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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.