- #1
TheNE
- 23
- 0
show that e-ikx is an eigenstate energy.
Do I start by multiplying the hamiltonian operator by ψ(x)?
So far I have ψ(x)(1/2m)(-i[STRIKE]h[/STRIKE]d/dx)2=e-ikx
Do I start by multiplying the hamiltonian operator by ψ(x)?
So far I have ψ(x)(1/2m)(-i[STRIKE]h[/STRIKE]d/dx)2=e-ikx