Hermitian adjoint of 4-gradient in Dirac equation

Expanding on my comment.

The basic idea is that what you mean by adjoint depends on the vector space being considered. For example, we might have $\mathbb{C}^n$ as our space, with the usual inner product; in that case, the adjoint of a vector or matrix is the transpose conjugate. Note that technically, taking adjoint of a vector doesn't return a vector, because row vectors and column vectors belong to different spaces.

We could also use $L^2(\mathbb{R}^n)$ as our vector space. Its elements are functions, and the inner product is defined by

$$(f,g) = \int d^n x\ f^* g$$

You can take adjoints here too, using the inner product defined above. It can be shown that the derivative operator, which is a linear transformation on $L^2$, is anti-Hermitian.

Now to the Dirac equation. The vector space (that is, spinor space) being considered here is $\mathbb{C}^4$, not $L^2$. That is, $\psi$ is a vector because it has four components, not because it's a function. The fact that its components are functions is irrelevant here. When we take adjoints, we transpose and conjugate vectors and matrices. The derivative is an operator if you think about what it does to functions, but it is not a $4\times4$ matrix; it does nothing to spinors. Therefore, the particular adjoint we're doing here doesn't affect it.

Perhaps the following argument is more convincing:

  1. The Dirac equation$^1$ $$ (i\gamma^{\mu}\stackrel{\rightarrow}{\partial}_{\mu}-m)\psi~=~0 \tag{A}$$ is by the fundamental lemma of variational calculus equivalent to $$ \forall \phi:\quad 0~=~\int d^4x~\bar{\phi}(i\gamma^{\mu}\stackrel{\rightarrow}{\partial}_{\mu}-m)\psi, \tag{B}$$ where $\phi$ is an arbitrary (off-shell) Dirac spinor.

  2. Hermitian conjugation in Dirac spinor space leads to $$ \forall \phi:\quad0~=~\int d^4x~\bar{\psi}(-i\stackrel{\leftarrow}{\partial}_{\!\mu}~\gamma^{\mu}-m)\phi, \tag{C}$$ which is equivalent to $$ \bar{\psi}(i\stackrel{\leftarrow}{\partial}_{\!\mu}~\gamma^{\mu}+m)~=~0,\tag{D}$$ cf. the above comment by Javier.

  3. On the other hand, if we also integrate (C) by parts, we get (after discarding boundary terms) $$ \forall \phi:\quad 0~=~\int d^4x~\bar{\psi}(i\gamma^{\mu}\stackrel{\rightarrow}{\partial}_{\mu}-m)\phi, \tag{E}$$ where the derivative now acts on $\phi$.


$^1$ We use the following conventions: $$ \bar{\psi} ~=~ \psi^{\dagger}\gamma^0 , \qquad\gamma^{\mu\dagger}~=~\gamma^0\gamma^{\mu}\gamma^0, \qquad (\gamma^0)^2~=~{\bf 1}_{4\times 4}. \tag{F}$$