Symmetric potential and the commutator of parity and Hamiltonian In one dimension -
How can one prove that the Hamiltonian and the parity operator commute in the case where the potential is symmetric (an even function)?
i.e. that $[H, P] = 0$ for $V(x)=V(-x)$
 A: You prove the equality of operators by applying them to a function, we have 
$$
H = - \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)
$$
Ergo:
$$
HP f(x) = H f(-x) = (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(-x) = - \frac{\hbar^2}{2 m} f''(-x) + V(x) f(-x)
$$
and
$$
PH f(x) = P (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(x) = P (- \frac{\hbar^2}{2 m} f''(x)) + P (V(x) f(x)) ...
$$
$$
... = - \frac{\hbar^2}{2 m} f''(-x) + V(-x) f(-x)
$$
When you use 
$$
V(-x) = V(x)
$$
you see that both expressions are equal.
A: $$[P,H]f(x)=(PH-Hp)f(x)$$
But 
$$H=P^2/2m+E(x)$$
$$ =PE(x)-Hf(x)$$
$$ =E(-x)-E(-x)$$
$$ =0 $$
The parity operator therefore commutes with Hamiltonian. 
A: While the accepted answer is very clear, I'll write an operator proof. The $\hat{p^2}$ in $\hat{H}$ commutes with $\hat{\mathbb{P}}$ (the parity operator). So, to show that $\hat{H}$ and $\hat{\mathbb{P}}$ commute, we have to show this:
$[\hat{V},\hat{\mathbb{P}}]=0$
Note that since $V(x)$ is a symmetric function i.e. even function, it is an eigenfunction of $\hat{\mathbb{P}}$.
$\hat{V}\hat{\mathbb{P}}-\hat{\mathbb{P}}\hat{V}$
$\Rightarrow V(x)\hat{\mathbb{P}}-V(-x)\hat{\mathbb{P}}=0$ (QED)
I did the last step keeping in mind that when you have a product of functions on which the parity operator needs to be applied, you can apply at one (i.e. change the $+x$ to $-x$) and transfer the Parity to the right.
P.S. As a consequence of this commutation, in one dimension, whenever you have a symmetric potential, the eigenstates are either even or odd, since, only even and odd functions are the eiegenstates of the parity operator.
