There are plenty of well-answered questions on Physics SE about the mathematical differences between gauge symmetries and global symmetries, such as this question. However I would like to understand the key differences between the transformations in terms of what they mean physically.

Say we have the Lagrangian for a scalar field interacting with the electromagnetic field,

\begin{equation} L = -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} + (D_{\mu} \phi)^*D_{\mu}\phi - m^2|\phi|^2, \end{equation}

where $D^{\mu} = \partial^{\mu} + ieA^{\mu}$. This is invariant under both a local gauge symmetry $A^{\mu} \rightarrow A^{\mu} + \partial^{\mu} \chi$ with $\phi \rightarrow e^{i\chi(x)} \phi$ and a global symmetry $\phi \rightarrow e^{i\chi} \phi$.

I am aware that by requiring the gauge symmetry we have introduced interaction terms coupling the scalar and vector boson fields, while the global symmetry gives us the conservation of particle number by Noether's theorem.

But now what do the local and global phase shifts mean physically? Or are their physical meanings defined purely by their introduction of field couplings and of particle conservation, respectively?


The first answer to such a question must always be: A gauge symmetry has no "physical" meaning, it is an artifact of our choice for the coordinates/fields with which we describe the system (cf. Gauge symmetry is not a symmetry?, What is the importance of vector potential not being unique?, "Quantization of gauge systems" by Henneaux and Teitelboim). Any gauge symmetry of the Lagrangian is equivalent to a constraint in the Hamiltonian formalism, i.e. a non-trivial relation among the coordinates and their canonical momenta.

In principle, any gauge symmetry may be eliminated by passing to the reduced phase space that has fewer canonical degrees of freedom. The gauge symmetry has no physical meaning in the sense that be may get rid of it by passing to a (classically) equivalent description of the system. A gauge transformation has no physical meaning because all states related by a gauge transformation are physically the same state. Formally, you have to quotient the gauge symmetry out of your space of states to get the actual space of states.

In contrast, a global symmetry is a "true" symmetry of the system. It does not reduce the degrees of freedom of the system, but "only" corresponds to conserved quantites (either through Noether's theorem in the Lagrangian formulation or through an almost trivial evolution equation in the Hamiltonian formalism). It is physical in the sense that states related by it may be considered "equivalent", but they are not the same.

Interestingly, for scalar QED, the global symmetry gives a rather inconvenient "Noether current" - one that depends on the gauge field (cf. this answer)! So the statement that "Noether's theorem" gives us charge/particle number conservation is not naively true in the scalar case (but it is in the Dirac case). Getting charge conservation from the gauge symmetry is also discussed in Classical EM : clear link between gauge symmetry and charge conservation.

Why then use such a "stupid" description in the first place, you might ask. The answer is that, in practice, getting rid of the superfluous degrees of freedom is more trouble than it's worth. It might break manifest invariance under other symmetries (most notably Lorentz invariance), and there can be obstructions (e.g. Gribov obstructions) to consistently fix a gauge. Quantization of gauge theories is much better understood in the BRST formalism where gauge symmetry is preserved and implemented in the quantum theory than in the Dirac formalism that requires you to be able to actually solve the constraints in the Hamiltonian formalism.

So the key difference between a gauge and a global symmetry is that one is in our theoretical description, while the other is a property of the system. No amount of shenanigans will make a point charge less spherically symmetric (global rotation symmetry). But e.g. the electromagnetic gauge symmetry simply vanishes if we consider electric and magnetic fields instead of the four-potential. However, in that case we lose the ability to write down the covariant Lagrangian formulation of electromagnetism - the current $J^\mu$ must couple to some other four-vector, and that four-vector is simply the potential $A^\mu$.

There is one further crucial aspect of gauge symmetries: Every massless vector boson necessarily is associated to a gauge symmetry (for a proof, see Weinberg's "Quantum Theory of Fields"). There is no other way in a consistent quantum field theory: You want massless vector bosons like photons - you get a gauge symmetry. No matter how "unphysical" this symmetry is - in the covariant framework of quantum field theory we simply have no other choice than to phrase such particle content in terms of a gauge field. This you might see as the true "physical" meaning of gauge symmetries from the viewpoint of quantum field theory. Going one step further, it is the spontaneous breaking of such symmetries that creates massive vector bosons. A theory of vector bosons is almost inevitably a theory of gauge symmetries.

As an aside: In principle, one might try to make any non-anomalous global symmetry into a gauge symmetry (cf. When can a global symmetry be gauged?). The question is whether gauging it produces any new physical states, and whether these states fit to observations.

  • 1
    $\begingroup$ Just a clarification to an otherwise correct and good answer: " it is the spontaneous breaking of such symmetries [gauge symmetries] that creates massive vector bosons. " This is not correct. It is the spontaneous breaking of the global part of the symmetry in a gauge theory that creates massive vector bosons. The local part cannot be spontaneously broken. $\endgroup$ – Diego Mazón May 23 '16 at 15:38
  • $\begingroup$ Could anybody comment on precisely where in Weinberg's QFT series the section which Mr. ACuriousMind♦ mentioned (proof that a massless vector boson implies a gauge symmetry) resides? $\endgroup$ – Arturo don Juan Dec 17 '18 at 17:20
  • 1
    $\begingroup$ @ArturodonJuan Chapter 5.9 shows that massless fields associated with particles of helicity $\pm 1$ cannot be described by an ordinary four-vector field but necessarily have a "gauge-like" transformation behaviour (eq. 5.9.31 and surrounding text). $\endgroup$ – ACuriousMind Dec 17 '18 at 18:53

This is a very broad question, so there are many ways to answer it. Here is one interpretation.

A principal distinction between gauge symmetries and global symmetries is that gauge symmetries lead to long-range interactions between charged particles; the gauge symmetry demands the existence of a massless field which can propagate over arbitrarily long distances (in a gapless phase).

The distinction is best characterized by the superselection rule associated to the gauge symmetry. Recall that a quantum field theory with Hilbert space $\mathcal{H}$ is said to obey a superselection rule if the Hilbert space may be written as a sum $$\mathcal{H} = \bigoplus_n \mathcal{H}_n,$$ such that for any local operator $\mathcal{O}(x)$, $$\langle n|\mathcal{O}(x)|m\rangle = 0$$ for $n\neq m$, where $|n\rangle$ and $|m\rangle$ are arbitrary states in $\mathcal{H}_n$ and $\mathcal{H}_m$. The $\mathcal{H}_n$ are called superselection sectors. Physically, superselection rules appear in gauge theories whenever there exist states with long-range influence that cannot be annihilated by local operators. They classify the long-range "hair" associated to physical states.

In the context of the scalar QED example that you mentioned, the Hilbert space breaks into superselection sectors labeled by the charge $Q$ associated with constant U(1) gauge transformations. Local gauge-invariant operators always have zero charge and cannot change the charge of a state. Physically, a charged particle gives rise to a long-range electric field, which cannot be destroyed (or created) by any local operator in an infinite spatial volume. This is the essential feature of gauge theories which I described above.

Turning off the gauge coupling, the corresponding global U(1) symmetry leads to no such superselection rule. The Hilbert space is still graded by the charge number, but local operators can be charged under the symmetry. There is no "long range hair" associated to a global symmetry.

At the risk of being too detailed, I should point out a subtlety: the above discussion assumed the gauge theory was in a gapless phase. (For the scalar QED example, this is called the "Coulomb phase"). Other phases may break the superselection rule. For example, if we deform the theory such that $\phi$ acquires a vev (in unitary gauge), the theory is then in a gapped phase (the "Higgs phase"), and the superselection rule breaks. Physically, charged particles are surrounded by a charge condensate which screens the long-range electric field. States no longer have long-range hair, and there is no superselection rule. On the other hand, non-abelian gauge theories are commonly found in the "confining phase," which is another kind of gapped phase in which non-perturbative effects force all states to carry zero charge. The superselection rule is broken and there are no long-range interactions.

For some nice discussion along these lines, see section 2 of this paper.


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