Reflectionless potential

V[x_,a_] = -a^2 Sech[a x]^2
Plot[V[x,1],{x,-5,5}]

Clear[a]

D[Sech[x],x]

Integrate[Sech[a x]^2,{x,-∞,∞},Assumptions->{A>0, a>0}]

ψ[x_, a_] = Sqrt[a/2]Sech[a x]

Integrate[ψ[x, a]^2 ,{x,-∞,∞},Assumptions->a>0]

Plot[ψ[x, 1] ,{x,-5,5}]

-1/2 D[ψ[x, a],{x,2} ]+ V[x,a]ψ[x, a]//FullSimplify

This tells us that :  a) ψ[x, a] is an eigenfunction and b) its energy (eigenvalue) is -(((a^(5/2)) )/(2 (Sqrt[2])))
Now, the solution to the TISE is known for this potential and it is

ψk[x_,a_, k_] = A(I k - a Tanh[a x])/(I k + a) Exp[I k x]

-1/2 D[ψk[x, a,k],{x,2} ]+ V[x,a]ψ[x, a,k]//FullSimplify

This is (k^2)/2 times ψk

Plot[Tanh[x],{x,0,10}]

So, for large x ψk is proportional to

Clear[x,k,a,A]
A[x_,k_,a_] = (k-a)/(k+a) Exp[I  k x]

Plot3D[Re[A[x,k,1]],{x,5,10},{k,0,4}]