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A scanning tunneling microscope brings a tip and surface, both electrical conductors, close enough to leave only a small vacuum gap. An applied electric potential difference makes occupied electron states on one side able to feed unoccupied states on the other. Although the gap is a classically forbidden potential barrier, quantum tunnelling through its finite width produces an electric current.

For an approximate rectangular potential barrier of gap width $d$, the transmission probability, and thus the electric current at fixed small bias, varies approximately as $e^{-2\kappa d}$. This strong distance dependence makes the device sensitive to changes in separation on the scale of a chemical atom. Feedback control moves the tip vertically to keep the electric current constant during a scan; the required height changes form an image of the surface, with contrast also depending on its electronic states. The finite gap permits transmitted probability current, whereas the semi-infinite step in part (b) does not.

For $V_0<0$, $k=\sqrt{2mE}/\hslash$ tends to zero as $E\to 0^{+}$ while $q=\sqrt{2m(E+|V_0|)}/\hslash$ stays finite. Matching gives probability amplitudes $r=(k-q)/(k+q)$ and $t=2k/(k+q)$, so quantum reflection probability $R=r^2\to 1$ and transmission probability $T=(q/k)|t{|}^2\sim 4k/q\to 0$. The final region is classically allowed; reflection results from the abrupt spatial change in the wavefunction's wavenumber. These are limits of positive-energy incoming states, not a flux ratio evaluated at exactly zero incident current.

For one-dimensional scattering from a real potential, write the left exterior wavefunction as $e^{ikx}+r{e}^{-ikx}$ with $k>0$. The reflected-to-incident probability-current ratio is $R=|r{|}^2$, since both components have the same speed and their currents have opposite signs. This probability differs from the complex reflection amplitude $r$. With transmission into a right exterior wavenumber $q$, the transmission probability is $T=(q/k)|t{|}^2$ and conservation of current gives $R+T=1$ when there is no absorption.

Quantum tunnelling is transmission through a potential barrier even when the particle energy is below the barrier height.

The wavefunction penetrates the classically inaccessible region; matching it to an outgoing scattering state on the far side gives a nonzero transmission probability. For a real, time-independent potential barrier, the transmitted energy equals the incident energy: quantum tunnelling does not require a temporary violation of conservation of energy. A tunnel junction turns this small transmission probability into a measurable electric current, whose sensitivity to separation makes the scanning tunnelling microscope possible.

For equal exterior potential energies and constant mass, the exact transmission probability isBelow the barrier, the hyperbolic sine is positive for $a>0$, so $0<T<1$. At $E=V_0$, the interior wavefunction is linear in $x$; either neighbouring formula has the displayed finite limit, since $\sinh z/z$ and $\sin z/z$ tend to one. Above the barrier, the interior wavefunction oscillates: the quantum reflection probability generally remains nonzero, but $qa=n\pi$ for a positive integer $n$ gives perfect transmission probability $T=1$ through constructive wave interference. These are scattering resonances, not below-barrier quantum tunnelling.

At half height, $E=V_0/2$, we have $k=\kappa$ and $T={\mathrm{sech}}^2(\kappa a)$, consistent with finite square barrier transmission at half barrier height after replacing this width $a$ by that example's width $2a$. In the illustration, the width is four times $\ell =\hslash /\sqrt{2m{V}_0}$ and $E/V_0=1/2$, giving $T\simeq 0.01388$. The plotted wavefunction comes from solving the four interface equations; the transmission probability curve uses the closed formula.

For a rectangular potential barrier of height $V_0$, width $a$, and equal constant potentials outside, a particle with $0<E<V_0$ has exterior wave number $k=\sqrt{2mE}/\hslash$ and interior decay constant $\kappa =\sqrt{2m(V_0-E)}/\hslash$. Write the three wavefunctions as $e^{ikx}+r{e}^{-ikx}$, $A{e}^{\kappa x}+B{e}^{-\kappa x}$, and $t{e}^{ikx}$. Continuity of the wavefunction and its derivative at both interfaces determines the transmission probability $T=|t{|}^2$ in the title. Equal exterior velocities are what permit the last equality.

For $\kappa a\gg 1$, ${\sinh }^2(\kappa a)\simeq e^{2\kappa a}/4$, soKeeping only $e^{-2\kappa a}$ estimates the exponentially dominant quantum tunnelling suppression; the interface prefactor can still change a numerical answer by a factor of order one.

For an attractive finite square well of width $a$, outside wave number $k$ and inside wave number $q$, matching the wavefunction and its first derivative at both endpoints gives the displayed transmission probability. It equals one when $qa$ is an integer multiple of $\pi$: the internal reflections cancel. For $q=2k$ the coefficient of ${\sin }^2(qa)$ is $9/16$.

For constant mass $m$, a finite jump from potential energy zero at $x<0$ to $V_0>0$ at $x>0$ requires continuity of the wavefunction and its first derivative. With incidence only from the left and $E>V_0$, put $k=\sqrt{2mE}/\hslash$ and $q=\sqrt{2m(E-V_0)}/\hslash$. Matching $A+B=C$ and $k(A-B)=qC$ gives reflected probability amplitude $B/A=(k-q)/(k+q)$ and transmission amplitude $C/A=2k/(k+q)$. The transmission probability is $(q/k)|C/A{|}^2$, conserving probability current.

For $0<E<V_0$, replace $q$ by $i\kappa$, where $\kappa =\sqrt{2m(V_0-E)}/\hslash$ is the tunnelling decay constant. The right-hand wavefunction is $C{e}^{-\kappa x}$, with zero probability current. Its probability density decays on length $1/(2\kappa )$, while its amplitude decays on length $1/\kappa$. The semi-infinite raised region gives total reflection, unlike a finite potential barrier allowing quantum tunnelling into a second allowed region.

For a thick rectangular barrier with equal exterior potentials, constant mass and $0<E<U$, the exact matched transmission probability has the displayed limit when $\kappa a\gg 1$. The exponential is fixed by the evanescent action; the interface prefactor remains energy-dependent. A leading-exponential estimate can therefore differ by an order-one multiplier from the matched result.

For one-dimensional quantum scattering, $t$ is the complex amplitude of the outgoing transmitted wave relative to the incoming wave. The transmission probability is $|t{|}^2$ times the ratio of transmitted and incident flux velocities; with equal asymptotic dispersion it is $|t{|}^2$. A reflectionless potential has $r=0$ and a unit-modulus transmission amplitude whose phase shifts continuum quantization. The supersymmetric factorization of the one-soliton potential gives $t(k)=(k+im)/(k-im)$ for its single-bound-state potential.

A tunnel diode is a p-n junction with sufficiently heavy semiconductor doping on both sides to make a thin depletion region and permit substantial interband tunnelling. The neutral regions are degenerate semiconductors. Its distinctive forward current-voltage characteristic contains a peak followed by a valley: over part of the curve, increasing the applied electric potential difference decreases the electric current. The Esaki tunnel-diode experiment made this behavior experimentally concrete.

The useful quantity is the local slope, not the ratio of total voltage to total electric current. With conventional electric current $I$ entering the positive-voltage terminal, defineNegative differential resistance means $g_d<0$, hence $r_d<0$ where defined. A forward operating point can still have $V>0$, $I>0$ and absorbed power $VI>0$. A small voltage increase then gives a negative current increment; this does not make the total electric current negative or turn the device into an unlimited source of energy. The external supply sets the operating point and supplies the energy available for an electronic amplifier.

An original idealized construction explains the shape. Put the n-type semiconductor on the left and the p-type semiconductor on the right. Take $q>0$ as the elementary charge and $V=V_p-V_n$. Increasing forward bias lowers the right-hand electron energy levels relative to the left. Denote the two reservoir electrochemical potentials by ${\mu }_n,{\mu }_p$, so that ${\mu }_n-{\mu }_p=qV$. At zero temperature, occupied left-hand conduction band states extend up to ${\mu }_n$; empty right-hand valence band states start at ${\mu }_p$. For transfer conserving electron energy, the available interval isIt is nonempty only when its upper endpoint exceeds its lower endpoint. Both an occupied initial quantum state and an empty final quantum state must exist at the same energy. The Pauli exclusion principle prevents transfer into an already occupied final quantum state.

For the original diagram, choose arbitrary energy units with $E_{c,n}=-0.5$, ${\mu }_n=0$, ${\mu }_p=-u$ and $E_{v,p}=0.5-u$, where $u$ is $qV$ in those units. The available interval has widthThus $L(0.25)=0.25$, $L(0.50)=0.50$, $L(0.70)=0.30$, and $L(1.15)=0$. Initially forward bias opens a transfer window; further forward bias closes it by moving the right-hand valence band edge below the left-hand conduction band edge. This is a state-counting illustration, not a formula for a measured electric current.

For reverse bias, the transfer direction changes: occupied right-hand valence band states can feed empty left-hand conduction band states. In the same construction, $u=-0.70$ opens the interval $0<E<0.70$ for this reverse transfer. A narrow depletion region allows a large reverse quantum tunnelling electric current at small electric potential difference, whereas the ordinary Shockley diode equation predicts only a small reverse saturation electric current before diode reverse breakdown. At zero biasing, equal reservoir occupations give zero net electric current; quantum tunnelling does not require a net equilibrium flow.

A finite-temperature refinement replaces sharp occupation cutoffs with the Fermi-Dirac distribution. An illustrative weak-transfer expression isHere $\mathcal{W}\ge 0$ describes allowed initial/final quantum states and their coupling, and $f_n,f_p$ are Fermi-Dirac distributions with the respective electrochemical potentials. The occupation difference follows by subtracting the reverse transfer factor from the forward one: $f_n(1-f_p)-f_p(1-f_n)=f_n-f_p$. An overlap interval alone does not determine $\mathcal{W}$: transmission probability, density of states, crystal momentum selection and phonons can matter. A complete germanium calculation is more demanding than multiplying two constant density of states factors. At larger forward bias, ordinary carrier injection and diffusion supply a rising current branch even after direct band overlap is lost.

For an opaque barrier between simple turning points, the leading transmission probability is $T\simeq e^{-2S_B/\hslash }$, where $S_B=\int \sqrt{2m[V(x)-E]}\mathrm{d}x$ over the forbidden interval. The exponent for probability is twice that for amplitude. Matching determines prefactors and the domain where this approximation applies.