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?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ Well, teh name of the variable is arbitrary, so there's nothing wrong with the potential energy being that function entirely. $\endgroup$– FGSUZCommented Oct 8, 2018 at 11:25
-
$\begingroup$ This form of potential energy in the 1D Schrodinger equation is a famous one. See e.g. physics.stackexchange.com/q/30807 and physics.stackexchange.com/q/246843 and en.wikipedia.org/wiki/P%C3%B6schl%E2%80%93Teller_potential (and references therein). $\endgroup$– user197851Commented Oct 8, 2018 at 11:29
-
$\begingroup$ @LonelyProf It probably wouldn't be hard to expand that into an answer $\endgroup$– David ZCommented Oct 8, 2018 at 11:33
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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).