1.3: Scattering From a 1D Delta-Function Potential

We are now ready to solve a simple scattering problem. Consider a 1D space with spatial coordinate denoted by $x$, and a scattering potential that consists of a “spike” at $x = 0$:

$$V(x) = \frac{\hbar^2\gamma}{2m} \,\delta(x).$$

The form of the prefactor $\hbar^2\gamma/2m$ is chosen for later convenience; the parameter $\gamma$, which has units of $[1/x]$, controls the strength of the scattering potential.

Figure $\PageIndex{1}$

If you are disturbed by the idea of a delta function potential, just regard it as the limiting case of a family of increasingly tall and narrow gaussian functions centered at $x=0$. For each non-singular potential, the applicability of the Schrödinger wave equation implies that the wavefunction $\psi(x)$ is continuous and has well-defined first and second derivatives. In the delta function limit, however, these conditions are relaxed: $\psi(x)$ remains continuous, but at $x=0$ the first derivative becomes discontinuous and the second derivative blows up. To see this, we integrate the Schrödinger wave equation over an infinitesimal range around $x = 0$:

$$\begin{align} \begin{aligned}\lim_{\varepsilon\rightarrow 0^+} \int_{-\varepsilon}^{+\varepsilon} dx\; \left[-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{\hbar^2\gamma}{2m} \delta(x)\right] \psi(x) &= \lim_{\varepsilon\rightarrow 0^+} \int_{-\varepsilon}^{+\varepsilon} dx\; E \psi(x) \\ = \lim_{\varepsilon\rightarrow 0^+} \left\{-\frac{\hbar^2}{2m} \left[\frac{d\psi}{dx}\right]_{-\varepsilon}^{+\varepsilon} \right\} + \frac{\hbar^2\gamma}{2m} \psi(0) &= 0 \end{aligned}\end{align}$$

Hence,

$$\lim_{\varepsilon\rightarrow 0^+} \left\{\; \left.\frac{d\psi}{dx}\right|_{x = +\varepsilon} - \left.\frac{d\psi}{dx}\right|_{x = -\varepsilon}\; \right\} = \gamma \,\psi(0). \label{delta_discontinuity}$$

To proceed, consider a particle incident from the left, with energy $E$. This is described by an incident state proportional to a momentum eigenstate $|k\rangle$, where $k = \sqrt{2mE/\hbar^2} > 0$. We said “proportional”, not “equal”, for it is conventional to adopt the normalization

$$|\psi_i\rangle = \sqrt{2\pi}\Psi_i |k\rangle \;\;\; \Leftrightarrow\;\;\; \psi_i(x) = \langle x|\psi\rangle = \Psi_i \, e^{ik x}.$$

The complex constant $\Psi_i$ is called the “incident amplitude.” Plugging this into the Schrödinger wave equation gives

$$\left[-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{\hbar^2\gamma}{2m}\delta(x)\right] \left(\Psi_i \, e^{ikx} + \psi_s(x) \right) = E \left(\Psi_i \, e^{ikx} + \psi_s(x) \right).$$

Taking $E = \hbar^2k^2/2m$, and doing a bit of algebra, simplifies this to

$$\left[ \frac{d^2}{dx^2} + k^2\right] \psi_s(x) = \gamma \delta(x) \left(\Psi_i \, e^{ikx} + \psi_s(x) \right),$$

which is an inhomogenous ordinary differential equation for $\psi_s(x)$, with the potential on the right hand side acting as a “driving term”.

To find the solution, consider the two regions $x < 0$ and $x > 0$. Since $\delta(x) \rightarrow 0$ for $x \ne 0$, the equation in each half-space reduces to

$$\left[\frac{d^2}{dx^2} + k^2\right] \psi_s(x) = 0.$$

This is the Helmholtz equation, whose general solution may be written as

$$\psi_s(x) = \Psi_i \left(f_1 \, e^{ik x} + f_2 \, e^{-ik x}\right).$$

Here, $f_1$ and $f_2$ are complex numbers that can take on different values in the two different regions $x < 0$ and $x > 0$.

We want $\psi_s(x)$ to describe an outgoing wave, moving away from the scatterer towards infinity. So it should be purely left-moving for $x < 0$, and purely right-moving for $x > 0$. To achieve this, let $f_1 = 0$ for $x < 0$, and $f_2 = 0$ for $x > 0$, so that $\psi_s(x)$ has the form

$$\psi_s(x) = \Psi_i \times \begin{cases}f_- \,e^{-ikx}, & x < 0 \\ f_+ \,e^{ikx}, & x > 0.\end{cases}$$

The complex numbers $f_-$ and $f_+$ are called scattering amplitudes. They describe the magnitude and phase of the wavefunction scattered backwards into the $x<0$ region, and scattered forward into the $x > 0$ region, respectively.

Recall from the discussion at the beginning of this section that $\psi(x)$ must be continuous everywhere, including at $x = 0$. Since $\psi_i(x)$ is continuous, $\psi_s(x)$ must be as well, so $f_- = f_+$. Moreover, we showed in Equation $\eqref{delta_discontinuity}$ that the first derivative of $\psi(x)$ is discontinuous at the scatterer. Plugging $\eqref{delta_discontinuity}$ into our expression for $\psi(x)$, at $x = 0$, gives

$$\Psi_i\big[ik(1+f_\pm) - ik(1-f_\pm)\big] = \Psi(1+f_\pm) \gamma.$$

Hence, we obtain

$$f_+ = f_- = -\frac{\gamma}{\gamma - 2ik}.$$

For now, let us focus on the magnitude of the scattering amplitude (in the next chapter, we will see that the phase also contains useful information). The quantity $|f_\pm|^2$ describes the overall strength of the scattering process:

$$|f_\pm|^2 = \left[1 + \frac{8mE}{(\hbar\gamma)^2}\right]^{-1}.$$

Its dependence on $E$ is plotted below:

Figure $\PageIndex{2}$

There are several notable features in this plot. First, for fixed potential strength $\gamma$, the scattering strength decreases monotonically with $E$—i.e., higher-energy particles are scattered less easily. Second, for given $E$, the scattering strength increases with $|\gamma|$, with the limit $|f|^2 \rightarrow 1$ as $|\gamma|\rightarrow \infty$. Third, an attractive potential ($\gamma < 0$) and a repulsive potential ($\gamma > 0$) are equally effective at scattering the particle.