Gauge covariant derivative of an adjoint action: $\psi(x) \to g \psi(x) g^{-1}$, instead of a left action $\psi(x)\to e^{iq\theta(x)} \psi(x)$

Question

In the case where the transformation on $\psi$ is applied from the left:

$$ \psi(x)\to e^{-iq\theta(x)}\psi(x). $$

The gauge covariant derivative is

$$ D_\mu = \partial_\mu - iqA_\mu \tag{1} $$

and the field is given as follows:

$$ F_{\mu\nu}=[D_\mu,D_\nu]. \tag{2} $$

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My question is; what are the equivalents to equation (1) and (2) if we have an adjoint action such as this

$$ \psi(x) \to g(x)\psi(x)g^{-1}(x) $$

where $g(x)$ could be arbitrary general linear transformations for instance. Does the use of a adjoint action transformation significantly changes (1) and (2)?

Answer 1

First, let me clear a possible confusion: as you've tagged the question as "electromagnetism", I assume you're going for an $U(1)$ symmetry. In this case, the adjoint does not transform, as the group is abelian (you can see it because if $\psi(x)$ is in an irreducible representation, then it's just a complex function, not a vector: $g$ and $\psi$ commute, so the transformation is the identity). So your question is trivial for electromagnetism: the covariant derivative of a field $\psi(x)$ in the adjoint representation is just the standard derivative.

But let's see the answer for a non abelian group anyway. A field $\psi(x)$ in the adjoint representation will be a linear combination of the generators $T^a$ of the adjoint representation (given by the structure constants). How do you get the covariant derivative of such a thing? Well, from the transformation itself!

As I'm following a book (Michele Maggiore's "A modern introduction to quantum field theory), forgive me for changing conventions. The stuff that I'm referring about is in section 10.4. I strongly suggest the book, it's very good.

What's the general transformation for a field in any representation, spanned by the generators $T^a$ of the adjoint representation? A general field in the representation can be written in components as $$ \psi(x)=\psi^a(x) T^a. $$ Component fields $\psi^a$ have by definition the following derivative: $$ (D_\mu\psi)^a(x)=\partial_\mu\psi^a(x)-\mathrm i g (A_\mu^c(x)T^c)^{ab}\psi^b(x). $$ The term on the right is basically a multiplication of the connection matrix $A_\mu^c(x)T^c$. I'm here using the fact that the adjoint representation has the same dimension as the dimension of the Lie algebra.I will keep all indexes up but for the spacetime index $\mu$.

Now, contract what you got with the generators, and let's keep all indexes out: you get $$ (D_\mu\psi)^a(T^a)^{bc}=(\partial_\mu \psi^a)(T^a)^{bc}-\mathrm i g A_\mu^d(T^d)^{ae}(T^a)^{bc}\psi^e. $$ Now, we use the fact that the generators are the structure constants, $(T^a)^{bc}=-\mathrm i f^{abc}$. You can rewrite the product of $T$'s as $$ (T^d)^{ae}(T^a)^{bc}=-f^{dae}f^{abc}=f^{dac}f^{eba}-f^{eac}f^{dba}=(T^e)^{ac}(T^d)^{ba}-(T^d)^{ac}(T^e)^{ba}=[T^d,T^e]^{cb}. $$ Here I've used antisymmetry and Jacobi identity on the structure constants. Please check my signs!

Let's return to our expansion. $$ (D_\mu\psi)^a(T^a)^{bc}=\partial_\mu\psi^a(T^a)^{bc}-\mathrm i g A_\mu^d[T^d,T^e]^{bc}\psi^e. $$ In matrix representation, we get $$ D_\mu\psi=\partial_\mu\psi-\mathrm i g[A_\mu,\psi]. $$ This is the generalization of (1) to fields in the adjoint representation. The definition of the field strength tensor remains unvaried.