1.3: Scattering From a 1D Delta-Function Potential
- Page ID
- 34590
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.
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:
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.