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

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/4700/2451 and physics.stackexchange.com/q/17939/2451 $\endgroup$
    – Qmechanic
    Commented Apr 2, 2012 at 17:23
  • 8
    $\begingroup$ This is easily one of the worst bounties I've seen - you cannot expect a technical result of QFT to have a "way to imagine". In particular, the ultimate answer to why the Higgs doesn't give the photon mass is because we don't observe the photon to have mass, so a theory which gives it mass would be pretty useless to describe reality, hence we have to build a theory that doesn't give the photon mass. $\endgroup$
    – ACuriousMind
    Commented Feb 18, 2016 at 16:08
  • 1
    $\begingroup$ Counterexample: arxiv.org/abs/hep-ph/0306245 $\endgroup$ Commented Feb 18, 2016 at 16:51
  • 2
    $\begingroup$ By construction. Higgs mechanism is a mechanism of breaking electroweak $SU(2)\times U(1)_Y$ to $U(1)_{EM}$, and the photon is the gauge boson of the unbroken $U(1)_{EM}$. $\endgroup$ Commented Feb 19, 2016 at 23:17

5 Answers 5


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.

  • 1
    $\begingroup$ Oh, I didn't know of the skullduggery by the spokesman. Thanks for the info! $\endgroup$ Commented Apr 2, 2012 at 9:08
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Apr 2, 2012 at 9:12
  • 1
    $\begingroup$ If I see nice unanswered questions about nice topics I sometimes feel very tempted to ping Lumo ;-) $\endgroup$
    – Dilaton
    Commented Apr 2, 2012 at 10:15
  • 2
    $\begingroup$ Lubos: the question has been edited, you may want to consider updating your answer accordingly. $\endgroup$
    – David Z
    Commented Apr 2, 2012 at 13:11
  • 2
    $\begingroup$ Thankyou for the elegant ans. I will split the questions accordingly $\endgroup$
    – Aritra De
    Commented Apr 3, 2012 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.


  • 1
    $\begingroup$ Hans: the question has been edited, you may want to consider updating your answer accordingly. $\endgroup$
    – David Z
    Commented Apr 2, 2012 at 13:12
  • 3
    $\begingroup$ 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. $\endgroup$
    – Ron Maimon
    Commented Nov 11, 2012 at 4:28
  • $\begingroup$ It's incredible that people can even comprehend this. $\endgroup$ Commented Feb 2, 2015 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 http://resonaances.blogspot.com/2012/07/h-day-3-how-to-pump-up-higgs-to-gamma.html.


Before saying anything about why photons don't interact with the Higgs field, i would like to emphasize the very meaning of mass

By Einstein's famous equation $E=mc^{2}$, "Mass and Energy are the same thing".

This connection between matter and energy was at the heart the solution to problem of how and from where particles (Guage bosons, except the photon) gets their mass, as proposed in 1964 by Peter Higgs and others who worked with him.

Now why and how do we know that photons are massless?

The gauge bosons a group of particles which are responsible for any of the fundamental interactions of nature, Along with other groups of particles (quarks and leptons) together form the Standard model of particle physics. Photons are one of the guage bosons which are responsible for the electromagnetic interactions.

Feynman diagram of an electron and a positron annihilation

In particular in order for the theory of electromagnetism to hold together in a mathematically self-consistent way the photon has to be exactly massless. if it did have a mass then if you try to do calculations with the theory with the massive photon included you would find that calculating the same quantity in different ways would give you different answers (inconsistent results)

Now mathematically for the other 3 guage bosons, say the WZ Boson, which is responsible for the weak nuclear interaction, should also have zero mass. But experimentally it seems that they're about a hundred times heavier than a proton

enter image description here

So in order to solve this mathematical inconsistency a radical new idea was introduced which states that

Space everywhere in the universe is filled with a quantum field which is called as the Higgs field.

The Answer

Some particles travel to this field without knowing it's there. For example the photon simply doesn't interact in any way with this field. Thus it doesn't have any mass.

Other particles interact highly with the higgs field thus slowing them down and this slowing down of the particles due to interactions cause the feeling of mass

Now why doesn't photon doesn't interact at all? well simply speaking, its just a property of photon, similar to properties of other particles like electrons or protons having a charge which is their property.

In order to describe this property we can use mathematics as other answers did.

just like the electromagnetic fields made of photons the Higgs field is made up the smaller particles are smaller bits for the field called Higgs bosons.

PS: Since a very simple but detailed answer(An answer for layman) was needed i didn't use any mathematics or very deep physics.


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".)


Not the answer you're looking for? Browse other questions tagged or ask your own question.