# Two approaches to a simple scattering problem

this is something that have puzzled me for some time and I couldn't figure it out. Consider the very simple first quantization $$1d$$ scattering problem ($$\hbar=1$$):

$$H = -iv_F \partial_x + c \delta(x)$$ where a main point here is the linearized dispersion relation. This can describe for example a scattering of an edge mode in a topological insulator off an impurity. We'll solve it in two ways.

# approach 1

A direct solution to the eigenproblem $$\left[-iv_F\partial_x + c\delta(x)\right]\psi(x)=v_F k \psi(x)$$. We have $$-iv_F\frac{d\psi}{\psi} = \left[v_Fk + c\delta(x)\right]dx$$ and by integrating we have $$\psi(x) = e^{i\frac{c}{v_F}\theta(x)}e^{ikx}$$ up to a normalization factor. We get the scattering phase $$c/v_F$$.

# approach 2

we write an Ansatz $$\psi(x) = \left[\theta(-x) + T \theta(x)\right]e^{ikx}$$ and by integrating an infinitesimal about $$0$$ $$\lim_{\epsilon\to 0}\int_{-\epsilon}^{\epsilon}H\psi(x) = \lim_{\epsilon\to 0}\int_{-\epsilon}^{\epsilon}v_F k \psi(x)$$ one gets $$iv_F \psi(x=-\epsilon) - iv_F\psi(x=\epsilon) + c\psi(0) = 0$$ we need to decide what is $$\psi(0)$$ and we choose $$\theta(0)=1/2$$ to get $$iv_F(1-T) = -\frac{c}{2}(1+T)$$ which gives $$\frac{1-T}{1+T} = -i\frac{c}{2v_F}$$ for $$T=\exp\left[2i \tan^{-1}(c/2v_F)\right]$$ so the scattering phase is $$2i\tan^{-1}(c/2v_F)$$ which is different than the previously found one.

# discussion and question

while the first approach is more direct, and the second one was done via an ansatz and an assumption on $$\theta(0)$$, I "like" the second result more. It makes no sense for $$c$$ to have unique values such that $$c=0$$ and $$c=2\pi n v_F$$ result in the same answer. Why the multiples of $$\pi$$? The second result is physically appealing, with $$c\to \pm \infty$$ resulting in scattering phase of $$\pm \pi$$, and an interpolation between.

The different results can also lead to physically meaningful different quantities. If we place everything on a line of length $$L$$ and enforce periodic boundary conditions, we get different quantization for the values of $$k$$ as $$k_n L + \varphi = 2\pi n$$ with $$\varphi$$ the scattering phase.

So which approach is the correct one (if any) and what is the source of the difference? I have a feeling that it has something to do with the linearized spectrum but I am really not sure about it.

• what earned me the downvote? i am open to suggestions for improvements
– user275556
Feb 23, 2023 at 15:15

The first method is the correct one. When in doubt, it’s best to revert to approximations of unity. I will use the usual one where I replace delta by: $$\delta_\epsilon(x)=\frac{f(x/\epsilon)}{\epsilon}$$ A usual choice is $$f=\text{rect}$$ the indicator function of the interval $$(-1/2,1/2)$$. But any $$f$$ of integral $$1$$ will do.
Up to a change of timescale, I'll set $$v_F=1$$. In this case, your first method still works: $$\psi(x) \propto \exp\left(ikx-ic\int_{-\infty}^x ds \delta_\epsilon(s)\right)$$
As $$\epsilon\to 0$$ this gives your first result, and the phase shift is exactly $$-c$$.
If you try to apply the second reasoning on this case, you'll find the problem. To make things easier, I'll assume $$f$$ to be supported in $$(-1,1)$$. You can again integrate over $$(-\epsilon,\epsilon)$$ your formula is still correct as $$\epsilon\to 0$$, except for the $$c\psi(0)$$ term. It is your choice of $$\theta(0)=1/2$$ which is problematic. You essentially impose: $$\frac{\psi(-\epsilon)+\psi(\epsilon)}{2} = \int_{-\epsilon}^\epsilon \delta_\epsilon(x)\psi(x)$$ But this is not true as you can check from the explicit formula. Form your perspective, $$\theta(0)$$ implicitly depends on $$\psi$$, so not only is the method wrong, but even when corrected does not provide resolution of the problem.