Transformation of Yang Mills Field Strength

Question

I am confused about the expression $$F_{\mu \nu} \to F_{\mu \nu}' = U F_{\mu \nu}U^{\dagger}.$$ I found related Phys.SE posts How would one show that a nonabelian field strength tensor transforms in a certain way under a local gauge transformation? and Gauge-covariance of the Yang-Mills field strength $F_{\mu\nu}^a$, but I would appreciate some clarification why that analysis makes sense.

To specify notation: Let $T^a$ be the generators of the gauge group. We let $U = \exp{i\theta^a T^a}$, and define $$D_{\mu}\psi = \partial_{\mu}\psi - ig A_{\mu}\psi.$$ By demanding that $$D_{\mu}\psi \to D_{\mu}'\psi' = U D_{\mu}\psi,$$ we can deduce that $$A_{\mu} \to A_{\mu}' = U A_{\mu}U^{\dagger} - \frac{i}{g}(\partial_{\mu}U)U^{\dagger}.$$

However, I don't know exactly what $U F_{\mu \nu}U^{\dagger}$ means. Since the $D_{\mu}$ is a differential operator, shouldn't it be defined by how it acts on something? If not, is the expression not ambiguous? How far to the right do operators act? The linked answer acts as if $D_{\mu}$ is a matrix.

Answer 1

I’m not quite sure if I fully understand your question, but I'll try to answer as much as possible.

It is true that $D_\mu$ should always act on some fields, but we should also note that this derivative is really flexible because it depends on the object it acts. If it acts on a field with non-abelian gauge symmetry, then it should be $\partial_\mu - igA_\mu$. If it acts on an operator $F$ living in the $SU(N)$ space and transforming as $UFU^{-1}$ and the covariant derivative should be $\partial_\mu - ig[A_\mu, F]$. And if it acts on a gauge scalar that is invariant under gauge transformation, the covariant derivative is just the ordinary derivative.

However, it should be noted that $A_\mu$ itself is not a covariant operator, since it doesn't transform as $UA_\mu U^{-1}$, but with an extra term.

As we have seen, the concept of covariant itself also depends on the objects. Covariant fields and covariant operators transform differently as you have mentioned. From this perspective, $F_{\mu\nu}$ is just an operator, and that's why we need to take the trace of it when we construct a pure gauge field action. The reason why operators transform differently is because if we want $F\Phi$ be covariant, then we require $F'\Phi'=UF\Phi=UFU^{-1}U\Phi$ which shows that if $\Phi$ transform as $U\Phi$, then $F$ must transform as $UFU^{-1}$ so that the net result transform again as a covariant field.

Now, let me say something about $F_{\mu\nu}$. By definition, it is $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu -g[A_\mu, A_\nu]$. Notice that this is not equal to $D_\mu A_\nu - D_\nu A_\mu$ because $A_\mu$ is neither a covariant field nor a covariant operator. You cannot replace $D_\mu$ by $\partial_\mu - igA_\mu$ because it makes no sense to define covariant derivative for an object that is not covariant at all. And the transformation rule of $F_{\mu\nu}$ comes solely from the transformation rule of $A_\mu$. There's nothing related to $D_\mu$ here. And to treat it as an independent gauge field, which is obviously different from field $\Phi$ seeing from perspective of the gauge transformation rules. And you can make action terms from gauge fields alone without acting on any $\Phi$, as long as the end result is a gauge scalar (for example, by taking a trace).

The core concept for gauge theory lies not on the redefinition of the derivative, but on how to preserve a gauge symmetry by combining two types of fields, one for what we normally perceived as real particles and one for interaction carriers. Hence, all other definitions should serve to this goal. For example, if we hope that $D_\mu (F\Phi) = (D_\mu F)\Phi + F (D_\mu\phi)$ holds, then we need $(D_\mu F)'=U(D_\mu F)U^{-1}$ so that the first term on the RHS transform as $U(D_\mu F) U^{-1}U\Phi= U(D_\mu F)\Phi$. Together with the second term which transform as $UFU^{-1}U(D_\mu\phi)=UF(D_\mu\phi)$, we get the whole thing on the LHS transform as a covariant field. Here, we ask that the derivative of a product of objects obeys the normal rule for ordinary derivative which seems not related to preserving gauge symmetry. But keep in mind that a Lagrangian that is gauge invariant will always be a gauge scalar which can be easily constructed by contract two gauge covariant fields (sandwiching some covariant operators between them is also allowed). That's why we care so much about covariant objects, because they are the lego pieces we need.