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Separate geometry from electronic response in an original one-dimensional toy tunnel junction. Write the geometric surface height as $h(x)$, tip height as $z(x)$, and positive gap as $s(x)=z(x)-h(x)$. Assume low fixed tip-sample electric potential difference: the sampled energy window is narrow enough that the tip density of states is featureless and the sample local density of states varies little across that window. Neglect thermal broadening, changes of tip shape and lateral averaging. Let $\rho (x)>0$ be the effective sample local density of states divided by a positive reference value, so it is dimensionless. Choose a spatially constant effective barrier energy $\Phi >0$, a positive coefficient $C$ with units of electric current, andHere $m_e$ is the electron mass, $\hslash$ the reduced Planck constant, and $\kappa$ the tunnelling decay constant. This is an illustrative low-bias model, not a universal calibration formula. At fixed $I=I_0>0$, taking a logarithm givesThe logarithms have dimensionless arguments. Only constant $\rho$ makes the recorded height differ from geometric height by a constant. Multiplying $\rho$ by $r>0$ raises the apparent height by $\ln r/(2\kappa )$ without moving the geometric surface.

For a concrete ambiguity, choose $\kappa =10{\mathrm{nm}}^{-1}$, $a=0.020\mathrm{nm}$, lateral period $L=1\mathrm{nm}$, $s_0=0.50\mathrm{nm}$, and $q(x)=\cos (2\pi x/L)$. CompareBoth produce exactly $z(x)=s_0+aq(x)$ at $I_0=C{e}^{-2\kappa s_0}=C{e}^{-10}$. Substitution checks this without identifying $\rho$ with geometric height: case A has constant gap $s_0$, while case B has gap $s_0+aq(x)$ and a compensating electronic factor. The chosen ${\rho }_B$ ranges from $e^{-0.4}$ to $e^{0.4}$, approximately $0.6703$ to $1.4918$; every gap remains positive. The chosen $\kappa$ corresponds to $\Phi \simeq 3.81$ electronvolts, a positive effective barrier.

At fixed tip height $z_f=s_0$, both cases instead giveTheir maximum-to-minimum electric current ratio is $e^{0.8}\simeq 2.226$. Thus even adding this fixed-height scan at the same electric potential difference cannot distinguish these deliberately chosen surfaces within the toy model. Independent structural information or additional electronic measurements are required. Changing electric potential difference can supply such information, but also changes the assumptions used to define $C$ and $\rho$.

Electron microscope resolution concerns distinguishable specimen detail under a specified contrast criterion. It depends on diffraction, lens defects, electron energy spread, specimen motion and detection. A smaller de Broglie wavelength permits finer detail in principle, but neither wavelength nor magnification is the instrument's measured resolution.

For an electron accelerated from negligible initial kinetic energy through a positive electric potential difference $U$, combine $K=eU$, the relativistic energy-momentum relation and the de Broglie wavelength:Here $h$ is the Planck constant, $m_e$ the electron mass, $e$ the elementary charge, and $c$ the speed of light. This includes the relativistic correction omitted by the low-voltage electron accelerating-voltage wavelength approximation. With $m_e\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ and SI defining values of $h,e,c$, calculated wavelengths are approximately $6.979$, $3.701$ and $1.969\mathrm{pm}$ at $30$, $100$ and $300\mathrm{kV}$ respectively. The rounded mass agrees with the recommended value.^[ref]

For a circular aperture admitting a small semi-angle $\alpha$, the Rayleigh criterion suggests a diffraction-limited resolution scale $d_D=0.61\lambda /\alpha$. At $100\mathrm{kV}$ and $\alpha =10\mathrm{mrad}$, this gives $0.226\mathrm{nm}$, already much larger than the $0.00370\mathrm{nm}$ wavelength. This is an ideal aperture estimate. Transmission electron microscopy with coherent illumination and a particular specimen requires its own image-contrast analysis; this scale alone is not a universal two-feature resolution test.

An original illustrative blur model adds competing effects:Define $C_s{\alpha }^3$ as a characteristic transverse blur from the electron-lens analogue of spherical aberration, and $C_c\alpha \Delta K/K$ as one from energy-dependent focusing, the analogue of chromatic aberration. The coefficients $C_s,C_c$ have units of length, $\Delta K$ is a chosen characteristic kinetic energy spread, and $b_0$ is a fixed residual blur. Adding their squares is a declared heuristic, not a measured point spread function or a rigorous resolution theorem. The model omits contrast transfer, specimen scattering and signal statistics, so its minimum is only an estimate within its assumptions.

Choose $C_s=C_c=1\mathrm{mm}$, $\Delta K=0.7\mathrm{eV}$ and $b_0=0.10\mathrm{nm}$ at $K=100\mathrm{keV}$. At $5\mathrm{mrad}$, the four contributions are about $0.452$, $0.125$, $0.035$ and $0.100\mathrm{nm}$, giving $b\simeq 0.480\mathrm{nm}$. At $10\mathrm{mrad}$, the diffraction contribution falls to $0.226\mathrm{nm}$ but the assumed spherical aberration contribution grows to $1.00\mathrm{nm}$. Thus opening the aperture indefinitely worsens this model's image. All these values are calculations with chosen parameters, not observations of any microscope.

An energy eigenstate is a quantum state represented by an eigenvector of the Quantum Hamiltonian. For a time-independent Quantum Hamiltonian, a normalized eigenfunction $\varphi$ with energy eigenvalue $E$ evolves according to the Schrödinger equation asHere $\hslash =h/(2\pi )$ is the reduced Planck constant. The changing factor is a global phase, so this is a stationary state: its position probability density is constant. For example, the second infinite square well mode has two density lobes separated by a node of a standing wave at $x=L/2$; both lobes stay fixed while its quantum phase rotates. An ideal energy measurement returns $E$ with probability one, but a position measurement still has a distribution. Stationary density does not describe a particle sitting at the peak of a lobe. Distinct energy eigenvalues accumulate different quantum phases, making their quantum superposition behave differently.

The equipartition theorem says that each independent quadratic term in a Hamiltonian described by classical mechanics has mean energy $k_{B}T/2$ in a canonical ensemble. Here $k_B$ is the Boltzmann constant and $T>0$ the absolute temperature. Count terms in the energy, rather than vaguely counting everything that can move: one harmonic oscillator has one position degree of freedom but TWO quadratic terms, one in momentum and one in displacement.

A precise sufficient setting is $H(y,\xi )=a{y}^2/2+H_0(\xi )$ with $a>0$, where $y$ is one of the canonical variables, $\xi$ denotes all remaining variables, and the phase space integration measure factors as $dyd\xi$. The range of $y$ is the whole real line, independently of $\xi$; there is no constraint coupling its allowed values to other variables. The partition function must be finite, and the relevant expected values must exist. The Boltzmann factor then givesThe probability density function ${\rho }_y$ is a normal distribution with variance $k_{B}T/a$. A larger coefficient narrows the fluctuations enough to cancel its greater energy cost. This is why the mean contribution is independent of mass or spring constant.^[ref]

For a proof outline, the one-variable Gaussian integral is $Z_y={\int }_{-\infty }^{\infty }{e}^{-\beta a{y}^2/2}dy=\sqrt{2\pi /(\beta a)}$, so $-{\partial }_{\beta }\ln Z_y=1/(2\beta )$. More generally, integration by parts in a canonical variable gives $⟨y{\partial }_{y}H⟩=k_{B}T$ when its boundary term $[y{e}^{-\beta H}]$ vanishes and the integrals converge. For $a{y}^2/2$ this is twice its mean energy. At a finite hard boundary the boundary term need not vanish; hard-wall correction to harmonic equipartition illustrates the resulting departure. A constraint, a nonquadratic potential energy, or a noncanonical integration measure therefore requires its own calculation. The theorem is an ensemble average, not equality of instantaneous energies, and applying it to a recorded trajectory requires equilibration and sufficient sampling.

For a classical one-dimensional harmonic oscillator, with positive mass $m$ and spring constant $k$,The origin is the equilibrium position. Its heat capacity is $k_B$ at fixed $m,k$, not $k_B/2$. Although the two mean energies agree, their measured amplitudes do not: $⟨p^2⟩=m{k}_{B}T$ and $⟨x^2⟩=k_{B}T/k$. In a harmonic optical trap, the position measurement observes the potential energy contribution; it does not measure both contributions by observing $x$ alone.^[ref]

For $N$ particles in a three-dimensional classical monatomic gas, neglect interactions and thermally accessible internal excitations. Its three translational momentum terms give internal energy $U=3N{k}_{B}T/2$, heat capacity at constant volume $C_V=3N{k}_B/2$, and heat capacity at constant pressure $C_p=C_V+N{k}_B=5N{k}_B/2$. Per mole, replace $N{k}_B$ by the molar gas constant $R$. For example, using stipulated $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the molar heat capacity at fixed volume is $12.471\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These results assume a dilute, nonrelativistic ideal gas with classical translational statistics, rather than a quantum degeneracy regime.

For an ideal gas of diatomic molecules, modeled as two chemical atoms with a bond, there are three translational quadratic terms, two accessible rotational kinetic energy terms for changes of bond orientation, and one bond-stretching harmonic oscillator with two quadratic terms. Rotation about the bond axis is not an additional active mode in this linear-molecule model. The three familiar limits of heat capacity at constant volume areCorresponding molar heat capacities are $3R/2$, $5R/2$ and $7R/2$. Counting the stretching harmonic oscillator only once as $k_{B}T/2$ would give the wrong last value.^[ref]

The frozen regimes require quantum mechanics. In the simple molecular rotation model, $E_j=k_B{\theta }_{r}j(j+1)$ with state degeneracy $2j+1$, ${\theta }_r={\hslash }^2/(2I{k}_B)$, and moment of inertia $I$. For the bond-stretching quantum harmonic oscillator, ${\theta }_v=\hslash \omega /k_B$ with angular frequency $\omega$ and $\hslash =h/(2\pi )$ from the Planck constant. When ${\theta }_r\ll {\theta }_v$, the three limits above correspond to $T\ll {\theta }_r$, ${\theta }_r\ll T\ll {\theta }_v$, and $T\gg {\theta }_v$, provided translational motion remains classical. They are smooth crossovers, not abrupt switches. Real molecules can have different allowed rotational populations; at high temperature, thermal dissociation, internal excitations and nonharmonic bond motion can invalidate the fixed-molecule model before its last limit is reached.^[ref]

For a quantum harmonic oscillator with fixed $\omega$, put $z={\theta }_v/T$. Its exact mean energy and heat capacity areAt $T\gg {\theta }_v$, the mean energy approaches $k_{B}T$ and $C_{\mathrm{osc}}$ approaches $k_B$. At $T\ll {\theta }_v$, thermal excitation is exponentially suppressed and $C_{\mathrm{osc}}$ tends to zero, although the zero-point energy $\hslash \omega /2$ survives. Frozen thus means little thermal response, not zero motion or zero energy.^[ref]

A motivating measurement is thermal calibration of an optical trap. A focused laser through a microscope objective confines a bead in water; imaging or a quadrant photodetector records a position time series. Gieseler and colleagues describe a polystyrene bead of radius $1.03\mu \mathrm{m}$ at $293.15\mathrm{K}$, a $1064\mathrm{nm}$ trapping laser, and photodetector acquisition at $10^5\mathrm{Hz}$. Their recorded bead continues undergoing Brownian motion; higher trapping power produces a narrower equilibrium position distribution. Infer the centered variance, not a squared distance from an arbitrary camera origin:The normal distribution position histogram and agreement with dynamical calibration methods support the local harmonic optical trap interpretation. For a stipulated example, $T=300\mathrm{K}$ and ${\sigma }_x=20\mathrm{nm}$ imply $k\simeq 1.04\times 10^{-5}N/m=0.0104pN/nm$. This calculated scale is not an experimental reading.^[ref]

Position calibration is essential: if detector output $v$ has conversion $x=s(v-v_0)$, its physical variance is $s^2\mathrm{Var}(v)$. An unknown scale $s$ cannot be determined together with $k$ from that one variance. A complementary calibration demonstrated by Tolić-Nørrelykke and colleagues combines the power spectrum of thermal motion with the bead response to known sinusoidal motion of a translation stage. The driven response supplies an additional length scale, and their method does not require bead size, liquid dynamic viscosity or wall distance as inputs. It still uses a thermal dynamical model; agreement between methods is a consistency check, not a theory-free proof of equipartition theorem.^[ref]

The readout itself can hide fluctuations. Wong and Halvorsen used a $1064\mathrm{nm}$ optical trap and adjustable-exposure CCD imaging of a bead in water. They varied exposure and trapping power, held the bead away from the nearest wall, adjusted illumination to maintain comparable images, and measured tracking noise using a stuck bead. The measured variance fell as exposure increased; corrected results agreed with their time-averaging model. For a stationary overdamped harmonic optical trap, with optical trap relaxation time $\tau$, an exposure of length $W$ measures ${\bar{x}}_W=W^{-1}{\int }_0^{W}x(t)dt$. Its ideal noise-free variance isFor $\alpha >0$, $S(\alpha )<1$: using the blurred variance in the uncorrected formula overestimates $k$ by $1/S(\alpha )$. At $W=\tau$, the calculated factor is $e/2\simeq 1.36$. Additive tracking noise instead raises the measured variance and biases the inferred $k$ downward if left uncorrected.^[ref]

Useful controls are to record over many optical trap relaxation times, compare blocks for drift, calibrate the position scale, measure fixed-bead noise, vary exposure, and check the position histogram for departures from a normal distribution. A sparse sequence of accurate instantaneous positions can estimate an equilibrium variance; a long exposure averages away motion even if many frames are collected. Flow, appreciable laser heating, escape from the harmonic region or nonconservative forces can break the assumed thermal equilibrium. Inferring $k$ by imposing equipartition theorem is a calibration, whereas testing the relation requires an independently established stiffness and position scale. The beautiful practical connection is that microscopic randomness can set a force scale: once calibrated, a small constant applied force displaces the mean by $f/k$. The same quadratic-term counting predicts macroscopic heat capacity and makes microscopic fluctuations useful measurements.

In the stationary-state terminology used here, an excited state is a normalized energy eigenstate with energy eigenvalue $E>E_0$, where $E_0$ is an attained ground-state energy. Its excitation cost above the ground state is $\Delta E=E-E_0>0$. Several energy eigenstates can share one excited energy eigenvalue through degenerate energy levels. An arbitrary quantum superposition with energy expectation value above $E_0$ need not be a single stationary excited state; its energy content is described by its Born rule probabilities or, for a continuum, its scalar spectral measure.

An excited state can be a stationary state: the second infinite square well mode is an immediate example. Spontaneous photon emission is a transition caused by coupling matter to the electromagnetic field; it is absent from an isolated matter-only Hamiltonian operator. The relevant energy eigenstate of that isolated model is therefore not automatically an energy eigenstate of the full interacting system. This distinction explains how an ideal stationary excited state can approximate a state with a finite experimental lifetime. See electric-dipole interaction for the coupling used in the simplest optical model.

For a one-photon transition between stationary internal states, neglecting recoil, conservation of energy requiresHere $h$ is the Planck constant and $\nu$ is the photon frequency. Energy matching alone does not guarantee a transition: in the electric-dipole interaction approximation its strength also involves the matrix element $⟨f|\mathbf{d}\cdot \mathbf{ϵ}|i⟩$, where $\mathbf{d}$ is the electric dipole moment operator and $\mathbf{ϵ}$ specifies photon polarization. A zero matrix element gives an electric-dipole selection rule. Thus an absent optical line does not prove the corresponding energy eigenvalue is absent; a different interaction can couple different states. ^[ref]. The Franck-Hertz experiment gives a particularly instructive comparison between collision excitation and photon emission.

The first excited state terminology assumes a least energy eigenvalue above $E_0$. Such a next level is not guaranteed by quantum mechanics: a continuous spectrum can start immediately above the spectral bottom, or energy eigenvalues can accumulate there. For example, a Hamiltonian operator with energy eigenvalues $0,\epsilon ,\epsilon /2,\epsilon /3,\dots$, where $\epsilon >0$, has a ground-state energy zero but no least positive energy eigenvalue: every candidate $\epsilon /n$ is exceeded downward by $\epsilon /(n+1)$. A free particle on the whole line has neither a ground state nor a first stationary excited state. Each use of a discrete excitation ladder therefore needs a specified Hamiltonian operator and its boundary conditions.

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

The Fermi energy is the highest occupied one-particle energy in the ground state of a noninteracting Fermi gas at zero temperature, measured here relative to the lowest one-particle energy. The Pauli exclusion principle allows only one fermion in each complete one-particle quantum state, including its spin, so adding more particles fills progressively higher energy eigenstates. A cold gas can therefore have substantial kinetic energy even when its temperature is zero. In the thermodynamic limit, with the same energy reference, $E_F=\mu (T=0)$ where $\mu$ is the chemical potential.^[ref]

For a uniform, three-dimensional gas of noninteracting electrons, with spin one-half, equal populations of the two spin states, no applied magnetic field, and nonrelativistic particles of mass $m=m_e$,Here $n$ is the total number density, $V$ the volume, $p_F$ the Fermi momentum, $\hslash =h/(2\pi )$ the reduced Planck constant, and $k_B$ the Boltzmann constant. The electron Fermi temperature $T_F$ expresses the same energy scale in kelvin; it is not the actual gas temperature. Strong degeneracy of a degenerate electron gas means $T\ll T_F$. The momentum formula counts states and survives changing the energy–momentum relation, whereas $E_F=p_F^2/(2m)$ requires $p_F\ll mc$.^[ref]

The reference matters: the displayed $E_F$ excludes rest energy and sets the zero at $p=0$. Including rest energy shifts both the one-particle energy and the zero-temperature chemical potential by $m{c}^2$, leaving occupations unchanged. More generally, adding a constant $C$ to every one-particle energy also adds $C$ to $\mu$; the Fermi-Dirac distribution depends only on their difference. In a finite quantum system the chemical potential at zero temperature can lie between the last occupied and first empty energy eigenstates; its equality with the filling edge is a continuum statement.

At positive temperature, $E_F$ remains the zero-temperature scale associated with the specified number density. The Fermi level $\mu (T)$ must instead satisfy the particle-number conditionwhere $g(E)$ is the extensive density of states, including both spin states. Smearing the Fermi-Dirac distribution while arbitrarily holding $\mu =E_F$ generally changes $N$. This distinction matters when interpreting electron heat capacity and any experiment that infers temperature from an occupation profile.^[ref]

For a two-dimensional electron gas in a perpendicular magnetic field of magnitude $B>0$, the filling factor is the electron number per unit area $n_s$ divided by the orbital degeneracy per unit area $eB/h$ of one spin branch of a Landau level. Here $e$ is the positive elementary charge and $h$ the Planck constant. The nominal filling factor varies continuously with $n_s/B$; it need not be an integer.

Spreading a point nuclear electric charge over a small spherical nuclear charge distribution weakens the central attraction and raises electronic energy levels. In nonrelativistic first-order time-independent perturbation theory, a slowly varying $s$-state contact density gives the displayed shift. For a uniform sphere, $⟨r^2{⟩}_{\mathrm{ch}}=3R^2/5$, and integration of the difference between spherical and point Coulomb potentials gives $\Delta E_{1s}=(4/5)Z^4\mathrm{Ry}(R/a_0{)}^2$ for a hydrogen-like ion. A nonrelativistic $p$ wavefunction vanishes at the origin, suppressing its leading shift to order $R^4$. Heavy atomic nuclei require relativistic and realistic-charge-distribution corrections. An observed transition frequency shifts by the difference of the two level corrections divided by the Planck constant.

Take a nonrelativistic particle of constant mass $m$ with energy $E>0$, and choose the same zero of potential energy on both sides of a rectangular potential barrier:The equality of the exterior potential energies matters: it gives the same exterior wavenumber $k=\sqrt{2mE}/\hslash$ and makes $T=|t{|}^2$. Here $\hslash =h/(2\pi )$ is the reduced Planck constant. This model has a finite jump in potential energy, no delta potential, and no absorbing term.

For $0<E<V_0$, the time-independent Schrödinger equation gives the spatial wavefunctionThe full stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. We set the incident probability amplitude to one and impose no incoming wave from the right; this scattering wavefunction is not a normalizable wavefunction. The tunnelling decay constant $\kappa$ describes exponential decay, rather than an oscillatory wavenumber in the barrier. Both exponential functions are needed on the finite interval: the growing term is not excluded by a condition at infinity.

The wavefunction and its derivative are continuous at each face, giving four linear matching conditions:Their solution supplies the transmission amplitudeA short derivation is to use the two right-face equations to express $A{e}^{\kappa a}$ and $B{e}^{-\kappa a}$ in terms of $t$, then substitute into the left-face equations. Taking the squared modulus gives the rectangular barrier transmission probability.

Assume a triangular array of straight, singly quantized Abrikosov vortices with nearest-neighbor distance $a$. Here $\overline{B}>0$ is the magnitude of the spatially averaged internal magnetic field, and ${\Phi }_0=h/(2e)$ is the superconducting flux quantum; $e>0$ is the elementary charge. Two lattice vectors of length $a$ meeting at $60^{\circ }$ span area $A_c=\sqrt{3}{a}^2/2$. One flux quantum per cell givesThus multiplying $\overline{B}$ by five decreases $a$ by $\sqrt{5}$; it does not multiply the flux of each vortex by five. Using the exact SI Planck constant and elementary charge, ${\Phi }_0\simeq 2.06783\times 10^{-15}\mathrm{Wb}$.^[ref]

For chosen internal fields $0.20\mathrm{T}$ and $1.0\mathrm{T}$, the spacings are approximately $109\mathrm{nm}$ and $48.9\mathrm{nm}$. These are geometric model calculations, not measurements from a particular sample. The local magnetic field varies within a cell; replacing $\overline{B}$ by the field at a core would not give this counting relation.

Consider an ideal uniform graphene sheet connected to an electron reservoir, with an insulated gate. Let $c_g>0$ be geometric capacitance per unit area, $e>0$ the elementary charge, and $V_0$ the neutral-point offset. Neglect temperature, trapped electric charge and spatial inhomogeneity. A change in gate voltage pays for both the electrostatic potential drop, $en/c_g$, and the change in the sheet's Fermi energy, $E_F/e$. Here $E_F$ is measured relative to the band crossing; it is negative for excess electron holes.

Combine this relation with the carrier density in a graphene cone. Write $u=|E_F|/e$, measured in volts, $V=|V_g-V_0|$, and $\alpha =e^3/[\pi (\hslash v_F{)}^2{c}_g]$. ThenThe positive root is unique because $u+\alpha u^2$ is strictly increasing for $u\ge 0$. The often useful geometric approximation $n\simeq c_g(V_g-V_0)/e$ neglects the Fermi energy term. It overestimates $|n|$ in this model, especially near neutrality where relatively few states are available.

For a deliberately chosen model with $c_g=0.010F/{\mathrm{m}}^2$, $v_F=10^{6}m/s$, $V_0=0$ and $V_g=1\mathrm{V}$, the formulas give $E_F\simeq 0.252\mathrm{eV}$ and $n\simeq 4.67\times 10^{12}\mathrm{c}{\mathrm{m}}^{-2}$. The purely geometric value would be $6.24\times 10^{12}\mathrm{c}{\mathrm{m}}^{-2}$. These are chosen-model predictions, not measurements of the historical devices. Numerical conversion uses the exact Planck constant and elementary charge.^[ref]

Use the quantum harmonic oscillator in units with reduced Planck constant $\hslash =1$, mass $m=1$, and angular frequency $\omega =1$. Its Hamiltonian operator and trial Gaussian wave packet areThis is the same Gaussian wave packet used for the Heisenberg uncertainty relation: $|{\psi }_{\sigma }{|}^2$ is a normal distribution with mean zero and variance ${\sigma }^2$. The Gaussian integral gives wavefunction normalization and $⟨x^2⟩={\sigma }^2$. These smooth, rapidly decaying trials lie in the one-dimensional harmonic oscillator form domain and even in the operator domain.

Since ${\psi }_{\sigma }^{\prime }=-x{\psi }_{\sigma }/(2{\sigma }^2)$, integration by parts with vanishing boundary terms gives the kinetic energy and potential energy:Narrow trials cost kinetic energy, while broad trials cost potential energy. The exact differenceshows that the unique optimal width is ${\sigma }^2=1/2$, with $E=1/2$. At $\sigma =1/2$ and $\sigma =1$, the trial energies are both $5/8=0.625$; the kinetic energy and potential energy contributions exchange places.

Minimizing within this trial family alone establishes $E_0\le 1/2$. To establish exactness, use the Heisenberg uncertainty relation to obtain $E_{\psi }\ge 1/2$ for every admissible quantum state, as in its existing oscillator example. Together the bounds give $E_0=1/2$. Direct substitution also yields $H{\psi }_{1/\sqrt{2}}={\psi }_{1/\sqrt{2}}/2$, agreeing with the Gaussian eigenstate for a quadratic potential. This is a useful distinction: the variational method supplies the upper bound, while the Heisenberg uncertainty relation supplies the complementary lower bound.

The gravitational constant $G$ sets the strength of Newton's law of universal gravitation. Two point masses $M$ and $m$ a center-to-center distance $r$ apart attract with force magnitudeFor fixed masses and separation, doubling $G$ would double the force. For bodies with spherical symmetry, the same exterior formula follows from the spherical shell theorem; for arbitrary extended bodies, add the attractions of their mass density elements instead. The gravitational constant connects a laboratory attraction to the Newtonian gravitational field controlling a Kepler orbit.

The gravitational constant $G$ and local gravitational acceleration $g$ are different quantities. Applying Newton's second law to a test particle gives $g=F/m=GM/r^2$ for one spherical source. The source's mass and the observer's position change $g$; $G$ remains the same parameter of the law. A local falling-body experiment therefore measures a field, not $G$ alone. On rotating Earth, an instrument fixed to the surface also responds to centrifugal acceleration, while surrounding mass density and altitude change the gravitational contribution.

Dimensional analysis explains the units. Denote the base physical dimensions of length, mass and time by $\mathsf{L},\mathsf{M},\mathsf{T}$. Since force has dimension $\mathsf{ML}{\mathsf{T}}^{-2}$,By contrast, $[g]=\mathsf{L}{\mathsf{T}}^{-2}$ and the standard gravitational parameter $\mu =GM$ has dimension ${\mathsf{L}}^3{\mathsf{T}}^{-2}$. A small numerical value of a dimensional constant depends on the units chosen: it is the resulting force for specified physical masses and separation that conveys the laboratory scale.

NIST's current complete listing givesfor the 2022 CODATA adjustment, the latest adjustment listed when checked on 14 October 2026. Parentheses indicate a standard uncertainty $u(G)=0.00015\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, approximately $22$ parts per million relative to $G$. The adjustment year is 2022; NIST's May 2024 data update is a different date.^{[ref] [ref]}

For an original scale calculation, stipulate two $1\mathrm{kg}$ point masses separated by $1\mathrm{m}$ and insert this recommended $G$: $F=6.67430\times 10^{-11}\mathrm{N}$. The corresponding acceleration of either mass is $6.67430\times 10^{-11}\mathrm{m}{\mathrm{s}}^{-2}$ at that instant. Reducing the separation to $0.10\mathrm{m}$ multiplies both by $100$. These are ideal model predictions, not observations; their tiny scale explains why stray forces matter.

Being a constant of a physical law does not make the numerical value exact. The SI fixes the numerical values of seven defining constants, including the speed of light and Planck constant; $G$ is determined experimentally. Exact unit definitions still require physical realizations with measurement uncertainty: calibrating a length or a mass does not become error-free.^[ref]

Quantizing the stretching coordinate gives a quantum harmonic oscillator with reduced mass $\mu$:Here $v$ is the vibrational quantum number and $\hslash =h/(2\pi )$ the reduced Planck constant. Consecutive energy eigenvalues have gap $\hslash \omega$; the ground state has zero-point energy $\hslash \omega /2$ above the well minimum. The offset $V_0$ cancels from every transition energy.

A short proof uses creation and annihilation operators: with $\ell =\sqrt{\hslash /(\mu \omega )}$, write $q=\ell (a+a^{†})/\sqrt{2}$. The Quantum Hamiltonian becomes $\hat{H}-V_0=\hslash \omega (a^{†}a+1/2)$. Nonnegativity of $a^{†}a$ gives the lowest level; repeated application of $a^{†}$ raises its energy by $\hslash \omega$ each time. This also gives $⟨q{⟩}_v=0$ andThe ground state therefore has a nonzero spread of bond lengths, even though its mean stretching displacement vanishes. A stationary energy eigenstate has a time-independent position probability density; the classical oscillation curve is not a trajectory of an atomic nucleus in that state.

The parabolic model formally extends $q$ to the whole real line, whereas an actual bond length satisfies $r>0$. It is useful for low levels concentrated near $r_e$, with typical stretching displacements much smaller than $r_e$. It fails near breakup of the chemical bond and when neglected anharmonicity becomes substantial.

Heisenberg's microscope is a thought experiment, not a report of a performed measurement. In Werner Heisenberg's Nobel lecture, The development of quantum mechanics, delivered on 11 December 1933, the imagined electron is illuminated with very short-wavelength light and observed through a high-resolution microscope. A scattered photon entering the observer's eye would, through Compton scattering, eject the electron from its original path. The passage challenges a classical trajectory picture; its microscope is hypothetical. The lecture date is 1933 even though the award is the Nobel Prize in Physics in 1932.^[ref]

The following idealized geometry makes the tradeoff visible. Let the collected light have wavelength $\lambda$ and let the microscope objective accept outgoing directions within a half-angle $\alpha$ about the vertical axis. Take $x$ transverse to that axis and hold the incoming photon momentum fixed. A diffraction-limited estimate of transverse resolution has scale $\delta x\sim \lambda /\sin \alpha$. The unobserved outgoing direction admits transverse photon momentum components between approximately $-(h/\lambda )\sin \alpha$ and $+(h/\lambda )\sin \alpha$, neglecting the angular variation of the scattered wavelength. By conservation of momentum, uncertainty in that component leaves uncertainty in the transverse recoil of the electron:Here $h=2\pi \hslash$ is the Planck constant; the two $\delta$ symbols denote heuristic resolution and recoil scales, not the standard deviations in the Heisenberg uncertainty relation. Increasing the acceptance angle improves the estimated resolution while broadening the allowed recoil range. Numerical factors depend on the resolution criterion and angular distribution; this scaling argument cannot supply the sharp constant $\hslash /2$.

The thought experiment makes the interaction needed to obtain an image memorable. The preparation uncertainty theorem has broader use: it applies to the prepared quantum state even when the two observables are measured on separate copies. No claim that Werner Heisenberg built this gamma ray apparatus is needed to understand either argument.

The Heisenberg uncertainty relation constrains the standard deviations of position and momentum in one preparation:On the real line, the position operator $x$ acts by multiplication, and the momentum operator is $p=-i\hslash d/dx$, with $\hslash$ the reduced Planck constant. Their canonical commutation relation $[x,p]=i\hslash I$ and the Robertson uncertainty principle give the bound on normalized quantum states satisfying the moment and product-domain assumptions above; Schwartz functions are a convenient sufficient class. Thus making a preparation more sharply localized in position forces a wider momentum distribution, even for ideal measurement in quantum mechanics.

An explicit saturating example is the normalized Gaussian wave packet, for $\sigma >0$,The wavefunction is a probability amplitude; its squared magnitude is the probability density function, by the Born rule. That density is a normal distribution of mean zero and standard deviation $\Delta x=\sigma$, rather than a density whose width is read directly from the amplitude's exponent. With the unitary, negative-exponent convention for the Fourier transform, its momentum amplitude isThe Gaussian integral evaluates this Fourier transform. Consequently $|{\tilde{\psi }}_{\sigma }(p){|}^2$ is another normal distribution, with mean zero and standard deviationFor example, in dimensionless units with $\hslash =1$, $\sigma =1/2,1,2$ give $(\Delta x,\Delta p)=(1/2,1),(1,1/2),(2,1/4)$. The Gaussian wave packets all saturate the same bound: narrowing one density broadens the other. More precisely, choose a reference length $L$ and set $X=x/L$, $P=pL/\hslash$; then $[X,P]=iI$, and the plotted widths are $\sigma /L$ and $L\Delta p/\hslash$.

The Heisenberg uncertainty relation connects localization to the cost in kinetic energy. In a quantum harmonic oscillator with mass $m>0$ and angular frequency $\omega >0$, the variance contribution to the expectation value of the energy is bounded using the arithmetic-geometric mean inequality:The mean-dependent contributions are nonnegative. The Gaussian eigenstate for a quadratic potential with ${\sigma }^2=\hslash /(2m\omega )$ and zero means reaches this ground state energy. This is why the tradeoff is more than a limitation on preparation: it predicts an unavoidable positive minimum energy.

Heisenberg's original article appeared in March 1927 as "Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik", Zeitschrift für Physik 43, pp. 172–198. Its abstract discusses simultaneous imprecision of canonically conjugate quantities and explanatory thought experiments. The precise standard deviation formulation here distinguishes preparation spreads from a slogan about measurement disturbance. ^[ref].

The zero-field energy levels here are averaged over hyperfine structure; $h$ is the Planck constant. Electron self-energy and vacuum polarization produce radiative corrections. Self-energy describes the bound electron's radiative interaction with its own electromagnetic field. Precision predictions also include nuclear recoil and proton charge radius effects; assigning recoil terms to the reference or Lamb shift is convention-dependent. A finite proton charge radius changes binding independently of radiative effects.^[ref]

A vacuum-electric field “jiggling” picture motivates one contribution, not every hydrogen atom energy correction.

Infrared is electromagnetic radiation with wavelengths longer than the red end of ordinarily visible light. A useful named convention is the CIE range $780\mathrm{nm}$ to $1\mathrm{mm}$; the approximate visible range $400$–$700\mathrm{nm}$ used in some teaching texts puts the boundary nearer $700\mathrm{nm}$. ^[ref]. Neither number is a sharp physical cutoff of human vision. The CIE divisions IR-A, IR-B and IR-C are respectively $0.78$–$1.4\mu \mathrm{m}$, $1.4$–$3\mu \mathrm{m}$ and $3$–$1000\mu \mathrm{m}$; the boundaries of near-, middle- and far-infrared vary between applications. ^[ref].

Use vacuum wavelength $\lambda$ when comparing these ranges. With frequency $\nu$, speed of light $c$, Planck constant $h$ and photon energy $E$,Thus the CIE infrared interval corresponds approximately to $3.00\times 10^{11}$–$3.84\times 10^{14}\mathrm{Hz}$ and $1.99\times 10^{-22}$–$2.55\times 10^{-19}\mathrm{J}$ per photon, with the smaller energy at the longer wavelength. A $940\mathrm{nm}$ infrared photon has $E\simeq 2.11\times 10^{-19}\mathrm{J}$, whereas a $254\mathrm{nm}$ ultraviolet photon has $E\simeq 7.82\times 10^{-19}\mathrm{J}$. These are calculations from $hc/\lambda$, not measurements of a particular source. A television remote is a familiar approximately $940\mathrm{nm}$ source; its information is carried by variations in the emitted radiant flux. ^[ref].

Detection and visual perception answer different questions. A photodetector can respond to infrared even when an observer perceives no light; a thermometer detects an increase in temperature after photon absorption. Conversely, a small readout can reflect a small spectral responsivity, rather than absence of incident electromagnetic radiation. Images made from infrared readouts use display colors chosen by the imaging system; those colors are not the colors of the incoming infrared.

The important microscopic mechanisms also differ. In infrared spectroscopy, photon absorption often increases the energy of a normal mode in which chemical bonds stretch or bend. The corresponding optical spectrum helps distinguish molecules by their motions. For the usual electric dipole mechanism, the motion must change the electric dipole moment; matching the frequency alone is not sufficient. ^[ref]. Electronic processes can also emit or absorb infrared, so it is not exclusively radiation from molecular motion. ^[ref]; ^[ref].

Infrared is especially useful for observing thermal radiation from objects near everyday temperatures. For an ideal $300\mathrm{K}$ blackbody, Wien's displacement law gives a wavelength peak near $9.66\mu \mathrm{m}$. A thermal camera's readout nevertheless depends on emissivity, reflected surroundings and transmission along the path, as well as temperature. ^[ref]. A source emitting infrared need not be hot: an infrared remote is useful precisely because a photodetector can distinguish its signal from a background. Visible light and ultraviolet can also cause heating when absorbed; “infrared” names a wavelength range, not a unique form of heat.

Herschel's 1800 discovery made the distinction between sight and detection experimentally concrete. His report describes a optical prism spreading sunlight across a white, ruled receiving surface, with three thermometers at different positions near the end of the visible optical spectrum. He observed heating beyond the red edge where no visible color appeared. The result established that sunlight contained invisible radiation less strongly refracted by that optical prism than the visible red; it did not measure a photon energy or establish today's infrared boundaries. ^[ref]; ^[ref].

A documented modern version by Caltech/IPAC uses a glass optical prism, white paper in a shaded box and three blackened liquid-filled thermometers. Initial readings in shade provide each thermometer's baseline; subsequent readings compare blue, yellow and just beyond red after the bulbs have had time to warm. The measured quantity is the rise in temperature, not a “temperature of a color”. The lesson reports the expected largest rise beyond visible red and specifically discusses matching the bulb coatings and thermometer calibration. Clouds and changes in the projected optical spectrum affect the comparison. ^[ref].

For a stronger modern interpretation, an additional shaded thermometer would monitor ambient drift, exchanging bulbs between positions would test differences between thermometers, and blocking the incident beam would distinguish beam-dependent heating from warmed supports or air. These are suggested controls, not claims about Herschel's historical procedure. Thermal conduction through the mounting and convection can change the readout. A broad bulb also samples a range of wavelengths; the optical prism's transmission and optical dispersion, the bulb's absorption and heat loss all affect the position of the largest rise. In particular, the result does not show that an infrared photon carries more energy than a visible photon.

A simple analytical illustration makes those limitations explicit. Let $\theta =T-T_{\mathrm{amb}}$ be the bulb's excess temperature, $C$ its heat capacity, $G$ its effective heat-loss coefficient in $\mathrm{W}{\mathrm{K}}^{-1}$, and $P_{\mathrm{abs}}$ the absorbed power. Assume one uniform bulb temperature, constant ambient temperature, constant $C,G$, and losses linear in $\theta$. Thenfor a beam switched on at $t=0$ with $\theta (0)=0$. After switching the beam off at $t=t_s$, $\theta (t)=\theta (t_s)e^{-(t-t_s)/\tau }$. The exponential function follows by solving this linear differential equation. The steady excess temperature measures $P_{\mathrm{abs}}/G$; it is not a direct measurement of the optical spectrum. With illustrative $C=0.020\mathrm{J}{\mathrm{K}}^{-1}$ and $G=0.0010\mathrm{W}{\mathrm{K}}^{-1}$, $\tau =20\mathrm{s}$. Absorbed powers of $1.0$ and $0.5\mathrm{mW}$ give steady rises of $1.0$ and $0.5\mathrm{K}$, regardless of which wavelengths supplied that absorbed power.

The upper schematic represents the logic of the optical prism experiment; its light ray positions are not a calibrated wavelength scale. The lower curves use the illustrative heat-balance model, with the beam blocked at $60\mathrm{s}$; they are not historical or modern measured data. The beauty of the experiment is that a simple change of detector extends observation beyond what the eye can perceive.

For a fixed amount of substance and composition of classical ideal gas, internal energy depends only on temperature: $U=U(T,n)$, with no independent volume dependence. This does not require constant heat capacity at constant volume. Between equilibrium endpoints,Here $C_V$ is total heat capacity at constant volume, $C_{V,\mathrm{m}}$ is molar heat capacity, and $U/m$ for sample mass $m$ is specific internal energy, whose change is ${\int }_{T_1}^{T_2}{c}_V(T)dT$. These endpoint relations apply even when the intervening process is not at fixed volume; they do not say that transferred heat always equals $\Delta U$. With constant $C_V$, the safe statement is $U(T)=U(T_{\ast })+C_V(T-T_{\ast })$. Writing $U=C_{V}T$ additionally chooses the reference energy; it does not prove that a physical gas remains classical and ideal down to zero temperature. ^[ref]

A thermodynamic proof outline uses the first law of thermodynamics in equilibrium form, $dU=TdS-pdV$, and the Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$:The last step uses the ideal gas law and the molar gas constant $R$, with fixed composition and amount of substance. Microscopically, separated ideal gas particles store no intermolecular interaction energy; their translational kinetic energy and any internal molecular energy are determined by temperature. Negligible intermolecular interactions still allow a molecule to have internal vibrational potential energy.

An original analytical model makes the distinction between an ideal gas and constant heat capacity concrete. Assume classical translation, two fully active rotational terms, and one independent quantum harmonic oscillator per diatomic molecule, with fixed angular frequency $\omega$. Define the vibrational temperature scale $\Theta =\hslash \omega /k_B$, using the reduced Planck constant $\hslash =h/(2\pi )$ and Boltzmann constant $k_B$. Summing its thermal occupations givesThe constant $U_{\ast }$ includes the chosen reference and zero-point energy, whose temperature derivative vanishes. When $T\ll \Theta$ the vibrational contribution to heat capacity is exponentially small; when $T\gg \Theta$ it tends to $R$ per mole. This model assumes rotations remain active throughout the interval and excludes dissociation, electronic excitation and real gas corrections. Its low-$T/\Theta$ edge is not a model of rotational freezing or of an actual substance at absolute zero.^[ref]