# Quantum Mechanics: Normalization Requirements (E>Vmin)

1. Oct 11, 2008

### njhajj

I have just begun using the 1-D Schrodinger Equation in my quantum mechanics course. We are assuming the potential, V, is solely a function of x (V(x)). I have been examining the solution to the differential equation through seperation of variables. Ψ(x,t) = ψ(x)φ(t)

(Assuming Ψ(x,t) is seperable) I understand the proof to show that the solutions of Ψ(x,t) produce probabilitiy densities that are invariant to time. And that |Ψ(x,t)|^2 = |ψ(x)|^2. I also understand why E must be real.

When the assumption of seperability is made we have:

-(h^2)ψ''(x)/2m +V(x)ψ(x) = Eψ(x) (h is hbar)

Which is equivalent to:

ψ''(x) = 2m(V(x) - E)(ψ(x))/h^2 (1)

My textbook (DJ Griffiths Introduction to Quantum Mechanics) claims that if E < Vmin => that Ψ(x,t) cannot be normalized. Because |Ψ(x,t)|^2 = |ψ(x)|^2 I can divert my attention to (1) and ignore φ(t).

Solving the differential equation for:

ψ''(x) = c^2(ψ(x), c is real

Gets me the general solution (I think!):

ψ(x) = Aexp(cx+b) + Bexp(-cx+d) A,B,b,d are constants

If we take a piecewise solution like ψ(x) = exp(-|x|) the integral of |ψ(x)|^2 over the whole real line is finite (and equal to 1, I believe). |ψ(x)|^2 is continuous as well.

The textbook also hints that it has to do with the the fact that the second derivative is always the same sign as the function itself. I can't formulate a proof as to why this should necessarily imply the function cannot be normalized. In fact, to the best of my knowledge I've provided a perfectly valid counterexample...

I've thought about it for an hour or so and I can't figure out why this answer is unacceptable. Can someone please tell me what I have done incorrectly?

I'm also curious about if there is something unphysical about having E<Vmin. Thanks for the help!

Last edited: Oct 11, 2008
2. Oct 11, 2008

### borgwal

Your "general solution" is not, in general, a solution to the correct equation, with a position-dependent V(x) [your c is not a function of x].

Also, as long as |V(x)| is finite, the wavefunction's derivative has to be continuous, too.

So, two reasons for your counter example to be incorrect.

Oh, and of course there is something unphysical about E being less than the minimum potential energy when E=T+V!