Basic qm derivation

1. Feb 6, 2005

broegger

Suppose that $$\psi (x)$$ is some solution to the time-independent Schrödinger equation;

$$-\frac{h^2}{2m}\frac{\partial^2\psi(x)}{\partial x} + V(x)\psi(x) = E\psi(x)$$.​

I want to show that if the potential V(x) is an even function, then $$\psi(-x)$$ is also a solution to same equation (same E and V).

I know I'm supposed to combine the facts that $$\psi(x)$$ is a solution and that V(x) = V(-x), but I can't see how. I've noted that

$$\frac{\partial^2\psi(-x)}{\partial x} = \frac{\partial^2\psi(x)}{\partial x}$$,​

but that's pretty much it

2. Feb 6, 2005

dextercioby

Your last formula contains 2 typo's.You may wanna repair it,because it's pretty important to the proof itself...

Daniel.

3. Feb 6, 2005

mikeu

Try simply making the substitution $$x\to-x$$ in the SWE, then using the fact that $$V(-x)=V(x)$$. The new form should then show directly that $$\psi(-x)$$ is a solution as well, since it satisfies the wave equation.

4. Feb 6, 2005

dextercioby

Yes,the way it's written and the condition imposed upon the potential energy,then the total Hamiltonian is parity invariant and of course the parity operator and the Hamiltonian commute,ergo they admit a complete set of eigenvectors...End proof...

Daniel.

5. Feb 6, 2005

Galileo

The proof also needs to show that the energies are nondegenerate.

6. Feb 6, 2005

dextercioby

What???Please,explain...I may be tired and i may not see it...

Daniel.

7. Feb 6, 2005

Galileo

Nevermind. I didn't read the actual question. I thought it said 'every solution to the SE is either even or odd'.

8. Feb 6, 2005

broegger

I've tried the substitution-thing - that was my first approach, I couldn't make it work. I'm too tired now, maybe I'll work it out tomorrow - thanks for your replies.

9. Feb 6, 2005

dextercioby

Weird,the way i see it,it's immediate... Anyway,i see that u didn't noticed.
$$\frac{\partial^{2} \psi}{\partial x}$$

is not correct.An essential "2" is missing...

Daniel.

10. Feb 6, 2005

broegger

Oh, yea, of course. My problem is that I don't know exactly what to end up with actually. Should I prove this:

$$\frac{\partial^2\psi(-x)}{\partial x^2} + V(x)\psi(-x) = E\psi(-x)$$​

or this:

$$\frac{\partial^2\psi(-x)}{\partial x^2} + V(-x)\psi(-x) = E\psi(-x)$$​

The essential difference being the minus in the potential function. Maybe it's because I'm tired (I am!), but I can't quite figure out?

11. Feb 6, 2005

dextercioby

They're equal,because the potential is parity invarint,viz.
$$V(x)=V(-x)$$

Daniel.

12. Feb 6, 2005

broegger

Oh, yea, I'm going to lie down now