Arrow and flow of charge in fermion propagator The momentum-space fermion propagator in the free Dirac theory is given by
 

The arrow on the fermion propagator is said to represent the flow of charge.
How can we derive this statement quantitatively from the Dirac Lagrangian?
What is the quantitative form of the charge being referred to here?
 A: The arrow is not really about charge. Neutrinos are neutral, and you need arrows in their propagators anyway. The more correct statement is that arrows represent the flow of fermion number.
Moreover, the arrows are not directly related to the Dirac Lagrangian, because you also need arrows for charged bosons. In fact, you need arrows whenever you have a particle with non-trivial quantum numbers.
In my opinion, the most intuitive way to understand the arrows is to go back to the basics: Wick contractions. When you want to evaluate the $n$-point function
$$
\langle\phi_1\phi_2\phi_3\cdots\phi_n\rangle
$$
where $\phi\in\{\psi,\psi^\dagger\}$, Wick's theorem tells you to calculate all possible contractions:
$$
\langle\phi_1\phi_2\phi_3\cdots\phi_n\rangle=\sum_\mathrm{contractions}\pm\langle\phi_i\phi_j\rangle\cdots\langle\phi_k\phi_\ell\rangle
$$
In principle, there are four types of contractions to consider:
$$
\langle\psi\psi\rangle,\qquad\langle\psi\psi^\dagger\rangle,\qquad\langle\psi^\dagger\psi\rangle,\qquad\langle\psi^\dagger\psi^\dagger\rangle
$$
but a simple calculation shows that
$$
\langle\psi\psi\rangle=\langle\psi^\dagger\psi^\dagger\rangle=0
$$
Indeed, if $\psi$ has any non-trivial quantum number, then so does $\psi|0\rangle$; and therefore $\langle0|\psi$ will have the opposite quantum number. This implies that these states are orthogonal and $\langle 0|\psi\psi|0\rangle=0$. For the same reason, $\langle 0|\psi^\dagger\psi^\dagger|0\rangle$ vanishes as well.
Therefore, we introduce the following diagrammatic rule: we always draw an arrow coming out of the $\psi$ fields, and one entering the $\psi^\dagger$ fields. Any diagram where the directions of the arrows is not conserved includes a factor of $\langle\psi\psi\rangle$ or $\langle\psi^\dagger\psi^\dagger\rangle$, and therefore it vanishes. This means that one need only consider diagrams where the direction of the arrows is conserved.
