quantum field theory - Higher-order gauge coupling terms in the Lagrangian

In QFT, one works with Lagrangians that are invariant with respect to a certain symmetry. Out of this invariance, one is able to write down interaction terms at first order in the gauge couplings. The invariance we are talking about is though only guaranteed up to the first order, in the expansion of the transformation operator.

My question is: What if we design Lagrangians invariant under higher orders, say the second (assuming this possible)? Would this lead to the correct form of the higher order interaction processes?

Answer

It's Lie theory.

When we have got a continuous symmetry, the means the symmetry group is a Lie group $G$. Saying that the Lagrangian $\mathcal{L}$ is invariant under $G$ means that $\rho(g)(\mathcal{L}) = \mathcal{L}\;\forall\; g \in G$, where $\rho$ is the representation the Lagrangian transforms in.

The "expansion to first order" physicists so often do is just switching from the action of $g$ to the action of its associated element in the Lie algebra. Every (well, not always every, but that's a technicality) element of the group is generated by the basis vectors $T^a$ of the Lie algebra by

$$ g = \mathrm{exp}(\phi^a T^a)$$

with one $\phi^a$ as coefficient for every $T^a$. This is, abstractly, not the exponential function we are used to, but for matrix groups, it turns out to be. The physicists approach is now to expand the exponential naively as

$$ \mathrm{e}^{\phi^a T^a} = 1 + \phi^a T^a + \mathcal{O}(\phi^2)$$

and henceforth consider the transformation under the infinitesimal symmetry $1 + \phi^a T^a$ instead of the group element. Every representation $\rho$ of the Lie group induces as representation $\mathrm{d}\rho$ of the Lie algebra, so we examine $(1 + \phi^a \mathrm{d}\rho(T^a))v$ for a vector $v$ whose transformation we are interested in. Invariance obviously means then that $\mathrm{d}\rho(T^a)v = 0$ for all $T^a$, since it must hold for arbitrary $\phi$. It follows that also the higher orders of the expansion would simply vanish when applied to $v$, you gain no new insights considering them, and you get $\rho(\mathrm{e}^{\phi^a T^a})v = v$ iff $\mathrm{d}\rho(T^a) v = 0$ from the first order alone.

More generally, the "higher order" terms in the expansion will not be elements of the Lie algebra, but of its universal enveloping algebra, which has not really something to do with our symmetry. The Lie algebra encodes all you need to know about infinitesimal transformations, and the Lie group encodes all you need to know about "finite" transformations.

It is very important to note that we are not doing an approximation when considering the Lie algebra - the connection between Lie algebra and Lie group is rigorous and does not need the power series of the exponential to be mathematically meaningful. The physicist's way of saying "expand to first order" is just a way to avoid all that pesky group theory while still getting the right results.