Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Coupling x potential width

  1. Dec 2, 2011 #1
    Hi,

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

    [itex]
    -\frac{d^2}{dx^2}+\lambda V
    [/itex]

    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. jcsd
  3. Dec 2, 2011 #2

    Ken G

    User Avatar
    Gold Member

    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
    [itex]
    -\frac{d^2}{dy^2}+\lambda a^2 V(y)
    [/itex],
    and we see that the equivalence class is defined by the value of [itex]\lambda a^2 V(x=a)[/itex], which characterizes the strength of the potential term relative to the kinetic energy term.
     
  4. Dec 3, 2011 #3
    Thank you for the answer
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook