Solving 1D Scattering Problem: Reflection & Transmission Coefficients

[Robin04]

TL;DR Summary

Calculate the transmission and reflection coefficients in the potential $V(x)=-\frac{1}{\cosh^2 (x)}$, given the wavefunction $\psi(x)=(\tanh(x)-ik)e^{ikx}$

We have the potential $$V(x)=-\frac{1}{\cosh^2 (x)}$$
Show that the Schrödinger equation has the solution
$$\psi(x)=(\tanh(x)-ik)e^{ikx}$$
and calculate the transmission and reflection coefficients for the scattering process.

It is easy to show that the given wavefunction indeed solves the Schrödinger equation and it has energy $$E=\frac{k^2}{2}$$
My problem is with the principle of calculating the transmission and reflection coefficients. So far, I have only solved problems for square potentials where the transfer matrices can be constructed by taking the ratios of the left and right moving coefficients of the wavefunction before and after the scattering, but I am confused about applying this to such a potential. I tried to calculated the probability current
$$j=\frac{1}{2}(\psi \psi'^*-\psi^*\psi')=k+k^3$$
because generally, the transmission and reflection coefficients are calculated from this, but I don't know how and also, the dimensions are a bit confusing.

 

[Paul Colby]

You need to write the total wave function as a sum of solutions, one for $k$ and the other $-k$. Select coefficients such that the asymptotic behavior goes to an incident wave at $-\infty$ plus a back scattered component while in the forward direction only a transmitted wave is present. This should give you two equations and two unknowns. Set the incident amplitude to 1. The remaining coefficients yield the forward and backscattered amplitudes.

 

[Robin04]

You need to write the total wave function as a sum of solutions, one for $k$ and the other $-k$. Select coefficients such that the asymptotic behavior goes to an incident wave at $-\infty$ plus a back scattered component while in the forward direction only a transmitted wave is present. This should give you two equations and two unknowns. Set the incident amplitude to 1. The remaining coefficients yield the forward and backscattered amplitudes.

So let the wavefunction be $$\psi(x)=A(\tanh(x)-ik)e^{ikx}+B(\tanh(x)+ik)e^{-ikx}$$
The limit in $x\rightarrow -\infty$ is
$$\psi(x\rightarrow -\infty)=-A(1+ik)e^{ikx}+B(ik-1)e^{-ikx}$$
and in $x\rightarrow \infty$
$$\psi(x\rightarrow \infty)=A(1-ik)e^{ikx}+B(ik+1)e^{-ikx}$$

What should the equations for the coefficients say?

 

[Paul Colby]

Well. good question. You're looking for the scattered field. Typically one separates the problem into an incident field plus a scattered one. Thus,

$\psi = \psi_\text{in} + \psi_\text{sc}$​

where, $\psi$ is the full solution and neither $\psi_\text{in}$ nor $\psi_\text{sc}$ are solutions of the full problem. I can say right off the bat,

$\psi_\text{in} = e^{-ikx}$.​

I can also say asymptotically that there is no $e^{ikx}$ at $\infty$. With this in mind one should get an inhomogeneous equation for A and B.

Oh, BTW you get the scattering coefficients from $\psi_\text{sc}$ which depend on A and B.

 

[vanhees71]

Well, this is a special case of the Pöschl-Teller potential ($\lambda=2$):

https://en.wikipedia.org/wiki/Pöschl–Teller_potential

Using the general ansatz given in #3:
$$\psi(x)=A(\tanh[x]-\mathrm{i} k) \exp(\mathrm{i} k x) + B(\tanh[x]+\mathrm{i} k) \exp(-\mathrm{i} k x),$$
you need to find the solution such that for $x \rightarrow \infty$ you only have a right-moving wave, which uniquely leads to $B=0$. The conclusion is that for this special choice of parameters you get a reflectionless scattering solution, because also for $x \rightarrow -\infty$ we only have a right-moving wave, which is also mentioned in the above quoted Wikipedia article.

 

[Paul Colby]

Something doesn't jive here I'd like to understand. In typical scattering calculations one writes the solution to the field equations as the sum of an incident or driving field plus a scattered wave. The scattering parameters are defined in terms of this scattered field, not the total field. Here we actually have all possible total fields so one might think the song and dance of splitting incident and scattered waves is not needed. Let's do it anyway.

The starting point is this split of incident and scattered waves,

$\psi = \psi_\text{in} + \psi_\text{sc}$​

The total field is known in terms of two unspecified coefficients while the incident field is simply $e^{-ikx}. Taking the limit$\tanh(x)\rightarrow 1$as the first equation (row) and$\tanh(x)\rightarrow -1## as the second yields the inhomogeneous matrix equation,

$\left( \begin{array}{cc} (1-ik) & (1+ik) \\ -(1+ik) & -(1-ik) \end{array} \right) \left( \begin{array} {c} A \\ B \end{array} \right) = \left( \begin{array} {c} 0 \\ 1 \end{array} \right)$​

Here the right hand side vector says the scattered wave has no incident component to the right and only a unit one to the left. The solution is

$A = -\frac{i(k^2+1)}{4k}$

$B = \frac{i(ik-1)}{4k}$​

Now, these may be substituted back to obtain the forward and backward scattered wave amplitudes, T and R

$T = B(1+ik)-1$

$R = -A(1+ik)$​

What did I miss?

I tried to fix the formatting on this and it simply won't behave. Sorry.

 

[vanhees71]

I don't understand this calculation. Here you just solve the time-independent Schrödinger equation for the given potential and then check the asymptotic behavior for $x \rightarrow \pm \infty$. You look for the case that for $x \rightarrow \infty$ you have only a right-moving wave, which corresponds to the asymptotic condition that a asymptotic free wave packet moves from the left toward to the potential (in-state). Then in general when it moves close to the region of the potential one part is reflected and one part runs through. One should always look at scattering problems in this way of moving wave packets. The energy-eigensolutions in the continuous part of the energy spectrum (the scattering states) are generalized eigenstates not representing physical states. Nevertheless you can use them to analyze scattering problems by using them in the sense that you can build all physical states with them (of course if there are bound states in addition you also need to take into account those, because only all eigenstates of the energy build a complete orthonormal set).

Now for some potentials there are special parameters (a bit like resonances), where there are no reflected parts, i.e., the entire wave goes forward (here moving to the right), and that's the case for the Pöschl-Teller potential as detailed in the above quoted Wikipedia article, and the solution becomes particularly simple, which I guess is why the problem was formulated for such a special case.

The solution for general parameters of the potential is a hypergeometric function, which is of the general form. Of course you have to choose again the right boundary conditions to describe the right-moving incoming state. Then you get for $x \rightarrow -\infty$ asymptotic waves moving both to the right and to the left, which represent the incoming wave and the part of the scattered wave due to reflection. By construction for $x \rightarrow \infty$ you have only a right-moving wave, which represents the part of the scattered wave going through the potential, and from the amplitudes of the reflection part for $x \rightarrow -\infty$ and the transmitted part for $x \rightarrow \infty$ you get the reflection and transmission coefficient. Due to unitarity (or equivalently the self-adjointness of the Hamiltonian) the sum is $R+T=1$.

 

[Paul Colby]

I don't understand this calculation.

Which is to say you don’t understand (more likely accept) the definition of scattered wave I’m using here. Ok, I need to reflect on it more .

 

[vanhees71]

I'm accepting the textbook definition of asymptotic states. In the sense, we've discuss this here, we solve the time-independent Schrödinger equation. For potentials which go sufficiently quickly to a constant for $x \rightarrow \pm \infty$ the naive asymptotic states are easily estimated.

For simplicity let's discuss potentials which go to 0 in both limits. Sufficiently quickly means that we assume
$$\lim_{x \rightarrow \pm \infty} x V(x)=0.$$
Then all energy eigenvalues $E>0$ are scattering states and they all are two-fold degenerate. Setting $E=\hbar^2 k^2/(2m)$ with $k>0$ the time-independent Schrödinger equation can be written as
$$\psi''(x)-\frac{2m}{\hbar^2} V(x)=-k^2 \psi(x).$$
Since $V(x)$ goes to 0 faster than $1/x^2$ for $x \rightarrow \pm \infty$ the general asymptotic forms are
$$\psi(x) \simeq A_{\pm \infty} \exp(\mathrm{i} k x) + B_{\pm \infty} \exp(-\mathrm{i} k x).$$
Now the usual scattering experiment is setup such that an asymptotic free particle moves from $x \rightarrow -\infty$ towards the potential. The corresponding energy-eigenstates are to be chosen such that $B_{+\infty}=0$. This determines the wanted energy eigenfunction uniquely, and $A_{-\infty}$ is the amplitude of the incoming wave, $B_{-\infty}$ is the amplitude of the reflected scattering wave, and $A_{\infty}$ that of the amplitude of transmitted scattering wave.

This behavior is just given by the solutions of the time-independent Schrödinger equation and not by just making some ansatz which doesn't solve it.

In the here discussed special case, as you can see immediately by just using
$$\tanh x \simeq \pm 1 \quad \text{for} \quad x \rightarrow \pm \infty$$
in this case the corresponding exact scattering state is such that $B_{-\infty}=0$ and thus you have a scattering process without reflection.

You can find such examples also for scattering on a finite rectangular potential well by choosing the corresponding energies. This is more simple to calculate, because you don't need the confluent hypergeometric functions but can deal with elementary exp and trig. functions only.

For some pictures/movies, see the following webpages I once prepared for a QM 1 lecture (it's however in German):

https://itp.uni-frankfurt.de/~hees/qm1-ss09/wavepack/node15.html

The "no-reflection case" is demonstrated here:

https://itp.uni-frankfurt.de/~hees/qm1-ss09/wavepack/node19.html

Of course for such a Gaussian initial asymptotic free wave packet you still get some small reflected parts, because the literally non-reflecting case is at a sharp energy value. It's similar to a resonance phenomenon.