5
$\begingroup$

In non-spontaneously broken QFT like QED the gauge bosons cannot have a mass due to gauge symmetry (follows from Ward identity). Also they have only 2 polarizations.

However in a spontaneously broken gauge theory the gauge boson becomes massive after the symmetry breaking. For example consider the standard U(1) example: $$L = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + D_\mu\phi^\dagger D^\mu\phi - V(\phi)$$ where $V(\phi)$ is a Mexican hat potential. After the symmetry breaking the photon is massive and has 3 polarizations.

What is the difference between both cases? Why is it relevant whether the theory is spontaneously broken or not (I guess it has something to do with the Ward identity)? And why are we so sure that this cannot happen in our 'normal' theories like QED (i.e. that in the end the photon becomes massive).

Edit: Maybe I should reformulate the question a bit. The definition of a spontaneous broken theory is that $Q|\Omega\rangle \neq 0$, where $Q$ is the charge operator as given by the symmetry and $|\Omega\rangle$ the groundstate. Why does the charge of the groundstate of my theory matter at all? I can formulate gauge invariance only as a Heisenberg operator symmetry: $\hat{A^\mu} \sim \hat{A^\mu} + \partial^\mu \hat{\alpha}$. Therefore the photon should still be massless. Obviously this cannot be true because we have plenty of examples where gauge bosons become massive (e.g. W boson or superconductors).

I guess that something goes wrong with the argument that the photon cannot be massive by gauge invariance. Since I don't know a nice proof of this (Only argumentations like this by resumming feynman diagrams) I guess something goes wrong there. The question is about what exactly goes wrong? And more important: Why is the mass of the photon protected in the normal case, but not in spontaneous symmetry breaking case. What's the crucial difference?

Edit 2: It took me a while to figure it out, but I think I found a suitable nonperturbative argumentation why the photon cannot become massive. Instead of unitarity gauge work in the usual Lorentz gauge (and Feynman gauge). Then the equal time commutation relations read $[A^\mu(x), A^\nu(0)] = 0$ and $[\dot{A}^\mu(x), A^\nu(0)] = i \eta^{\mu\nu}$. Also the equation of motion is given by $\partial_\alpha\partial^\alpha A^\mu = J^\mu$. Therefore apply the equation of motion on the propagator $\langle 0|TA^\mu(x)A^\nu(0)|0\rangle$:

\begin{align} \partial_\alpha \partial^\alpha \langle 0|TA^\mu(x)A^\nu(0)|0\rangle &= \partial_\alpha \langle 0|T(\partial^\alpha A^\mu(x))A^\nu(0)|0\rangle + \partial_\alpha \left(\delta(t) \delta_{0\alpha} \langle 0|[A^\mu(x),A^\nu(0)]|0\rangle\right)\\ &= \langle 0|T(\partial_\alpha\partial^\alpha A^\mu(x))A^\nu(0)|0\rangle + \delta(t) \langle 0|[\partial_0 A^\mu(x),A^\nu(0)]|0\rangle\\ &= \langle 0|TJ^\mu(x)A^\nu(0)|0\rangle + i\eta^{\mu\nu}\delta^4(x) \end{align}

The commutator come from time ordered product and can be replaced by there equal time value because of the $\delta(t)$ in front. After a Fourier transformation this reads (call the Fourier transformed propagator $D^{\mu\nu}(p)$):

$$-p^2 D^{\mu\nu}(p) = \int d^4x e^{-ipx} \langle 0|TJ^\mu(x)A^\nu(0)|0\rangle + i\eta^{\mu\nu}$$

For $J^\mu = 0$ (free case) this simply gives the free photon propagator $D^{\mu\nu}(p) = \frac{-i\eta^{\mu\nu}}{p^2}$. Now if $J^\mu \neq 0$ multiply both sides with $p_\mu$ (which translates to $\partial_\mu$ for $J^\mu$)

\begin{align}-p^2 p_\mu D^{\mu\nu}(p) &= -\int d^4x e^{-ipx} \langle 0|T(\partial_\mu J^\mu(x))A^\nu(0)|0\rangle - \int d^3x e^{-ipx} \langle 0|[J^0(x),A^\nu(0)]|0\rangle + ip^\nu\\ &= - \int d^3x e^{-ipx} \langle 0|[J^0(0,\vec{x}),A^\nu(0)]|0\rangle + ip^\nu \end{align}

here I used $\partial_\mu J^\mu = 0$. Now as long as $[J^0(0,\vec{x}),A^\nu(0)] = 0$ (this is for example the case in the theory above since $J^0$ does not contain any $\partial_0 A^0$) we have $$-p^2 p_\mu D^{\mu\nu}(p) = ip^\nu$$ or $$p_\mu D^{\mu\nu}(p) = -i\frac{p^\nu}{p^2}$$ which means that $D^{\mu\nu}(p)$ must have a pole at $p^2 = 0$ (i.e. the photon is massless). Clearly this it not satisfied by the propagator for a massive particle $\sim \frac{1}{p^2 - m^2}$

Note that this argumentation never used any property of the ground state of our theory. In the Higgs mechanism one expands the field around its vev and ignores higher orders. After that the photon is massive (in lowest order of perturbation theory) which is a clear contradiction to my derivation above. So my first guess would be that the mass of the photon will again cancel when taking into account all feynman diagrams.

$\endgroup$
  • $\begingroup$ I very much does happen in QED, inside a superconductor where the U(1) group is SSBroken , and the photon becomes effectively massive through the Higgs mechanism: the resulting magnetic flux exclusion is called the Meissner effect. $\endgroup$ – Cosmas Zachos Feb 8 at 23:55
  • $\begingroup$ Near duplicate. $\endgroup$ – Cosmas Zachos Feb 8 at 23:57
  • $\begingroup$ Viola that's a quite nice example. So let's say I have a superconductor in the non superconducting phase. There I have a massless photon (right?). Then I cool it below the critical temperature. From the QFT principles the photon cannot obtain mass during the whole process (because it is protected by the ward identity). But if I just treat the final state with the spontaneous symmetry breaking formalism I find a massive photon. So in order for the symmetry breaking formalism to be correct there must be something that gives mass to the photon at some point during the process. $\endgroup$ – toaster Feb 9 at 0:31
  • $\begingroup$ Whether or not a particle is massive or not cannot depend on the particular choice of gauge. During performing the spontaneous symmetry breaking formalism one expands the Higgs field around the VEV. What tells me that I am not doing something wrong here? In particular basic QFT principles tell me that the photon cannot obtain mass whatever interaction I introduce! On the other hand side I have the Higgs mechanism which is an approximation but gives me a massive photon. So either the first or the second statement must be wrong in general! I guess it's the first one but I would like to know why. $\endgroup$ – toaster Feb 9 at 1:18
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos Feb 9 at 1:47
4
+50
$\begingroup$

Why does the charge of the groundstate of my theory matter at all?

Because it dictates the coupling form of all states with vanishing v.e.v. to it, and hence the realization of the relevant symmetry, from which all else follows. It makes all the difference. Gauge invariance is always there.

You may be falling victim to chronic abuse of language plausibly invented for this very purpose. "Protection" has a technical connotation involving arbitrary radiative corrections, irrelevant in this context.

The assumption behind the Higgs mechanism, or the lack thereof, is that all quantum calculations have been carried out or guessed, to produce a low energy effective theory. This theory has or lacks SSB, the latter case being your "normal" theory.

Now, I could not, of course, improve on Kibble's near-perfect explanation of his work, beyond vulgarizing it with cartoons and annotations, possibly steering you to "follow the math".

If the theory has no SSB, as in conventional QED, the vacuum is symmetric, so U(1) rotations leave it alone, $e^{i\alpha Q} |0\rangle = |0\rangle$. Consequently $Q|0\rangle = 0$.

By contrast, inside a medium such as a superconductor's, $Q|0\rangle \neq 0$.

Kibble illustrate's Goldstone's celebrated (1961) sombrero potential, a gimmick allowing ready grasp of the group theory (but Landau & Ginzburg did that earlier, 1950, without covariance). All fields are defined with zero v.e.v. s as they should, at the true vacuum, at the minimum of the sombrero.

Specifically, it enforces that the bottom of the potential, the vacuum, has a massive "Higgs mode", $\varphi _1$; and a massless "Goldstone mode", $\varphi_2$.

The conserved EM current (13) is then $$ j_\mu= \frac{v}{2} \partial_\mu \varphi_2 + ..., $$ ignoring irrelevant multi-field terms. $v$ is a dimension-one parameter quantifying symmetry breaking in the sombrero, historically deducted from the higgs to ensure $\langle 0|\varphi_1|0\rangle=0$. The linear term, which makes all the difference, is the hallmark of the SSB, or "Nambu-Goldstone" realization of the symmetry.

Firstly, it ensures that, given (14), $$ \langle \varphi_2 | \int d^3x \frac{v}{2} \partial_0 \varphi_2(x) |0\rangle \neq 0, $$ that is Q cannot annihilate the vacuum, but, semiclassically, pumps zero energy goldstons in or out of it, to yield another vacuum degenerate with it.

By contrast, in the normal, "unbroken" phase, for $v=0$, there is no such chance, and the vacuum is annihilated.

The second, related, phenomenon, is that a crucial piece of the covariantly completed kinetic term for the goldston and higgs may be declared as a B field mass term, $$\tfrac{1}{2} \left ( \partial_\mu \varphi_2 + ev A_\mu \right )^2\equiv \frac{e^2 v^2 }{2} B_\mu^2 .$$

In the $\varphi_2=0$ gauge, where the goldston was all pumped into the "higgs" $\varphi_1$, we can afford to ignore it. (This is easier in polar field coordinates.)

You may then, if you are inclined, call the B a "massive photon", or whatever. In the above gauge, it identifies with the photon. The crucial physics, however, is that, at the level of this effective lagrangian, there are no massless modes, photon or Goldstone, but only massive modes (B), instead. You might go wild changing gauges and fixing them and identifying modes, but that's just language games. (The SM pros, of course, cannot live without them.)

In the "normal" case without SSB, v=0, no such luck and the photon is visibly massless, no matter what you do, while $\varphi_2$ is visibly massive and not a goldston. The current starts with the usual bilinear rotation term, dubbed "Wigner-Weyl" realization.

The theory is gauge invariant for vanishing or non-vanishing v, and you may well have escaped from the clutches of slippery "protection" talk. There has been lots of work anticipating whether a theory will, as a result of quantum interactions, end up as a zero or nonzero v effective lagrangian, but there are no foolproof guidelines.

$\endgroup$
  • $\begingroup$ Thanks for your answer. I have read your answer and Kibbles paper several times now and I am still not convinced. In the first orders of perturbation theory the photon looks massive. But what tells me that there is no cancellation of the photon mass at higher orders? I think I found a argumentation why the photon is massless in general (The only difference is a different gauge). I added it to my question, maybe you could take a look at it. $\endgroup$ – toaster Feb 12 at 17:11
  • $\begingroup$ That was the basic starting assumption here: that, instead of connecting tree level with loops, one has computed all loops, and encoded them in an effective action with or without nontrivial vacuum. Promoting the argument to a Ward-identity or Slavnov-Taylor identity, better, is a messy story, and I thought, irrelevant here. I suspect you are heading headlong for Coleman-Weinberg technology, where radiative corrections do all the damage, but then you should express your question in its language, surely with an answer in its reviews. $\endgroup$ – Cosmas Zachos Feb 12 at 18:37
  • $\begingroup$ Possibly related , if that's where you really wish to go. $\endgroup$ – Cosmas Zachos Feb 12 at 19:45
1
$\begingroup$

I'll try to give my undergrad understanding of spontaneous symmetry breaking. Additions and corrections are welcome.

The term spontaneously broken is a bit misleading. One of my lecturers even calls it "the worst terminology in physics"... The whole theory is still symmetric even after spontaneous symmetry breaking. Only in the ground state the symmetry is let's call it "hidden" such that seemingly non gauge invariant terms like in your case a photon mass term can appear when we expand the scalar field $\phi$ around its minimum.

Regarding QED. We are 100% sure that the same thing could occur in some extension of QED. There exist several Toy models that lead to massive Photons hence "breaking" the U(1) gauge symmetry. However experiments tell us that if the photon has a mass it is very small...

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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