michaeld wrote:

In QM a parity transformation is given by an operator P acting on the
Hilbert space H of the theory. The fact that P^2 = 1 implies that the
Hilbert space splits up as a direct sum of eigenspaces of P with
eigenvalues +-1. A state in one of these eigenspaces is said to have
parity charge +-1 depending on its eigenvalue.

A system has parity symmetry iff the Hamiltonian is invariant under
parity, i.e.:

PHP^(-1) = H

This is equivalent to [P,H] = 0. If this holds then the Heisenberg
equation for P is:

dP/dt = [P,H] = 0

i.e. P is a conserved quantity.

Or if you prefer the Schrodinger picture, note that [P,H] = 0 is
equivalent to the statement that e^(-iHt) P e^(iHt) = P and so given a
state |psi(t_0):

e^(-iHt) P e^(iHt) |psi(t_0) = P |psi(t_0)

i.e.

P |psi(t_0+t) = e^(iHt) P |psi(t_0)

In particular if |psi(t_0) has parity charge n = +-1 then:

P |psi(t_0+t) = e^(iHt) n |psi(t_0)
= n |psi(t_0+t)

so the final state |psi(t_0+t) also has charge n. Therefore if a given
reaction starts with say a state with parity charge 1 and ends with
parity charge -1 then you can rule out this reaction (as long as your
system is parity invariant).

Thank you for the breath of fresh air.