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I wonder what the QED Lagrangian would look like if you carefully write out all units of the terms and make sure they are consistent. I think there is something missing about Coulomb.
Can you write down the QED Lagrangian such that the physical units of all terms agree and state the units of all variables and terms?
The Wikipedia article you linked to gives the action, not the Lagrangian or Lagrangian density, so I’ll do the same.
Note that since $x^0 \equiv ct$, we have $d^4x = c \, dt \, d^3x$ and therefore
$$S \equiv \int dt \, L \equiv \int dt \int d^3x \, \mathscr L = \int \frac{d^4x}{c} \mathscr L$$
where $L$ is the Lagrangian and $\mathscr L$ is the Lagrangian density. Thus $\mathscr L$ is the quantity inside the square brackets below, and $L$ is the integral of this density over the spatial coordinates.
$S$ must have units of action, which is the same as energy times time. The Lagrangian $L$ therefore must have units of energy, and the Lagrangian density $\mathscr L$ must have units of energy density.
The units of electromagnetic quantities such as $e$, $\phi$, $\mathbf A$, $\mathbf E$, $\mathbf B$, $A^\mu$, and $F^{\mu\nu}$ are the same in QED as in classical electromagnetism. Both the classical theory and the quantum theory can be done in various unit systems. I discuss four of them below.
The Minkowski metric $\eta_{\mu\nu}$ is dimensionless in Cartesian coordinates, and few people do QED calculations in anything other than Cartesian coordinates. In any case, since $\mathscr L$ must be a Lorentz scalar, if $\mathscr L$ has the right dimensions in Cartesian coordinates, it will have the right dimensions in any coordinates.
$$S = \int\frac{d^4x}{c}\left[ -\frac{1}{4\mu_0}F^{\mu\nu}F_{\mu\nu} + \bar\psi ( i \hbar c \gamma^\mu D_\mu - m c^2) \psi \right]$$
$$D_\mu \equiv \partial_\mu + \frac{ie}{\hbar}A_\mu$$
$$x^\mu \equiv (ct, \mathbf x)$$
$$A^\mu \equiv \left( \frac1c \phi, \mathbf A \right)$$
Base units: kilograms, meters, seconds, amperes
Table of $\text{[quantity]} = \text{kg}^a \text{m}^b \text{s}^c \text{A}^d$
$$\begin{array}{c|c|c|c|} \text{quantity} & \text{kg} & \text{m} & \text{s} & \text{A} \\ \hline S & 1 & 2 & -1 & 0 \\ \hline d^4x & 0 & 4 & 0 & 0 \\ \hline c & 0 & 1 & -1 & 0 \\ \hline \mu_0 & 1 & 1 & -2 & -2 \\ \hline F^{\mu\nu} \text{ or } F_{\mu\nu} & 1 & 0 & -2 & -1 \\ \hline\ \psi \text{ or } \bar\psi & 0 & -3/2 & 0 & 0 \\ \hline \hbar & 1 & 2 & -1 & 0 \\ \hline \gamma^\mu \text{ or } \gamma_\mu & 0 & 0 & 0 & 0 \\ \hline m & 1 & 0 & 0 & 0 \\ \hline \partial_\mu \text{ or } \partial^\mu & 0 & -1 & 0 & 0 \\ \hline e & 0 & 0 & 1 & 1 \\ \hline A^\mu \text{ or } A_\mu & 1 & 1 & -2 & -1 \\ \hline g_{\mu\nu} \text{ or } g^{\mu\nu} & 0 & 0 & 0 & 0 \\ \hline \end{array}$$
$$S = \int\frac{d^4x}{c}\left[ -\frac{1}{16\pi}F^{\mu\nu}F_{\mu\nu} + \bar\psi ( i \hbar c \gamma^\mu D_\mu - m c^2) \psi \right]$$
$$D_\mu \equiv \partial_\mu + \frac{ie}{\hbar c}A_\mu$$
$$x^\mu \equiv (ct, \mathbf x)$$
$$A^\mu \equiv \left(\phi, \mathbf A \right)$$
Note: These are sometimes called "unrationalized" Gaussian units. The $F^2$ term has an extra factor of $1/4\pi$; this introduces factors of $4\pi$ into the field equations but removes them from solutions. For example, in these units the classical potential of a point charge is simply $q/r$.
In this system, $\mu_0$ and $\epsilon_0$ do not exist. This is because Gaussian units don't have a electromagnetic base unit; charge is defined in terms of the mechanical base units.
Base units: grams, centimeters, seconds
Table of $\text{[quantity]} = \text{g}^a \text{cm}^b \text{s}^c$
$$\begin{array}{c|c|c|c|} \text{quantity} & \text{g} & \text{cm} & \text{s} \\ \hline S & 1 & 2 & -1 \\ \hline d^4x & 0 & 4 & 0 \\ \hline c & 0 & 1 & -1 \\ \hline F^{\mu\nu} \text{ or } F_{\mu\nu} & 1/2 & -1/2 & -1 \\ \hline\ \psi \text{ or } \bar\psi & 0 & -3/2 & 0 \\ \hline \hbar & 1 & 2 & -1 \\ \hline \gamma^\mu \text{ or } \gamma_\mu & 0 & 0 & 0 \\ \hline m & 1 & 0 & 0 \\ \hline \partial_\mu \text{ or } \partial^\mu & 0 & -1 & 0 \\ \hline e & 1/2 & 3/2 & -1 \\ \hline A^\mu \text{ or } A_\mu & 1/2 & 1/2 & -1 \\ \hline g_{\mu\nu} \text{ or } g^{\mu\nu} & 0 & 0 & 0 \\ \hline \end{array}$$
$$S = \int\frac{d^4x}{c}\left[ -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + \bar\psi ( i \hbar c \gamma^\mu D_\mu - m c^2) \psi \right]$$
The rest is the same as for Gaussian units.
These are sometimes called "rationalized" Gaussian units. The $F^2$ term has a prefactor of $-1/4$ instead of $-1/16\pi$. This removes factors of $4\pi$ from the field equations.
$$S = \int d^4x \left[ -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + \bar\psi ( i \gamma^\mu D_\mu - m) \psi \right]$$
$$D_\mu \equiv \partial_\mu + ieA_\mu$$
$$x^\mu \equiv (t, \mathbf x)$$
$$A^\mu \equiv \left(\phi, \mathbf A \right)$$
Base units: GeV
Table of $\text{[quantity]} = \text{GeV}^a$
$$\begin{array}{c|c|} \text{quantity} & \text{GeV} \\ \hline c\equiv 1 & 0 \\ \hline \hbar\equiv 1 & 0 \\ \hline S & 0 \\ \hline d^4x & -4\\ \hline F^{\mu\nu} \text{ or } F_{\mu\nu} & 2 \\ \hline\ \psi \text{ or } \bar\psi & 3/2 \\ \hline \gamma^\mu \text{ or } \gamma_\mu & 0\\ \hline m & 1 \\ \hline \partial_\mu \text{ or } \partial^\mu & 1 \\ \hline e & 0 \\ \hline A^\mu \text{ or } A_\mu & 1 \\ \hline g_{\mu\nu} \text{ or } g^{\mu\nu} & 0\\ \hline \end{array}$$
