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I'm having issues deriving the Dirac adjoint equation, $$\overline{\psi}(i\gamma^{\mu}\partial_{\mu}+m)=0.\tag{1}$$ I started by taking the Hermitian adjoint of all components of the original Dirac equation, giving me $$\psi^{\dagger}(-i\gamma^{\mu\dagger}\partial_{\mu}^{\dagger}-m)=0.\tag{2}$$ The adjoint of the gamma matrices is defined to be $\gamma^{\mu\dagger}=\gamma^0\gamma^{\mu}\gamma^0$, so no issues there. Now intuitively, I would think that the adjoint of the 4-gradient would be $\partial_{\mu}^{\dagger}=-\partial_{\mu}$. In non-relativistic quantum mechanics, it can be shown that first derivative operators are anti-Hermitian, so for example, $\frac{d}{dx}^{\dagger}=-\frac{d}{dx}$. So I would think this would the same case for the 4-gradient, but apparently it isn't. Among the many derivations I've gone over, for example, on page 77 here, it claims that the 4-gradient is self-adjoint. Could someone please explain why my intuition is incorrect?

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    $\begingroup$ Depends on the space. If you're taking adjoints in $L^2$, then yes, it's anti-Hermitian. But here you're taking adjoints in spinor space (that is, transposing and conjugating spinors and matrices), so the derivative is unaffected. $\endgroup$ – Javier Aug 13 '16 at 23:02
  • $\begingroup$ @Javier Could you explain why exactly the derivative is unaffected? I'm not very familiar with spinor space. Thanks. $\endgroup$ – connorp Aug 14 '16 at 16:42
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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.

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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 $\phi$.

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

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