Using EOM in QED Lagrangian [duplicate]

Question

Let's have the QED Lagrangian.

$$\mathcal{L} = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} + \bar{\Psi}(i\partial_\mu \gamma^\mu - m)\Psi + g\bar{\Psi}A_\mu \gamma^\mu \Psi.\tag{1}$$

The equations of motion are:

$$ \partial_\mu F^{\mu\nu} = g\bar{\Psi}\gamma^\nu \Psi \tag{2}$$

and

$$(i\partial_\mu \gamma^\mu - m)\Psi = -g A_\mu \gamma^\mu \Psi.\tag{3}$$

If we use the first EOM and replace the current (in the third term of the $\mathcal{L}$) by the derivative of $F_{\mu\nu}$ we find the following Lagrangian.

$$\mathcal{L} = \frac{1}{2} F_{\mu\nu}F^{\mu\nu} + \bar{\Psi}(i\partial_\mu\gamma^\mu - m)\Psi.\tag{4}$$

We can see that the interacting term between the fermion and the gauge boson disappeared. The main question is How is possible that two equivalent lagrangians (we can go from one to the other by EOM) have different Feynman diagrams?

Answer 1

Let's take a step back and look at a Lagrangian from classical mechanics. For example, $$L(x,\dot x)=\tfrac 1 2 m\dot x^2-U(x),$$ but we could choose any Lagrangian here. We can also define an action $S[x(t)]=\int dt\, L(x(t),\dot x(t))$; the action is a functional, which takes a function as input and outputs a number. Functionals are often indicated with square brackets. In contrast, the Lagrangian is a function. For the Lagrangian $x$ is just an argument (a number) but for the action $x(t)$ is a function (a parametrization). For example, $L(y,z)=\tfrac 1 2m y^2-U(z)$ is the same Lagrangian as in the first equation.

Any relevant Lagrangian obeys a stationary-action principle: whenever the action is minimized it produces equations of motion. We can minimize the action by taking a functional derivative and then setting it to zero. This will result in a function $x_0(t)$ which minimizes the action given some boundary conditions. Formally $$\frac{\delta S}{\delta x(t)}[x_0(t)]=0.$$

Now you are basically asking why $L(x_0(t),\dot x_0(t))$ gives a different result. The answer is that it doesn't make sense to plug $x_0(t)$ back into the Lagrangian and you should not expect the same result at all.