18 Tunnelling

Scattering on a step

The scattering regime \(E>V_0\) corresponds to the case where classically an incoming particle would be automatically transmitted. We covered this in the last lecture so we summarise the results here.

The Hamiltonian eigenfunctions take the form \[\phi(x) = \begin{cases} e^{ikx} + r e^{-ikx} & \quad x < 0 \\ t e^{ik'x} & \quad x > 0 \end{cases} \, ,\] where \[k = \sqrt{2mE/\hbar^2} \qquad k' = \sqrt{2m(E-V_0)/\hbar^2} \, .\] The coefficients \(r\), \(t\) are found by requiring that the wave function and its derivative are continuous at \(x = 0\), with the result \[r = \frac{k-k'}{k+k'} \qquad t = \frac{2k}{k+k'} \, .\]

The wave functions are not square-normalizable and to extract physical information we instead compute ratios of probability currents. The probability current is \[J(x) = \begin{cases} J_I - J_R& \quad x < 0 \\ J_T & \quad x >0 \end{cases}\] where \[J_I = \frac{\hbar k}{m} \qquad J_R = \frac{\hbar k}{m} |r|^2 \qquad J_T = \frac{\hbar k'}{m} |t|^2 \, .\] The reflection and transmission probabilities are then \[\begin{aligned} R & = \frac{J_R}{J_I} = |r|^2 = \left( \frac{k-k'}{k+k'} \right)^2 \\ \nonumber T & = \frac{J_T}{J_I} = \frac{k'}{k}|t|^2 = \frac{4kk'}{(k+k')^2} \, . \end{aligned}\] As a consistency check, we have \(R + T = 1\). The reflection and transmission coefficients are sketched below for \(V_0 < E < \infty\).

Tunnelling into a step

Now consider \(0<E<V_0\). In the region \(x>0\), where \(E-V_0\) has changed sign, the Hamiltonian eigenfunction now has an exponential decay \[\phi(x) = \begin{cases} e^{ikx} + r e^{-ikx} & \quad x < 0 \\ t e^{-\kappa x} & \quad x > 0 \end{cases} \, ,\] where now \(\kappa = \sqrt{2m(V_0-E)/\hbar^2}\). We discard the other potential solution \(e^{\kappa x}\) which would diverge at \(x \to +\infty\).

Notice that the above wave function can be obtained from the scattering problem by replacing \(k' \to i \kappa\). This means we can immediately write down the solution for the coefficients \(r\), \(t\), \[r = \frac{k-i\kappa }{k+i\kappa} \qquad t = \frac{2k}{k+i\kappa} \, .\] This dramatically changes the computation of the probability currents, which involve both the wave function and its conjugate. In particular, the reflected probability current is now equal to the incoming probability current, \[J_R = \frac{\hbar k }{m}|r|^2 = \frac{\hbar k }{m}\left| \frac{k-i\kappa }{k+i\kappa} \right|^2 = \frac{\hbar k}{m} = J_I \, .\] Meanwhile, the transmitted probability current now vanishes, \[\begin{aligned} J_T & = \frac{\hbar}{2mi}(\overline{\phi(x)}\partial_x \phi(x) - \phi(x)\partial_x \overline{\phi(x)}) \qquad x>0 \\ \nonumber & =\frac{\hbar}{2mi}(e^{-\kappa x}\partial_x e^{-\kappa x} - e^{-\kappa x}\partial e^{-\kappa x}) \\ & = 0 \, , \end{aligned}\] because \(e^{-\kappa x}\) is a real function of \(x\).

In summary, we conclude that \[R = 1 \qquad T = 0\] in the tunnelling regime \(0<E<V_0\). Note the following points:

• This is consistent with the limit \(E \to V^+_0\) in the scattering regime \(E>V_0\).

• It coincides with the classical expectation that the particle is always reflected when \(0<E<V_0\).

Despite the fact that \(T = 0\), the probability density is non-zero in the region \(x>0\) so there is therefore a non-vanishing probability to find the particle with \(x>0\). The probability density is sketched below. In contrast, the particle is forbidden from the region \(x>0\) in classical mechanics

This has important consequences if we were to add a step down to \(V(x) = 0\) at some finite distance \(x = L\). The wave function would then be expected to decay exponentially for \(0<x<L\), but then become trigonometric again for \(x>L\) and there is a possibility for the particle to escape to \(x \to + \infty\). This is known as “quantum tunnelling". We explore this in more detail now.

Scattering off a barrier

The scattering regime corresponds again to \(E>V_0\). The Hamiltonian eigenfunctions are trigonometric in all regions, \[\phi(x) = \begin{cases} e^{ikx} + r e^{-ikx} & \quad x < 0 \\ A e^{ik' x} + B e^{-ik' x} & \quad 0<x<L \\ t e^{ i k x} & \quad x > L \end{cases} \, ,\] where the wavenumbers \(k\), \(k'\) are defined as above.

Since the potential remains finite at \(x = 0\) and \(x = L\), we need to impose that the wave function and its derivative are continuous there. This gives the constraints \[\begin{aligned} 1+r & = A+B \\ k(1-r) &= k'(A-B) \end{aligned}\] from \(x=0\) and \[\begin{aligned} Ae^{i k' L} + Be^{- i k'L} & = t e^{ikL} \\ k' (Ae^{ i k' L} - Be^{- ik' L}) & = k t e^{ikL} \, . \end{aligned}\] from \(x=L\). We have four linear equations for four variables \(r\), \(t\), \(A\), \(B\). The solution is found by elementary but tedious linear algebra that I will not reproduce here. The important output is the reflection / transmission probabilities \[\begin{aligned} R & = |r|^2 = \frac{(k^2-k'^2)^2 \sin^2(k'L)}{\left(k^2+k'^2\right)^2\sin^2(k'L) +4 k^2 k'^2 \cos^2(k'L)}\\ T & = |t|^2 = 1-R\, . \end{aligned}\] The important features are summarized below.

• The limit \(E\to \infty\) corresponds to \(k,k' \to \infty\) with \(k/k' \to 1\). We find \(R \to 0\), \(T\to1\) reproducing the classical expectation for \(E>V_0\). The potential barrier is negligible compared to the energy and the particle is transmitted with probability \(1\).

• The limit \(E \to V_0^+\) corresponds to \(k' \to0^+\), with \[R \to \frac{\gamma}{1+\gamma} \qquad T \to \frac{1}{1+\gamma}\] where \(\gamma = mL^2V_0 / 2\hbar^2\) is a dimensionless parameter.

• The function has trigonometric dependence on \(k'\). In particular, the transmission probability becomes \(1\) whenever \(k'L = n\pi\) or equivalently \[E = V_0 + \frac{\hbar^2}{2m}\left(\frac{n\pi}{L} \right)^2\, .\] These “transmission resonances" correspond to a standing wave in \(0<x<L\). If we remember that the wavelength is \(\lambda = 2\pi / k'\), the condition becomes \(2L = n\lambda\) so the distance from \(x=0\) to \(x=L\) and back is an integer number of wavelengths. Intuitively, there is constructive interference between incident wave at \(x = 0\) and the standing wave in the region \(0<x<L\).

Tunnelling through a barrier

The tunnelling regime is \(0<E<V_0\). The Hamiltonian eigenfunctions are now \[\phi(x) = \begin{cases} e^{ikx} + r e^{-ikx} & \quad x < 0 \\ A e^{\kappa x} + B e^{-\kappa x} & \quad 0<x<L \\ t e^{ i k x} & \quad x > L \end{cases} \, ,\] where \(k\) and \(\kappa\) are defined as above.

A Jupyter notebook (on Colab) to compute the time dependence of a wave packet incident on a finite-height barrier, and see the effect of tunnelling.

As before, the coefficients \(r\), \(t\) are found by replacing \(k' \to i\kappa\) in the scattering regime. This modifies the reflection and transmission coefficients to \[\begin{aligned} R & = \frac{\left(k^2+\kappa ^2\right)^2 \sinh^2(\kappa L)}{\left(k^2-\kappa ^2\right)^2\sinh^2(\kappa L) +4 \kappa ^2 k^2 \cosh^2(\kappa L)} \\ T & = 1-R \, . \end{aligned}\] Note that there is a non-vanishing probability for the particle to be transmitted through the potential barrier and reach \(x = +\infty\), which is forbidden in classically. This is known as “quantum tunnelling".

• The limit \(E \to 0\) corresponds to \(k \to 0\) with \(\kappa\) fixed. We find \(R\to 1\), \(T \to 0\) reproducing the classical expectation for \(E<V_0\).

• The limit \(E \to V^-_0\) in the tunnelling regime coincides with the limit \(E \to V_0^+\) in the scattering regime.

• Note the exponential rather than trigonometric dependence on \(\kappa\) in \(0<x<L\). In particular, there are no “resonances" like in the scattering regime.

Finally, we can combine the results in the scattering and tunnelling regimes to sketch the reflection / transmission probabilities across the entire range \(0<E<\infty\).