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Let ${\sigma }^2=\hslash /(2m\omega )$ and $g(x)=(2\pi {\sigma }^2{)}^{-1/2}{e}^{-x^2/(2{\sigma }^2)}$ be the quantum harmonic oscillator ground-state probability density. Put $p(a)={\int }_{-a}^{a}g(x)dx$. The first excited-state density is $x^{2}g(x)/{\sigma }^2$. Integration by parts, using $g^{\prime }=-xg/{\sigma }^2$, gives the first-order differenceThe bracket grows from zero to one, since its derivative is $2a^{2}g(a)/{\sigma }^2$. Thus the first-order predicted level gap for $H_0+\lambda H^{\prime }$ is $\hslash \omega [1-\lambda \{p(a)-2ag(a)\}/2]$. For every $\lambda >2$ it becomes negative at sufficiently large $a$.

For $\lambda >1$ the full potential has a central barrier and symmetric minima at $\pm a$. The unperturbed even ground state samples the central barrier more strongly than the odd excited state, whose node reduces its shift. This explains the perturbative tendency, but the negative predicted gap is a failure diagnostic of the first-order truncation. The exact one-dimensional ground state remains nodeless and below the first excited state; in a high symmetric barrier the actual pair has a small positive double-well tunneling splitting.

For a diatomic molecule, let $V(r)$ be the electronic potential energy at bond length $r$, with the atomic nuclei treated as fixed while finding the electronic energy. Choose zero at infinitely separated chemical atoms in the specified electronic quantum states. If the minimum occurs at $r_e$, the bond potential-well depth isThe bottom of this curve describes an equilibrium bond length, not an attainable motionless vibrational quantum state. Changing the zero of potential energy shifts both endpoints equally and leaves $D_e$ unchanged. A quantum harmonic oscillator approximates the curve near its minimum but rises without limit, so that local approximation alone cannot supply a dissociation asymptote. A Morse potential supplies both a minimum and a finite separated-fragment limit.^[ref]

Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

A real centered Gaussian wave packet with initial position variance $1/(2\alpha )$ and momentum variance ${\hslash }^2\alpha /2$ evolves under a quantum harmonic oscillator byThe covariance matrix obeys $\Sigma (t)=M\Sigma (0)M^{\mathsf{T}}$ and $\det \Sigma ={\hslash }^2/4$. Thus the Robertson-Schrödinger uncertainty relation stays saturated, although the position-momentum variance product can exceed ${\hslash }^2/4$ because of nonzero quantum covariance. The widths are stationary only for $\alpha =m\omega /\hslash$; otherwise they breathe with frequency $2\omega$.

A canonical quantum harmonic oscillator has geometric number probabilities and mean occupation $(e^{\beta \hslash \omega }-1{)}^{-1}$. The position operator then gives the displayed variance. It interpolates between zero-point fluctuations $\hslash /(2m\omega )$ and the classical variance $k_{B}T/(m{\omega }^2)$. The kinetic and potential energies are equal at every temperature, although classical equipartition describes their values only at high temperature.

In the isotropic three-dimensional quantum harmonic oscillator, the first excited Cartesian energy eigenstates are proportional to $x,y,z$ times the common Gaussian $e^{-r^2/(2a_0^2)}$. If ${\varphi }_x,{\varphi }_y,{\varphi }_z$ are normalized with the same real phase convention, their combinations in the title are normalized angular momentum eigenstates with $l=1$ and $m=1,0,-1$.

Indeed, $z=r\cos \theta$ and $x\pm iy=r\sin \theta e^{\pm i\phi }$ identify a common radial factor times the spherical harmonics $Y_{10},Y_{1,\pm 1}$. The transformation is unitary, so it changes the basis rather than the threefold orbital degeneracy.

For noninteracting particles in a three-dimensional isotropic quantum harmonic oscillator, measure the one-particle ground energy as zero. The single-particle partition function is $Z_1=(1-e^{-\beta \hslash \omega }{)}^{-3}$, from the product of three geometric series. In the dilute Maxwell–Boltzmann statistics regime, $\ln \Xi =e^{\beta \mu }{Z}_1$, so the grand potential is the displayed expression and $N=e^{\beta \mu }{Z}_1$. For $k_{B}T\gg \hslash \omega$,The same approximation follows from integrating the density of states $g(E)\simeq E^2/[2(\hslash \omega {)}^3]$ against $e^{-\beta E}$. Extrapolating the classical formula until the fugacity is one gives the scale of Bose condensation in a three-dimensional harmonic trap, but the quantitative transition temperature requires the Bose occupation law rather than Maxwell-Boltzmann statistics.

Suddenly shifting a quantum harmonic oscillator equilibrium from zero to a signed position $x_0$ leaves the old ground state unchanged at the instant of the quantum quench. The new annihilation operator is $a^{\prime }=a-\lambda$, with $\lambda =\sqrt{m\omega /(2\hslash )}{x}_0$, so $a^{\prime }|0⟩=-\lambda |0⟩$. In the translated energy eigenstates $|n^{\prime }⟩=e^{-ip{x}_0/\hslash }|n⟩$, the expansion isThe sign follows by translating the old vacuum by $-x_0$ before expanding it in the old Fock states. A Baker--Campbell--Hausdorff formula with central commutator separates the resulting exponential into creation and annihilation parts. Evolution with the shifted Hamiltonian preserves this coherent state, with $\alpha (t)={\alpha }_0{e}^{-i\omega t}$. Its mean occupation is $|{\alpha }_0{|}^2$ and its mean energy above the new potential minimum is $\hslash \omega (|{\alpha }_0{|}^2+1/2)$. Omitting $(-\lambda {)}^n$ in favour of $|\lambda {|}^n$ changes the state when $x_0>0$.

A normalized oscillator coherent state obeys $a|\alpha ⟩=\alpha |\alpha ⟩$, so its mean occupation number is $N=|\alpha {|}^2$. Under the quantum harmonic oscillator Hamiltonian operator $\hslash \omega (a^{†}a+1/2)$, its evolution is $|\alpha (t)⟩=e^{-i\omega t/2}|\alpha e^{-i\omega t}⟩$. Inserting $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ and $p=-i\sqrt{m\hslash \omega /2}(a-a^{†})$ givesThese mean values obey the classical harmonic oscillator equations. The displacement amplitude is $\sqrt{2\hslash N/(m\omega )}$, whose classical orbit has energy $\hslash \omega N$. The full quantum mean energy has the additional zero-point energy $\hslash \omega /2$. Indeed the variances are $\hslash /(2m\omega )$ and $m\hslash \omega /2$, and their kinetic and potential contributions sum to $\hslash \omega /2$.

For the dimensionless quantum harmonic oscillator $H=-d^2/d{x}^2+x^2$, use the compactly supported trial ${\psi }_{\gamma }=\cos (\gamma x)$ on $|x|\le \pi /(2\gamma )$ and zero elsewhere, with $\gamma >0$. It is continuous and belongs to the oscillator energy quadratic form domain: its weak derivative is square-integrable even though its ordinary derivative jumps at the endpoints. Exact integration givesThe minimum occurs at ${\gamma }^4=({\pi }^2-6)/12$, giving the displayed upper bound, approximately $1.13572$, compared with exact ground state value $1$. This illustrates why Rayleigh-Ritz variational principle trials need the energy form domain rather than the smaller operator domain.

For the dimensionless quantum harmonic oscillator $H=-d^2/d{x}^2+x^2$, the normalized compact trial function is ${\psi }_a=\sqrt{15/(16a^5)}(a^2-x^2)$ on $|x|\le a$ and zero elsewhere. Its kinetic and potential contributions to the energy quadratic form are $5/(2a^2)$ and $a^2/7$. Minimizing their sum gives $a^4=35/2$ and the displayed upper bound, approximately $1.19523$, compared with exact ground state eigenvalue $1$. The weak derivative is square-integrable because the trial is continuous at its endpoints. Jumps in the ordinary first derivative do not invalidate the Rayleigh-Ritz variational principle: it requires the energy form domain, rather than the smaller operator domain.

For two identical fermions of mass $m$ in the same one-dimensional quantum harmonic oscillator, use center of mass $X=(x_1+x_2)/2$ and separation $r=x_1-x_2$. The corresponding oscillator masses are $2m$ and reduced mass $m/2$, with the same frequency $\omega$. If $N,n$ are their occupation numbers, the unperturbed energy is $\hslash \omega (N+n+1)$. The fermionic exchange symmetry requires even $n$ with a spin-one-half singlet state and odd $n$ with a spin-one-half triplet state.

For a delta potential $\lambda \delta (r)$ the first-order shift is $\lambda |{\phi }_n^{\mathrm{rel}}(0){|}^2$. Every odd relative wavefunction vanishes at contact, so these states remain exact unaffected eigenstates of the spatial problem. For the even relative ground state, the oscillator length is $\sqrt{2}\ell$, where $\ell =\sqrt{\hslash /(m\omega )}$, and the shift is $\lambda /(\sqrt{2\pi }\ell )$, independent of $N$. This separates interaction-sensitive singlets from contact-insensitive triplets.

In the plane wave basis of a ring, $V(x)=-\lambda \cos (2\pi x/L)$ has matrix elements $V_{mn}=-(\lambda /2)({\delta }_{m,n+1}+{\delta }_{m,n-1})$. Near $\alpha =1/2$, degenerate perturbation theory on $n=0,1$ givesup to corrections from remote states. Thus the avoided crossing gap is $|\lambda |$ to first order when $|\lambda |\ll E_L$.

For $|\lambda |\gg E_L$, a parabolic approximation about a potential minimum instead gives a local quantum harmonic oscillator spacing $\sqrt{2|\lambda |E_L}$. A local gauge transformation removes the flux from this approximation. Actual ring levels retain exponentially small flux dependence through quantum tunnelling, while nonquadratic corrections to their mean energies are algebraic in $\sqrt{E_L/|\lambda |}$.

A creation operator raises the occupation number operator eigenvalue of a quantum harmonic oscillator by one. Repeated application builds Fock states.

Repulsive trapped condensates change smoothly from a nearly ideal quantum harmonic oscillator ground state for $\chi \ll 1$ to a broad Thomas–Fermi approximation for a condensate for $\chi \gg 1$. Comparing the Gaussian interaction energy of a trapped condensate with the total ideal energy gives $E_I/E_0=2\chi /(3\sqrt{2\pi })$. Thus $N\sim a_0/a$ is a crossover scale, not a sharply defined critical particle number. Total interaction energy must be compared with total ideal energy; comparing it with one oscillator quantum confuses total and per-particle scales.

Adding $-Fx$ to the quantum harmonic oscillator potential gives $\frac{1}{2}m{\omega }^2(x-d{)}^2-F^2/(2m{\omega }^2)$, where $d=F/(m{\omega }^2)$. Translation preserves the oscillator spectrum, apart from the displayed constant shift. Eigenfunctions are translated by $d$ and the energy spacing is unchanged.

The dynamic electric polarizability relates the induced electric dipole moment at a driving frequency to the applied electric field at that frequency in linear response theory. A complex coefficient can encode both a dispersive phase and absorption. For a charged one-dimensional quantum harmonic oscillator with charge magnitude $e$, mass $m$, and natural frequency ${\omega }_0$, the undamped response isOne derivation uses the oscillator electric-dipole selection rule and first-order time-dependent perturbation theory. The same result follows from the classical Lorentz oscillator model: $m\ddot{z}+m{\omega }_0^{2}z=-eE$ and $d=-ez$ imply $\ddot{d}+{\omega }_0^{2}d=e^{2}E/m$. At exact resonance there is no bounded periodic response without damping; a finite switch-on produces secular growth rather than a finite steady-state value of the pole expression.

The Einstein model of a solid represents the vibrations of $N$ chemical atoms by $3N$ independent one-dimensional quantum harmonic oscillators, all with the same positive angular frequency $\omega$. Equivalently, each chemical atom has three independent oscillator directions. The common angular frequency is a simplifying model assumption, not a claim about every vibration of a real solid. Hold the volume, $N$ and $\omega$ fixed. Define the characteristic temperature ${\theta }_E=\hslash \omega /k_B$ and the dimensionless ratio $x={\theta }_E/T$. The vibrational internal energy isThe first term is the zero-point energy $3N\hslash \omega /2$. It is constant at fixed $\omega$, so it contributes no heat capacity at constant volume. The temperature-dependent term givesThis is the vibrational heat capacity at constant volume of the model, with no expansion or other contributions included. ^[ref]

For a worked comparison, choose the model parameter ${\theta }_E=300\mathrm{K}$. The normalized heat capacity at constant volume depends only on $T/{\theta }_E$:

At $T={\theta }_E$, one mole of chemical atoms therefore has vibrational heat capacity at constant volume $3R(0.92067)\simeq 22.965\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, using the unrounded model value. Even at twice the characteristic temperature, the model remains about $2.06\%$ below the Dulong-Petit law limit. These are calculations for the ideal Einstein solid, not material measurements.

A short derivation uses the thermal partition function of a quantum harmonic oscillator. The energy eigenvalues $E_n=\hslash \omega (n+1/2)$ and their Boltzmann factors form a geometric series, giving the one-oscillator canonical partition function $z=e^{-x/2}/(1-e^{-x})$. The mean occupation is $1/(e^x-1)$, hence the displayed $U$ after summing over $3N$ independent quantum harmonic oscillators. Taking its derivative with $dx/dT=-x/T$ gives the displayed heat capacity at constant volume. ^[ref]

The two limits explain the shape. At high temperature, a Taylor series in small $x$ givesAt low temperature, $1-e^{-x}\to 1$, soThe zero-point energy remains even when the thermal response is tiny: freeze-out means that changing temperature scarcely changes the occupations, not that the chemical atoms have zero energy. This is the useful lesson of the Einstein model of a solid: the classical counting rule reappears when many levels are accessible, while discrete excitation spacings explain its failure upon cooling.

The exponential low-temperature decay is also the model's limitation. Real three-dimensional crystalline solids have acoustic phonon branches with arbitrarily small angular frequencies in the bulk limit. The Debye model includes these collective normal modes and predicts a low-temperature vibrational contribution $C_V\propto T^3$, which the single-frequency Einstein solid cannot reproduce. This qualification concerns the vibrational contribution; additional excitations can change a material's total heat capacity. ^[ref]

Einstein introduced the quantum treatment of solid vibrations in his 1907 paper Die Plancksche Theorie der Strahlung und die Theorie der spezifischen Wärme. Its historical importance is that quantum excitation energies explained why the classical Dulong-Petit law could fail; the modern quantum harmonic oscillator formulation above includes the zero-point energy explicitly. ^[ref]

An electric dipole transition is a transition driven, in the long-wavelength approximation, by the electric-dipole interaction $H_{\mathrm{int}}=-\mathbf{d}\cdot \mathbf{E}(t)$. Its leading amplitude is proportional to a matrix element of the electric dipole moment between the initial and final states. A vanishing element gives an electric-dipole selection rule; a nonzero element alone does not ensure a large probability, since the drive frequency, duration and strength also matter. For a charged quantum harmonic oscillator driven along its axis, $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ gives the single-quantum rule $\Delta n=\pm 1$.

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.