Parity of wave function corresponding to even potential

[theneedtoknow]

Homework Statement

Using time independent 1D Shrodinger equation, show that if V(x) is even and Psi(x) is a solution, Psi(-x) is also solution. Then, assume Psi(-x) and Psi(x) differ only by a constant, show that the constant is either +1 or -1.

Homework Equations

The Attempt at a Solution

Can I get some help just on where to start the first part for now?
I tried solving the time independent shrodinger equation for V(x) and setting V(x) = V(-x)
which gave me:
Psi(x)/Psi(-x) = Psi''(-x)/Psi''(x)
but I can't figure out where to go from here or even if I'm doing the right thing.

 

[anathema]

To start, let's recall the time independent 1D Schrodinger equation:

H(psi) = E(psi)

Where H is the Hamiltonian operator, psi is the wave function, and E is the energy of the system. In this case, we are dealing with a 1D system, so our Hamiltonian operator can be written as:

H = -h^2/2m * d^2/dx^2 + V(x)

Where h is Planck's constant, m is the mass of the particle, and V(x) is the potential energy function. Now, let's consider the given condition that V(x) is an even function, meaning that V(x) = V(-x). This implies that the potential energy at position x is equal to the potential energy at position -x, which makes sense since the potential energy should be symmetric about the origin.

Now, let's substitute this condition into the Schrodinger equation:

H(psi) = E(psi)

-b^2/2m * d^2/dx^2 + V(x) (psi) = E(psi)

-b^2/2m * d^2/dx^2 + V(-x) (psi) = E(psi)

We can see that the only difference between these two equations is the potential energy function, which is simply the sign of x. This means that if psi(x) is a solution to the first equation, then psi(-x) is a solution to the second equation. This is because the wave function is not affected by the sign of x, so if it satisfies the first equation, it will also satisfy the second equation.

For the second part of the problem, we are asked to show that if psi(x) and psi(-x) differ only by a constant, then that constant must be either +1 or -1. To do this, we can write out the two equations for psi(x) and psi(-x) and set them equal to each other:

-b^2/2m * d^2/dx^2 + V(x) (psi) = E(psi)

-b^2/2m * d^2/dx^2 + V(-x) (psi) = E(psi)

Now, let's assume that psi(x) and psi(-x) differ only by a constant, so we can write:

psi(x) = c * psi