Let us revisit the finite step potential,
\begin{equation}
V(x) = \begin{cases}
0 & \quad x \lt{} 0 \\
V_0 & \quad x \geq 0
\end{cases} \, ,
\end{equation}
where we assume \(V_0 \gt{}0\). We consider incoming particles of definite energy \(E\gt{}0\). There are two regimes to consider:
18.1.1.Scattering on a step
In the scattering regime of the finite step problem, the
wave function is oscillatory on both sides of the step. There is an
incoming part, as well as a reflected and transmitted part, all
described by plane waves.
The scattering regime \(E\gt{}V_0\) 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
\begin{equation}
\phi(x) = \begin{cases}
e^{ikx} + r e^{-ikx} & \quad x \lt{} 0 \\
t e^{ik'x} & \quad x \gt{} 0
\end{cases} \, ,
\end{equation}
where
\begin{equation}
k = \sqrt{2mE/\hbar^2} \qquad k' = \sqrt{2m(E-V_0)/\hbar^2} \, .
\end{equation}
The coefficients \(r\), \(t\) are found by requiring that the wave function and its derivative are continuous at \(x = 0\), with the result
\begin{equation}
r = \frac{k-k'}{k+k'} \qquad t = \frac{2k}{k+k'} \, .
\end{equation}
The wave functions are not square-normalizable and to extract physical information we instead compute ratios of probability currents. The probability current is
\begin{equation}
J(x) = \begin{cases}
J_I - J_R& \quad x \lt{} 0 \\
J_T & \quad x \gt{}0
\end{cases}
\end{equation}
where
\begin{equation}
J_I = \frac{\hbar k}{m} \qquad J_R = \frac{\hbar k}{m} |r|^2 \qquad J_T = \frac{\hbar k'}{m} |t|^2 \, .
\end{equation}
The reflection and transmission probabilities are then
\begin{align}
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{align}
As a consistency check, we have \(R + T = 1\). The reflection and transmission coefficients are sketched below for \(V_0 \lt{} E \lt{} \infty\).
18.1.2.Tunnelling into a step
In the tunnelling regime of the finite step problem, the
wave function is oscillatory to the left of the step, but
exponentially damped (but nonzero) to the right of the step.
Now consider \(0\lt{}E\lt{}V_0\). In the region \(x\gt{}0\), where \(E-V_0\) has changed sign, the Hamiltonian eigenfunction now has an exponential decay
\begin{equation}
\phi(x) = \begin{cases}
e^{ikx} + r e^{-ikx} & \quad x \lt{} 0 \\
t e^{-\kappa x} & \quad x \gt{} 0
\end{cases} \, ,
\end{equation}
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 \(r\), \(t\),
\begin{equation}
r = \frac{k-i\kappa }{k+i\kappa} \qquad t = \frac{2k}{k+i\kappa} \, .
\end{equation}
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,
\begin{equation}
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 \, .
\end{equation}
Meanwhile, the transmitted probability current now vanishes,
\begin{align}
J_T & = \frac{\hbar}{2mi}(\overline{\phi(x)}\partial_x \phi(x) - \phi(x)\partial_x \overline{\phi(x)}) \qquad x\gt{}0 \\ \nonumber
& =\frac{\hbar}{2mi}(e^{-\kappa x}\partial_x e^{-\kappa x} - e^{-\kappa x}\partial e^{-\kappa x}) \\
& = 0 \, ,
\end{align}
because \(e^{-\kappa x}\) is a real function of \(x\).
In summary, we conclude that
\begin{equation}
R = 1 \qquad T = 0
\end{equation}
in the tunnelling regime \(0\lt{}E\lt{}V_0\). Note the following points:
Despite the fact that \(T = 0\), the probability density is non-zero in the region \(x\gt{}0\) so there is therefore a non-vanishing probability to find the particle with \(x\gt{}0\). The probability density is sketched below. In contrast, the particle is forbidden from the region \(x\gt{}0\) 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 \(0\lt{}x\lt{}L\), but then become trigonometric again for \(x\gt{}L\) 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.
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