I am reading Sidney Coleman's QFT ch. 27 (in particular Eq. (27.73)) where he said that we can use integration by parts to write the term in the action \begin{equation} \int d^4x (\mathcal{D}^{\mu} \bar{\psi}) \sigma_{\mu \nu} (\mathcal{D}^{\nu} \psi) \to -\int d^{4}x \bar{\psi} \sigma_{\mu \nu} (\mathcal{D}^{\mu} \mathcal{D}^{\nu} \psi), \end{equation} where \begin{equation} \mathcal{D}^{\mu} = \partial^{\mu} + ie A^{\mu} \end{equation} is the U(1) gauge covariant derivatives for QED. $A^{\mu}$ is the electromagnetic gauge fields. $\bar{\psi}$ and $\psi$ are Dirac 4-component spinors. Notice that $(\mathcal{D}^{\mu} \bar{\psi})$ means that $\mathcal{D}^{\mu} $ only acts on $\bar{\psi}$ and $(\mathcal{D}^{\nu} \psi)$ means that $\mathcal{D}^{\nu} $ only acts on $\psi$. I try to prove that we can do integration by part using straightforward calculation in follows: \begin{align} (\mathcal{D}^{\mu} \bar{\psi}) \sigma_{\mu \nu} (\mathcal{D}^{\nu} \psi) & = (\partial^{\mu}\bar{\psi} + ieA^{\mu}\bar{\psi}) \sigma_{\mu \nu} (\mathcal{D}^{\nu} \psi) \\ & = (\partial^{\mu} \bar{\psi}) \sigma_{\mu \nu} (\mathcal{D}^{\nu} \psi) + ie A^{\mu} \bar{\psi} \sigma_{\mu \nu} (\mathcal{D}^{\nu} \psi) \\ &\to - \bar{\psi} \sigma_{\mu \nu} \partial^{\mu} (\mathcal{D}^{\nu} \psi) + \bar{\psi} \sigma_{\mu \nu} (ie A^{\mu} ) \mathcal{D}^{\nu} \psi \\ & = -\bar{\psi} \sigma_{\mu \nu} (\partial^{\mu} - ieA^{\mu}) (\mathcal{D}^{\nu} \psi). \end{align} But this does not equal to $-\bar{\psi} \sigma_{\mu \nu} \mathcal{D}^{\mu}\mathcal{D}^{\nu} \psi$ since my definition of covariant derivative defined above is \begin{equation} \mathcal{D}^{\mu} = \partial^{\mu} + ie A^{\mu}. \end{equation} I have used that since $A^{\mu}$ is just a number field so that I can move it to any place I want. I also only use the integration by parts for moving the $\partial^{\mu}$ acting on $\bar{\psi}$ to act on the entire $\mathcal{D}^{\nu} \psi$ and then add a minus sign. I would like to know where does my calculation went wrong. This is just the simplest U(1) gauge field case but I just can't sort out how to do integration by part with U(1) gauge covariant derivative, though I know that we should be able to do so even in the non-Abelian gauge theory. Thanks in advance!

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    $\begingroup$ $\bar\psi$ carries the opposite representation as $\psi$, so $\mathcal D$ doesn't act on it the same way it acts on $\psi$. There is a crucial minus sign. $\endgroup$ – AccidentalFourierTransform Jan 31 at 3:15
  • $\begingroup$ Thanks a lot, I now understand! $\endgroup$ – ocf001497 Jan 31 at 3:22

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