# Coupling x potential width

1. Dec 2, 2011

### intervoxel

Hi,

I'm studying a theorem that is valid for a small coupling constant λ in the Schrodinger operator below

$-\frac{d^2}{dx^2}+\lambda V$

The potential has a parameter α which defines its width.

My question is: Starting with λ=1 and α=1, can I claim that if I increase the value of α (α >> 1) is equivalent to decrease λ (λ << 1)?

2. Dec 2, 2011

### Ken G

Actually, I would have said that you can prove that an increase in a is equivalent to an increase in lambda. The way to do it is define a new energy variable, call it F = a2 E, and a new position variable, call it y = x/a. Then in terms of the new energy F, the Shroedinger operator is
$-\frac{d^2}{dy^2}+\lambda a^2 V(y)$,
and we see that the equivalence class is defined by the value of $\lambda a^2 V(x=a)$, which characterizes the strength of the potential term relative to the kinetic energy term.

3. Dec 3, 2011