Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2022 / Paper 2 / Section B / 21

One consistent five-postulate formulation is the following; different textbooks group the same content differently.

The position-dependent probability density is $|\psi {|}^2$, and its flux is the probability current. These distinguish the wave amplitude from the measurable particle flow in the following scattering problem.

Place the interface at $x=0$ and let the incident beam travel in the positive x direction. Set the vacuum potential to zero and the solid potential lower by $V_0>0$:Equivalently, $V(x)=-V_0\Theta (x)$. The incident kinetic energy is $E=m{v}^2/2$. This ideal potential step assumes a sharp, planar interface, constant particle mass and no absorption, inelastic processes or lateral structure. It models the prescribed decrease in potential without asserting that every real solid has the same electronic interface potential.

For a stationary scattering state, the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)+V\psi =E\psi$. The left side contains the incident and reflected waves, while the solid has no incident wave arriving from positive infinity:Multiplying by $e^{-iEt/\hslash }$ gives the time-dependent wavefunctions. The kinetic energy rises inside the lower-potential solid, so $k_2>k_1$ and the transmitted wavelength is shorter. A downward step permits a propagating transmitted wave for every $E>0$, but still generally reflects some incident flux. There is no forbidden region merely because $E<V_0$; the relevant transmitted kinetic energy is $E+V_0$.

Classically every particle crosses a downward step and accelerates, so this partial reflection has no classical counterpart. If the step is instead upward to $U>0$, $E>U$ gives oscillatory transmission with possible reflection, while $E<U$ gives a decaying evanescent wave. A semi-infinite forbidden region carries no transmitted stationary flux; a finite barrier can permit quantum tunnelling to a propagating region beyond it.

For a finite step and equal masses, continuity of the wavefunction and its derivative givesHence the reflected and transmitted amplitude ratios areA travelling wave has probability current $j=(\hslash k/m)|A{|}^2$, so the flux coefficients areThey satisfy $R+T=1$. Using $|t{|}^2$ alone as the transmission probability would omit the different transmitted velocity.

For $E=2\mathrm{eV}$ and a downward step of $2\mathrm{eV}$, the transmitted kinetic energy is $4\mathrm{eV}$ and $k_2/k_1=\sqrt{2}$. ThereforeThe nonzero reflection is compatible with current conservation even though the potential drops.

Here the step is upward by $U=3\mathrm{eV}$, so the forbidden-region energy difference is $U-E=1\mathrm{eV}$. The bounded wavefunction isThe requested amplitude penetration depth is $\delta =1/\kappa$. Using the supplied mass and $h$,The probability density decays as $e^{-2\kappa x}$, so its $1/e$ depth is half this value. For an ideal lossless semi-infinite step below threshold, the evanescent tail does not imply net stationary particle current deep into the material; the reflected flux is unity.

Scanning tunneling microscopy uses electron tunnelling between a conducting tip and a sample across a narrow vacuum gap. At fixed bias, the strong distance dependence of the current enables sensitive height or electronic-structure measurements; the result depends on local electronic states as well as topography.

A tunnel diode uses tunnelling between occupied and available states across a very thin, heavily doped semiconductor junction. The alignment of those states changes with bias and can produce a region of negative differential resistance, useful in fast switching or oscillation circuits. Both examples require a finite barrier with states available on the other side; an infinitely extended forbidden region is not an equivalent transport geometry.