- #1
robotopia
- 13
- 0
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 Schrödinger 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<br /> \qquad \text{and} \qquad<br /> \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
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 Schrödinger 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<br /> \qquad \text{and} \qquad<br /> \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