Tunnelling

Figure 20.8:

A particle of energy $E$ is incident on a step potential of hight $V>E$ as shown in Figure 20.8. The step potential extends from $x=0$ to $x=a$, the potential is zero on either side of the step. The particle has wavefunction

$$\psi_I(x,t) = e^{-iEt//\hbar} \left[ A_I e^{ i p x/\hbar} + B_I e^{-ipx/\hbar} \right]$$ (20.64)

in region $I$ to the left of the step. The first pat $A_I e^{ipx/\hbar}$ represents the incident particle ie. travelling along $+x$ direction and the second part $B_I e^{-ipx/\hbar}$ the reflected particle travelling along the $-x$ axis.

In classical mechanics there is no way that the particle can cross a barrier of height $V>E$ In quantum mechanics the particles's wave function penetrates inside the step and in region $II$ we have

$$\psi_{_{II}} (x,t) = e^{-iEt/\hbar} \left[ A_{II} e^{-qx/\hbar} + B_{II} e^{qx/\hbar}\right]$$ (20.65)

In region $III$ the wave function is

$$\psi_{III} (x,t) = e^{-iEt/\hbar} \left[ A_{III} e^{ipx/\hbar}+ B_{III}e^{-ipx/\hbar}\right]$$ (20.66)

where the term $A_{III} e^{ipx/\hbar}$ represents a particle travelling to the right and $B_{III} e^{-ipx/\hbar}$ represents a particle incident from the right. In the situation that we are analyzing there no particles incident from the right and hence $B_{III}=0$.

In quantum mechanics the wave function does not vanish in region $II$. As shown in Figure 20.8 the incident wave function decays exponentially in this region , and there is a non-zero value at the other boundary of the barrier. As a consequence there is a non-zero wavefunction in region $III$ implying that there is a non-zero probability that the particle penetrates the potential barrier and gets through to the other side even though its energy is lower than the height of the barrier. This is known as quantum tunnelling. It is as if the particle makes a tunnel through the potential barrier and reaches the other side. The probability that the incident particle tunnels through to the other side depends on the relative amplitude of the incident wave in region $I$ and the wave in region $III$. The relation between these amplitude can be worked out by matching the boundary conditions at the boundaries of the potential barrier.

The wave function and its $x$ derivative should both be continuous at all the boundaries. This is to ensure that the Schrodinger's equation is satisfied at all points including the boundaries.

Matching boundary conditions at $x=0$ we have

$$\psi_I(0,t) = \psi_{II} (0,t)$$ (20.67)

and

$$\left( \frac{\partial \psi_{I}}{\partial x} \right)_{x=0} = \left( \frac{\partial \psi_{II}}{\partial x} \right)_{x=0}$$ (20.68)

We also assume that the step is very high $V \gg E$ so that

$$q = \sqrt{2m (V-E)} \approx \sqrt{2mV}$$ (20.69)

and we also know that

$$p= \sqrt{2mE} , \, \hspace{1cm} \mbox{so} \hspace{1cm} \,\frac{p}{q} = \sqrt{\frac{E}{V}} \ll 1$$ (20.70)

Applying the boundary conditions at $x=0$ we have

and

(20.72)

The latter condition can be simplified to

$$A_I-B_I = \frac{iq}{p} \left( A_{II}-B_{II}\right)$$ (20.73)

Applying the boundary conditions at $x=a$ we have

$$A_{II} e^{-qa/\hbar} + B_{II}e^{qa/\hbar}= A_{III} e^{ipa/\hbar}$$ (20.74)

and

$$-q \left[A_{II} e^{-qa/\hbar} - B_{II}e^{qa/\hbar}\right] = ip A_{III} e^{ipa/\hbar} \,.$$ (20.75)

The latter condition can be simplified to

$$A_{II} e^{-qa/\hbar} - B_{II}e^{qa/\hbar}= \frac{-ip}{q} A_{III} e^{ipa/\hbar}$$ (20.76)

Considering the $x=a$ boundary first and using the fact that $p/q \ll 1$ we have

$$A_{II} e^{-qa/\hbar}-B_{II} e^{qa/\hbar}=0$$ (20.77)

which implies that

(20.78)

Using this in equation (20.74) we have

$$A_{III}=2 e^{-ipa/\hbar} e^{-qa/\hbar} A_{II}$$ (20.79)

Considering the boundary at $x=0$ next, we can drop $B_{II}$as it is much smaller than the other terms. Adding equations (20.71) and (20.73) we have

$$\left( 1+ \frac{iq}{p} \right) A_{II} = 2 A_I$$ (20.80)

and as $q/p \gg 1$, this gives us

$$A_{II} = - \frac{i p}{q} 2 A_I\,.$$ (20.81)

Using this in equation (20.79) we have

$$A_{III} = - 4 i \frac{p}{q}\, e^{-ipa/\hbar} \, e^{-qa/\hbar}\, A_I \,.$$ (20.82)

The transmission coefficient

$$T = \frac{\left\vert A_{III} \right\vert^2}{\left\vert A_I \right\vert^2} = 16 \frac{p^2}{q^2} e^{-2qa/\hbar}$$ (20.83)

gives the probability that an incident particle is transmitted through the potential barrier. This can also be expressed in terms of $E$ and $V$ as

$$T= 16 \frac{E}{V} e^{-2 a\sqrt{2mV}/ \hbar}$$ (20.84)

The transmission coefficient drops if either $a$ or $V$ is increased. The reflection coefficient $R=1-T$ gives the probability that an incident particle is reflected.