I am following a course about gauge theories in QFT and I have some questions about the physical meaning of what we are doing.

This is what I understood:

When we write a Lagrangian $\mathcal{L}(\phi)$, we are looking for its symmetries. Its symmetries are the transformation we apply on the fields that let the Lagrangian unchanged.

It means we are acting with an operator $U$ on the field $\phi$ and we will have: $\mathcal{L}(\phi'=U \phi)=\mathcal{L}(\phi)$.

And the operators $U$ belongs to a group.

Symmetries are very important because according to Noether theorem we can find the current conserved by knowing the symmetries.

In gauge theories, we allow the transformation $U$ to act "differently" on each point of the space. Then we have $U(x)$ (x dependance of the group element).

Thus, in my class the teacher did the following:

He remarked that this quantity:

$$ \partial_{\mu} \phi $$ doesn't transform as:

$$ \partial_{\mu} \phi'=U(x) \partial_{\mu} \phi $$ (because of the $x$ dependance of $U$).

And then he said "we have a problem, let's introduce a covariant derivative $D_{\mu} \phi$ that will allow us to have:

$$D_{\mu} \phi'=U(x)D_{\mu} \phi $$

My questions are the following:

Why do we want to have this "good" law of transformation? I am not sure at all but this is what I understood and I would like to check.

  • First question: please tell me if I am right in this following paragraph

I think it is because we want to write the Lagrangian as invariant under gauge transformation. To do it we don't start from scratch: we start from a term that we know should be in the Lagrangian: $\partial_\mu \phi^{\dagger} \partial^{\mu} \phi$. We see that this term is not gauge invariant, so we try to modify it by "changing" the derivatives: $\partial_\mu \rightarrow D_\mu$. We see that if we have $D_{\mu} \phi'=U(x)D_{\mu} \phi$ we will have the good law of transformation. And finally, after some calculation we find the "good" $D_\mu$ that respect $D_{\mu} \phi'=U(x)D_{\mu} \phi $.

So: Am I right in my explanation?


Why do we want a Lagrangian invariant under gauge transformations? Is there a reason behind it or it is just a postulate? I could understand that we want Lagrangian invariant under global transformation (if we assume the universe isotropic and homogenous it makes sense), but for me asking a local invariance is quite abstract. What is the motivation behind all this?

I know that if we have lagrangian invariant under all local symmetries then it will be invariant under global symmetries, but this "all" is "problematic" for me.

  • Next question in the following two lines:

Why should the lagrangian be invariant under all local symmetries? It is a very strong assumption from my perspective.

I would like a physical answer rather than too mathematical one.


3 Answers 3


We do not start from the assumption that the Lagrangian "should" be invariant under gauge transformations. This assumption is often made because global symmetries are seen as more natural than local symmetries and so writers try to motivate gauge theory by "making the global symmetry local", but this is actually nonsense. Why would we want a local symmetry just because there's a global one? Do we have some fetish for symmetries so that we want to make the most symmetrical theory possible? One can derive gauge theory this way but as a physical motivation, this is a red herring.

The actual point is not that we "want" gauge symmetry, but that it is forced upon us when we want to describe massless vector bosons in quantum field theory. As I also allude to in this answer of mine, every massless vector boson is necessarily described by a gauge field. A Lagrangian gauge theory is equivalently a Hamiltonian constrained theory - either way, the number of independent degrees of freedom that are physically meaningful is less that the naive count, since we identify physical states related by gauge transformations.

The true physical motivation for gauge theories is not "we want local symmetries because symmetries are neat". It's "we want to describe a world with photons in it and that can only covariantly be done with a gauge theory".

A non-quantum motivation of gauge theory can also be given: If you write down the Lagrangian of free electromagnetism, motivated because its equations of motion are the Maxwell equations, not because we like gauge symmetry, then you find it comes naturally with a $\mathrm{U}(1)$ gauge symmetry, corresponding to the well-known fact that adding a gradient to the vector 4-potential is physically irrelevant. Now, if you want to couple other fields to this free electromagnetism, you need to make the additional terms gauge invariant, too, else the theory is no longer "electromagnetism coupled to something else" in any meaningful sense since suddenly adding gradients can change the physics. Once again, gauge symmetry is something one discovers after physically motivating the Lagrangian from something else, not some sort of a priori assumption we put in.

  • 2
    $\begingroup$ So you're saying that the empirical motivation for local gauge invariance is the masslessness of the photon? I heard that that local gauge invariance is the reason we know the photon is exactly massless (and not just very, very light), but it sounds like you're saying that explanation is backwards. Do I understand you correctly? $\endgroup$
    – JohnK
    Mar 17, 2019 at 19:13
  • 1
    $\begingroup$ very well written, particularly the last example $\endgroup$
    – Gravity_CK
    Jul 21, 2021 at 21:48

One of the underlying ideas in gauge invariance is that the physics exist independently of the coordinate parameterization (and thus gauge) we apply to the system.

For unit-preserving changes in coordinates, the Lagrangian (as a measure of the "energy" in the system") should thus have a value independent of which coordinates we choose.

The motivation for covariant derivatives, etc., is then to account for disagreements between coordinate systems (and the underlying physical manifold) of what makes a "straight line" or a "right angle".


The free Lagrangian is invariant under global gauge symmetries. For this we don't need to introduce covariant derivatives or gauge fields.

We then artificially impose the condition that even local gauge changes are symmetries. But this is just a mathematical trick. A redundancy in our description. So the Lagrangian and equations of motion should not depend on our choice of gauge, because a local gauge change is not "physically real". Hence we use the covariant derivative to make sure the Lagrangian remains gauge independent.

For example in the case of U(1). Suppose we make a local change of phase of the fermion field. But this change is not real. So we must have a corresponding term that also changes in such a way that it cancels the affect of this artificial local phase change. This is done by the gauge field $A_{\mu}(x)$.

If you want a good intuitive understanding of Gauge theory I would strongly suggest the book "An elementary primer for gauge theory" by K. Moriyasu


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