# Why define $D_\mu = \partial_\mu -ieA_\mu$ with the electric charge $e$?

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?

• 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! Aug 23 '19 at 15:23

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.

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.

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