When is emitting a particle (e.g. neutrino) the same as absorbing an antiparticle? I've read that Pauli first proposed the neutrino to explain energy and momentum conservation in this equation:
$$n \to  p^+ + e^- + \overline\nu_e$$
But it is also written differently here:
$$\nu_e + n  \to  p^+ + e^-$$
Is there a difference between the two equations? If not, does it mean we ascribe a negative energy to one of $\overline\nu_e$ or $\nu_e$?
 A: Moving a particle from one side of a reaction to the other while flipping to it matterness keeps all the necessary particle quantum numbers in balance, but it changes the kinematics.
For instance, in your example the top equation necessarily has a broad energy spectrum (one of the main reasons for Pauli's proposal), which the lower one can have a discrete spectrum (must have a discrete spectrum if the initial particles are prepared in well defined momentum states).
Nor can you even guarantee that this kind of reversal preserve the possibility of the reaction. For instance
$$ p^- \to \bar{n} + e^- + \bar{\nu}_e $$
has the same isospin and lepton quantum numbers quantum numbers as your top reaction but it is strictly forbidden by energy considerations (the final state is heavier than the initial state).
Also things like
$$ \bar{p} + n \to e^- + \bar{\nu}_e $$
are allowed by quantum numbers but are vanishingly unlikely in practice. It would take some cleverness just to exhibit a Feynmann diagram at the quark-gluon-lepton level that totals up to the process exhibited here are the hadron-lepton level.
The neutrino moving case that you exhibit is unusually clear-cut because it involves a fundamental particle and a very small fraction of the mass in the system.
