Does time evolution preserve parity? Let $\psi(t_0, \cdot)$ be the state of a quantum system corresponding to a Hamiltonian $H$ in the position representation at time $t_0$. Assume $\psi(t_0, -x) = \psi(t_0,x)$, that is $\psi(t_0, \cdot)$ is symmetric. Is then $\psi(t, \cdot)$ symmetric for every time $t$? Or in other words: Is $$(e^{-itH}\psi)(t_0, -x)=(e^{-itH}\psi)(t_0, x)?$$ If not, can there be some conditions on $H$ such that this holds?
 A: To answer your question, let's first define the parity flip operator $P : \mathcal H \rightarrow \mathcal H$, such that it flips the sign of $x$ for any position eigenvector:
$$P |x \rangle \equiv |-x \rangle\qquad (\mathrm{I})$$
Note that $P$ is a Hermitian operator, since the matrix elements $\langle x'| P | x \rangle  = \langle x' | -x \rangle = \delta(x+x')$ are invariant under a conjugate transpose. It's also in fact unitary, but that's not very important here.
Now the assumption of the initial state $|\psi(t_0) \rangle$ being symmetric simply translates to:
$$P|\psi(t_0)\rangle = |\psi(t_0)\rangle \qquad (\mathrm{II})$$
This is because:
$$P|\psi(t_0)\rangle = P \int_{\mathbb R} dx \ \psi(t_0,x) \ |x \rangle = \int_{\mathbb R} dx \ \psi(t_0,x) \ P|x \rangle=\int_{\mathbb R} dx \ \psi(t_0,x) \ |-x \rangle=\int_{\mathbb R} dx \ \psi(t_0,-x) \ |x \rangle$$ 
which together with $(*)$ implies:
$$\int_{\mathbb R} dx \ \psi(t_0,-x) \ |x \rangle=\int_{\mathbb R} dx \ \psi(t_0,x) \ |x \rangle \Leftrightarrow \psi(t_0,-x)=\psi(t_0,x)$$
With that equivalent statement in mind, let's now look at the Schrodinger equation:
$$i \hbar \frac{d}{dt} |\psi(t) \rangle = H |\psi(t) \rangle$$
Now transform both sides with $P$:
$$i \hbar \frac{d}{dt} P|\psi(t) \rangle = PH |\psi(t) \rangle \qquad (\mathrm{III})$$
where I exchanged the order of $d/dt$ and $P$, since $P$ is a time-independent operator. Now inserting $\mathbb I =P^{-1} P$ into the left side of (III) gives:
$$i \hbar \frac{d}{dt} P|\psi(t) \rangle = PH \ \mathbb I \ |\psi(t) \rangle$$
$$i \hbar \frac{d}{dt} P|\psi(t) \rangle = (PHP^{-1}) \ P |\psi(t) \rangle \qquad (\mathrm{IV})$$
Eq.(IV) tells you how the symmetry properties of the system's state evolve with time under a specific Hamiltonian. For instance, if the Hamiltonian commuted with the parity flip operator, we would have:
$$H P = P H$$
i.e.
$$PHP^{-1} = H$$
Under this assumption, eq.(IV) gives:
$$i \hbar \frac{d}{dt} P|\psi(t) \rangle = H  P |\psi(t) \rangle$$
Now since the initial state is symmetric $P|\psi(t_0) \rangle = |\psi(t_0) \rangle$, and the Schrodinger equation has a unique solution for a specific initial state, we conclude that:
$$P |\psi(t) \rangle = |\psi(t) \rangle \qquad t \in [t_0,+\infty)$$
Meaning that:
the symmetry of the initial state is preserved throughout the evolution if the Hamiltonian commutes with the generator of that symmetry.
As an aside, also note the very deep observation that since $[P,H] = 0$, the expectation value of $P$ (which is an observable, since it's Hermitian) is a constant of motion! This is because by the Ehrenfest theorem:
$$i \hbar \frac{d}{dt} \langle P \rangle = \langle [P,H] \rangle = 0$$
This is a very deep and general result, which is not at all limited to the particular symmetry that was considered here. Any symmetry in the dynamics of a quantum system corresponds to a symmetry generating operator that commutes with the Hamiltonian, and is associated with an observable that is conserved throughout the evolution.
