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Why don’t photons interact with the Higgs field and hence remain massless?

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Could you please split this into two separate questions? Your questions are unrelated, so each deserves a second thread. – Manishearth Apr 2 '12 at 8:50
@LuboMotl: Alright, makes sense. Though it may be confusing for future visitors :/ – Manishearth Apr 2 '12 at 9:09
We generally split unrelated questions unless there is some real reason not to do so. – Colin K Apr 2 '12 at 11:56
I fail to see the link between the two questions – Frédéric Grosshans Apr 2 '12 at 12:31
I agree with Manishearth, this should be split. Since OPERA has been covered by a number of other questions on this site, I'll remove that portion and just leave this as a question about the Higgs field. Aritra, please see our other questions about OPERA and then if your question is still not answered by them, you may repost it as a separate question (but do make sure to indicate why the existing questions aren't enough for you). – David Z Apr 2 '12 at 13:08

Massless photon

Photons interact with the "Higgs doublet" but they don't interact with the "ordinary" component of the Higgs field whose excitations are the Higgs bosons.

The reason is that the Higgs vacuum expectation value is only nonzero for the component of the Higgs field whose total electric charge, $Q=Y+T_3$ where $Y$ is the hypercharge and $T_3$ is the $z$-component of the $SU(2)_w$ weak isospin gauge group, is equal to zero, i.e. for $Y=\pm 1/2$ and $T_3=\mp 1/2$. That's why the coefficient of the $(h+v) A_\mu A^\mu$ term is zero.

In other words, the vacuum condensate of the Higgs field that fills the space is charged under the weak charges, including the hypercharges and the weak $SU(2)$ charge, but exactly under the right combination of these charges, the electric charge, the condensate is neutral. It would be "bad" if the vacuum carried a nonzero electric charge. It doesn't.

So the $A_\mu A^\mu$ interaction, whose coefficient is proportional to the electric charge of the Higgs field, isn't there. The photon remains massless and the electromagnetic interaction remains a long-range force, dropping as a power law at long distances (instead of the exponential decrease for short-range forces: W-bosons and Z-bosons do interact with the Higgs condensate and they get massive and their forces get short-range).

OPERA anomaly

The OP's question used to have two parts but this second part has been deleted. But I won't delete the answer because the votes and other things may have already reacted to this part as well etc.

Yes, the anomaly of the OPERA neutrino speed measurement has been resolved. First, ICARUS, using directors in the very same cave, measured the speed as well and got $v=c$ within the error margin (the same error margin as OPERA's).

Second, a few months ago, OPERA found out that they had a loosely connected fiber optical cable to a computer card. Using some independent data OPERA recorded, it was possible to determine that the cable error (plus another source of error whose mean value is much smaller) shifts the timing by $73\pm 9$ nanoseconds in the right direction (it's the right direction because the cable problem had delayed some older neutrino-free measurements of the time but was fixed once the neutrinos were being measured), see

so when the error is corrected, the "neutrinos by $60\pm 10$ nanoseconds too fast" become "neutrinos coming $13\pm 15$ nanoseconds after light" which is consistent with $v=c$. Note that relativity with light but massive neutrinos predicts $c-v\sim 10^{-20} c$ for these neutrinos, experimentally indistinguishable from $v=c$.

The spokesman of the experiment and the physics coordinators have already resigned; the spokesman resigned first: before another no-confidence vote but after some preparation votes for the no-confidence vote. It seems that they have known the mistake since December 8th, 2011, but they were hiding it for a few months (it was leaked to Science News by someone else in February) and they wanted to do experiments for additional months, even in May 2012, even though the error has been known to eliminate the anomaly for quite some time. They apparently enjoyed the unjustifiable fame.

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Oh, I didn't know of the skullduggery by the spokesman. Thanks for the info! – Manishearth Apr 2 '12 at 9:08
Oh, by the way, since you're not always online, you may want to consider adding yourself here. That way we can send any interesting questions which you may be able to answer better your way even if they get buried by other questions. – Manishearth Apr 2 '12 at 9:12
If I see nice unanswered questions about nice topics I sometimes feel very tempted to ping Lumo ;-) – Dilaton Apr 2 '12 at 10:15
Lubos: the question has been edited, you may want to consider updating your answer accordingly. – David Z Apr 2 '12 at 13:11
Thankyou for the elegant ans. I will split the questions accordingly – Aritra De Apr 3 '12 at 9:24

The massless photon:

The zero mass is not due to a special value of the Weinberg angle, the angle which determines the mass of the other three bosons $W^+$, $W^-$ and $Z$ The mass is zero because the vacuum expectation value of the Higgs field doublet is single valued rather than two valued. This means it can in principle always be expressed by.

$\langle \phi \rangle ~=~ \left(\begin{array}{c} 0 \\ v \end{array}\right)$

It's the 0 here which leaves one of the four bosons massless. Just to show a bit of the math:

The gauge transform of the Higgs field is defined with $\beta$ corresponding to an Abelian field and the three $\alpha$ corresponding to Non Abelian fields.

$\phi \longrightarrow ~~\exp \,\frac{i}{2}\left\{\beta \left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right) + \alpha^1\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right) + \alpha^2\left(\begin{array}{cc} 0 & \!-\!i\\ i & 0 \end{array}\right) + \alpha^3\left(\begin{array}{cc} 1 & 0\\ 0 & \!-\!1 \end{array}\right) \right\}~\phi$

The $\beta$ corresponds to the hypercharge Y (see also Luboš Motl's post) and $\alpha^1$, $\alpha^2$ and $\alpha^3$ correspond to the three components of the iso-spin T. Now the combination $\beta=\alpha^3$ results in the matrix.

$\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right) + \left(\begin{array}{cc} 1 & 0\\ 0 & \!-\!1 \end{array}\right) = \left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right)$

So it is this combination which doesn't interact with the vacuum expectation value,

$\left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right)\left(\begin{array}{c} 0 \\ v \end{array}\right)\equiv\left(\begin{array}{c} 0 \\ 0 \end{array}\right)$

and it is this combination which represents the massless photon.


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Hans: the question has been edited, you may want to consider updating your answer accordingly. – David Z Apr 2 '12 at 13:12
You should use a different terminology other than "single valued" and "two valued", it's confusing, because you mean "can be rotated to have one component zero". But this is not necessary, there is a massless photon for any spinor coordinates you choose of the Higgs value. – Ron Maimon Nov 11 '12 at 4:28
It's incredible that people can even comprehend this. – Inquisitive Feb 2 '15 at 23:09

There is an aspect to this question that nobody seems to have addressed and that is, although the higgs (the 'radial' component of the field) is neutral, and therefore doesn't interact with the photon at 'tree level' we still see the decay $h \rightarrow \gamma \gamma$. This is because, roughly, by quantum effects a higgs will fluctuate into a particle/ anti-particle pair (electrons, quarks etc) which can they produce photons. So while the higgs does not strictly interact with the photon, at low energies we can parameterize a low-energy effective interaction where the higgs does interact with the photon. This is diagrammatically expressed in the Feynman diagrams: higgs to gamma gamma

which I have borrowed from

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First of all, photons are observed to be massless, and the $W^\pm$ and $Z$ are observed to have mass. So we have to build a model that agrees with this.

Mathematically (at a non-rigorous but intuitive level), the $SU(2)_W\times U(1)_Y$ electroweak gauge symmetry is a 4-dimensional Lie group ($=U(2)$). Given the experimental facts, we want to find a representation under which the orbit of the non-zero vev under this Lie group action is 3-dimensional, so that the 3-dimensional orbit contributes to the longitudinal components of $W^\pm, Z$ (3 of them); since the original Lie group is 4-dimensional, so there is 1-dimension that acts trivially on the vev, that is the photon. What representation satisfies this? The simplest choice is 2-component complex vectors (i.e. doublet) transforming under $U(2)$ matrices. The orbit of $U(2)$ acting on a non-zero complex doublet is $S^3$, leaving one dimension in the $U(2)$ acting trivially. (This can be attributed to the fact that by knowing two component vectors $v, u$ and $U v= u$, you cannot uniquely determine $U$.) On the other hand, a complex doublet takes value in an $\mathbb{R}^4$; this $\mathbb{R}^4$ mod out the $S^3$ orbits leave one dimension that is the higgs.

So the rough idea is, given the symmetry group and the representation:

$$dim(\mbox{symmetry group})-dim(\mbox{orbit of action})=dim(\mbox{residual symmetry})$$ $$dim(\mbox{rep space})-dim(\mbox{orbit of action})=dim(\mbox{physical degrees of freedom})$$ (The dimension of orbit of action in global symmetry is the number of massless Goldstone bosons; in gauge symmetry the longitudinal components of massive gauge fields. If the symmetry is global, the "physical degrees of freedom" is replaced by "massive particles".)

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protected by Qmechanic Mar 13 '13 at 15:41

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