# 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

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