In the Standard Model, the Higgs boson is expected to have spin 0 and even parity. I know how to get the spin-0 approach, but how do I argue for the even parity? Could you give a simple and a more detailed explanation for this even parity expectation?
2 Answers
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It is not an assumption; both $0^+$ and $0^-$ were considered as possible Higgs states. The angular distribution of decay products (like in $h\to ZZ$, $h\to WW$, $h\to f\bar{f}$, $h\to \gamma\gamma$ or in Higgstrahlung) is dependent on the parity of the Higgs particle. Alternatively, you can measure the helicities of the outgoing photons (in the $h\to\gamma\gamma$ case); the observed distribution is consistent with an even parity Higgs.
This workshop has a good overview.
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$\begingroup$ I changed the wording here. Put "expectation" instead of "assumption" and then my question remains. The information you provide does not answer my question. Also, in your linked material, on slide 4/49 it says "In SM, the only fundamental neutral scalar is a JPC= 0++". My question is - why? $\endgroup$– madCommented Jun 30, 2014 at 9:08
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$\begingroup$ When a signal excess was first found, we were unsure whether it was "the" Higgs boson that would be responsible for spontaneous symmetry breaking. A particle that can acquire a VEV and give us the usual Higgs mechanism must have the same quantum numbers as the observed vacuum, or such a VEV would break observed invariances: the vacuum has no charge, the vacuum is rotationally invariant, the vacuum is invariant under parity, etc. Such a particle would have to have $J^{PC}=0^{++}$. Only afterwards did we measure the properties of the new boson to be consistent with the Higgs. $\endgroup$ Commented Jul 1, 2014 at 5:35
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$\begingroup$ Thanks for your comment, I really appreciate it. But that's clear to me. I want to know WHY the (theoretical implementation of the) vacuum is invariant under parity. Which term in the Lagrangian represents this even parity? I am not referring to the observed state but to the theoretical model. A handwaving argument is okay, I just want to motivate the even parity of the Higgs in the SM. $\endgroup$– madCommented Jul 1, 2014 at 8:43
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1$\begingroup$ There are terms in the SM Higgs Lagrangian that are trilinear: $hhh$, $hZZ$, $hW^+W^-$ and the like. We know that the intrinsic parity of $Z$ and $W^{\pm}$ is $-1$, so for the whole Lagrangian to be a true scalar (not a pseudoscalar), we must have the parity of $h$ be $+1$. Otherwise, we would need an extension of the SM (like the MSSM) to generate CP-odd fields. $\endgroup$ Commented Jul 1, 2014 at 19:42
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1$\begingroup$ I was a little too cavalier in my previous statement: the bosonic portion of the Lagrangian separately conserves $P$ and $C$, while adding fermions only conserves $CP$ (at least approximately: ignore the CKM phase for now). In a purely bosonic theory, we MUST have $h=0^{++}$, while in a theory with fermions, we can have $h=0^{\pm\pm}$. This is why we "assume" that the Higgs is a true scalar: it works in both regimes, so it's easier to work with. Ultimately, experiment needs to verify that it's not a pseudoscalar. (Excellent review is pp. 197-199 in The Higgs Hunters Guide) $\endgroup$ Commented Jul 2, 2014 at 19:28
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One argument could be the Yukawa coupling, which is responsible for the coupling to the fermions.
In the Yukawa coupling term in the Lagrangian, $\mathcal{L}_{\text{Yukawa}}$ , there are no terms that contain a $\gamma^5$ matrix, defined as $$\gamma^5 := i\gamma^0 \gamma^1 \gamma^2 \gamma^3$$ This publication states how terms in the Lagrangian transform under parity operation, namely (giving only the relevant information here) $$ \Psi \bar \Psi \;\;\; \scriptsize transforms\, as \normalsize \;\;\;\text{scalar (parity = +1)}$$ $$ \Psi \gamma^5 \bar \Psi \;\;\; \scriptsize transforms\, as \normalsize \;\;\;\text{pseudoscalar (parity = -1)}$$
and therefore the Yukawa coupling term gives direct hint to the expectation $\text{P}(\text{Higgs})=+1$. However, as the Yukawa coupling theory could be the wrong model, experiments are supposed to check for the parity sign, too.
Thanks go to my university professor for pointing that out in his script.