Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
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?
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 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.
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.
