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Based on my C grade of QFT (technically F in grad), I learned that a complex scalar field can give mass to a maseless gauge boson by SSB, with real component becoming a massive higgs, and orthogonal (complex) becoming a maseless Goldstone boson and "eaten up" by maseless gauge boson to give gauge boson's mass.

If Higgs can give gauge boson a mass with singlet model, why is SM higgs a double? Is it because we have 3 massive gauge bosons? or something else?

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  • $\begingroup$ No, not something else. There are three gauge bosons hankering for a mass, and so you need 3 Goldstone bosons, one neutral and two oppositely charged. In addition to the surviving "debris" Higgs particle, associated with the v.e.v. In all, a complex doublet. Is this your question, or, alternatively why we have three gauge bosons? $\endgroup$ Feb 13 '20 at 17:15
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There is only one standard model Higgs particle.

In this general article

In the Standard Model, the Higgs particle is a boson with spin zero, no electric charge and no colour charge. It is also very unstable, decaying into other particles almost immediately. The Higgs field is a scalar field, with two neutral and two electrically charged components that form a complex doublet of the weak isospin SU(2) symmetry. The Higgs field has a "Mexican hat-shaped" potential. In its ground state, this causes the field to have a nonzero value everywhere (including otherwise empty space), and as a result, below a very high energy it breaks the weak isospin symmetry of the electroweak interaction. (Technically the non-zero expectation value converts the Lagrangian's Yukawa coupling terms into mass terms.) When this happens, three components of the Higgs field are "absorbed" by the SU(2) and U(1) gauge bosons (the "Higgs mechanism") to become the longitudinal components of the now-massive W and Z bosons of the weak force. The remaining electrically neutral component either manifests as a Higgs particle, or may couple separately to other particles known as fermions (via Yukawa couplings), causing these to acquire mass as well

So there is one Higgs particle in the standard model as seen in the table, the one measured at th LHC experiments. The doublet language refers to the Higgs field , which is a different story.

There are extensions of the standard model where there are more than one Higgs particles, and experiments at LHC are looking to see if one could find them , with no success up to now.

At present no Higgs boson particle beyond the standard model has been seen.

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  • $\begingroup$ What is the theoretical reason that the SM higgs field should be a doublet, not a singlet? $\endgroup$ Feb 13 '20 at 19:08
  • $\begingroup$ Because in the SU(3)xSU(2)xU(1) of the standard model, to be able to give the mass to what become the W+, W- ,and Z, and photon the SU(2)xU(1) the mathematics needs the complex doublet , so as to fit the experimental data.. SM is a model that fits data, it is not a theory of everything. $\endgroup$
    – anna v
    Feb 13 '20 at 20:16
  • $\begingroup$ I see..... that means If I want new gauge groups introduced in SM, per say U(1)dark and U(1)bright, and make U(1)dark and U(1) bright's gauge boson to obtain mass from Higgs Mechanism, then HIggs field should be at least triplet?... $\endgroup$ Feb 13 '20 at 22:07
  • $\begingroup$ @trentacoollime I am not a theorist to be able to answer that, but it should be more fields for sure. $\endgroup$
    – anna v
    Feb 14 '20 at 5:24
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In electroweak theory, neither three guage bosons have a mass term nor does the fermions, as they would break the guage invariance (for example a term like, $m^2 W_\mu W^\mu$). But standard model fermions do have mass and thus we say $SU(2)_W * U(1)_Y$ symmetry is not exact. Higgs mechanism or Spontaneous Symmetry Breaking gives masses without violating guage invariance. Since Higgs field couple to $SU(2)_W * U(1)_Y$ bosons (in the covariant kinetic term), it is included as a doublet.

[The weak isospin $T^3$ (= $ \pm 1/2$ for left handed fermions and =0 for right handed as they are singlets), hypercharge $Y_W$ and charge $Q$ are related by $Q = T^3 + 1/2 Y_W$. Now, theoretically we could consider a charged higgs field but then we wold expect a mass term for photons, which is not realistic (The electromagnetism symmetry $U(1)_{EM}$ is not broken through higgs mechanism). Hence the conventional way of defining higgs field as a charge 0 field is justified. Higgs can have $Y_W = \pm 1/2$.]

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In electroweak physics you have not one but four spin-1 particles: $\gamma$, $Z$ and $W^\pm$. All of these interact with each other. To describe this you have to employ the non-Abelian Yang-Mills theory. In this each gauge boson corresponds to a certain gauge group generator and their interaction vertices are given by the structure constants of the gauge group. This allows you to determine the gauge group. It is $U(1)\times SU(2)$ with $\gamma$ and $Z$ corresponding to a certain orthogonal superpositions of the $U(1)$ generator and $T^3$ whereas $W^\pm$ correspond to $T^\pm=\frac{T^1\pm i T^2}{\sqrt{2}}$.

To break this gauge symmetry you need a field that would interact with $SU(2)$ part of the gauge theory. This means that it has to be in the nontrivial representation of $SU(2)$ - to have many components that rotate in a certain way under the gauge transformations. The simplest such representation is a doublet - two-component complex column on which the $T^k$ are represented simply as properly normalized Pauli matrices.

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