Symmetry of QED interaction Lagrangian

[joneall]

TL;DR

Is it all U(1)?

I am trying to get a foothold on QFT using several books (Lancaster & Blundell, Klauber, Schwichtenberg, Jeevanjee), but sometimes have trouble seeing the forest for all the trees. My problem concerns the equation of QED in the form
$$
\mathcal{L}_{Dirac+Proca+int} =
\bar{\Psi} ( i \gamma_{\mu} \partial^{\mu} - m ) \Psi
+ g A_{\mu} \bar{\psi} \gamma^{\mu}\psi
-\frac{1}{2} (
\partial^{\mu} A^{\nu} \partial_{\mu} A_{\nu} -
\partial^{\mu} A^{\nu} \partial_{\nu} A_{\mu}
)
$$
consisting of a Klein-Gordon scalar term, an interaction term and a massless vector. The interaction term may be found by insisiting that the scalar term be invariant under a local U(1) transformation $e^{i\alpha (x)}$ . One then notes that The EM vector potential A must have internal symmetry $$
A_{\mu} \rightarrow A_{\mu}^{\prime} =
A_{\mu} + \partial_{\mu}a(x)
$$,
which can be done explicitly or by including it in a so-called covariant derivative.

So I have two questions:
(1) Is not the transform of A a separate requirement from U(1) symmetry for the KG Lagrangian? It has nothing to do with U(1)?
(2) Is there any correspondence between this covariant derivative and the one in GR with Christoffel symbols?

My goal is to understand what this is all about, not just to be able to derive the equations.

Thanks in advance for any help. Please note that my level of expertise in this domain is as an advanced beginner.

 

[Gaussian97]

Well, first of all, you are writing the QED Lagrangian, not the sQED, so there's no Klein-Gordon field here, $\Psi=\psi$ is a Dirac field, also in the Dirac term you forgot to put the bar in the first field.

Now I don't really understand what's your problem with (1), if you do a gauge transformation you will see that you need the $A$ transformation to have you Lagrangian invariant, and by the way, the functions $\alpha(x)$ and $a(x)$ that you have must be related by $a(x)=-g\alpha(x)$.

For the second question, well, for sure there's an analogy, because in both cases you define a transformation that depends on the space-time point, therefore the derivative, which compares the value of the function for two different points ($f(x+\varepsilon)-f(x)$) will not transform nicely (because one will transform with the transformation law of the point $x$ and the other with the transformation law of $x+\varepsilon$).
You can solve this by defining some kind of parallel transport to move the two functions to the same point, so when you compare the difference and make a transformation, the transformation will factor out without problems.
To define such parallel transport you need to introduce connections, in GR these connections are the Christoffel symbols, in QED this connection is the $A$ field.

 

[joneall]

Sorry, I meant Dirac, not K-G. As for the bar, I think it is there, it's just hard to see on the capital $\Psi$. At least, I meant to put it there. And, yes, I should have used the same $\Psi$ everywhere.

I guess my first question might have been posed differently. If transformation by $e^{i\alpha(x)}$ is a member of a U(1) Lie group, to what group does $A_{\mu} \rightarrow A_{\mu}^{\prime} =A_{\mu} + \partial_{\mu}a(x)$ belong? (Can the added derivative be part of a Lie generator?) If it's not U(1), then the QED equation is not just due to that particular symmetry.

 

[Gaussian97]

Well, I'm not a mathematician, but from the introductory course in group theory that I took, let me define a map $T_\alpha$ such that:
$$T_\alpha (A^\mu) = A^\mu(x) - \partial^\mu \alpha(x)$$
then the following map, $\Phi$:
$$e^{i\alpha(x)} \to T_\alpha$$
fulfills that for the identity (i.e. $\alpha(x)=0$) we have
$$T_0(A^\mu) = A^\mu - \partial^\mu 0 = A^\mu$$
so $T_0$ is the identity, so $\Phi$ maps the identity to the identity, furthermore because
$$T_{\alpha+\beta}(A^\mu)=A^\mu - \partial^\mu (\alpha(x)+\beta(x)) = (A^\mu - \partial^\mu \alpha(x)) - \partial^\mu \beta(x) = T_{\beta}(T_{\alpha} (A^\mu))=(T_\beta \circ T_\alpha)(A^\mu)$$
$$\Phi(e^{i\alpha(x)}e^{i\beta(x)})=\Phi(e^{i(\alpha(x)+\beta(x))})=T_{\alpha+\beta} = T_\alpha \circ T_\beta = \Phi(e^{i\alpha(x)})\circ\Phi(e^{i\beta(x)})$$
Then it is a representation of $U(1)$.
So simply, a Dirac field transform under the $U(1)$ group by the fundamental representation $$\psi'=e^{i\alpha}\psi$$
while a Proca-Maxwell field will transform under the representation
$$A'_\mu=T_{\alpha}(A_\mu)$$

Again, I'm not a mathematician so there may be non-rigorous steps or maybe other errors.

 

[joneall]

Your equations look good to me. (I envy you. When I did grad work in <1965, there was no course offered in group theory, which is why I'm struggling some now.) Interesting that I have never seen in any of these books what you just went thru. So it all is U(1), it just looks different in the two representations, Dirac and Proca.

Thanks a lot.

 

[joneall]

Uh, isn't there a way to mark a thread "Answered" or "Resolved"?