Can increasing the potential width α decrease the coupling constant λ?

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SUMMARY

The discussion centers on the relationship between the potential width parameter α and the coupling constant λ in the context of the Schrödinger operator defined as -d²/dx² + λV. The user posits that increasing α (with α >> 1) may be equivalent to decreasing λ (with λ << 1). They propose a transformation using new variables F = α²E and y = x/α, leading to a modified Schrödinger operator -d²/dy² + λα²V(y). This transformation reveals that the strength of the potential term is characterized by λα²V(x=α), establishing a clear relationship between the parameters.

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Hi,

I'm studying a theorem that is valid for a small coupling constant λ in the Schrödinger 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)?
 
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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.
 
Thank you for the answer
 

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