# Higgs boson quantum numbers

by Jochent
Tags: boson, higgs, numbers, quantum
 P: 2 Hi everyone, I have studied QFT, the SM and the Higgs mechanism when I was in university and after reading an article from CMS (CERN) about the spin-parity measurement of the HZZ channel, which shows that $J^{P}=0^+$ is favoured versus $J^{P}=0^-$, I went back to the theory of the Higgs boson. I am struck by the fact that the theory books do not explain how the quantum numbers of the Higgs boson are deduced. (in particular parity P, charge conjugation C and time reversal T) The books that I have just cover that the SM Higgs boson has $J^{PC}=0^{++}$. I would like to understand how one deduces this. Is this extracted from the Higgs Lagrangian and if so, how does one do it? Thanks a lot !
 Sci Advisor Thanks P: 4,160 Since the vacuum state contains a Higgs condensate, the Higgs boson must also have the quantum numbers of the vacuum, namely 0++.
 Emeritus Sci Advisor PF Gold P: 16,465 Bill is correct, provided that you consider only the simplest Higgs models. If you put in a second Higgs doublet, you get two charged Higgses, and three neutrals. Two of the three neutrals are 0+ and the third is 0-. In the simplest model, you need hhh and hhhh couplings. If the Higgs held any conserved parity quantum number that was odd, you could not have hhh couplings. So they are all even.
 P: 2 Higgs boson quantum numbers Thank you both for the quick answers. So technically one could say that the quantum numbers are fixed since the minimum is, or more correct can be, defined as $\phi_0=\frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ v\end{pmatrix}$ with v the vacuum expectation value. Since the fact that H is introduced as the expansion around this minimum as $\phi(x)=\frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ v+H(x)\end{pmatrix}$ , H has to comply with the minimum itself and therefore its quantum numbers should be compatible with the minimum. Is this a correct reasoning ? Then secondly, after the reparametrization of the Lagrangian due to the expansion around the chosen minimum, a $H^3$ self-coupling term arrives in the Lagrangian, which comfirms the fact that all the quantum numbers have to be even. Thanks again ! It feels like my QFT is getting rusty.
Thanks
P: 4,160
 Quote by Vanadium 50 If you put in a second Higgs doublet, you get two charged Higgses, and three neutrals. Two of the three neutrals are 0+ and the third is 0-.
Which one is the condensate?
P: 1,020
 Quote by Jochent Hi everyone, I have studied QFT, the SM and the Higgs mechanism when I was in university and after reading an article from CMS (CERN) about the spin-parity measurement of the HZZ channel, which shows that $J^{P}=0^+$ is favoured versus $J^{P}=0^-$, I went back to the theory of the Higgs boson. I am struck by the fact that the theory books do not explain how the quantum numbers of the Higgs boson are deduced. (in particular parity P, charge conjugation C and time reversal T) The books that I have just cover that the SM Higgs boson has $J^{PC}=0^{++}$. I would like to understand how one deduces this. Is this extracted from the Higgs Lagrangian and if so, how does one do it?
In Standard model Higgs boson is a scalar particle having constructed from potential which respects ##\phi→-\phi##. It is the only scalar particle in SM ( not to be confused with some pseudoscalar particles which we know), hence it transforms even under parity transformation.

It is a scalar particle hence zero value of J. Also to see the C-parity, you can look at one possible decay which is it's decay to two photons. C-parity of a photon is -1, hence C-parity of Higgs boson is ##(-1)^2=1## .

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