Deriving Even Function Solutions to TISE

[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

 

[dextercioby]

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

Daniel.

 

[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.

 

[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.

 

[Galileo]

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

 

[dextercioby]

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

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

Daniel.

 

[Galileo]

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

Daniel.

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

 

[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.

 

[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.

 

[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?

 

[dextercioby]

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

Daniel.

 

[broegger]

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