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In a semiclassical escape model, an alpha-particle assault frequency is the interior rate scale multiplying the barrier transmission probability. Confinement alone gives $p\sim \hslash /L$ and hence $\nu \sim \hslash /(m_{\alpha }{L}^2)$ up to geometric factors. Counting encounters with either wall differs by a factor of two from returns to one wall. A stationary bound wavefunction does not itself follow a classical collisional trajectory.

The alpha preformation probability is a model-dependent nuclear-structure factor multiplying an interior assault frequency and barrier transmission probability in an alpha-decay rate estimate. It accounts for the availability of an alpha-like cluster at the matching surface. It is not fixed by the exterior Coulomb tunnelling exponent alone.

For a real one-dimensional potential, a real-axis transmission probability is a modulus square. Its algebraic complex continuation need not have only physical bound-state poles. In a square well, $t=e^{-ika}/D(k)$ while the continued probability is $1/[D(k)D(-k)]$. Zeros of the second factor can encode a different boundary condition. In units ${\hslash }^2/(2m)=1$, a well of width two and depth $1+{\cot }^{2}1$ has a continued-probability pole at $k=i\cot 1$ which satisfies neither decaying even nor decaying odd matching. The physical bound-state criterion uses the outgoing amplitude on its decaying sheet.

Tunneling through a controllable vacuum gap by Gerd Binnig and others made barrier width experimentally adjustable. Their apparatus used a pointed electrical conductor opposite a conducting plate, with fine positioning by piezoelectricity. Rubber-tube support and magnetic levitation above a cooled bowl exhibiting superconductivity suppressed vibrations; the junction itself remained near room temperature.^[ref]

The readout was electrical resistance versus displacement at a fixed electric potential difference of 60 mV of either polarity. They observed an exponential function dependence spanning four orders of magnitude. Cleaning and better vacuum increased the inferred barrier from approximately 0.6-0.7 eV to 3.2 eV. Reproducible curves and the cleaning response argued against deforming contamination layers; an appropriate slope also helped exclude an ordinary conducting contact. The paper reports residual contamination, electronics-limited range and a jump attributed to release of thermal stress; its curve origins generally do not determine absolute gap widths.^[ref]

The ideal tunnel junction model explains the slope. At fixed small electric potential difference, fixed tip geometry and approximately unchanged density of states, take an effective barrier energy $\Phi$ above the relevant electron states and approximate the current magnitude byThe coefficient $C$ includes the applied electric potential difference, available occupied and empty quantum states, and tip geometry. A single transmission probability is not an electric current: a junction combines contributions from many electron states in both directions, with the electric potential difference producing a net imbalance. Its tilted, material-dependent barrier is only approximated by the equal-exterior rectangular potential barrier; the exact single-particle formula above is not a complete junction model.

For the chosen model value $\Phi =4$ electronvolts, $\kappa \simeq 10.25{\mathrm{nm}}^{-1}$. Increasing separation by $0.1\mathrm{nm}$ multiplies $I$ by about $0.129$, a reduction by about $7.8$. The following diagram chooses $I(0.4\mathrm{nm})=1\mathrm{nA}$ solely for illustration; it contains no measured data and is not a fit to the 1982 curves.

Distance calibration is essential. In Scanning Tunneling Microscopy: From Birth to Adolescence by Gerd Binnig and others, the authors describe an initial inferred barrier near 1 eV. Recalibration found only half the quoted piezo displacement sensitivity and raised the inferred barrier above 4 eV. Thus an exponential trace alone was insufficient evidence for tunnelling through vacuum. The model makes the sensitivity explicit: the inferred $\Phi ={\hslash }^2(d\ln I/ds{)}^2/(8m_e)$ depends quadratically on the distance scale.^[ref]

For scanning tunnelling microscopy, feedback control adjusts the vertical position to hold electric current fixed during a lateral scan; the recorded height follows separation only when the surface is electronically homogeneous. Changes in the local density of states can also change apparent height. This limitation is already explicit in the authors' account of constant-current imaging.^[ref]

A nonzero probability density inside a barrier does not by itself imply transport. In one dimension, the probability current of a wavefunction is $J=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$. A single evanescent wave $C{e}^{-\kappa x}$ has ${\psi }^{\ast }{\psi }^{\prime }=-\kappa |C{|}^2{e}^{-2\kappa x}$, a real quantity, hence $J=0$. This is what happens on the inaccessible side of a potential barrier extending to infinity with $E<V_0$.^[ref]

For the finite-barrier wavefunction $A{e}^{\kappa x}+B{e}^{-\kappa x}$, the cross terms instead giveThe relative quantum phase of the two probability amplitudes permits a nonzero, spatially constant probability current, even though neither exponential function separately carries it. The probability continuity equation gives ${\partial }_{x}J=0$ for this stationary state. Matching the wavefunction and its derivative also makes $J$ continuous at both faces, soHere $R=|r{|}^2$ is the quantum reflection probability, and $T=J_{\mathrm{tr}}/J_{\mathrm{inc}}$ is the transmission probability. The net exterior probability current equals the interior probability current; it is not lost as the wavefunction decreases. A physical incident wave packet supplies finite total probability, while the stationary state describes the energy component used to compute its transmitted flux.

For fixed $0<E<V_0$ and $\kappa a\gg 1$, the exact rectangular barrier transmission probability givesThis is the leading term of an asymptotic expansion: the wavefunction amplitude scale is $e^{-\kappa a}$, and the transmission probability scale is $e^{-2\kappa a}$, with an interface prefactor. Increasing width by $\Delta a$ multiplies this approximation by $e^{-2\kappa \Delta a}$. At fixed incident energy, increasing barrier height increases the tunnelling decay constant as $\sqrt{V_0-E}$; heavier nonrelativistic particles also have stronger suppression. For the numerical example in rectangular barrier transmission probability, the approximation gives $0.01397$, compared with the exact $0.01388$.

The regime condition matters. Near $E=V_0$ at fixed width, $\kappa a$ tends to zero and the opaque barrier approximation fails; use the exact threshold value. For $\kappa a\ll 1$, expansion of the hyperbolic sine givesThis tends to one as $a\to 0$ at fixed $E,V_0$, but the condition $\kappa a\ll 1$ alone does not imply nearly perfect transmission if the remaining factor is large. At $E>V_0$, use the oscillatory formula and its scattering resonances, rather than an exponential decay law.

The WKB approximation describes a wave whose local wavelength changes slowly compared with one wavelength. For the one-dimensional stationary Schrödinger equation, put $p(x)=\sqrt{2m[E-V(x)]}$. In a classically allowed region,The phase is the classical action divided by $\hslash$. The amplitude varies as $p^{-1/2}$ so that a single travelling component has constant probability current. A useful local validity condition is $|\hslash p^{\prime }/p^2|\ll 1$.

For $V>E$, write $\kappa =\sqrt{2m(V-E)}/\hslash$. The wave instead has growing and decaying components proportional to ${\kappa }^{-1/2}\exp (\pm \int \kappa \mathrm{d}x)$. At a simple turning point, $p$ vanishes and neither local WKB expression is valid. Linearizing the potential gives an Airy-function solution that connects the two sides and supplies the turning-point phase shift. It is not legitimate to carry the divergent WKB amplitude straight through the turning point.^[ref]

A useful application is tunnelling through a smooth opaque barrier. With turning points $x_1,x_2$, the leading transmission probability isThe exponent is twice the forbidden-region action in units of $\hslash$, because transmission is a probability rather than a wave amplitude. For an ideal rectangular barrier of height $V_0>E$ and width $d$, the same leading exponential is $\exp [-2d\sqrt{2m(V_0-E)}/\hslash ]$; matching at the abrupt edges supplies an additional prefactor.

For a bound state between two smooth turning points, the related WKB quantization rule is ${\int }_{x_1}^{x_2}p\mathrm{d}x=(n+1/2)\pi \hslash$. In a harmonic potential the integral equals $\pi E/\omega$, recovering $E_n=(n+1/2)\hslash \omega$. This exact agreement of energies for the harmonic oscillator does not make WKB wavefunctions uniformly exact near its turning points.

The classical transmission probability is $P_{\mathrm{cl}}(E)=\Theta (E-V_0)$; the convention exactly at the threshold does not affect a thermal integral. Below the barrier it is zero, whereas the quantum result permits tunnelling. Above the barrier it is one, whereas quantum reflection is generally nonzero except at transmission resonances. Both descriptions tend to unit transmission at arbitrarily high energy, and both have zero transmitted probability in the zero-energy limit for this positive barrier.

Assume $a>0$ and put $\mu =mg/{\hslash }^2$, $q=k/\hslash$, and $\gamma =\mu /q$. At a delta potential located at $x_0$, integration of the Schrödinger equation gives the matching conditionsWriting the adjacent plane wave coefficients as $(A_L,B_L)$ and $(A_R,B_R)$ givesSolving these two linear equations yields the transfer matrix of a delta potential:For the symmetric double-delta potential, use separate plane wave pairs in $x<-a$, $-a<x<a$, and $x>a$. Successive matching gives $M=D(a)D(-a)$ between the two exterior pairs. For real $q$, multiplication givesThe determinant identity $\det M=1$ reads $|d{|}^2-F^2=1$. Left incidence means $(t,0{)}^T=M(1,r{)}^T$, whenceHere the first relation also follows from $t=M_{11}-M_{12}{M}_{21}/M_{22}=\det M/M_{22}$. Therefore the requested transmission probability and quantum reflection probability, as functions of the printed momentum $k$, areThey obey $R+T=1$. As checks, $g=0$ gives $T=1$, and the coincident limit $a\to 0$ gives $T=q^2/(q^2+4{\mu }^2)$, the answer for a single delta potential of strength $2g$.

For a one-dimensional wavefunction, the probability density and probability current areThe Time-dependent Schrödinger equation with a real potential is $i\hslash {\psi }_t=-{\hslash }^2{\psi }_{xx}/(2m)+V\psi$. Multiply it and its complex conjugate by ${\psi }^{\ast }$ and $\psi$, respectively. Subtracting cancels the real-potential terms and givesThis derives the continuity equation ${\rho }_t+j_x=0$.

For $k>0$, the specified free-region wavefunction combines a right-moving incoming wave of unit amplitude and a left-moving reflected wave with complex reflection amplitude $R$. Applying the free Schrödinger operator to either plane wave gives $E={\hslash }^2{k}^2/(2m)$. Its time factor has modulus one, soThe density includes the stationary interference pattern; $|R{|}^2$, rather than $R$, is the reflection probability.

For the finite square well, let $k=\sqrt{2mE}/\hslash$ outside and $q=\sqrt{2m(E+V_0)}/\hslash$ inside. Write the stationary parts as $e^{ikx}+R{e}^{-ikx}$ to the left, $A{e}^{iqx}+B{e}^{-iqx}$ inside, and $T{e}^{ikx}$ to the right. At both finite potential steps, the wavefunction and its first spatial derivative are continuous. Integrating the stationary Schrödinger equation across a shrinking step proves derivative continuity; a jump of the wavefunction would create an unmatched distributional derivative and is excluded.

Propagate the boundary values at $a$ back through the well. The values at zero areMatching them to $1+R$ and $ik(1-R)$, and adding the two equations after division of the second by $ik$, givesSince the exterior wave numbers agree, the transmission probability is the current ratio $|T{|}^2$, soWhen $V_0=3E$, $q=2k=\sqrt{8mE}/\hslash$, yieldingClassically an incoming particle of positive energy crosses the attractive well and emerges with its original energy, so transmission is one. The quantum result agrees exactly when $qa$ is an integer multiple of $\pi$:These are widths with resonant transmission through a square well; the degenerate width zero also has unit transmission.

PutThe time-independent Schrödinger equation has the formsThe first region contains the incident and reflected waves, while the final region contains only the transmitted wave.

Continuity of $\psi$ and ${\psi }^{\prime }$ at both edges of the finite square well gives the standard transmission coefficientWhen $V_0=3E$, one has $K=2k$, and thereforeHence the transmission probability is

For a finite potential step and constant mass, $\psi$ and ${\psi }^{\prime }$ are continuous. A finite jump of ${\psi }^{\prime }$ would require a delta-function contribution in the integrated equation, while a jump of $\psi$ would be still more singular. Matching the preceding amplitudes givesThe transmission probability is $|F/A{|}^2$ because the incident and transmitted regions have equal wave vector and mass; in unequal leads a velocity ratio is also required. These matching conditions do not apply unmodified to infinite walls, delta barriers or mass discontinuities.

Use the one-way density per unit length $g_{+}/L=2/(h|v|)$. The electron-flux magnitude in an interval is charge times velocity times line density:If $g$ is instead defined for a finite length $L$, the traversal frequency is $|v|/L$; the same cancellation follows. The printed $I=Nev$ therefore requires $N$ to be a line density, not an unscaled total particle number.

Subtract the opposite reservoir's incoming flux. With the stated electron electrochemical potentials, conventional positive current is opposite to electron-number flux, soAt low temperature and small bias, each fully transmitted occupied spin-degenerate subband contributes $2e^2/h$. Hence $\boxed{G=N_{b}2e^2/h}$. More generally, the Landauer formula weights each mode by its transmission probability; imperfect transmission removes exact integer plateaux.

In a classically forbidden region the local wavevector becomes imaginary. For a slowly varying barrier, the WKB transmission probability has leading exponentialwhere $r_1,r_2$ are turning points and $\mu$ is the reduced mass. This is a probability exponent; the wave amplitude has only half that exponent. Near a turning point one uses matching formulas rather than the unmodified local WKB expression.

For alpha decay, an attractive nuclear interior supports a quasibound state behind the repulsive Coulomb barrier of the daughter nucleus. For zero orbital angular momentum, write the exterior potential as $C/r$, $C=2Z_d{e}^2/(4\pi {ϵ}_0)$, with nuclear matching radius $R$ and outer turning point $b=C/Q>R$. The exterior action can be integrated:The decay rate is approximately $\lambda =\nu P_{\mathrm{pre}}{e}^{-2G}$ and the half-life is $\ln 2/\lambda$. Here $\nu$ is the interior assault-frequency scale and $P_{\mathrm{pre}}$ is the alpha preformation probability. They are model-dependent; barrier penetration alone does not determine them. Nonzero orbital angular momentum adds a centrifugal barrier.

For $R/b\ll 1$, the leading dependence is $G\propto Z_d/\sqrt{Q}$. Thus a modest increase in decay energy can cause an enormous increase in decay rate, giving the Geiger-Nuttall law relating logarithmic half-life to inverse square-root energy along a suitable isotope sequence. Tunnelling conserves the total energy: the escaping alpha particle does not acquire energy from the barrier.

Single travelling wave. Differentiation of the plane wave gives ${\partial }_x\psi =ik\psi$, so its probability current isThis is probability density times the free-particle velocity $\hslash k/m$.

Counterpropagating waves. Remove the common time phase and set $z=B^{\ast }A{e}^{2ikx}$. Direct multiplication givesConsequentlyThe interference term $z-z^{\ast }$ is purely imaginary in the last real-part operation. The probability density can have interference fringes even though this net probability current is spatially constant.

Scattering amplitudes. Locate the step at $x=0$ and first take $E>V$. With $k_1=\sqrt{2mE}/\hslash$ and $k_2=\sqrt{2m(E-V)}/\hslash$, useThe common factor $e^{-iEt/\hslash }$ has been suppressed. A finite step with the same mass on both sides requires continuity of $\psi$ and ${\psi }^{\prime }$, giving $1+r_a=t_a$ and $k_1(1-r_a)=k_2{t}_a$. HenceIf $0<E<V$, choose $k_2=i\kappa$, $\kappa =\sqrt{2m(V-E)}/\hslash$, so the transmitted wave decays. The same amplitude formulae continue to hold, but the semi-infinite forbidden region carries no outgoing probability current: $R=1$ and $T=0$.

Flux balance. In the propagating case,In particular,The velocity ratio in $T$ is essential: the squared transmission amplitude alone is not the transmission probability when the two velocities differ.

Numerical transmission. The mass cancels from the ratio, and the energies can remain in the same electronvolt units:Thus about $92.8\%$ is transmitted and $7.18\%$ reflected, despite the incident energy exceeding the step.

Equation and boundary conditions. The delta potential has strength $V_0$ with units of energy times length, because the Dirac delta function has units of inverse length. The time-independent Schrödinger equation isThe wavefunction is continuous at the origin. A finite jump in $\psi$ would produce a derivative of a Dirac delta function in ${\psi }^{\prime \prime }$, with no corresponding term in this potential. A jump in ${\psi }^{\prime }$ does produce a delta contribution and is allowed. Integrate through a symmetric interval:For continuous finite $\psi$, the right-hand side vanishes as $ϵ\to 0$. ConsequentlyThis derivative jump is the boundary condition associated with the attractive delta potential.

Unique bound state. For $E<0$, write $\kappa =\sqrt{-2mE}/\hslash >0$. Away from the origin, ${\psi }^{\prime \prime }={\kappa }^2\psi$. Normalizability excludes exponentials growing at either infinity, leavingContinuity gives $A=B=C$. The derivative jump then becomes $-2\kappa C=-2m{V}_{0}C/{\hslash }^2$. For a nonzero state,There is exactly one positive value of $\kappa$, so exactly one negative-energy bound state. An odd solution would have $\psi (0)=0$; the allowed decaying exponentials would then both vanish identically. Normalize the even solution usingChoosing the overall phase real and positive gives the bound state in a delta potential:It peaks at the origin, has a downward cusp there, and decays exponentially on both sides. Its value, but not its slope, is continuous.

Scattering amplitudes and fluxes. For $E>0$, put $k=\sqrt{2mE}/\hslash$ and retain $\kappa =m{V}_0/{\hslash }^2$ as a strength parameter. A unit-amplitude wave incident from the left hasContinuity gives $t=1+r$. The jump condition givesSolving,The probability current of a plane wave of amplitude $A$ is $\hslash k|A{|}^2/m$ in its propagation direction. Since the asymptotic wavenumber is the same on both sides, reflection probability and transmission probability are just $|r{|}^2$ and $|t{|}^2$:They satisfy $R+T=1$, expressing conserved probability current in this real potential.

The right-hand plot uses $\eta =\kappa /k=V_0\sqrt{m/(2E)}/\hslash$, proportional to $V_0$ at fixed incident energy. Then $R={\eta }^2/(1+{\eta }^2)$ rises monotonically from zero to one, while $T=1/(1+{\eta }^2)$ falls from one to zero; they are equal at $\eta =1$. Despite the absence of a classical forbidden barrier, an abrupt attractive interaction reflects a quantum wave. Increasing attraction strengthens the derivative mismatch and can approach complete reflection at fixed positive energy. At fixed $V_0$, increasing incident energy instead makes the well increasingly transparent. The negative-energy bound state and the positive-energy scattering states are different stationary solutions, not an absorptive loss channel.

From the second-edge amplitude in part (c),This uses ${\psi }_{\mathrm{R}}=F{e}^{i{k}_2(x-a)}$. If the transmitted wave is instead written $T{e}^{i{k}_{2}x}$, its coefficient is $\boxed{T/A=i\sqrt{k_0/k_2}{e}^{-i{k}_{2}a}}$; the extra factor only changes the quantum phase convention.

Using the probability current of a plane wave,Thus the flux transmission probability is one. The larger transmitted density $|F{|}^2>|A{|}^2$ compensates for the smaller transmitted velocity $\hslash k_2/m$; equal currents need not have equal wave amplitudes.

Inside the barrier, the stationary Schrödinger equation gives an exponentially decaying wavefunction with inverse penetration lengthThe amplitude falls approximately as $e^{-\kappa a}$, so the leading exponential estimate for the transmission probability isFor comparison, wavefunction matching at a finite potential step at both edges gives the rectangular-barrier tunnelling probabilityThe difference is the interface prefactor: for an opaque barrier it is $16E(V_0-E)/V_0^2=2.56$. Both give a transmission of order $10^{-6}$; the first is the requested exponential estimate.

The probability current of the incident wave is $j_{\mathrm{in}}=\hslash k_1|A{|}^2/m$, and that of the reflected wave is $j_{\mathrm{ref}}=-\hslash k_1|B{|}^2/m$. The quantum reflection probability is the positive ratio of reflected to incident flux magnitudes:For comparison, the transmitted probability current is $j_{\mathrm{tr}}=\hslash k_2|C{|}^2/m$, so the transmission probability isThese satisfy $R+T=1$, as required by conservation of probability current for a real, nonabsorbing potential energy. In particular, $|C/A{|}^2$ alone is not the transmitted flux fraction when $k_2\ne k_1$.

The time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.