Solutions to schrodinger equation with potential V(x)=V(-x)

[partyday]

Homework Statement

Write down the time independent one dimensional Schrodinger
equation for a particle of mass m in a potential V (x). Show that if V (x) = V (−x) and real, then the
solutions ψn(x) have the property that ψn(x) = ±ψn(−x).

Relevant Equations

V(x) = V(-x)

$C \psi''(x) + V(x)\psi(x)=E\psi(x)$

C is just the constant by $\psi''$

My initial attempt was to write out the schrodinger equation in the case that x>0 and x<0, so that

$$ \frac {\psi'' (x)} {\psi (x)} = C(E-V(x))$$
and

$$ \frac {\psi'' (-x)} {\psi (-x)} = C(E-V(-x))$$

And since V(-x) = V(x) I equated them and rearranged them so

$$\frac {\psi'' (x)} {\psi (x)} = \frac {\psi '' (-x)} {\psi(-x)} $$ I feel like this is where I'm supposed to be in the problem, but I'm struggling to connect this result to ψn(x) = ±ψn(−x). Any suggestions?

 

[timetraveller123]

fyi i no nothing about qm but this just seems to be about even odd functions
so the question is asking to show wavefunction is purely odd or purely even
i think you might you want to use the fact that any function can be written as a sum of even and odd functions
also if a function is purely odd or even what can you tell about the second derivative.
if this wrong please forgive me.

 

[partyday]

fyi i no nothing about qm but this just seems to be about even odd functions
so the question is asking to show wavefunction is purely odd or purely even
i think you might you want to use the fact that any function can be written as a sum of even and odd functions
also if a function is purely odd or even what can you tell about the second derivative.
if this wrong please forgive me.

I think I see the solution now. Bit of a brainfart for me. The ratios of the first and second derivative are the same for an odd or even function at x and -x, and that means that by that fact it's either $\psi(x) = \psi(-x)$ (even) or $\psi(x) = - \psi(-x)$ (odd) right?

 

[timetraveller123]

i think you might also need to show that any wave function is necessarily of that form

i am slightly confused why can't the wavefunction be like zero for negative x and x=zero. and non-zero for positive values of x or the other way around.

 

[PeroK]

I think I see the solution now. Bit of a brainfart for me. The ratios of the first and second derivative are the same for an odd or even function at x and -x, and that means that by that fact it's either $\psi(x) = \psi(-x)$ (even) or $\psi(x) = - \psi(-x)$ (odd) right?

I think you may be trying to prove something that is not necessarily true. What is generally true is as follows:

If $V$ is an even function, then for each eigenvalue you can find an eigenfunction that is either even or odd.

If you make the further assumption that the eigenspace associated with an eigenvalue is one-dimensional, then any eigenfunction must be odd or even - this follows from the above result.

But, if you have an eigenspace corresponding to a single eigenvalue that is of higher dimension, then you can combine odd and even eigenfunctions to get an eigenfunction that is neither odd nor even.