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Coupling x potential width

  1. Dec 2, 2011 #1

    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. jcsd
  3. Dec 2, 2011 #2

    Ken G

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    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
    -\frac{d^2}{dy^2}+\lambda a^2 V(y)
    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
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