I am confused about the expression $$F_{\mu \nu} \to F_{\mu \nu}' = U F_{\mu \nu}U^{\dagger}.$$ I found this answer on StackExchange, 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.

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.

I am confused about the expression $F_{\mu \nu} \to F_{\mu \nu}' = U F_{\mu \nu}U^{\dagger}$. I found this answer on StackExchange, 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.

I am confused about the expression $$F_{\mu \nu} \to F_{\mu \nu}' = U F_{\mu \nu}U^{\dagger}.$$ I found this answer on StackExchange, 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.