# Normal Base for Hilbert Space of delta Potential Well

I'm interested in the problem of an attractive $$\delta$$ potential. The Hamiltonian is given by

$$H = - \frac{\partial_x^2}{2m} - V \delta(x).$$

Solving this typically entails looking at scattering and bound states separately (see, for example, Wikipedia). The bound state

is

$$\psi_\mathrm{B}(x) = \sqrt{mV} \mathrm{e}^{-mV |x|}$$

while the scattering solutions are for a wave coming in from the left of the form

$$\psi_{k,\mathrm{l}}(x) = \begin{cases} A \mathrm{e}^{\mathrm{i}kx} + r\mathrm{e}^{-\mathrm{i}kx}& \text{if x<0} \\\\ t \mathrm{e}^{\mathrm{i}kx} & \text{otherwise,} \end{cases}$$

and correspondingly, for a wave from the right. Up to a global prefactor, $$A$$, $$r$$ and $$t$$ are fixed and functions of $$V$$, $$m$$ and $$k$$.

What does a complete, normal base for the problem look like? The bound states is as is orthogonal to the scattering states, but the same cannot be said for the scattering states of different $$k$$ or direction. And do I need to include left and right moving states?

More concretely, I need to evaluate terms of the form

$$\sum_m \left c_m$$

for a (time-dependent) perturbation.

• You can construct eigenstates with periodic boundary conditions as linear combinations of the scattering states. Commented May 27 at 14:14
• I don't really want periodic boundary conditions as, as an end goal, I want to move the potential. And changing coordinates to a moving frame would be quite a bit more involved with periodic boundary conditions. Commented May 28 at 9:03

A complete set of solutions of the eigenvalue problem $$-\frac{1}{2m} \phi^{\prime \prime} (x) -V \delta(x) \phi(x) =E \phi(x), \quad V\gt0,$$ is given by \begin{align} \phi_B(x)&=\sqrt{mV} e^{-mV|x|}, \\[5pt] \phi_k(x) &= \begin{cases} \frac{1}{\sqrt{2\pi}}\left[\left(e^{ikx}+\frac{imV}{k-imV}e^{-ikx}\right)\Theta(-x)+\left(1+\frac{imV}{k-imV}\right)e^{ikx}\Theta(x)\right] \; \,\text{for} \, k\gt 0 \\[5pt] \frac{1}{\sqrt{2\pi}}\left[\left(e^{ikx} -\frac{imV}{k+imV} e^{-ikx} \right)\Theta(x)+\left(1-\frac{imV}{k+imV} \right)e^{ikx}\Theta(-x) \right] \; \, \text{for} \; k\lt 0 \end{cases} , \end{align} with energy eigenvalues $$E=-mV^2/2$$ for the bound state $$\phi_B$$ and $$E=k^2/2m$$ for the scattering states $$\phi_k$$.

Checking orthogonality ($$\phi_B \perp \phi_k$$, $$\langle \phi_k |\phi_{k^\prime}\rangle =\delta(k-k^\prime)$$) and the completeness relation $$\phi_B(x)\phi_B(y)^\ast+\int_\limits{-\infty}^\infty \! dk \, \phi_k(x)\phi_k(y)^\ast=\delta(x-y)$$ is left as a straightforward (but somewhat tedious) homework exercise.

• Thanks a lot for the answer! That's basically what I've tried, but I can't get $\left<\phi_k|\phi_k'\right>=\delta(k-k')$ as I'm always getting some additional constant term. So you're saying that this is just because I made some computational mistakes that I should find? Commented May 28 at 14:06
• @DanielHauck Hint: $\lim\limits_{\epsilon \downarrow 0} \int\limits_0^\infty\! dx \, e^{ipx-\epsilon x}=i \mathcal{P} \frac{1}{p}+ \pi \delta(p)$ and $\lim\limits_{\epsilon \downarrow 0} \int\limits_{-\infty}^0 \! dx \, e^{ipx +\epsilon x} = -i \mathcal{P} \frac{1}{p} +\pi \delta(p)$. Commented May 29 at 5:20
• Now I got it to work, thank you so much! In the end, I was hung up on some quirkiness of Mathematica when evaluating Fourier transforms. Commented Jun 3 at 14:38

The usual approach for scattering problems is to construct a basis consisting of a continuous part (scattered waves) and a discrete part (the bound states). You will then of course get combinations of sums and integrals in the calculations.

The states in the continuous part can be "in" states, like the $$\psi_{k,l}$$ and $$\psi_{k,r}$$, but you can also construct "out" states, which are essentially the complex conjugate. Look for instance at [question 635540] to see if you need them.

For the "in" states, you could also consider symmetric and antisymmetric combinations of $$\psi_{k,l}$$ and $$\psi_{k,r}$$, if that makes it easier to solve the system. (A bit like how a plane wave from one direction in 3D is decomposed in waves with a certain simplifying symmetry, the $$Y_l^m$$'s, but here you do not need a complicated Bauer formula to do that!)

• Thank you for the answer! Appart from the out states, that's what I'm trying to do. But for some reason, no matter what I try I can't get the scattering states to be orthogonal for different $k$. Does this mean I just made some calculation mistakes? For me, $\left<k,\mathrm{l}|k',\mathrm{l}\right>$ (or the corresponding right moving part) isn't just proportional to $\delta(k-k')$, but has a constant part. Here I mean with the notation, that $\Psi_{k,\mathrm{l}}=\left<x|k,\mathrm{l}\right>$. Commented May 28 at 9:01
• Orthogonality seems a bit non-trivial with infinite integrals, I'd have to give it a closer look... Commented May 28 at 9:32
• Is there even an outgoing eigenstate here? I tried an ansatz of $\psi_\mathrm{out}(x)=A\theta(x) \mathrm{e}^{\mathrm{i}kx}+B\theta(-x)\mathrm{e}^{-\mathrm{i}kx}$ with the Heaviside theta function. And this can't be an eigenstate as there is no way to satisfy $\psi_\mathrm{out}(0^+)=\psi_\mathrm{out}(0^-)$ and $\psi_\mathrm{out}'(0^+)-\psi_\mathrm{out}'(0^-)=-2mV\psi_\mathrm{out}(0)$ for a real $k$. Commented May 28 at 11:37