Interpretation of the adjoint Dirac equation

Adjoint spinors $\bar{\psi}$ satisfy the adjoint Dirac equation $$\bar{\psi} \left( \gamma^{\mu} \, p_{\mu} - mc \right) = 0 \;$$ (I'm using the terminology and units of Griffiths' "Intro. to Elementary Particles"). What puzzles me here is that $p_{\mu}$ is presumably a (four-) vector operator (as indicated below), and by normal convention one expects the operand (i.e. that which it the operator operates upon) to stand on its right side... but here there is nothing on its right side to operate upon. Is it implicit, then, that here $p_{\mu}$ "acts" to its left?

Secondarily, and under the convention $x^{\mu}=(ct, \mathbf{x})$ with diagonal components of the metric tensor being $$(g_{00}, g_{11}, g_{22}, g_{33}) = (1,-1,-1,-1) \; ,$$ is it correct that since $$p_{\mu}=i \hbar \partial_{\mu} = i \hbar (\frac{\partial}{\partial (ct)}, \nabla)$$ one must have $$p^{\mu}=i \hbar (\frac{\partial}{\partial (ct)}, - \nabla) \; ?$$ (Note: in my original post I had unintentionally reversed these two definitions, for the co- and contravariant cases.)

Perhaps the resolution of my question is that $p_{\mu}$ is here to be interpreted here not as an operator, but as $p_{\mu}=\hbar k_{\mu}$, where $k_{\mu}$ is the wavenumber vector?

• It really depends on whether you are working in position or momentum space. In momentum space, your first equation is just a matrix equation, and there is no need to talk about how $p$ acts. In position space one often adds an arrow above $p$ that indicates that $p$ acts to the left. – user178876 Sep 18 '18 at 17:34