1. ## Parity

If $V(-x) = V(x)$, prove $[H,P] = 0$, where $H$ is the hamiltonian operator and $P$ is the parity operator.

..

We know that this is an even function (so even parity..) and has behavior like cosine. Not sure how to prove this- maybe using the general $HP - PH$ idea?

2. If someone doesn't get back to you before tonight, I'll do this when I'm online again.

-Dan

3. Originally Posted by DiscreteW
If $V(-x) = V(x)$, prove $[H,P] = 0$, where $H$ is the hamiltonian operator and $P$ is the parity operator.

..

We know that this is an even function (so even parity..) and has behavior like cosine. Not sure how to prove this- maybe using the general $HP - PH$ idea?
The parity operator in 1 dimension is the same as simply replacing x with -x. The easiest way to calculate the commutator is to apply it to an arbitrary wavefunction. Since the wavefunction can be decomposed into a linear combination of energy eigenstates we might as well talk about an energy eigenstate.

Call this state $\psi (x)$. Now we know that
$H \psi (x) = (K(x) + V(x) ) \psi (x) = E \psi (x)$
where K(x) is the kinetic energy operator.

Well, we know that $K(x) \propto \frac{d^2}{dx^2}$, which is an even function, and we know that V(x) is given to us as an even function, so the only possible kind of eigenfunction we can have is an even function.

So the eigenstate $\psi (x)$ is an even function. (We may easily extend this argument to an arbitrary function that satisfies the Schrodinger equation with an even V(x).)

On to the commutator:
$[ H, P ] \psi (x) = HP \psi (x) - PH \psi (x)$

$= H \left ( \psi (-x) \right ) - P \left ( E \psi(x) \right )$

But we know that $\psi (-x) = \psi (x)$, so
$= H \psi (x) - E \left ( P \psi (x) \right )$
(the last term since E is simply a constant.)

$= E \psi (x) - E \psi (-x)$
and, again, $\psi (-x) = \psi (x)$, so

$= E \psi (x) - E \psi (x) = 0$

Since $[ H, P ] \psi (x) = 0$ for an arbitrary function $\psi (x)$, thus [H, P] = 0.

-Dan