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The spatial wavefunction $\chi (x)$ is the position amplitude of a stationary quantum state, so $|\chi (x){|}^2$ is its probability density after normalization. The real function $U(x)$ is the potential energy, $E$ is the energy eigenvalue, $m$ is the particle mass, and $\hslash$ is the reduced Planck constant.

For a particle of rest mass $m_0$ and velocity $\mathbf{u}$, the relativistic energy and momentum areFor a photon of wavelength $\lambda$ travelling along $\hat{\mathbf{n}}$,Here $h$ is Planck constant and $c$ is the speed of light.

Let $\epsilon$ and ${\epsilon }^{\prime }$ be the incident and scattered photon energies. The stationary electron initially has energy $m_e{c}^2$. By four-momentum conservation, its final energy and momentum areUse the electron's relativistic energy-momentum relation, rather than assuming its recoil is nonrelativistic:Cancelling common terms givesSince $\lambda =hc/\epsilon$, the Compton wavelength shift isThe word elastic here means that the final electron has the same rest mass as initially and the complete photon-electron system conserves total energy; the photon alone loses energy to recoil.

For backscattering, $1-\cos \theta =2$, and henceChoose the incident direction as $+x$, so that the scattered photon travels in the $-x$ direction. An observer moving with velocity $+\beta c\hat{\mathbf{x}}$ meets that photon head-on. The Lorentz transformation of four-momentum gives the observed energyThis is the approaching-observer relativistic Doppler effect. Set $D=\epsilon /{\epsilon }^{\prime }=2.511/0.511$ and solve:The observer must move opposite to the scattered photon, along the original incident direction, at approximately $0.9205c$. Moving with the scattered photon would redshift it instead.

For a sinusoidal plane electromagnetic wave in vacuum, write the maximum electric field as $E_0$. The magnetic field maximum is $B_0=E_0/c$, so the cycle-averaged Poynting vector magnitude, or light intensity, isEach photon carries energy $hf$, where $h$ is the Planck constant. Therefore the photon number flux density, namely the number per unit normal area per unit time, isMultiplying by an illuminated normal area gives the photon arrival rate. Each absorbed photon transfers momentum $hf/c$ at normal incidence, giving the radiation pressure on a perfectly absorbing surfaceThese are time-averaged quantities. At oblique angle of incidence $\theta$, the normal radiation pressure is $I{\cos }^2\theta /c$; the simple displayed expression assumes normal incidence.

The one-dimensional time-independent Schrödinger equation isHere $m$ is the particle mass, $V(x)$ its potential energy, $E$ the stationary energy eigenvalue, and $\hslash =h/(2\pi )$ the reduced Planck constant. The spatial wavefunction is $\Psi$; the full stationary state is $\Psi (x)e^{-iEt/\hslash }$. Physical bound states satisfy the boundary conditions and wavefunction normalization $\int |\Psi {|}^{2}dx=1$. The operator on the left is the Hamiltonian operator, the sum of kinetic and potential energies.

For a photon, $E=hc/\lambda$. Using Planck constant $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, speed of light $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$ givesThe Rayleigh criterion for a circular aperture of diameter $D$ is ${\theta }_R=1.22\lambda /D$, set by the first zero of its Airy disk. Thus the ideal diffraction-limited resolution isThe conversion to transverse separation uses the small-angle approximation. An order-of-magnitude estimate $\lambda /D$ instead gives $2.07\times 10^{-7}\mathrm{rad}$ and $4.13\times 10^{10}\mathrm{m}$; the distinction is the chosen resolution criterion. These are aperture limits, before instrumental aberrations and detector sampling.

After the photon has thermalized, its energy $E_{\gamma }=hc/\lambda$ belongs to the macroscopic body's internal energy. The body's heat capacity makes the resulting temperature change negligible. Its equilibrium entropy therefore increases byThe Boltzmann entropy $S=k_B\ln \Omega$ then gives the microstate multiplier from photon absorption:For scale, explicitly choosing a representative visible wavelength $\lambda =500\mathrm{nm}$ and room temperature $T=300\mathrm{K}$ gives $\Delta S=1.32\times 10^{-21}\mathrm{J}{\mathrm{K}}^{-1}$, or $\Delta S/k_B=95.9$, and a multiplier approximately $4.5\times 10^{41}$. These are illustrative choices, not additional supplied data: the exact multiplier depends exponentially on the particular visible wavelength and temperature. The exact Planck constant, speed of light, and Boltzmann constant fix the conversion.^[ref]

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.

Classically, a particle with $0<E<V_0$ is completely reflected and never enters the raised region. In quantum mechanics the wavefunction penetrates the classically forbidden side as an evanescent wave. Define the tunnelling decay constantBoundedness at positive infinity excludes $e^{+\kappa x}$, soMatching the wavefunction and its derivative givesThe reflected probability amplitude has unit modulus, hence quantum reflection probability one. The decaying tail carries zero probability current: this semi-infinite potential step has no transmitted beam at infinity, unlike a finite potential barrier. The sketch shows an exact matched example rather than attempting to assign a single real value to a complex wavefunction.

Because the probability density decays as $|C{|}^2{e}^{-2\kappa x}$, the requested $1/e$ penetration distance isFor a proton, take $m_p=1.67262\times 10^{-27}\mathrm{kg}$, $V_0-E=10\mathrm{MeV}=1.60218\times 10^{-12}\mathrm{J}$ and $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe wavefunction amplitude decay length $1/\kappa =1.44\mathrm{fm}$ is twice this value. The nonrelativistic equation supplied in the paper is appropriate for this estimate since $10\mathrm{MeV}$ is small compared with the proton rest energy, approximately $938\mathrm{MeV}$. If the original PDF's unbarred $h$ is instead read literally as the Planck constant, the displayed differential equation gives $d_{\mathrm{literal}}=2\pi d=4.53\mathrm{fm}$. This is a valid calculation for that literal coefficient, but $0.720\mathrm{fm}$ is the physical estimate using the standard Schrödinger equation.

The time-independent Schrödinger equation is the energy eigenvalue equation for the Hamiltonian operator:Here $\hslash$ is the reduced Planck constant. A corresponding stationary wavefunction evolves as $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, which follows by substitution in the Time-dependent Schrödinger equation. Its position probability density is time independent because the time factor has unit modulus. Physical energy eigenstates must also satisfy the system's boundary conditions; for a bound state, wavefunction normalization requires ${\int }_{-\infty }^{\infty }|\psi (x){|}^{2}dx=1$.

Choose the null affine parameter so that the photon tangent is its four-momentum $P^{\mu }$. A fundamental observer has $U^{\mu }=(1,0,0,0)$ in cosmic-time coordinates, so the measured energy is $E=P_{\mu }{U}^{\mu }=c^2{P}^t$. The null normalization and conserved radial quantity in part (b) giveThus redshifting of peculiar momentum gives $E\propto a^{-1}$. Since $E=h\nu$, with $h$ the Planck constant, the received-to-emitted frequency ratio is ${\nu }_R/{\nu }_E=a(t_E)/a(t_R)$. By the definition of cosmological redshift,

The constants are the Planck constant $h$, photon frequency $\nu$, speed of light $c$, Boltzmann constant $k_B$ and absolute temperature $T$. The supplied $U_{\nu }$ is a spectral energy density, with units $\mathrm{J}{\mathrm{m}}^{-3}\mathrm{H}{\mathrm{z}}^{-1}$. It describes equilibrium radiation per volume and frequency interval, rather than emitted power itself. For isotropic cavity radiation the corresponding surface spectral radiant exitance is $M_{\nu }=c{U}_{\nu }/4$.

At low frequency the curve rises as ${\nu }^2$; it has one positive maximum and an exponentially decaying high-frequency tail. The left panel below shades the area $u(T)={\int }_0^{\infty }{U}_{\nu }d\nu$ under one curve. Put $x=h\nu /(k_{B}T)$; thenThe remaining dimensionless integral is independent of temperature, soThus doubling temperature doubles the characteristic frequency scale, raises the spectrum's height by eight and increases its area by sixteen. Total emitted energy also depends on emitting area and elapsed time.

Here $h$ is the Planck constant, $\nu$ is the radiation frequency, $c$ is the speed of light in vacuum, $k_B$ is the Boltzmann constant, and $T$ is the absolute temperature in kelvin. The quantity $U_{\nu }$ is spectral energy density per unit frequency, so $U_{\nu }d\nu$ is energy per unit volume in the interval $[\nu ,\nu +d\nu ]$, and $h\nu /(k_{B}T)$ is dimensionless.

Photocurrent is the electric current attributable to illumination. Here $I_{\mathrm{ph}}\ge 0$ denotes its magnitude after background subtraction. For a photodiode without internal multiplication, let $P$ be incident radiant flux concentrated at wavelength $\lambda$, and let $\eta$ be external quantum efficiency, including losses before collection. An incident photon has energy $E_{\gamma }=hc/\lambda$, with Planck constant $h$ and speed of light $c$. Dividing energy per second by energy per photon gives the incident photon rate ${\Phi }_{\gamma }$. Multiplying by $\eta$ gives the mean collected pair rate ${\dot{N}}_{\mathrm{pair}}$; multiplying by elementary charge $q$ givesThe factor is $q$: each collected pair replenishes an electron at one contact and removes an electron at the other. A cross-section of the external wire therefore counts one transferred electron per completed pair, consistent with charge conservation. The electron and electron hole are two internal charge carriers, not two independent transfers through that wire.

Choose an illustrative photodiode model with $\lambda =650\mathrm{nm}$, $\eta =0.75$ and $P=1.00\mu \mathrm{W}$. Using the exact SI values $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ givesA mean fraction $0.75$ does not require each fourth photon to be lost in a repeating pattern. It specifies a mean conversion rate. Nor does $0.75$ mean $0.75\mathrm{A}{\mathrm{W}}^{-1}$: quantum efficiency is dimensionless, whereas spectral responsivity compares electric current with power.

At fixed radiant flux, longer-wavelength photons have less energy, so more arrive per second. This explains the rising straight line when quantum efficiency is held constant. A real photodiode's quantum efficiency varies with wavelength; the plotted assumption cannot be extended past its band gap absorption cutoff. At fixed incident photon rate instead, $I_{\mathrm{ph}}=\eta q{\Phi }_{\gamma }$ has no additional factor of $\lambda$.

The photoelectric effect is the release of electrons from matter following photon absorption. Here the focus is emission from a metal surface into vacuum: a photon supplies energy, the work function pays the escape cost, and the remainder can become the emitted electron's kinetic energy. A photocurrent counts collected electric charge; a stopping potential probes the upper kinetic energy limit. These are different measurements.

In 1905, Albert Einstein proposed that an individual light quantum could transfer its whole energy to one electron. His §8 allowed losses before escape, distinguished the fastest emitted electrons from the rest, and predicted a straight stopping potential versus frequency relation with a material-independent slope. He separately predicted that changing light intensity changes the number emitted without changing their velocity distribution under otherwise identical conditions. He also discussed incomplete transfer and limits to the account.^[ref]

The useful modern starting model is a clean, homogeneous metal with independent single-photon events, negligible initial thermal broadening, negligible lowering of the surface barrier by an applied electric field, and negligible electric fields from the emitted electrons themselves. Specify the surface and its work function; it is not enough to name the bulk material. The Einstein photoelectric equation describes the upper edge of the emission energy distribution. It neither predicts that every absorbed photon liberates an electron nor supplies the whole photocurrent curve. Its appeal is that a complicated surface process still exposes the universal Planck constant through a simple slope.

Let a conducting sample and conducting analyzer be in good electrical contact, unbiased, at equilibrium, and without appreciable charging during illumination. Their Fermi levels then coincide. Choose that common Fermi level as zero. Define $E_B>0$ as the binding energy relative to this zero: removing an electron into the electron reservoir at the Fermi level costs $E_B$. It is not the cost of removing the electron all the way to the sample's vacuum level.

Write the sample and analyzer work functions as ${\varphi }_s$ and ${\varphi }_a$. Their local vacuum levels lie at ${\varphi }_s$ and ${\varphi }_a$ above the common Fermi level. Conservation of energy for an electron emitted without additional losses gives total outgoing energy $h\nu -E_B$, where $h$ is the Planck constant and $\nu$ the photon frequency. Immediately outside the sample its kinetic energy is $K_s=h\nu -E_B-{\varphi }_s$. The contact-related electric potential changes this toThus the sample work function cancels when the measured kinetic energy is referenced to the analyzer. Substituting ${\varphi }_s$ for ${\varphi }_a$ in the final conversion would mix two different energy references. An instrument's calibrated binding energy scale can incorporate its effective analyzer work function, so the correction need not appear as a separate operator-entered number.

As an original numerical example, stipulate $h\nu =1486.6\mathrm{eV}$, ${\varphi }_a=4.5\mathrm{eV}$ and a line at $E_B=284.0\mathrm{eV}$. The predicted analyzer kinetic energy is $1198.1\mathrm{eV}$. If ${\varphi }_s=5.0\mathrm{eV}$, the surface value is instead $K_s=1197.6\mathrm{eV}$; the contact-related increase is $0.5\mathrm{eV}$. A second line at $286.0\mathrm{eV}$ gives $1196.1\mathrm{eV}$ at the analyzer. A $+2.0\mathrm{eV}$ chemical shift in photoelectron spectroscopy therefore appears as a $-2.0\mathrm{eV}$ shift in measured kinetic energy. These are chosen values, not a measured carbon spectrum.

Charging breaks the assumed reference. If a uniformly charging sample rises by electric potential $V>0$ relative to the analyzer, an escaping electron loses an additional energy $eV$, with $e$ the positive elementary charge. Applying the unbiased conversion consequently returnsTwo features from the same stable potential retain their separation. Features from regions at different potentials acquire different offsets, which can mimic chemical differences or broaden a line. A conducting holder alone does not establish electrical contact through an insulating surface layer. Compare repeat scans, electrical contact and charge-compensation settings before interpreting an offset as a change in chemical bonds.

Measurement calibration is a separate issue: a known reference line tests the scale, while charge compensation controls the specimen's electric potential. In an original offset model $E_{B,\mathrm{read}}=E_{B,\mathrm{true}}+b$, one known line determines $b$. An unknown scale factor in $E_{B,\mathrm{read}}=a{E}_{B,\mathrm{true}}+b$ requires two distinct known lines to determine both $a$ and $b$. Re-reading a reference after a sample checks drift. Treating an unidentified contamination peak as a universally fixed reference would assume the chemical identity and electrical behavior that the measurement should test.

The energy carried by one photon is proportional to its frequency $\nu$, with proportionality constant the Planck constant $h$. The equivalent forms use angular frequency $\omega =2\pi \nu$ and vacuum wavelength $\lambda =c/\nu$. Different observers can measure different photon energies because they measure different frequencies; the photon remains massless. In signature $(+---)$, an observer with four-velocity $u^{\mu }$ measures $E=p_{\mu }{u}^{\mu }$, where $p^{\mu }$ is the photon's four-momentum and $u^{\mu }{u}_{\mu }=c^2$.

A photon gas is a collection of photons described statistically as radiation. Here the equilibrium model is homogeneous, isotropic electromagnetic radiation in vacuum at temperature $T>0$: each photon has energy $E=c|\mathbf{p}|=h\nu$, with momentum $\mathbf{p}$ and frequency $\nu$, and there are two independent transverse states of electromagnetic polarization. An absorbing cavity with walls at a common temperature provides a way to establish this thermal equilibrium. The bulk formulas assume cavity dimensions much larger than the typical wavelength $hc/(k_{B}T)$, so the electromagnetic mode density in a box can replace individual cavity modes by a continuum. ^[ref]

The crucial constraint is whether photon emission and photon absorption, or other number-changing processes, can equilibrate the photon population. When they can, photon number is not an independently conserved quantity and the equilibrium photon chemical potential is ${\mu }_{\gamma }=0$. For each polarization and frequency, the mean occupation number then follows the Bose-Einstein distribution:If number-changing processes are ineffective but number-conserving interactions redistribute energy, a constrained Bose-Einstein distribution can instead have $f=[\exp ((h\nu -{\mu }_{\gamma })/(k_{B}T))-1{]}^{-1}$ with chemical potential ${\mu }_{\gamma }\le 0$ fixed by the number constraint. Thus thermal equilibrium with respect to energy exchange alone does not establish ${\mu }_{\gamma }=0$. The zero value can also persist in an already prepared Planck photon distribution during suitable subsequent evolution; continual photon emission is not required to preserve it. ^[ref]

Combining $f$ with the two-polarization density of states, $g_{\nu }/V=8\pi {\nu }^2/c^3$, gives the Planck photon distribution and the spectral photon energy density:Here $B_{\nu }$ is the spectral radiance in Planck's law; $u_{\nu }$ is stored energy density per unit frequency, not outgoing power per area. Their different units are connected by the full-sphere solid angle and the speed of light $c$. The Boltzmann constant is $k_B$, and $h$ is the Planck constant. ^[ref]

The equilibrium equation of state is particularly simple:The radiation constant $a_R$ is the same constant called $a_{\mathrm{rad}}$ in black-body radiation; $\sigma$ is the Stefan-Boltzmann constant. An outline of the energy density calculation substitutes $x=h\nu /(k_{B}T)$ and uses ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation pressure follows because the angular average of the squared directional component of photon momentum is $1/3$ for isotropy; the thermodynamic derivation of blackbody radiation pressure supplies an independent route. These formulas describe the radiation contribution, not the cavity walls. The isotropic massless relation $p=u/3$ survives a different isotropic spectrum, whereas the coefficient in $u=a_R{T}^4$ requires the zero-chemical potential Planck photon distribution.

For a computed example with stipulated $T=2.725\mathrm{K}$, the photon energy density is $u\simeq 4.17174\times 10^{-14}\mathrm{J}{\mathrm{m}}^{-3}$ and the radiation pressure is $p\simeq 1.39058\times 10^{-14}\mathrm{Pa}$. The photon number density in the International System of Units isHere $\zeta$ is the Riemann zeta function. The mean photon energy differs from the energy at the peak of a spectral curve: counting photons and weighting them by energy emphasize different frequencies. The calculations use the exact International System of Units values of the Planck constant, Boltzmann constant and speed of light. ^[ref]

A motivating observation is the cosmic microwave background spectrum measured by the Far Infrared Absolute Spectrophotometer (FIRAS) aboard COBE. Its apparatus used a polarizing Michelson interferometer, with one input viewing a roughly seven-degree sky patch and the other an internal blackbody. Scanning mirrors changed the optical path length difference; heat-sensitive detectors recorded the varying power, and Fourier transform spectroscopy recovered the spectral difference. Operation outside the terrestrial atmosphere removed atmospheric emission and absorption, while shielding and cooling the optics in a helium cryostat suppressed the instrument's own radiation. ^[ref]

Absolute calibration came from moving an external blackbody into the sky aperture, filling the same beam used for the sky. This was an iron-loaded epoxy absorber shaped as a re-entrant cone. Repeated reflections made it a close approximation to a blackbody, while thermometers supplied its temperature. Mather and colleagues explicitly evaluated reflections, temperature gradients, radiation leaking through the material and leakage around its edge. These are limits on how faithfully the calibrator realizes thermal equilibrium radiation, rather than details that can be eliminated merely by calling it black. ^[ref]

During sky observations the internal blackbody temperature was adjusted to nearly null the spectral difference, reducing gain and drift errors. Fixsen and colleagues' 1996 analysis used calibrated spectra over wavenumbers $2$ to $21\mathrm{c}{\mathrm{m}}^{-1}$, about $60$ to $630\mathrm{GHz}$. They separated the sky's spatially uniform component, its dipole and Galactic emission, and varied sky cuts and foreground templates as controls. The resulting cosmic microwave background spectrum agreed with a single Planck function, with reported RMS deviations below 50 parts per million of its peak brightness. This is a residual statistic over the analyzed spectrum, not a pointwise relative error bound at every frequency. ^[ref]

The observation motivates using a photon gas as a physical model rather than merely a formal counting exercise. Its interpretation has a clear boundary: agreement with Planck's law constrains departures from an equilibrium spectral shape within the calibrated band and foreground assumptions; it does not directly measure radiation pressure or entropy density of a photon gas. Nor does it establish that today's cosmic microwave background is continually re-equilibrating with matter. A previously thermal spectrum can retain its Planck function shape as expansion of the universe lowers every photon frequency in the same proportion. The usefulness of the model is that one measured spectral temperature, together with its equilibrium assumptions, predicts energy density, radiation pressure, photon number density and entropy density of a photon gas consistently.

Planck's law gives the spectral radiance of an ideal blackbody in thermal equilibrium at absolute temperature $T>0$. It predicts how much power is carried in each small frequency or wavelength interval, per projected area and per solid angle. The whole curve is determined by $T$: no material-specific parameter enters this ideal Planck function.

Writing $\nu$ for frequency in hertz and $\lambda$ for wavelength in metres, the two forms of the Planck function areHere $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light in vacuum. Their exact values in the International System of Units are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. ^[ref]

The units matter: $B_{\nu }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, whereas $B_{\lambda }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Both are spectral radiance, rather than radiant exitance or energy density. To report $B_{\lambda }$ per micrometre, multiply its per-metre value by $10^{-6}$; a micrometre-wide interval is $10^{-6}$ metres wide.

For a computed example, at $T=6000\mathrm{K}$ and $\lambda =0.500\mu \mathrm{m}$, the exponent is $hc/(\lambda k_{B}T)\approx 4.79592$. Substitution into Planck's law gives $B_{\lambda }\approx 3.17569\times 10^{13}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$, or $3.17569\times 10^7\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ per micrometre. At the same wavelength, the illustrative temperatures $3000\mathrm{K}$ and $4500\mathrm{K}$ give $2.60268\times 10^5$ and $6.37793\times 10^6$ in the latter units. These are ideal predictions for stipulated temperatures.

The key change of variables is $\nu =c/\lambda$. Corresponding positive intervals carry the same power, so $B_{\lambda }d\lambda =B_{\nu }|d\nu |$ andEqual-width wavelength intervals have unequal widths in frequency; the factor $c/{\lambda }^2$ compensates for this. Merely replacing the horizontal-axis labels of $B_{\nu }$ by $c/\nu$ does not produce $B_{\lambda }$. This explains the frequency and wavelength peaks of Planck's law.

A short derivation uses the Boltzmann distribution for a single electromagnetic mode with thermal energies $nh\nu$, $n=0,1,\dots$. Put $q=e^{-h\nu /(k_{B}T)}$. The geometric series gives $Z={\sum }_{n\ge 0}{q}^n=(1-q{)}^{-1}$ and mean thermal energy $h\nu q/(1-q)$. The electromagnetic mode density in a box, including both independent polarization of an electromagnetic wave states of a photon, is $8\pi {\nu }^2/c^3$ per volume per unit frequency. Multiplication gives spectral energy density $u_{\nu }=8\pi h{\nu }^3/[c^3(e^{h\nu /(k_{B}T)}-1)]$; isotropy gives $B_{\nu }=c{u}_{\nu }/(4\pi )$. This counts thermal excitation energy, excluding zero-point energy. ^[ref]

Planck's law connects quantum mechanics to a measurable radiation curve. Its small-$h\nu /(k_{B}T)$ limit is the Rayleigh-Jeans law, $B_{\nu }\approx 2k_{B}T{\nu }^2/c^2$; its large-$h\nu /(k_{B}T)$ limit is the Wien approximation, $B_{\nu }\approx (2h{\nu }^3/c^2)e^{-h\nu /(k_{B}T)}$. The suppression at high frequency makes the total energy density finite. Integrating the curve gives the Stefan–Boltzmann law, while locating its peak gives Wien's displacement law. Thus one formula predicts both the spectral shape and the total emission.

The quantum Hall effect connects an electrical measurement to a quantum counting rule: a two-dimensional electron gas in a perpendicular magnetic field can exhibit plateaus of transverse electrical resistance. The integer case supplies a resistance scale built from the Planck constant and elementary charge, rather than the sample's dimensions. The fractional quantum Hall effect involves additional many-electron physics.

The quantum harmonic oscillator describes a particle of mass $m>0$ in a quadratic potential energy with angular frequency $\omega >0$. In the position representation its Quantum Hamiltonian isThe model is useful far beyond a literal spring: a Taylor series about a stable equilibrium with positive potential curvature has this quadratic term as its leading nonconstant approximation. The pure quartic oscillator has zero curvature at its minimum, so that particular approximation does not apply. ^[ref].

In units with $\hslash =1$, creation and annihilation operators giveHere $N$ is the number operator. Restoring the reduced Planck constant, the energy eigenvalues are $E_n=\hslash \omega (n+1/2)$ for $n=0,1,2,\dots$: neighboring levels are separated by $\hslash \omega$, and the ground-state energy is the zero-point energy $\hslash \omega /2$.

With length scale $\ell =\sqrt{\hslash /(m\omega )}$ and dimensionless coordinate $u=x/\ell$, the ground state is the normalized Gaussian wave packetFor $\hslash =m=\omega =1$, this is precisely the $a=1$ trial wavefunction in the quantum variational principle. The balance of kinetic energy and potential energy therefore recovers the exact ground-state energy using only Gaussian integrals.