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 \(E > V_0\).
Comments
Post a Comment