Let's consider a complex $\phi$ coupling minimally to $U(1)$ gauge field: $$ \mathcal{L} = - \frac{1}{4} F_{\mu\nu} F^{\mu\nu} + (D_\mu\phi)^*(D^\mu\phi) - m^2 \vert\phi\vert^2 + \dots $$
For now, I want to look at the 1-loop correction to the scalar propagator due to the gauge sector only. There are two Feynman diagrams:
Before we proceed, we need to fix a gauge, and the gauge field propagator is quite different in different gauges: $$ -i \left(\frac{g_{\mu\nu} - (1-\xi) k_\mu k_\nu/k^2}{k^2}\right), $$ where the value of $\xi$ corresponds to a particular gauge choice.
Question 1:
Diagram (1) by right is quadratically divergent. In either gauge, the loop integral is proportional to $$ \int \frac{d^dk}{k^2} $$ but the coefficient is obviously gauge-dependent: $$ (g_{\mu\nu} - (1-\xi) k_\mu k_\nu/k^2) \, g^{\mu\nu} = (d-1) + \xi $$ appearing on the numerator.
Now I know the particle physics folks will just define the integral to be zero using dimensional regularization, and tell me there's no problem. Am I right in saying that, if we believe that the UV theory is still gauge-invariant, then by some unknown black magic it should go away by itself and dropping it is supposedly the right thing to do here?
Question 2:
Now look at diagram (2). If I write down the loop integrals in Landau gauge ($\xi = 0$) and Feynman gauge ($\xi = 1$), and take the difference, I get something like:
$$ \int d^dk \frac{(k^2 - 2 \, p \cdot k)^2}{k^4[(k-p)^2 - m^2]}. $$
This is quite ugly and I have not worked through it. But the point is that the integral
- is obviously non-zero
- is quadratically diverging. Ok let's say that goes away by itself...
- it still contains a logarithmic divergence
- and the finite part has to depend on $p$ and $m$, and cannot be subtracted off cleanly in any renormalization scheme
So what happens to gauge invariance here?
