Question:
Find the coefficients of transmission and reflection for a particle incident on the potential shown. The particle has energy, \(E\), which is lager than the potential energy, \(V_0\).
Answer:
This is a quantum system, so we can use the Schroedinger equation.
\[ H\psi = E\psi \]
where \(H\), \(\psi\), and \(E\) are the hamiltonian, the wave function, and the energy eigenvalue, respectively. The hamiltonian operator consists of the kinetic and potential energies. In this case, the incident beam is a free particle, so there is zero potential energy before incident on the potential. Thus, we need to consider two cases of the equation:
When \(x<0\),
\[ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} \psi = E\psi \]
\[ \psi” + \frac{2mE}{\hbar^2} \psi = 0 \]
The momentum can be expressed as \(p=\hbar k\) where \(k\) is the wave number. Thus,
\[ E = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m} \]
We can derive
\[ k^2 = \frac{2mE}{\hbar^2} \]
The equation (2) will become
\[ \psi” + k^2 \psi = 0 \]
The solution is
\[ \psi_I = Ae^{ikx} + Be^{-ikx} \]
When \(x>0\),
\[ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} \psi = (E-V_0)\psi \]
\[ \psi” + \frac{2m(E-V_0)}{\hbar^2} \psi = 0 \]
Let us put \(k’=\frac{2m(E-V_0)}{\hbar^2}\). The solution is
\[ \psi_{II} = Ce^{ik’x} + De^{-ik’x} \]
The term \(De^{-ik’x}\) represents the wave emanating from the right side, which is none. Therefore, we can let \(D=0\). Namely,
\[ \psi_{II} = Ce^{ik’x} \]
Due to the consistency of quantum theory, the wave function and its first derivative must be continuous at \(x=0\). This means
\[ \psi_I(0) = \psi_{II}(0) \]
\[ \frac{d}{dx}\psi_I(0) = \frac{d}{dx}\psi_{II}(0) \]
Use the previous results \(\psi_I = Ae^{ikx} + Be^{-ikx}\) and \(\psi_{II} = Ce^{ik’x}\).
\[ A + B = C \]
\[ A – B = \frac{k’}{k}C \]
Take the ratios \(C/A\), which is related to the transmission, and \(B/A\), which is related to the reflection.
\[ \frac{C}{A}=\frac{2}{1+k’/k}, \quad \frac{B}{A}=\frac{1-k’/k}{1+k’/k} \]
Now, the current density \(J\) for the Schroedinger equation can be derived from the continuity equation, \(\frac{\partial \rho}{\partial t}+\frac{\partial J}{\partial x}=0\):
\[ J = \frac{\hbar^2}{2mi}\left(\psi^* \frac{d\psi}{dx} – \psi \frac{d\psi^*}{dx}\right) \]
We can obtain the current density for the incident, transmission, and reflection by using equation \(J = \frac{\hbar^2}{2mi}\left(\psi^* \frac{d\psi}{dx} – \psi \frac{d\psi^*}{dx}\right)\); and \(\psi_{\mathrm{inci}}=Ae^{ik}\), \(\psi_{\mathrm{trans}}=Ce^{ik’}\), and \(\psi_{\mathrm{reflec}}=Be^{ik}\)
\[ J_{\mathrm{inci}} = \frac{\hbar^2}{2mi}2ik|A|^2 \]
\[ J_{\mathrm{trans}} = \frac{\hbar^2}{2mi}2ik’|C|^2 \]
\[ J_{\mathrm{reflec}} = \frac{\hbar^2}{2mi}2ik|B|^2 \]
Let us define transmission and reflection coefficients, \(T\) and \(R\).
\[ T \equiv \left| \frac{J_{\mathrm{trans}}}{J_{\mathrm{inci}}} \right| = \left|\frac{C}{A}\right|^2 \frac{k’}{k} \]
\[ R \equiv \left| \frac{J_{\mathrm{reflec}}}{J_{\mathrm{inci}}} \right| = \left|\frac{B}{A}\right|^2 \]
Use the result of the ratios \(C/A\) and \(B/A\); then, we have
\[ T=\frac{4k’/k}{(1+k’/k)^2}, \quad R=\left|\frac{1-k’/k}{1+k’/k}\right|^2 \]