0 which turned out to be consistent.
More about Landau-Yang theorem: https://en.wikipedia.org/wiki/Landau–Yang_theorem
H $\rightarrow \gamma\gamma$ first measurements: https://cds.cern.ch/record/2140979/files/HIG-15-005-pas.pdf
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Because Hogs Boson are only related with any particle mass. Not with angular moment or dipole moment. So it is needless for it to have any spin. and particles mass is never changed whether it rotate (spin) or not. So Hgs Boston need not any spin value.
