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?
Also:
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.
Best Answer
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.