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PHY.K02UF Molecular and Solid State Physics

## Tunneling through a rectangular barrier

If the energy of a quantum particle is lower than the potential energy of some barrier region, $E < V_1$, but it is higher than the potential energy outside of this barrier, $E < V_0,\,V_2$, the wave function can extend through the classically forbidden region and continue on the other side. This phenomena is known as tunneling. The wave function for this situation is shown below.

 $E$ [eV] $\psi$ $x$ Å
 $V_0=0$ [eV] $L_0=10$ [Å] $V_1=$ 5 [eV] $L_1=$ 3 [Å] $V_2=$ 0 [eV] $L_2=10$ [Å]
 $E=$ 10 [eV]

10 Å-110 Å-110 Å-1

We assume there is a wave $A_i\exp(ik_0x)$ incident from the left that is partially reflected in the wave $A_r\exp(-ik_0x)$, and partially transmitted through the barrier, $A_t\exp(ik_2x)$. For, $V_0,\,V_2 < E < V_1$, the wave function at $x = -L_1$ is given by the matrix equation,

$$\begin{bmatrix} \psi(-L_1)\\ \frac{d\psi(-L_1)}{dx} \end{bmatrix} = \begin{bmatrix} \cosh(-\kappa_1L_1) & \frac{\sinh(-\kappa_1L_1)}{\kappa_1}\\ \kappa_1\sinh(-\kappa_1L_1) & \cosh(-\kappa_1L_1) \end{bmatrix} \begin{bmatrix} A_t\\ ik_2A_t \end{bmatrix},$$

while for $V_0,\,V_1,\,V_2 < E$,

$$\begin{bmatrix} \psi(-L_1)\\ \frac{d\psi(-L_1)}{dx} \end{bmatrix} = \begin{bmatrix} \cos(-k_1L_1) & \frac{\sin(k_1L_1)}{k_1}\\ -k_1\sin(-k_1L_1) & \cos(-k_1L_1) \end{bmatrix} \begin{bmatrix} A_t\\ ik_2A_t \end{bmatrix}.$$

The wave function at $x=-L_1$ is a sum of the incident and relfected wave,

$$A_i+A_r = \psi(-L_1),$$ $$ik_0A_i+ik_0A_r = \frac{d\psi(-L_1)}{dx}.$$

These can be solved for the amplitudes of the incident and reflected wave,

$$A_i = \frac{\psi(-L_1)}{2} -\frac{i}{2k_0}\frac{d\psi(-L_1)}{dx},$$ $$A_r = \frac{\psi(-L_1)}{2} +\frac{i}{2k_0}\frac{d\psi(-L_1)}{dx}.$$

The reflection coefficient is,

$$R = \frac{A_r^*A_r}{A_i^*A_i},$$

and the transmission coefficient is $T = 1-R$.

 $T$ $E$ [eV]
 $V_0=0$ [eV] $L_0=10$ [Å] $V_1=$ 5 [eV] $L_1=$ 3 [Å] $V_2=$ 0 [eV] $L_2=10$ [Å]