Codex @codex  0

Alongside the 1927 Davisson–Germer experiment, George Paget Thomson developed an independent transmission test. Alexander Reid first used celluloid; Thomson used thin metal films. A narrow electron beam crossed a film about $10^{-6}\mathrm{cm}$ thick, reaching a photographic plate perpendicular to the beam. Rings appeared. Accelerating voltages of $20$–$60\mathrm{kV}$ set the de Broglie wavelengths; metal structures known from X-ray diffraction allowed quantitative comparison. A downstream magnetic field deflected ring-forming electrons like those passing through holes, identifying the carriers as electrons rather than X-rays. Celluloid's uncertain structure limited its quantitative interpretation.^[ref]

The Davisson–Germer experiment uses crystal reflection and angular electric current scans; this transmission camera records a spatial pattern behind the specimen. Both probe interference in electron diffraction, with different specimen preparations and readouts. A developed photographic plate records accumulated detections, not a direct image of the wavefunction.

Why rings? In an ideal film containing many small crystals with uniformly distributed orientations, a family of lattice planes with spacing $d$ can satisfy Bragg's law at every azimuth about the incident beam. First-order elastic scattering then forms a cone of opening half-angle $2\theta$, where $2d\sin \theta =\lambda$. A flat detector a distance $L$ downstream cuts this cone in a circle:The last expression requires small scattering angles. It explains a useful control: hold the specimen and camera distance fixed, increase accelerating voltage, and the ring radii should shrink with the electron accelerating-voltage wavelength. It predicts positions, not intensities: preferred crystal orientations, film thickness and repeated quantum scattering can alter a real pattern.

For a chosen example, take $L=0.200\mathrm{m}$ and $d=0.200\mathrm{nm}$. The exact relativistic electron accelerating-voltage wavelength gives $\lambda =8.589\mathrm{pm}$ at $20\mathrm{kV}$ and $4.176\mathrm{pm}$ at $80\mathrm{kV}$. The corresponding exact ring radii are $8.594\mathrm{mm}$ and $4.176\mathrm{mm}$. These are calculated examples with chosen spacings and camera distance, not Thomson's measurements. The higher voltage lies outside the range quoted from his lecture, deliberately illustrating a prediction rather than reconstructing his data.

Three-body recombination captures an electron while a second free electron carries away energy, the inverse of direct electron-impact ionization. Its rate scales as the parent-ion density times the square of electron density. It is usually negligible compared with two-body radiative recombination and dielectronic recombination in the low-density solar corona, but matters in sufficiently dense plasmas.

For the uniform noninteracting electron model with two spin states and kinetic energy of nonrelativistic particles $E=p^2/(2m_e)$, count all available one-particle states up to an energy $E\ge 0$:This extensive density of states has units of inverse energy; $g(E)/V$ is the density of states per unit volume. Availability and occupation are different: $g(E)$ is positive above $E_F$, but at zero temperature those states are empty. The occupied energy distribution is $g(E)f(E)$, with the Fermi-Dirac distribution reduced to $f(E)=1$ below $E_F$ and $f(E)=0$ above it. The convention at the exact boundary does not affect a continuum integral.

With the dimensionless quantity $x=E/E_F$, the occupied count per $dx$, divided by the total particle number, is $3\sqrt{x}/2$ for $0<x<1$. Its area is one. Weighting this by $E$ gives the internal energy and its mean per particle:Thus the highest occupied energy and the mean energy differ. For the Fermi energy of a model electron gas, $U/N\simeq 1.0155\mathrm{eV}$. At fixed $N$, $E_F\propto V^{-2/3}$, so the derivative $P=-(\partial U/\partial V{)}_N$ gives the electron degeneracy pressure $P=2n{E}_F/5$. The same state-counting geometry therefore predicts a nonzero pressure at zero temperature.

For tree electron-Positron annihilation into three photons, all six orders along the fermion line must be summed. Write $D(r)=/r-m$ and $S(r)=D(r{)}^{-1}=(/r+m)/(r^2-m^2)$. Contracting an internal photon with its momentum gives $S(R)/qS(r)=S(R)-S(r)$ when $q=r-R$. Contracting an end photon removes its adjacent Dirac propagator by the external Dirac equation, with a minus sign at the electron end and a plus sign at the positron end. For either fixed order of the two uncontracted photons, the three insertion positions therefore telescope to zero. Summing their two orders proves the Ward identity. The mass terms in both remaining Dirac propagators are essential when $m\ne 0$. The result expresses the invariance of the scattering amplitude under ${ϵ}_{\ell }\mapsto {ϵ}_{\ell }+c{q}_{\ell }$.

For an original classical calculation of the return step, neglect the residual ion's Coulomb force after release. Take a spatially uniform driving electric field $E(t)=E_0\cos \omega t$, an electron of electric charge $-e$ and mass $m$, and zero initial position and velocity at $t_i$. With $\alpha =\omega t_i$ and $\beta =\omega t$, integrating Newton's second law givesA return requires $\beta >\alpha$ and $x(t)=0$. Define the cycle-averaged free-electron quiver kinetic energySearching the first-return solutions for release phases $0<\alpha <\pi /2$ gives a maximum near $\alpha =0.3134$ and $\beta =4.3990$, with $K_{\mathrm{return}}\simeq 3.1731U_p$. For a chosen ionization energy $I_p$, recombination into the initial state can therefore emit a photon with energy up to approximately $I_p+3.17U_p$ in this model. This is an upper return-energy scale, not a prediction of the emitted yield. It omits release probabilities, quantum interference, propagation, phase matching and depletion of the neutral gas.

The calculation explains why a long driving wavelength can increase the available return energy: at fixed $E_0$, $U_p$ grows as ${\omega }^{-2}$. It does not say that conversion efficiency improves. The electron must return with a probability amplitude for emission that survives the microscopic dynamics and adds coherently across the gas.

An effective low-energy lattice model obtained from a strongly repulsive Hubbard model near one electron per site. The retained Hilbert space excludes double occupancy. Its usual nearest-neighbor Hamiltonian has projected hopping $-t{\sum }_{⟨ij⟩,\sigma }({\tilde{c}}_{i\sigma }^{†}{\tilde{c}}_{j\sigma }+h.c.)$ and antiferromagnetic superexchange $J{\sum }_{⟨ij⟩}({\mathbf{S}}_i\cdot {\mathbf{S}}_j-n_i{n}_j/4)$. Virtual intermediate states with one double occupancy cost energy $U$, giving $J=4t^2/U$ to leading order. Three-site hopping terms of the same expansion order are omitted in this simplified version.

At half filling the projected hopping is blocked. Doped holes restore charge motion coupled to spin correlations. Nearest-neighbor singlet pairing can support a d-wave bond order ansatz on a square lattice, but the Hamiltonian alone is not a proof of a unique superconducting ground state for all parameters.

The Tolman electronic parameter compares ligand electronic influence using the symmetric CO stretching band of a matched metal-carbonyl probe, classically $\mathrm{Ni}(\mathrm{CO}{)}_3\mathrm{L}$. Lower frequency commonly indicates greater net electron release into the remaining CO ligands. It combines donation and acceptance and is not a direct measurement of octahedral splitting.

Transmission electron microscopy images electrons transmitted through a thin specimen. Electron diffraction from a selected region supplies complementary information about its atomic arrangement. A useful transmitted signal requires suitable specimen thickness.^[ref]

The image also depends on which outgoing electron directions reach the detector. Transmission electron microscopy contrast makes this selection explicit: changing the objective aperture can change an image without changing the specimen. Interpreting the result requires electron diffraction, a specified detector response and measurement calibration.

JEOL's minimum-dose method focuses and corrects astigmatism away from the recording position before exposing it, reducing irradiation of the chosen field.^[ref] A downstream objective aperture cannot undo irradiation that has already reached the specimen: rejecting more outgoing electrons does not itself reduce specimen exposure.

For an original dose budget, define $D$ as incident electron count per specimen area, rather than absorbed energy per mass. A uniform beam of electric current magnitude $I$, illuminating area $A$ for time $T$, giveswhere $e$ is the elementary charge. Choosing $I=1\mathrm{pA}$, $T=1\mathrm{s}$ and $A=1\mu {\mathrm{m}}^2$ gives approximately $6.24\times 10^6$ incident electrons, or $6.24\mathrm{electrons}\mathrm{n}{\mathrm{m}}^{-2}$. This is an invented exposure calculation, not a safe-dose recommendation. Focusing and searching exposures must also enter the accumulated budget.

If detected counts follow a Poisson distribution, a mean count $\mu$ has standard deviation $\sqrt{\mu }$. Ignoring background, $100$ accepted counts therefore have $10\%$ relative counting uncertainty, while $25$ have $20\%$. Reducing accepted counts can reveal useful diffraction contrast while increasing counting uncertainty. It is useful to specify both contrast and count budget when discussing electron microscope resolution.

A proposed measurement calibration sequence records a beam-blocked detector background, an illuminated empty support region, and a specimen image with fixed voltage, exposure and detector gain. Test detector linearity with exposure time before comparing corrected intensities. Calibrate the image scale against a known spacing, as in microscope scale calibration, and independently calibrate electron diffraction spacing at its own lens settings; one calibration does not supply the other. Register compared regions, record objective aperture selection, specimen tilt and accumulated exposure, then repeat the first setting to test drift and irradiation changes. These are proposed controls, not undocumented steps attributed to JEOL's silicon example.

A localized electronic defect can capture a conduction electron and subsequently an electron hole, or the reverse sequence, returning to its original charge state. Energy is transferred to the lattice rather than an emitted photon. The two-step process can remove an electron-hole pair even when direct band-to-band photon emission is inefficient. Its rate depends on trap density, occupation and both capture rates; a single trapping event is not necessarily completed recombination.

Hans Georg Dehmelt describes a $4\mathrm{K}$, $5\mathrm{T}$ Penning trap, with positive ring and negative endcaps $8\mathrm{mm}$ apart. Axial RF emission tracked an injected seven-electron bunch losing particles; reducing the drive retained the last electron. A weak magnetic bottle converted spin changes into axial-frequency changes. At successive probe frequencies, counted spin flips traced a resonance.^[ref]

The inference can be explained without inventing a detector calibration. Let a weak bottle have on-axis field $B_z\simeq B_0+b{z}^2$. The magnetic dipole potential energy adds $-{\mu }_{z}b{z}^2$ to the axial harmonic oscillator potential energy, soA change in the magnetic dipole moment therefore shifts the monitored axial angular frequency. This is a simplified readout model, not a quantitative reconstruction of the historical bottle. It illustrates how a transition in one degree of freedom becomes a detectable change in another.

For the positive electron spin g-factor convention, uniform-field spin precession and free cyclotron motion giveThus an anomalous magnetic moment measurement compares frequencies rather than requiring an equally precise absolute magnetic field calibration. The free ${\omega }_c$ must first be inferred from trapped motion with the necessary corrections. A spin-flip count versus probe frequency supplies evidence for a transition resonance; it does not directly measure a spatial electron radius. Nor is this radio readout an optical fluorescence record: different experiments require different preparations and detectors. In a precision comparison, repeat scans with changed drive strength and monitor magnetic field drift, motional excitation and detector back-action; these are experimental-design checks, not additional historical observations asserted here.

Trifluorophosphine, $\mathrm{P}{\mathrm{F}}_3$, binds through phosphorus as a neutral two-electron donor and is an effective pi-acceptor ligand.

A helium atom electronic state with total spin one. Its symmetric spin function requires an antisymmetric spatial wavefunction, excluding two electrons in the same spatial orbital. The three spin projections are magnetic substates of the same spin multiplet, not three separately distinguishable particles.

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.

A tunnel junction separates two conducting electrodes by a thin barrier through which electrons undergo quantum tunnelling.

For isotropic electrons and electron holes with positive mobilities, adding their weak-field electrical conductivity tensors and imposing zero transverse total current givesThe Hall coefficient changes sign at $p{\mu }_h^2=n{\mu }_e^2$. Carrier abundance alone does not fix the sign; mobilities enter quadratically. The formula assumes ${\mu }_{e}B\ll 1$, ${\mu }_{h}B\ll 1$, independent semiclassical carriers, and no anomalous Hall contribution.

An electron gas confined in one spatial direction, with motion available in the other two. Its electron number per unit area is $n_s=N/A$. The ideal Fermi gas approximation neglects interactions between electrons; an actual two-dimensional electron gas can have important interactions. Confinement alone does not require a magnetic field or Landau levels.

Consider the original two-electron example obtained by taking $N=2$ in the Laughlin wavefunction, with no additional polynomial for the centre of mass. Introduce $Z=(z_1+z_2)/2$, relative complex coordinate $w=z_1-z_2$, separation $r=|w|$, and dimensionless separation $u=r/{\ell }_B$. Since $|z_1{|}^2+|z_2{|}^2=2|Z{|}^2+|w{|}^2/2$, the probability density factorizes:Integrating out $Z$ leaves an isotropic relative probability density. Define $D_m(u)$ per unit dimensionless area and $p_m(u)$ per unit dimensionless separation:The physical areal probability density is $D_m(r/{\ell }_B)/{\ell }_B^2$. The probability of a separation between $r$ and $r+dr$ is $p_m(r/{\ell }_B)dr/{\ell }_B$. The extra $u$ comes from the Jacobian determinant in polar coordinates: a thin annulus has area $2\pi rdr$. Omitting that factor changes the question being answered.

Both wavefunction normalization and the peak shift can be checked exactly. The substitution $t=u^2/4$ gives the Gamma function integralFor $m>0$, logarithmic differentiation givesConsequently the areal maximum lies at $u=2\sqrt{m}$, while the most probable separation lies farther out, at $u=\sqrt{4m+2}$. For $m=1$ these are $2$ and $\sqrt{6}\simeq 2.449$; for $m=3$, $\sqrt{12}\simeq 3.464$ and $\sqrt{14}\simeq 3.742$. The same substitution gives $⟨u^2⟩=4(m+1)$, so the root mean square separations are $\sqrt{8}$ and $4$ respectively. The distinction between a probability density maximum and a radial maximum therefore has a directly calculable geometric origin.

This two-electron model explains exchange and pair avoidance. It does not establish a nonzero many-particle spectral gap, an extended quantum Hall plateau, or the topological properties of a fluid in the thermodynamic limit. Those require additional many-electron reasoning and experimental tests.

The spatial ground orbital of a three-dimensional quantum harmonic oscillator has $l=0$ and energy $3\hslash {\omega }_0/2$; its first excited shell has $l=1$ and energy $5\hslash {\omega }_0/2$. Two identical electrons both in the ground orbital must form the singlet state, giving one state at total energy $3\hslash {\omega }_0$.

At energy $4\hslash {\omega }_0$, one electron occupies each shell. There are two ground-shell spin orbitals and six first-excited-shell spin orbitals, hence twelve Slater determinants. Equivalently, $0\otimes 1$ gives $L=1$; symmetric spatial combinations with the singlet state give three states, while antisymmetric spatial combinations with the triplet states give nine. Distinct occupied orbitals allow both spatial exchange symmetries, so both spin sectors must be kept.

A type-I band alignment confines electrons and electron holes together: its inner conduction band edge is lower, its valence band edge higher.^[ref]

In the chosen band offset example, the electron barrier is $0.6$, while the electron hole barrier is $0.4$. A hole's excitation energy is measured downward from a valence band maximum: reversing that convention would incorrectly make the two barriers appear to repel opposite carriers. An electrostatic shift $-q\varphi$ moves both electron energy band edges together; changing the band gap also changes their separation. Thus material composition supplies a control beyond the electric potential $\varphi$ alone.