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What is the electric charge of Higgs particle?

  1. Jan 20, 2015 #1
    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?
     
    Last edited by a moderator: Jan 20, 2015
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  3. Jan 20, 2015 #2

    mfb

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    Please use the edit function instead of making multiple posts.

    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.
     
  4. Jan 20, 2015 #3

    Orodruin

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    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.
     
  5. Jan 20, 2015 #4

    ChrisVer

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    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:
    [itex] H^+ = Re [H^+] + i Im[H^+] [/itex]
    [itex] H^0 = Re [H^0] + i Im[H^0] [/itex]
    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:
    [itex]H_{phys} (x)= <H> + h(x) [/itex]
     
  6. Jan 20, 2015 #5

    vanhees71

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    The clever way to do this are 't Hooft's ##R_{\xi}## gauges. You use the usual Faddeev-Popov path-integral formalism. You can go to the unitary gauge as the limit ##\xi \rightarrow \infty##. The point is that the proper vertex functions are renormalizable only for finite ##\xi##. So if you want to do loop calculations you should use a finite ##\xi##. The S-matrix elements of physical processes are gauge invariant and thus independent of ##\xi##. This shows that the particle content is as given in the unitary gauge, particularly there are no would-be Goldstone bosons as observable particles left but only one Higgs boson.

    A great book, explaining this very well, is

    J. C. Taylor, Gauge Theories of Weak Interactions, Cambridge University Press (1976)

    It's one Higgs boson in the usual formulation of the standard model with a minimal Higgs sector; you can invent more sophisticated Higgs fields, where you have more then one kind of Higgs boson left, and it's an interesting question, if the "Higgs like particle" declared as discovered on July 4, 2012 is the one described with the minimal Higgs sector or whether there are more Higgs bosons in nature or whatever it is what's behind the "Higgs mechanism".
     
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