Scattering

Free Particle

In the time-independent Schrodinger's equation, when there is no potential $$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial^2x}\phi_E = E \phi_E$$ Let $k = \frac{\sqrt{2mE}}{\hbar}$ The general solution is $$\phi_E(x, 0) = Ae^{ikx} + B e^{-ikx}$$ which is not normalizable. So, a free particle with single plane wave as wave function is not physical. The wave function can only be a wave packet. $$\Psi(x, 0) = Ne ^{-\frac{(x-x_0)^2}{2a^2}}e^{ikx}$$ which is a superposition of momentum eigenfunctions $$\Psi(x, 0) = \int dk \frac{e^{ikx}}{\sqrt{2\pi}}f(k)$$ where $$f(k) = e^{-\frac{a^2(k-k_0)^2}{2}}\tilde{N}e^{-ikx_0}$$ So, the wave function should be $$\Psi(x,t) = \int dk f(k) \frac{1}{\sqrt{2\pi}} e^{i(kx - \omega t)}$$

Potential Step

When the energy of the particle $E$ is smaller than $V_0$ The general solution is $$\phi_E(x) = \begin{cases} Ae^{ikx}+B^{-ikx}, \text{ $$x<0$$ }\\ Ce^{-\alpha x} + De^{\alpha x}, \text{ $$x>0$$ } \end{cases}$$ where $$k = \frac{\sqrt{2mE}}{\hbar}$$ and $$\alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar}$$ Let it be scattering from left to right, so $D = 0$ $$\phi_E(x) = \begin{cases} Ae^{ikx}+B^{-ikx}, \text{ $$x<0$$ }\\ Ce^{\alpha x}, \text{ $$x>0$$ } \end{cases}$$ $\phi_E$ and $\phi_E'$ are continuous at $x=0$, so $$\begin{cases} A + B = C \\ ikA - ikB = -\alpha C \end{cases}$$ So, $$B = \frac{ik + \alpha}{ik -\alpha}A = \frac{k -i\alpha}{k + i\alpha}A$$ and $$C = \frac{2ik}{ik - \alpha}A = \frac{2k}{k+i\alpha}A$$ Define $$t \equiv \frac{B}{A} = \frac{k -i\alpha}{k + i\alpha}A$$ and $$r \equiv \frac{C}{A} = \frac{2k}{k+i\alpha}A$$ to illustrate the ratio of the amplitude of the transmitted and reflected wave to the incident wave. The transmission probability is $$T = \frac{|J_C|}{|J_A|} = \frac{|C|^2\frac{\alpha}{m}}{|A|^2\frac{k}{m}} = |t|^2\frac{\alpha}{k} = \frac{4\alpha k}{(k+\alpha)^2}$$ and the reflection probability is $$R = \frac{|J_B|}{|J_A|} = \frac{|B|^2\frac{k}{m}}{|A|^2\frac{k}{m}} = |r|^2 = \left(\frac{k - \alpha}{k + \alpha}\right)$$ Similarly for $E > V_0$.