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!