Tunneling effect and one dimensional barriers in Quantum Mechanics

Quantum mechanical tunneling effect or barrier penetration the the process in which a particle is transmitted through a potential barrier of finite width and height even when its energy is less that that of the barrier height. This is possible due to wave nature of particle and the particle appears to cross the barrier without going over the top. It is used to explain $\alpha - decay$ , field emission, field ionization and tunnel diodes.

The transmission of the particle is measured as a probability of the particle transmission through the barrier and is given by

$T=\dfrac{\text{Transmitted probability current density}}{\text{Incident probability current density}}$

Let us take a potential step as $V(x)=0; x<a \text{ and } V(x)=V_0;x \ge a$ and incident particle $E>V_0$ .

Consider the beam of particles with energy E incident from left to right. In the case of $E>V_0$ we have the Schrodinger equations in region I and II as;

Region I:

$-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi_I}{dx^2}=E\psi_I$

$\psi_I + K_0^2\psi_I=0 \text{ where } k_0^2=\dfrac{2mE}{\hbar^2}$ …… (1)

Region II:

$-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi_{II}}{dx^2}+V_0\psi_{II}=E\psi_{II}$

$\psi_{II} + K^2\psi_{II}=0 \text{ where } k^2=\dfrac{2m}{\hbar^2}(E-V_0)$ …….. (2)

The solution of (1) and (2) are

$\psi_I = Ae^{ik_0x} + Be^{-ik_0x}$

$\psi_{II} = Ce^{ikx} + De^{-ikx}$

Since no particle coming from the right D = 0;

$\psi_{II} = Ce^{ikx}$

The boundary condition are:

$\psi_I = \psi_{II} \text{ at } x=a$

and $\psi_I^{'} = \psi_{II}^{'} \text{ at } x=a$

Applying boundary condition we get,

$Ae^{ik_0a}+Be^{-ik_0a}=Ce^{ika}$ …….. (3)

and $ik_0Ae^{ik_0a}-ik_0Be^{ik_0a}=(ik)Ce^{ika}$

or $Ae^{ik_0a}-Be^{ik_0a}=Ce^{ika}(\dfrac{k}{k_0})$ ……… (4)

Adding equation (3) and (4) we get,

$\dfrac{C}{A}=\dfrac{2k_0}{k+k_0}e^{ia(k-k_0)}$

Subtracting equation (3) and (4), we get,

$\dfrac{B}{A}=\dfrac{k_0-k}{k_0+k}e^{2ik_0a}$

The reflection coefficient

$R_E=\dfrac{\text{Reflected probability current density}}{\text{Incident probability current density}}$
$=\dfrac{\hbar k_0}{m} \dfrac{B^2}{A^2}$
$=[\dfrac{B}{A}][{\dfrac{B}{A}}]^{\ast}$
$=\dfrac{(k_0-k)^2}{(K_0+k)^2}e^{2ika}e^{-2ika}$
$=\dfrac{[1-(\dfrac{k}{k_0})^2]}{[1+(\dfrac{k}{k_0})^2]}$
$R_E = [\dfrac{1-\mu}{1+\mu}]^2$ here $\mu=\dfrac{k}{k_0}$