Show that E must exceed the minimum value for V(x) for every normalizable solution to the TISE, Hψ = Eψ
Hint: rewrite the TISE as d2ψ/dx2 = (2m/ħ)(V(x) - E)
I understand how to get the TISE Hψ = Eψ into the form of the hint by using the hamiltonian and the differential form of the momentum operator. That's all fine
The Attempt at a Solution
The solution to the differential equation given in the hint is:
ψ = Ae√(2m/ħ)(V(x) - E) + Be-√(2m/ħ)(V(x) - E)
For the solution to be normalizable, it has to go to zero at infinity. This means it can't be oscillatory, therefore the powers of the exponents have to be real.
But doesn't this mean that V has to be greater than E and not the other way round?
If E is greater than V anywhere, then in that region there will be an imaginary exponent and the solution will be oscillatory.