# Proof that E > V(x) in TISE?

1. May 27, 2013

### robotopia

I'm looking for feedback on my answer to the following question (if my proof has holes in it, etc.) In particular, I've tacitly assumed that ψ is real. Does this extend naturally to complex ψ? Or don't I have to worry about that?

Problem 2.2 from Griffiths' Intro to QM:

Show that E must exceed the minimum value of V(x) for every normalizable solution to the time-independent Schrodinger equation. What is the classical analog to this statement? Hint: Rewrite Equation 2.4 in the form
$$\frac{d^2\psi}{dx^2} = \frac{2m}{\hbar^2} [V(x) - E]\psi;$$
if $E < V_\text{min}$, then ψ and its second derviative always have the same sign—argue that such a function cannot be normalized.

My attempt:
Assume both $E < V_\text{min}$ and ψ is normalizable. Then, as stated in the question, ψ and $\frac{d^2\psi}{dx^2}$ always have the same sign. Moreover, normalizability requires that
$$\lim_{x \rightarrow \infty} \psi = 0 \qquad \text{and} \qquad \lim_{x \rightarrow -\infty} \psi = 0.$$
ψ cannot be zero everywhere (which is not normalizable), and must therefore either have a non-zero global maximum or a non-zero global minimum (or both). Moreover, since ψ asymptotes to zero at both ends, we can say that ψ must have either a positive global maximum, or a negative global minimum (or both).

Suppose ψ has a positive global maximum at $x=x_\text{max}$ . Then $\psi(x_\text{max})$ is positive, and because $E<V_\text{min}$, so is $\frac{d^2}{dx^2}\psi(x_\text{max})$. But $\frac{d^2}{dx^2}\psi(x_\text{max})$ must be negative, since it is a local (as well as a global) maximum. Contradiction.

Similarly (by exchanging all ‘maximum’s with ‘minimum’s and all ‘positive’s with ‘negative’s), ψ cannot have a negative global minimum. This contradiction allows us to say that if $E<V_\text{min}$, then ψ is not normalizable. QED

2. May 28, 2013

### aim1732

The proof automatically generalizes to complex ψ because the real and complex parts of the wave function must separately go to zero at ±∞ while satisfying the condition of signs you talk about so well separately.So it should hold.An interesting thing to note probably is that this is true only because the TISE is linear.