Scattering on potential $$V(x) = -\frac{(\hbar a)^2}{m}\text{sech}^2(ax)$$ with 1D equation of Schrodinger is famous problem. It is dealt with in Problem 2.48 of Griffiths book or online here.

It is called reflectionless potential because every incident wave has perfect transmission $T = 1$ "regardless of its energy".

I understand for some energies for $E<0$ that $T=1$. I also understand resonant scattering means at some values of $E$ when $E>0$ that $T=1$.

I do not understand how "regardless of its energy" particles may pass the barrier with $T=1$.

Question: How can this phenomena be explained?


2 Answers 2


TL;DR: The supersymmetric partner potential to OP's potential is the constant potential, which is clearly reflectionless.

Define for later convenience the constant $\kappa:=\hbar/\sqrt{2m}$. The constant potential and OP's potential are just the two first cases ($\ell=0$ and $\ell=1$) in an infinite sequence of reflectionless attractive$^1$ potentials

$$ V_{\ell}(x)~:=~-\frac{(\kappa a)^2 \ell(\ell+1)}{\cosh^2 ax}, \qquad \ell~\in~\mathbb{N}_{0}. \tag{1}$$

Let us next consider a sequence of two superpartner potentials

$$ V_{\pm,\ell}(x)~:=~ (\kappa a)^2\left( \ell^2 -\frac{\ell(\ell \mp 1)}{\cosh^2 ax}\right). \tag{2}$$

Note that reflection properties are not altered by simultaneously shifting the potentials $V_{\ldots}$ and the energy-level $E$ upwards or downwards with an overall constant. That's just a matter of choosing a zero-point level. So from now on we will identify two potentials iff they differ by an overall constant. E.g. the three potentials

$$ V_{+,\ell +1} ~\sim~ V_{-,\ell}~\sim~V_{\ell} \tag{3}$$

only differ by overall constants.

The two superpartner potentials (2) satisfy

$$ V_{\pm,\ell} ~=~ W_{\ell}^2 \pm \kappa W_{\ell}^{\prime} , \tag{4}$$


$$ W_{\ell}(x)~:=~\ell \kappa a \tanh ax \tag{5} $$

is the superpotential. One may show under fairly broad assumptions$^2$ that two superpartner TISEs$^3$

$$ -\kappa^2 \psi^{\prime\prime} +V_{\pm,\ell}\psi~=~E \psi \tag{6}$$

share bound state spectrum (except for the ground state for $V_{-,\ell}$), and (absolute value of) the reflection and transmission coefficients, cf. Ref. 1. Hence we have linked all the considered potentials

$$\begin{align} 0~\sim~& V_{-,0}~\sim~ V_{+,1}~\stackrel{\text{SUSY}}{\longleftrightarrow}~ V_{-,1}\cr ~\sim~& V_{+,2}~\stackrel{\text{SUSY}}{\longleftrightarrow}~ V_{-,2}~\sim~ V_{+,3}~\stackrel{\text{SUSY}}{\longleftrightarrow}~\ldots \end{align}\tag{7}$$

to the constant potential. This explains why the sequence (1) consists of reflectionless potentials for a non-negative integer $\ell\in\mathbb{N}_{0}$.

Finally, if $\ell \notin \mathbb{Z}$ is not an integer, the superpotential (5) still makes sense. However, the potential (1) cannot be linked via the use of supersymmetric partners and constant shifts to the trivial potential, and in fact, the potential (1) is not reflectionless if $\ell \notin \mathbb{Z}$.


  1. F. Cooper, A. Khare, & U. Sukhatme, Supersymmetry and Quantum Mechanics, Phys. Rept. 251 (1995) 267, arXiv:hep-th/9405029; Chapter 2.


$^1$ For completeness, let us mention that there is also a sequence of reflectionless repulsive potentials. The analogues of eqs. (1), (2) and (5) read $$ \tag{1'} V_{\ell}(x)~:=~\frac{(\kappa a)^2 \ell(\ell+1)}{\sinh^2 ax}, \qquad \ell~\in~\mathbb{N}_{0},$$ $$\tag{2'} V_{\pm,\ell}(x)~:=~ (\kappa a)^2\left(\ell^2 +\frac{\ell(\ell \mp 1)}{\sinh^2 ax} \right), $$ $$\tag{5'} W_{\ell}(x)~:=~\ell \kappa a \coth ax , $$ respectively. For $a\to 0$, this becomes $$ \tag{1''} V_{\ell}(x)~:=~\frac{\kappa^2 \ell(\ell+1)}{x^2}, \qquad \ell~\in~\mathbb{N}_{0},$$ $$\tag{2''} V_{\pm,\ell}(x)~:=~ \frac{\kappa^2\ell(\ell \mp 1)}{x^2} , $$ $$\tag{5''} W_{\ell}(x)~:=~\frac{\ell \kappa}{x} , $$ respectively. Also we have for notational simplicity suppressed a freedom to shift the potential profiles along the $x$-axis $x\to x-x_0$ .

$^2$ For starter, one has to assume that both the limits $\lim_{x\to\pm\infty} W(x)$ exist and are finite, which hold in OP's case.

$^3$ The TISE (6) can be transformed into the associated Legendre differential equation, which famously describes spherical harmonics and angular momentum states in QM with azimuthal quantum number $\ell\in\mathbb{N}_{0}$.

  • $\begingroup$ "...reflection properties are not altered by shifting the potentials upwards or downwards with an overall constant." What does "reflection properties" mean? Is it just that whether the particle reflects isn't affected, or that the reflection coefficient doesn't change at all? $\endgroup$ Commented Jul 8, 2023 at 15:10
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Jul 9, 2023 at 15:56

I will give an answer to this old question since I just saw this question referenced in Reddit, and since Qmechanic's very comprehensive answer above did not really answer OP's question.

I understand for some energies for E<0 that T=1. I also understand resonant scattering means at some values of E when E>0 that T=1. I do not understand how "regardless of its energy" particles may pass the barrier with T=1. Question: How can this phenomena be explained?

This is a valid question but the answer is simply that the statement "regardless of its energy" is not correct. As far as I understand from the answer by Qmechanic, this potential in question has some (infinite number) discrete values of energies (corresponding to bound states) where resonant transmission happens, just as OP was expecting. (The same happens in the usual square potential barrier.) It would be very hard to imagine a non-trivial potential, for which a continuum of states would show resonant transmission. With the resonant transmission - quasi bound states -correspondence, such a potential would have to have a continuum of quasi bound states.


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