# What is the electric charge of Higgs particle?

What is the electric charge of Higgs particle?Then if it is charge particle why must we give that particle the charge?

In GWS model we introduce a scalar field in the spinor representation of SU(2):(H^+,H^0) .Then why must we choose positive charge for H^+?

Now,I think that Q=T^3+Y/2,and we choose Y=1 so that H^+ has +1e and H^0 has 0e?

So Higgs particle is neutral particle?

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mfb
Mentor

Which Higgs particle do you mean? The standard model particle clearly has no electric charge, other models predict charged Higgs bosons in addition.

Every positively charged particle also has a negatively charged antiparticle.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
To expand a bit on mfb's reply: The scalar field that you introduce is a complex SU(2) doublet and you can introduce it as it is generally introduced or as the complex conjugate of how it is generally introduced. This is essentially just bookkeeping and will result in the same physics, you could just as well have chosen a doublet with a negative component and a neutral component, it is just a matter of what you are choosing to call different things.

As the Higgs field acquires a vev and EW symmetry is broken, the two components of the charged field are eaten by the W bosons and the imaginary part of the neutral component is eaten by the Z boson. You are left with one real scalar field, which is the Higgs boson, which has charge zero.

vanhees71
ChrisVer
Gold Member
None of these H+ or H0 are the Higgs particles... Because both of these are complex fields... In particular both of them together have 4 degrees of freedom:
$H^+ = Re [H^+] + i Im[H^+]$
$H^0 = Re [H^0] + i Im[H^0]$
So 4 fields....

The Higgs is just one of these components- the other 3 are "eaten" by Ws and Zs... In particular you can do that, in an introductory level, by rotating out the Higgs field and making those unphysical degrees of freedom vanish...
However if anyone has any reference for how this is done in a Path Integral formalism, I'd be happy to see it :)
What happens is that one of these components gets a non-vanishing vev, and Higgs physical field is said to be the perturbations around that vev:
$H_{phys} (x)= <H> + h(x)$

vanhees71