I've been trying to think of some approximate way of solving quantum scattering problem for a single electron obeying Schrodinger equation with a locally supported potential:

$$\nabla^2 \Psi( \mathbf{r})+(k^2-u( \mathbf{r}))\Psi( \mathbf{r})=0$$

$$k^2=\frac{2mE}{\hbar^2}, \qquad u( \mathbf{r})=\frac{2mU( \mathbf{r})}{\hbar^2}$$

The usual asymptotic conditions apply, i.e. we can represent $\Psi( \mathbf{r})$ as a sum of incoming plane wave and a scattered wave, with the scattered wave obeying Sommerfeld radiation condition. Let $x$ be the direction of the incoming wave:

$$\Psi( \mathbf{r})=e^{i kx}+\psi( \mathbf{r})$$

Substituting this, we obtain inhomogeneous equation:

$$\nabla^2 \psi( \mathbf{r})+(k^2-u( \mathbf{r}))\psi( \mathbf{r})=u( \mathbf{r}) e^{i kx}$$

I'm interested in the general form of $u( \mathbf{r})$, not spherically (or axially) symmetric, and I do not want to use finite elements or other discretization methods.

So I have tried to think of some suitable approximation methods. And I had the following idea:

Let's represent the scattered wavefunction as an infinite sum, which we assume converges at least at large distances from the potential.

$$\psi=\psi_1+\psi_2+\psi_3+\dots $$

Let all the terms obey Sommerfeld radiation condition as well.

Now let them obey the following system of equations:

$$\nabla^2 \psi_1+k^2\psi_1=u e^{i kx}$$

$$\nabla^2 \psi_2+k^2\psi_2=u \psi_1$$

$$\nabla^2 \psi_3+k^2\psi_3=u \psi_2$$

$$ \dots $$

It seems to me that if we sum all of the equation above the series then obey the original inhomogenous Schrodinger equation.

However, solving each of the equations above is relatively straightforward, through the Green's function, for example:

$$\psi_1( \mathbf{r})=-\int_{ \mathbb{R} ^3} G( | \mathbf{r}- \mathbf{r'} |) u( \mathbf{r'}) e^{i kx'} dV'$$

Naturally, I have attempted this method for the most simple problem: 1D finite rectangular potential well:

$$u=-u_0 [a \leq x \leq b]$$

The 1D Green's function for the Sommerfeld condition is:

$$G(x)=\frac{i}{2k} e^{i k |x|}$$

Then I obtain the following series for $\psi(x)$ for $x>b$:

$$\psi(x)=\left( \frac{i u_0}{2k}(b-a)+\frac{i^2 u_0^2}{4k^2}(b-a)^2+\frac{i^3 u_0^3}{8k^3}(b-a)^3+ \dots \right) e^{ikx}$$

Or, using geometric series and summing also the incoming wave, we obtain for the transmitted wave:

$$\Psi_t (x)=\frac{1}{1-i \frac{u_0(b-a)}{2k}} e^{i kx}$$

Here our main assumption is that $$ \left| \frac{u_0(b-a)}{2k} \right| <1$$

The transmission probability then is:

$$T(k)=\frac{4k^2}{4k^2+u_0^2 (b-a)^2}$$

Meanwhile, exact solution of the Schrodinger equation for this problem gives us the following formula:

$$T(k)=\frac{4k^2}{4k^2+u_0^2 \frac{\sin^2 \left((b-a)\sqrt{k^2+u_0} \right)}{k^2+u_0}}$$

To obtain our approximate expression we should use the following assumption: $$|(b-a)\sqrt{k^2+u_0}| << 1$$

My question is: why does the assumption needed to obtain this approximate solution different and stronger than the assumption I used to sum the series? What are the hidden assumptions I seem to have made which influenced this result?

It seems to me that since in the approximate expression for $T(k)$ we do not get any resonanses (i.e. virtual bound states) and the second condition looks to be that the wavelength of the particle is much larger than the width of the well, then I probably can't use Sommerfeld conditions for all the functions $\psi_k$ as I did. Is this right?


1 Answer 1


The condition that your series converges at large distances may not be sufficient. Every element of the series should have the same large distance asymptotics such that you can recover a physical scattering amplitude. You assert that this is the case but this assertion is left dangling - you don't know if that's the case and hence you can't justify that your $\psi(x)$ is correct.

I'm a little confused as to why you would not just expand G in a pertubation series instead. You're doing something similar in spirit, why not do what is conventional?

  • $\begingroup$ Thank you for your answer! I wanted to do something new. As for the condition, I was sure in the example I used the series converges for every x>b. Is this not sufficient? Should I also check the convergence for every x<b? I mean the condition for convergence $\endgroup$
    – Yuriy S
    Commented Nov 22, 2017 at 20:37
  • $\begingroup$ @YuriyS another issue is that the recursive relation you write down doesn't really follow, at least intuitively. You really should be expanding E (or $k^2$) as a series too for consistency which would just make this perturbation theory. The overarching problem, I think, is that your series is physically inconsistent at each order which is what I meant in my answer. $\endgroup$
    – BB681
    Commented Nov 22, 2017 at 21:01
  • $\begingroup$ @ BB681, thank you again, I'll try to wrap my head around all this. Unfortunately, perturbation theory is not what I'm looking for, but I will look into something else, maybe I'll have to use some discretization scheme $\endgroup$
    – Yuriy S
    Commented Nov 22, 2017 at 22:07
  • 1
    $\begingroup$ @YuriyS To save you some trouble, $\nabla^2 \psi( \mathbf{r})+(k^2-u( \mathbf{r}))\psi( \mathbf{r})=u( \mathbf{r}) e^{i kx}$ does not have a general solution for arbitrary u(r). What you have to do is either numerics or some perturbative technique or just try ansatz solutions and optimize. $\endgroup$
    – BB681
    Commented Nov 22, 2017 at 22:14
  • $\begingroup$ If I may ask, what do you mean by "consistency" in this case? Is there a self-contradiction involved in the recursive relation I set? I thought it had a physical interpretation as each wave-like term acts as a source for the previous wave-like term, modulated by the potential $\endgroup$
    – Yuriy S
    Commented Nov 24, 2017 at 12:36

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