Let us consider the following scattering problem in 1 dimension, with potential

\[\begin{equation} V(x) = \begin{cases} \pm V_0 & 0 < x < a \\ 0 & \text{else} \end{cases} \end{equation}\]

where \(V_0 > 0\). The Schrödinger equation to solve is

\[\begin{equation} \left[-\frac{\hbar^2}{2m}\frac{\d^2}{\d x^2} + V(x)\right]\psi = E\psi \end{equation}\]

If \(V_0 > 0\), this is a potential barrier, and if \(V_0 < 0\), this is a potential well. For the case \(V_0 > 0\), there are no bound states at all, and if \(V_0 < 0\), there are bound states with negative energy, and scattering states with positive energy. We will solve both of these and contrast the solutions.

Potential well

Let us now turn to studying the potential well. The potential well is depicted below. The Schrödinger equation is given by

\[\begin{equation} \begin{cases} -\frac{\hbar^2}{2m}\psi" - V_0\psi = E\psi, & |x| < a \\ -\frac{\hbar^2}{2m}\psi" = E\psi, & |x| > a \end{cases} \end{equation}\]

Function

Potential well with depth \(V_0\).

To solve the problem, we solve the Schrödinger equation in each domain. If we are interested in scattering states, we only consider scattering states with \(E > 0\). Since \(E > 0\) means that the particle’s energy is greater than the potential in each domain, they are sinusoidal in each region. The wave function can be defined in three different domains,

\[\begin{equation} \psi(x) = \begin{cases} Ae^{ik_0x} + Be^{-ik_0x}, & x < -a \\ Ce^{ikx} + De^{-ikx}, & -a < x < a \\ Fe^{ik_0x}, & a < x \end{cases} \end{equation}\]

Here, \(k_0 = \sqrt{2mE}/\hbar\), and \(k = \sqrt{2m(E+V_0)}/\hbar\). The relation between the constants is found by applying the boundary conditions. We have 4 equations:

\[\begin{align} Ae^{-ik_0a} + Be^{ik_0a} ={}& Ce^{-ika} + De^{ika} \\ k_0(Ae^{-ik_0a}-Be^{ik_0a}) ={}& k(Ce^{-ika}-De^{ika}) \\ Ce^{ika} + De^{-ika} ={}& Fe^{ik_0a} \\ k(Ce^{ika} - De^{-ika}) ={}& k_0Fe^{ik_0a} \end{align}\]

First, we eliminate either \(C\) or \(D\) from both equations to get

\[\begin{align} A(k+k_0)e^{-ik_0a} + B(k-k_0)e^{ik_0a} ={}& 2Cke^{-ika} \\ A(k-k_0)e^{-ik_0a} + B(k+k_0)e^{ik_0a} ={}& 2Dke^{ika} \\ 2kCe^{ika}={}& (k+k_0)Fe^{ik_0a} \\ 2kDe^{-ika}={}& (k-k_0)Fe^{ik_0a} \end{align}\]

Then, we can rewrite these as

\[\begin{align} A(k+k_0)e^{i(k-k_0)a} + B(k-k_0)e^{i(k_0+k)a} ={}& 2kC \\ A(k-k_0)e^{-i(k_0+k)a} + B(k+k_0)e^{i(k_0-k)a} ={}& 2kD \\ 2kC={}& (k_0+k)Fe^{i(k_0-k)a} \\ 2kD={}& (k-k_0)Fe^{i(k_0+k)a} \end{align}\]

and finally eliminate both \(C\) and \(D\),

\[\begin{align} A(k+k_0)e^{i(k-k_0)a} + B(k-k_0)e^{i(k_0+k)a} ={}& (k_0+k)Fe^{i(k_0-k)a} \\ A(k-k_0)e^{-i(k_0+k)a} + B(k+k_0)e^{i(k_0-k)a} ={}& (k-k_0)Fe^{i(k_0+k)a} \end{align}\]

Lastly, we need \(B/A\) and \(F/A\), which are found by solving

\[\begin{equation} \begin{pmatrix} (k_0-k)e^{i(k_0+k)a} & (k_0+k)e^{i(k_0-k)a} \\ -(k_0+k)e^{i(k_0-k)a} & (k-k_0)e^{i(k_0+k)a} \end{pmatrix} \begin{pmatrix} B/A \\ F/A \end{pmatrix} = \begin{pmatrix} (k+k_0)e^{i(k-k_0)a} \\ (k-k_0)e^{-i(k_0+k)a} \end{pmatrix} \end{equation}\]

Thus, we get

\[\begin{align} \frac{B}{A} ={}& \frac{ie^{-2ik_0a}\sin(2ka)(k_0^2-k^2)/(2kk_0)}{\cos(2ka) - \frac{k_0^2+k^2}{2kk_0}i\sin(2ka)} \\ \frac{F}{A} ={}& \frac{-e^{-2ik_0a}}{\cos(2ka) - \frac{k_0^2+k^2}{2kk_0}i\sin(2ka)} \end{align}\]

The transmission coefficient is

\[\begin{align} T^{-1} ={}& \bigg|\frac{A}{F}\bigg|^2 \\ ={}& \cos^2(2ka) + \left(\frac{k_0^2+k^2}{2kk_0}\right)^2\sin^2(2ka) \\ ={}& 1 + \left[\left(\frac{k_0^2+k^2}{2kk_0}\right)^2 - 1\right]\sin^2(2ka) \\ ={}& 1 + \left(\frac{k_0^2-k^2}{2kk_0}\right)^2\sin^2(2ka) \end{align}\]

Now plugging in the wave vectors

\[\begin{equation} k_0 = \frac{\sqrt{2mE}}{\hbar}, \qquad k = \frac{\sqrt{2m(E+V_0)}}{\hbar} \end{equation}\]

we find the transmission coefficient to be

\[\begin{align} T^{-1} ={}& 1 + \frac{V_0^2}{4E(E+V_0)}\sin^2\left(\frac{2a}{\hbar}\sqrt{2m(E+V_0)}\right) \end{align}\]

We see that \(T^{-1}\) is minimized to 1 (so \(T\) is maximized to 1) when the sine is zero, or, when \(2ka = n\pi\). When this occurs, the wave in the potential region is commensurate with the size of the well, namely \(k = n\pi/2a\) (\(2a\) is the width of the well). When the transmission is \(T=1\) for this particular wave vector, we call this a resonance. In general, the energy of the particles which pass unimpeded through the potential are

\[\begin{equation} E = -V_0 + \frac{\hbar^2}{2m} \left(\frac{n\pi}{2a}\right)^2 \end{equation}\]

Since scattering states have \(E > 0\), the first resonance occurs when \(2a\sqrt{2mV_0}/\hbar\pi = n\).