# Integration by parts of covariant derivatives in QED

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 $$$$\int d^4x (\mathcal{D}^{\mu} \bar{\psi}) \sigma_{\mu \nu} (\mathcal{D}^{\nu} \psi) \to -\int d^{4}x \bar{\psi} \sigma_{\mu \nu} (\mathcal{D}^{\mu} \mathcal{D}^{\nu} \psi),$$$$ where $$$$\mathcal{D}^{\mu} = \partial^{\mu} + ie A^{\mu}$$$$ is the U(1) gauge covariant derivatives for QED. $$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$$ since my definition of covariant derivative defined above is $$$$\mathcal{D}^{\mu} = \partial^{\mu} + ie A^{\mu}.$$$$ I have used that since $$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!

• $\bar\psi$ 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. – AccidentalFourierTransform Jan 31 at 3:15
• Thanks a lot, I now understand! – ocf001497 Jan 31 at 3:22