Barrier penetration, tunneling

In the previous example, the quantum particle could penetrate for some distance through walls which were infinitely thick. What would happen if a quantum particle were to propagate towards walls which were not infinitely thick ? The particle in fact has a finite probability of passing through such a wall. This is known as the tunneling effect, and it is a purely quantum phenomenon. Figure 16 depicts a particle incident from the left approaching a barrier. The height of the barrier $U$ is larger than the energy $E$ of the particle, $E<U$. Some of the amplitude of the quantum particle is reflected, while some passes through the barrier (non-classical behaviour), and emerges to continue travelling towards the right on the other side.

Figure 16: A particle with an energy below the barrier height approaches the barrier. Some amplitude is reflected, and some amplitude tunnels through to the other side.

The potential is :

$$U(x) = \left\{ \begin{array}{lll} 0 & x < 0 &\qquad \mbox{r... ... 0 & x > L &\qquad \mbox{region III} \end{array} \right. .$$ (87)

It is clear, as in the previous problem, that in the potential free areas we will have oscillatory solutions, while in the area where the particle energy is less than the barrier height $E<U$, we will have exponential solutions.

Using the identities

$$e^{i\theta} = \cos\theta + i\sin\theta$$ (88)

$$e^{-i\theta} = \cos\theta - i\sin\theta$$ (89)

we can write the oscillatory solution as

$$\psi_{I} = Ae^{ik_1 x} + Be^{-ik_1x}$$ (90)

$$\psi_{III} = Fe^{ik_1x} + Ge^{-ik_1x}$$ (91)

where

$$k_1 = i\frac{\sqrt{2mE}}{\hbar}.$$ (92)

This is a more suitable form to describe wave functions moving in a particular direction. For example,

$$\psi_{I+} = Ae^{ik_1 x} \qquad \mbox{incident wave}$$ (93)

is a wave function in region I incident from the left. At $x=0$, the wave $\psi_{I+}$ strikes the barrier and is partially reflected, as the wave function

$$\psi_{I-} = Be^{-ik_1 x} \qquad \mbox{reflected wave}.$$ (94)

A classical particle would rebound off the barrier completely, not partially. However, we already know that some of the amplitude of the wave function will penetrate into the barrier in the classically non-allowed region II. Inside the barrier (region II from $0 \le x \le L$), we know that $U>E$, and therefore the Schrödinger wave equation will have not have oscillatory solutions, but exponential solutions.

$$\psi_{II} = Ce^{-k_2x} + De^{-k_2x}$$ (95)

where

$$k_2 = \frac{\sqrt{2m(U-E)}}{\hbar}.$$ (96)

If the barrier is not infinitely thick, then some amplitude may still remain on the far side of the barrier. On the far side of the barrier, in region III ($x>L$), the transmitted wave would again be moving towards the right.

$$\psi_{III+} = Fe^{ik_1 x} \qquad \mbox{transmitted wave}$$ (97)

Applying the boundary conditions, we demand that the wave functions in all three regions match smoothly to each other, in both value and slope.

$$\left . \begin{array}{r c l} \psi_{I} &=& \psi_{II} \\ ... ...}{\partial x}\psi_{II} \\ \end{array} \right \} \quad x = 0$$ (98)

at the left hand side of the barrier and

$$\left . \begin{array}{r c l} \psi_{II} &=& \psi_{III} \ ... ...l}{\partial x}\psi_{III} \end{array} \right \} \quad x = L.$$ (99)

Specifying these conditions, we find we must solve the equations

$$A + B = C + D$$ (100)

$$ik_1A - ik_1B = - ik_2C + ik_2D$$

$$Ce^{-k_2 L} + De^{k_2 L} = Fe^{ik_1 L}$$

$$-k_2Ce^{-k_2 L} + k_2De^{k_2 L} = ik_1Fe^{ik_1 L}$$

Considering figure 16, we can see that the interesting quantity to calculate is the amount of transmitted wave amplitude relative to the incident amplitude. This is know as the transmission coefficient $T=\vert\frac{F}{A}\vert^2$. (The steps to realise the transmission coefficient from here on are presented for your interest, and are not examinable.)

$$\frac{A}{F} = \left [ \frac{1}{2} + \frac{i}{4}\left(\frac{k_2}{k_1} - \frac{k_1}{k_2} \right) \right] e^{(ik_1+k_2)L}$$ (101)

$$+ \left [ \frac{1}{2} - \frac{i}{4}\left(\frac{k_2}{k_1} - \frac{k_1}{k_2} \right) \right] e^{(ik_1-k_2)L}.$$

If the barrier is much higher than the particle energy, then $\frac{k_2}{k_1}>\frac{k_1}{k_2}$ so that

$$\frac{k_2}{k_1} - \frac{k_1}{k_2} \approx \frac{k_2}{k_1}.$$ (102)

Suppose also the barrier is wide enough so that the transmitted amplitude is weak $k_2L \gg 1$

$$e^{k_2 L} \gg e^{-k_2 L}.$$ (103)

We can now simplify the expression for the transmission probability

$$\frac{1}{T}=\left \vert \frac{A}{F}\right\vert^2 = \left\ver... ...}{2} +\frac{ik_2}{4k_1} \right) e^{(ik_1-k_2)L} \right\vert^2.$$ (104)

finally

$$T=\left\vert\frac{F}{A}\right\vert^2 = \left ( \frac{1}{4} -\frac{k_2^2}{16k_1^2}\right)^{-1} e^{-2k_2L}.$$ (105)

As the bracketed quantity is more slowly varying than the exponential, we can simplify the transmission probability further to

$$T \approx e^{-2k_2L}.$$ (106)

The transmission of a quantum particle through a barrier is shown in figure 17 below.

Figure 17: A quantum particle tunneling through a barrier.