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.

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

1 Answer 1


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.

Hope this helps.


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