Why don’t photons interact with the Higgs field and hence remain massless?
[Physics] Why don’t photons interact with the Higgs field
higgsmassphotonsquantum-field-theorystandard-model
Related Solutions
The Higgs field is a scalar field, $h$, so as far as the Lorentz symmetry goes, it is allowed and expect to interact with any other field. Whatever is the Lorentz-invariant term in the Lagrangian for other fields may be multiplied by $h$ to get an allowed interaction term, often a renormalizable one.
In particular, the Higgs field interacts with all the fermionic fields via the so-called Yukawa interaction, schematically $y\cdot h\bar\psi \psi$, where $\psi$ is a fermion field. The (classically) dimensionless Yukawa couplings $y$ may be large or small. The top quark Yukawa coupling is very large, of order one, which makes the top quark heavy. On the contrary, the Yukawa couplings with the neutrinos is (almost) zero which keeps the neutrinos massless.
In the previous paragraph, I ignored the fact that the Higgs is really a component of the Higgs doublet and the interaction terms has to be gauge-invariant. So a doublet must be contracted with another doublet, and so on. The vev produces masses for one component of the doublet only. There are interactions with $h$ (doublet) and its complex conjugate that make all quarks massive. However, it's not possible to write down the simplest interaction that would give the neutrinos masses in the simplest model. In the simplest model, the neutrinos are left-handed and it's exactly the left-handed part of the neutrinos that can't get the masses. But the neutrinos are still interacting with the Higgs field.
The gauge invariance in fact dictates that the Higgs field has to interact with other fields, the gauge fields. Because the Higgs field carries no color charges like quarks, it's neutral under $SU(3)_{QCD}$ which means that it doesn't have any interactions with gluons.
However, the Higgs field has nonzero charges under the electroweak $SU(2)_W$ because the Higgs field is a doublet; and under the hypercharge $U(1)_Y$ part of the electroweak gauge symmetry. These charges of the Higgs fields mean that the derivatives in its kinetic term have to be replaced by the covariant derivatives and this produces interactions with the electroweak gauge fields.
In general, that makes the gauge bosons for $SU(2)\times U(1)$ massive, too, when the Higgs gets a vacuum expectation value. However, the vev is such that the classical vacuum configuration is invariant under the electromagnetic $U(1)_{em}$ generated by the electric charge $Q = (1/2)Y+T_3$. So under this particular generator, the component of the Higgs field that has a nonzero vev is neutral which is why the Higgs field doesn't directly interact with the corresponding gauge field, the electromagnetic field, and that's why the photon stays massless and the electromagnetic force remains a long-range force. The other three generators of the electroweak group produce gauge bosons $W^\pm, Z^0$ if an orthonormal basis is chosen which is why these three gauge bosons are massive due to the Higgs mechanism.
This blog post is a good one to get a feeling of the Higgs mechanism.
One should definitely separate the Higgs field from the Higgs boson. The Higgs boson is an elementary particle , attendant to the existence of the Higgs field as its excitation, and acquires its mass as all the other particles in the standard model table.
The Higgs field that gives the elementary particle masses in contrast to all the other fields posited by field theory, which are zero everywhere unless a "particle" exists as an excitation of the field, is nonzero everywhere.
Please read the link to understand that the masses are not acquired by interaction of the type "Higgs+electron" or "Higgs + quark" Feynman vertices, but by the change in the basic vacuum expectation value through which the other fields, photon, electron etc , exist and move.
So chirality has nothing to do with the acquisition of mass from the quarks and the other fermions through the Higgs mechanism. (In any case chirality is definite only for zero mass fermions , whether quarks or leptons.)
Best Answer
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.