Derivation of tunneling probability through barrier

Tunnelling is the quantum effect in which a particle with energy $E$ passes through a potential barrier of height $V > E$, which is impossible in classical mechanics. It happens because the wave function does not drop to zero inside the barrier but decays exponentially; if the barrier is thin, some of the wave reaches the far side. Examples: tunnel diode, scanning tunnelling microscope, field emission, conduction through thin oxide layers on contacts.

 V(x)
  ^        __________
  |   V   |          |
  | E-----|- - - - - |-----
  | ~~~~~ | \_       | ~~~
  |  I    |  II \__  | III
  |_______|__________|_____> x
          0          L

Wave functions in three regions

Let $k = \dfrac{\sqrt{2mE}}{\hbar}$ and $\alpha = \dfrac{\sqrt{2m(V - E)}}{\hbar}$.

• Region I ($x < 0$, $V = 0$): $\psi_I = A_1 e^{ikx} + B_1 e^{-ikx}$ (incident + reflected)
• Region II ($0 < x < L$, potential $V$): $\psi_{II} = A_2 e^{\alpha x} + B_2 e^{-\alpha x}$
• Region III ($x > L$, $V = 0$): $\psi_{III} = A_3 e^{ikx}$ (transmitted only)

Boundary conditions

$\psi$ and $d\psi/dx$ are continuous at $x = 0$ and $x = L$:

$$\begin{aligned} A_1 + B_1 &= A_2 + B_2 \\ ik(A_1 - B_1) &= \alpha(A_2 - B_2) \\ A_2 e^{\alpha L} + B_2 e^{-\alpha L} &= A_3 e^{ikL} \\ \alpha(A_2 e^{\alpha L} - B_2 e^{-\alpha L}) &= ik A_3 e^{ikL} \end{aligned}$$

Transmission coefficient

The tunnelling probability is the ratio of transmitted to incident flux. Since the speed is the same in regions I and III:

$$T = \frac{|A_3|^2}{|A_1|^2}$$

Eliminating $B_1, A_2, B_2$ from the four equations gives the exact result

$$T = \left[1 + \frac{V^2 \sinh^2(\alpha L)}{4E(V - E)}\right]^{-1}$$

Wide or high barrier ($\alpha L \gg 1$)

Then $\sinh(\alpha L) \approx \tfrac{1}{2}e^{\alpha L}$, so $\sinh^2(\alpha L) \approx \tfrac{1}{4}e^{2\alpha L} \gg 1$ and the 1 can be neglected:

$$T \approx \frac{16E(V - E)}{V^2}\, e^{-2\alpha L} = T_0 \exp\left[-\frac{2L\sqrt{2m(V - E)}}{\hbar}\right]$$

with $T_0 = \dfrac{16E(V-E)}{V^2}$.

Conclusions

1. $T$ falls exponentially with barrier width $L$; doubling $L$ greatly reduces tunnelling.
2. $T$ falls as $\sqrt{V - E}$ rises, so a higher barrier or a lower electron energy reduces tunnelling.
3. Heavier particles (larger $m$) tunnel far less, which is why tunnelling is noticed for electrons and only for barriers of about a nanometre or less.
4. Reflection probability is $R = 1 - T$.