# What evidence is there for the electroweak higgs mechanism?

The wikipedia article on the Higgs mechanism states that there is overwhelming evidence for the electroweak higgs mechanism, but doesn't then back this up. What evidence is there?

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LOL...it has been proved to be true... –  user7757 Jul 4 '12 at 13:29

## 2 Answers

The evidence for the Higgs mechanism is that it predicts the W and Z boson mass ratio from the coupling constants of the electroweak interaction. The coupling of SU(2) and U(1) to the Higgs determine the mass of the W's and Z's, and since you know the Higgs charge and SU(2) representation, it predicts the ratio of the masses of the two in terms of the dimensionless couplings. This relation is satisfied to 5 decimal places, and cannot be a coincidence. This establishes the Higgs mechanism occurs in nature as Weinberg said it does beyond reasonable doubt.

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What is the higgs charge, do you know when & where the experiment was carried out? Presumably, although the mass ratio is correct, so indirectly verifying the higgs boson for the electroweak field, direct identification of the higgs boson has still not happened? –  Mozibur Ullah Jun 22 '12 at 2:42
@MoziburUllah . The charge of the standard model higgs is zero. The Higgs itself may have been seen experimentally in the latest LHC data to be announced in July. It will take some work to establish that the resonance seen at 125GeV is the Higgs but we shall soon know. To establish it one has to measure coupling constants in decays in all channels where this resonance is seen to see consistency with the standard model predictions. –  anna v Jun 22 '12 at 4:46
@MoziburUllah: The Higgs field (not the Higgs particle, this is only the neutral component we call the Higgs--- the Higgs field includes parts which are eaten by the Ws and Zs to make them massive) is an SU(2) doublet with U(1) charge 1/2 (this charge depending on your normalization convention). It is the same as a bosonic version of the electron/neutrino doublet. The ratio of the W and Z mass is predicted to be $g_2/\sqrt{g_1^2 + g_2^2}$, something which is unfortunately usually called the Weinberg angle, even though it isn't an angle at all, it's just a ratio of couplings. –  Ron Maimon Jun 22 '12 at 6:58

The simplest way to see that we need a Higgs is note first right down the Standard model without the Higgs particle, in particular the W and Z boson lagrangian which will contain masses terms for these particles like:

$\mathcal{L} = -\frac{1}{4}W_{\mu \nu}W^{\mu \nu} - m_w^2 W_{\mu }W^{\mu }-\frac{1}{4}Z_{\mu \nu}Z^{\mu \nu} - m_z^2 Z_{\mu }Z^{\mu }$

On the surface of things, this isn't so bad, we appear to have a renormalizable (all terms in the Lagrangian are dimension 4 or less) field theory. However, intuitively at high energies we know that we should be able to ignore the mass of the W and Z bosons. In order to do this we could rewrite the lagrangian above as

$\mathcal{L} = -\frac{1}{4}W_{\mu \nu}W^{\mu \nu} -\frac{1}{4}Z_{\mu \nu}Z^{\mu \nu} + v^2 Tr (D_\mu e^{i \frac{ \pi^a t^a}{v} } D^\mu e^{i \frac{ \pi^a t^a}{v} } )$

where $D_\mu$ is the appropriate covariant derivative and I have essentially 'reversed' the eating of the would-be goldstone bosons. Now we don't even have to bother with taking the massless limit of the W and Z particles since there are no more massless to concern ourselves with. The thing is this way of writing things shows clearly we were ignorant to the fact that we are in actuality dealing with a nonrenormalizable mode! (Since the the above Lagrangian contains terms with dimension greater than 4 - to see this, just expand out the exponentials). Now nonrenormalizable theories are perfectly fine except for the fact that predict their own demise - at scales near to v the theory will become strongly coupled (those higher order terms in the Lagrangian we can ignore at low energies become important). We could throw our hands up at this point, but we could also try and look for some model that holds up to higher energies (UV completes) the above Lagrangian. The simplest model that does this is by including the higgs boson in the standard model. This entails realizing the above is just the Lagrangian for a scalar doublet in the broken phase and taking:

$e^{i \frac{ \pi^a t^a}{v} }(o,v)^T \rightarrow e^{i \frac{ \pi^a t^a}{v} } (o,h)^T$.

Now this is a just an arbitrary SU(2) transformation acting on a vector so we can rewrite it as

$e^{i \frac{ \pi^a t^a}{v} } (o,h)^T = ( \phi_1 +i \phi_2 , h +i \phi_3)^T$

Now we can write the above Lagrangian as

$\mathcal{L} = -\frac{1}{4}W_{\mu \nu}W^{\mu \nu} -\frac{1}{4}Z_{\mu \nu}Z^{\mu \nu} + D_\mu H D^\mu H$

and the theory is renormalizable (which you can see by expanding out in terms of the h and $\phi_i$'s). We could have done this a handful of other ways - namely Technicolor or Little higgs but electroweak precision (tests of beyond the standard model higgs, see http://en.wikipedia.org/wiki/Peskin%E2%80%93Takeuchi_parameter) imply that a light perturbative higgs is strongly favored. In this sense 3 of the 4 degress of freedom of the Higgs have already been discovered!

In short: the Higgs one of handful of ways of rendering the Standard Model minus the Higgs satisfactory up to higher energies and of all these ways, experiments so far favor the Higgs over alternatives.

See Nima explain it in better detail:

http://streamer.perimeterinstitute.ca/Flash/b962bd27-28b7-4f93-b10e-5b10ed5744ca/viewer.html

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