# Does classical electrodynamics have $U(1)$ symmetry? If yes, how?

Quantum electrodynamics (QED) is based on $U(1)$ symmetry. What happens to this symmetry in classical electrodynamics?

Addendum The books on classical electrodynamics such as J. D. Jackson, does not mention about $U(1)$ symmetry in the context of gauge invariance (as far as I know). Gauge invariance is simply understood, in classical electrodynamics books, as the invariance of Maxwell's equations under $A_\mu\to A_\mu+\partial_\mu\chi(x)$. There is no sign of U(1) invariance that I can discover here. On the other hand, when something like Dirac equation or Dirac field is brought into the scene then the implementation U(1) transformation is clear. But that is always discussed in quantum field theory books. It appears that it is essential to have a Dirac field to understand U(1) symmetry. So the question is whether it is possible to understand the existence of U(1) symmetry in classical electrodynamics without bringing in the Dirac field into the picture?

• Yes, it's the same. – Qmechanic Aug 19 '17 at 8:00
• Why do you think anything about the $\mathrm{U}(1)$ symmetry is quantum? Generally, we obtain a quantum field theory by quantizing a classical one, how do you think new symmetries could appear in this process? (As opposed to disappearing, cf. quantum anomalies) – ACuriousMind Aug 19 '17 at 10:39
• I'm guessing your difficulty might be that you've only seen $U(1)$ symmetry derived from a minimal coupling condition with Schrödinger's / Dirac equation (i.e. that the EM field gauge conditions can be made to absorb the extra terms that appear in the uncoupled Schrödinger / Dirac equation when the quantum state $\psi$ is multiplied by an arbitrary phase term $e^{i\,\phi(\mathbb{r})}$)? If this answers @ACuriousMind's rhetorical question, then this tells us the answer you need. – WetSavannaAnimal Aug 19 '17 at 10:51
• @ACuriousMind I'm not quite thinking about anomalies. I'm asking how do I convince myself that U(1) invariance is there in classical electrodynamics. In Classical Electrodynamics books such the one by J. D. Jackson there is no mention of $U(1)$ symmetry in the context of gauge invariance (as far as I know). This symmetry is mentioned only in QFT books in the context of gauge invariance. Moreover, in classical electrodynamics, the fields are real fields where as the elements of $U(1)$ are complex, in general. – SRS Aug 26 '17 at 12:32
• @WetSavannaAnimalakaRodVance Yes! – SRS Aug 26 '17 at 12:33

1. A free "$\mathrm{U}(1)$" gauge theory can never tell whether the gauge group is $\mathrm{U}(1)$ or $\mathbb{R}$ because the only field in the theory, the gauge potential $A$, transforms as $$A\mapsto A + \partial_\mu \chi,$$ where $\chi$ is just a real-valued function, and the real numbers are the Lie algebra of both $\mathrm{U}(1)$ and $\mathbb{R}$. This is not a classical or quantum property, you simply cannot tell the difference. So, in a sense, asking whether this theory has $\mathrm{U}(1)$ symmetry or not is meaningless - it has $\mathfrak{u}(1)$ symmetry, and there is no meaningful notion of the symmetry group.

2. Electromagnetism coupled to an external conserved current still cannot tell what the gauge group is, since the current is gauge-invariant.

3. Electromagnetism coupled to other fields can tell what the gauge group is, since part of coupling it to other fields is specifying how these fields transform under gauge transformations. There we have a choice between (infinitesimally) $\psi \mapsto \psi + \chi \psi$ and $\psi\mapsto \psi + \mathrm{i}\chi \psi$, which lead to finite transformations $\psi\mapsto \mathrm{e}^{\chi}\psi$ and $\psi\mapsto \mathrm{e}^{\mathrm{i}\chi}\psi$, respectively. The former corresponds to a gauge group $\mathbb{R}$, the latter to $\mathrm{U}(1)$. Again, none of this is classical or quantum.

The reason you likely think that the $\mathrm{U}(1)$ is a quantum feature is that it much more natural in quantum field theory than in classical field theory to have complex-valued fields, but in fact we can consider e.g. classical electromagnetism coupled to a classical complex scalar field and then we are likewise forced to specify the gauge group.

• If I understood it correct, you're saying in point 1 of your answer that a free “U(1)” gauge theory is associated with an unambiguous Lie algebra but no unambiguous Lie group. How are you sure that there is an unambiguous Lie algebra? How does the algebra look like? Thanks. @ACuriousMind – SRS Aug 26 '17 at 14:44
• @SRS $\chi$ is real-valued, so the Lie algebra $\mathfrak{u}(1)$ is simply $\mathbb{R}$ with the trivial Lie bracket. – ACuriousMind Aug 26 '17 at 14:45
• Well. By a Lie algebra, I understand the algebra of the generators of a Lie group. If there is no Lie group, what does it mean to have a Lie algebra? @ACuriousMind – SRS Aug 26 '17 at 14:54
• @SRS Come to think of it, I've even heard people talk about "noncompact $U(1)$ gauge symmetry", on this very site even, which does my head in, because clearly(well, maybe!!) they mean $(\mathbb{R},\,+)$- that's an awfully confusing way to talk about things. – WetSavannaAnimal Aug 26 '17 at 15:00
• @SRS A Lie algebra is simply a vector space with a Lie bracket. – ACuriousMind Aug 26 '17 at 15:14

Using compact notation of differential forms, the differential operator $D=d+A$ is a covariant operator with the gauge potential $1$-form $A$. The field $2$-form is $F~=~D\wedge D$ $=~(d~+~A)\wedge(d~+~A)$. This acting on a unit with $d\wedge d~=~0$ (boundary of boundary is 0) gives $$F~=~dA~+~A\wedge A,$$ The $U(1)$ group simply means that $A\wedge A~=~0$. We of course know that a $1$-form in an exterior product with itself is zero. However, if there are internal color or charge indices you can have a number of these. In $SU(2)$ there are $3$ gauge field and in $SU(3)$ there are $8$. With $U(1)$ there is only one gauge potential with no charge index.

The field $2$-form has components $F_{\mu\nu}$ and with some effort you can show that $$F_{0i}~=~\frac{\partial A_i}{\partial t}~-~\frac{\partial A_0}{\partial x^i}~=~E_i$$ $$F_{ij}~=~\frac{\partial A_i}{\partial x_j}~-~\frac{\partial A_j}{\partial x^i}~\rightarrow~B_i~=~-(\nabla\times A)_i.$$ These are standard electromagnetic calculations. This would all be very different if there were the nonlinear terms $A\wedge A$ included. So classical electromagnetism is abelian.

• But is the symmetry still $U(1)$, or something like $\mathbb{R}$? After all, there are no complex numbers around, all that changes is the potential. – Javier Aug 19 '17 at 12:19
• Sigh! that is really a non-issue. Abelian symmetry is $U(1)$ and the $\pm 1$ weights correspond to the $\pm e$ charges. Further, classical solutions can have $exp(-ikx + i\omega t)$ terms. – Lawrence B. Crowell Aug 19 '17 at 13:33
• I think @Javier's worried about the topology of the group: $U(1)$ is not quite all of Abelian symmetry: its universal cover is $(\mathbb{R}, +)$. Of course they are the same at the algebra level: I'm getting the feeling that both the OP and Javier really want to know how the compactness of the symmetry group arises and what role it plays. – WetSavannaAnimal Aug 20 '17 at 6:01
• @LawrenceB.Crowell The 'form notation' I'm not comfortable with. Can you be a little liberal and add an answer without using them (if possible)? – SRS Aug 26 '17 at 12:36
• @LawrenceB.Crowell a good overview, but it doesn't fully answer the question of $ℝ$ vs U(1), since $ℝ$ is also Abelian and $A ∧ A$ would still be 0 if the group were $ℝ$ rather than U(1). – alexchandel Jan 2 '19 at 2:46