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If $D_\mu = \partial_\mu - ieA_\mu$ then the QED Lagrangian is invariant under $$A_\mu \to A_\mu + \frac{1}{e}\partial_\mu\alpha(x)$$ $$\psi \to e^{i\alpha(x)}\psi$$ However if $D_\mu = \partial_\mu -iA_\mu$, the transformation needed for $A_\mu$ is simpler: $$A_\mu \to A_\mu + \partial_\mu\alpha(x)$$

The lagrangian is still left invariant by this transformation.

What is the reason to add the electric charge to the definition of the gauge covariant derivative?

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  • $\begingroup$ Please do not leave answers in the comments -- particularly if you think the question isn't on-topic. Comments are for clarifying the question and not for discussion/answers. Thanks! $\endgroup$ – tpg2114 Aug 23 at 15:23
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The answers in the comments are gone and so in case someone else has the same question I will type it here but not accept the answer as it is mostly not my own.

Benefits of defining $D_\mu = \partial_\mu - i e A_\mu$ include:

  • The coupling constant between electron and photon $e$ is written explicitly
  • The conserved current becomes $j^\mu = e \bar{\psi}\gamma^\mu\psi$, which gives an electric charge operator proportional to $e * (Number\,\, Operator)$

The conventions with and without an explicit $e$ are both used: Without the $e$ in more abstract settings which do not require explicit calculation, and with the $e$ in (for example) phenomenology.

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In your QED Lagrangian you have $$ \mathcal{L} = \bar{\psi} ( i \gamma^{\mu} D_{\mu} - m ) \psi - \tfrac{1}{4} F_{\mu\nu} F^{\mu\nu} $$ where $F_{\mu\nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$. The field $A_{\mu}$ are your photons, and the field $\psi$ are your electrons/positrons.

$\mathcal{L}$ tells you the rules for calculating things in QED. Focus on the following term in $\mathcal{L}$ $$ \bar{\psi} i \gamma^{\mu} D_{\mu} \psi = \bar{\psi} i \gamma^{\mu} ( \partial_{\mu} - i e A_{\mu})\psi = \bar{\psi} i \gamma^{\mu} \partial_{\mu} \psi + e\; \bar{\psi} \gamma^{\mu} A_{\mu} \psi $$ You see that there is now a combination of fields $e \bar{\psi} \gamma^{\mu} A_{\mu} \psi $ appearing. This describes how the photons and electrons interact with each other $\to$ so a way of thinking about why $e$ is there, is that it is a measure of how strong these particles interact with each other.

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  • $\begingroup$ I think you misinterpreted my question. I might rephrase it as, "why not define the covariant derivative without the electric charge?" $\endgroup$ – doublefelix Aug 22 at 19:37
  • $\begingroup$ I see, I think @AccidentalFourierTransform has the answer you want then. $\endgroup$ – Greg.Paul Aug 22 at 19:40

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