## Gauge Invariance

You can see the original post here.

I’m learning about the Dirac field, and I’ve been working my way through Ch. 3 of Peskin & Schroeder, which starts off by considering what kind of Lorentz invariant theory of spin 1/2 particles we can construct, eventually leading to the Dirac equation. I’ve reached the point where we are about to couple the complex Dirac field $\psi (x)$ to the vector potential $A_{\mu}$ of electromagnetism. In an effort to approach this from a different direction, I skipped to Ch 15.1 The Geometry of Gauge Invariance, and I’m glad I did.

When I first took quantum field theory (and even more so particle phenomenology) in graduate school, lots of emphasis was placed on the importance of symmetry in these theories. To be honest, I never really saw how important it was, I was more worried about working through some math to find solutions to my homework assignment. Now that I’m approaching this for the second time (with no homework due), I’m spending a lot more time trying to understand the concepts, in this case symmetry and gauge invariance.

Let’s start with the complex field $\psi (x)$, and demand that the dynamics of the field be invariant under the transformation

$\psi (x) \rightarrow e^{i\alpha (x)} \phi (x)$,

i.e., a position dependent phase rotation by the angle $\alpha$. Exactly why we are demanding this is unclear to me right now, but let’s just say that we don’t want our theory to be messed up by someone throwing a weird phase field at us, because who needs that. We next look at what kind of terms in the Lagrangian are invariant under this transformation. The mass term is easy. That is to say, terms of the form $m\bar{\psi}\psi$ are invariant, where $m$ turns out to be the mass of the particle. If we want the field to be dynamic, were going to need a kinetic term involving derivatives of the field. If we take the derivative of the field in some direction defined by the vector $n^{\mu}$, it’s defined by

$n^{\mu}\partial_{\mu} \psi (x) = \lim_{\eta \to 0}\frac{1}{\epsilon}[\psi(x+\epsilon n)-\psi (x)]$.

It is clear that this behaves differently than $\psi$ does under the above transformation, given the position dependence of $\alpha (x)$. Those familiar with General Relativity will recognize this problem, as it’s similar to the issue of the metric tensor $g_{\mu\nu}$ being a function of position. In any case, like in GR we need a derivative $D_{\mu}\psi(x)$ that will transform like $\psi (x)$. If we had a scaler function $U(y,x)$ that transformed like $U(y,x) \rightarrow e^{i\alpha(y)} U(y,x) e^{-i\alpha(x)}$, we would be able to define a covariant derivative like so

$n^{\mu}D_{\mu} \psi (x) \equiv \lim_{\epsilon \to 0} \frac{1}{\epsilon} [\psi (x+\epsilon n) - U(x+\epsilon n, x)\psi (x) ]$.

With this definition, $D_{\mu}\psi$ transforms in the same way as $\psi$, so terms like $\overline{D^{\mu}\psi}D_{\mu}\psi$ will be invariant, and we have our kinetic term for the Lagrangian.

Now we need the form of $U(y,x)$ and then $D_{\mu}$. Since we don’t want the physics of our field to get messed up by the derivative (and for other reasons I don’t quite know), $U(y,x)=exp(i\phi(y,x))$, and since we need it only for the derivative, we can expand it about the displacement $\epsilon n^{\mu}$, i.e.,

$U(x+\epsilon n, x) = 1-ie\epsilon n^{\mu} A_{\mu}(x) + \mathcal{O}(\epsilon^2 )$.

Here (as in P&S), we pulled out a constant $e$ from the vector field $A_{\mu}(x) = \frac{1}{e}\partial_{\mu}\phi(y,x)\vert_{y=x}$… Note that P&S don’t say that $A_{\mu}$ is equal to this, but it seems to me that it must given their definition of the form of $U(y,x)$ given above. Given this, the covariant derivative takes the form

$D_{\mu}\psi(x)=\partial_{\mu}\psi(x)+ieA_{\mu}\psi(x)$,

and the vector field will transform according to

$A_{\mu} \rightarrow A_{\mu}-\frac{1}{e}\partial_{\mu}\alpha(x)$

According to P&S, $U(y,x)$ is called the comparator, and fields like $A_{\mu}$, which arise as “the infinitesimal limit of a comparator of local symmetry transformations” are called connections…sounds familiar from parallel transport of a vector in GR.

Let me recap for a second. We started from a complex field $\psi(x)$, and after demanding that the Lagrangian (i.e., its dynamics) be invariant under a spatially varying phase rotation, we found out that we need the presence of a vector field $A_{\mu}$ in order for there to be a kinetic term. To complete the theory, we need a term in the Lagrangian that determines the dynamics of $A_{\mu}$, i.e., that includes only the vector field and not the Dirac field. We can find this by noticing that if the covariant derivative of $\psi$ transforms as the field itself, so does the second derivative. Since that is true, so must the commutator of the covariant derivative, i.e.,

$[D_{\mu},D_{\nu}]\psi \rightarrow e^{i\alpha(x)}[D_{\mu},D_{\nu}]\psi(x)$,

so $[D_{\mu},D_{\nu}]$ must be invariant… but this commutator is not a differential operator at all, it instead is equal to

$[D_{\mu},D_{\nu}]\psi=ie(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})\psi=ieF_{\mu\nu}\psi$.

So kinetic terms in the Lagrangian for the vector potential (we can call it that now) can be constructed from $F_{\mu\nu}$, which is the familiar tensor for the electromagnetic field! This is amazing, and I can see why it is said that the symmetry groups are what define the Standard Model. With this one symmetry, we got most of the form of the QED Lagrangian, and we will get all of it once we also impose Lorentz invariance (which is how they proceed in Ch. 3).