Show that psi(-x) is a solution with the same eigenvalue as psi(x)

Sep 2018
2
0
I have been given the assignment to show that ψ(-x) is another solution with the same energy eigenvalue as ψ(x), given that the potential is even around x=0. The solutions are one-dimensional. I have tried to get my head around it but I am not sure what to look for. My idea would be to divide through by the wave function in the wave equation, and we would then be left with the energy-eigenvalue E as a function of the potential, which we know is even around the origin:

-ℏ^2/2m (d^2 ψ(x))/(dx^2 )+V(x)ψ(x)=ψ(x)E.

Dividing by ψ(x):

-ℏ^2/2m (d^2)/(dx^2 )+V(x)=E

And the eigenvalue is now a function of the potential only. Is this an illegal move? Or where do I go from here?
 
Aug 2010
434
174
I have been given the assignment to show that ψ(-x) is another solution with the same energy eigenvalue as ψ(x), given that the potential is even around x=0. The solutions are one-dimensional. I have tried to get my head around it but I am not sure what to look for. My idea would be to divide through by the wave function in the wave equation, and we would then be left with the energy-eigenvalue E as a function of the potential, which we know is even around the origin:

-ℏ^2/2m (d^2 ψ(x))/(dx^2 )+V(x)ψ(x)=ψ(x)E.

Dividing by ψ(x):

-ℏ^2/2m (d^2)/(dx^2 )+V(x)=E

And the eigenvalue is now a function of the potential only. Is this an illegal move? Or where do I go from here?
Yes, that's an illegal move! What do you think "d^2/x^2", without a function, even means? d^2ψ(x)/dx is NOT "d^2/dx^2" multiplied by
ψ(x), it is the operator applied to the function. You cannot just "divide" by ψ(x).

Can you use the "chain rule" to determine dψ(-x)/dx and d^2ψ(-x)/dx^2?
 
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