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I am reading a book where it write the one-dimensional stationary Schrodinger equation as $$ [-\frac{\hbar^2}{2m}\frac{d^2}{dζ^2}-Γ{\rm sech}^2(bζ)]ψ(ζ)=Eψ(ζ). $$ It is known that the equation is usually written as $$ [-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)]ψ(x)=Eψ(x). $$ I speculate that the author may have $bζ$ instead of $x$. But I am confused with the component of potential energy $-Γ{\rm sech}^2(bζ)$. My guess is that we may have $2mΓb^2(2π)^2/h^2 $ instead of the $V$. Isn't it?

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closed as unclear what you're asking by ZeroTheHero, John Rennie, Kyle Kanos, A.V.S., AccidentalFourierTransform Oct 28 '18 at 2:28

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On prompting from a moderator, I'm offering my comment as an answer!

It seems most likely that the potential energy function given in your equation was a deliberate choice, as it is a famous potential: the Pöschl-Teller potential. It can be solved exactly, giving a finite number of bound states. Also, for suitable values of the strength parameter $\Gamma$, it is reflectionless, meaning that incident waves transmit perfectly through it.

The value of $b$ establishes a length scale for the potential, and your speculation about the combination of $\Gamma$ and factors of $\hbar^2/2m$ (where $\hbar=h/2\pi$) probably relates to making a convenient choice of energy scale so as to reduce the equation to a dimensionless form (see the pages I referenced, as well as this one).

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