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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.

Add the random thermal force to the deterministic optical trap relaxation time model. In one stationary coordinate of a harmonic optical trap, the overdamped Langevin equation isHere $\xi$ is Gaussian white noise understood through its integrals, $\delta$ is the Dirac delta function, and $\zeta$ is the linear friction coefficient. The covariance coefficient has units ${\mathrm{N}}^2\mathrm{s}$; multiplying by $\delta (t-t^{\prime })$, of units ${\mathrm{s}}^{-1}$, gives squared force. This fluctuation-dissipation relation for a Langevin particle makes the stationary variance ${\sigma }^2=k_{B}T/k$, consistent with the equipartition theorem. The solution is an Ornstein-Uhlenbeck process with autocovariance functionIndeed, at positive lag the noise has zero correlation with the initial position, so $\zeta C_x^{\prime }=-k{C}_x$; its initial value is ${\sigma }^2$. This supplies both a decay time and an amplitude, making a stronger calibration test than a position histogram alone.

Specify the Fourier transform convention before using a power spectral density. For frequency $f$ in hertz, define $S_x^{(2)}(f)={\int }_{-\infty }^{\infty }{C}_x(t)e^{-2\pi ift}dt$ and the one-sided density $S_x^{(1)}(f)=2S_x^{(2)}(f)$ for $f>0$. Integration of the exponential givesThe position power spectral density has units ${\mathrm{m}}^2/Hz$; its integral is a variance, not a standard deviation. At $f=f_c$ it is half its zero-frequency plateau. Angular frequency conventions require a corresponding $2\pi$ conversion in both axis and integration measure. If $x=\beta (V-V_0)$ with position sensitivity $\beta$ in $m/V$, then $S_V^{(1)}=S_x^{(1)}/{\beta }^2$ in ${\mathrm{V}}^2/Hz$ before independent detector noise is added.

For an original invented calibration example, choose $a=0.500\mu \mathrm{m}$, dynamic viscosity $\eta =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, $T=300\mathrm{K}$, $k=0.0200pN/nm=2.00\times 10^{-5}N/m$, and $\beta =100nm/V$. The unbounded no-slip Stokes drag law givesThe corresponding voltage standard deviation is $0.14391\mathrm{V}$, and the position power spectral density plateau is $0.390369\mathrm{n}{\mathrm{m}}^2/Hz$, or $3.90369\times 10^{-5}{\mathrm{V}}^2/Hz$ in the detector. Recovering both $k=k_{B}T/{\sigma }^2$ and $k=2\pi \zeta f_c$ is an internal model check; actual agreement also tests the assumed temperature, drag and detector response.

The generator uses the exact stationary update $x_{j+1}=\rho x_j+\sigma \sqrt{1-{\rho }^2}{Z}_j$, with $\rho =e^{-\Delta t/\tau }$ and independent standard normal random variables $Z_j$. For a demeaned block of $N$ samples and its unnormalized discrete Fourier transform $X_j$, the two-sided periodogram is $\Delta t|X_j{|}^2/N$; doubling positive bins except the zero and Nyquist endpoints gives a one-sided density. Summing those bins times $\Delta f=1/(N\Delta t)$ returns the block mean square by discrete Parseval identity. Independent detector noise, finite exposure and aliasing would alter a measured spectrum and must enter a fit before interpreting $f_c$.

For a bead in a stationary viscous liquid, linear drag often dominates inertia on the observed timescale. Neglecting both inertia and the random thermal force, the deterministic force balance givesHere $\zeta >0$ is the drag coefficient and $\tau$ the optical trap relaxation time: the time to reduce the deterministic displacement by a factor $e$. The bead approaches equilibrium without oscillation in this overdamped model. The approximation requires times long compared with the inertial time $m/\zeta$ and, for separated timescales, $m/\zeta \ll \zeta /k$, where $m$ is the bead's mass. For a no-slip spherical bead of radius $a$ in an unbounded liquid of dynamic viscosity $\eta$, Stokes drag law gives $\zeta =6\pi \eta a$. Nearby walls change that drag coefficient.

For example, $a=0.5\mu \mathrm{m}$, $\eta =10^{-3}\mathrm{Pa}\mathrm{s}$ and $k=0.01pN/nm=10^{-5}N/m$ give $\zeta \approx 9.42\times 10^{-9}\mathrm{N}s/m$ and $\tau \approx 0.942\mathrm{ms}$. A displacement of $100\mathrm{nm}$ gives a restoring force of magnitude $1\mathrm{pN}$, and its deterministic displacement drops to $100/e\approx 36.8\mathrm{nm}$ after one optical trap relaxation time. These are predictions for the stated model, not measured trajectories.

The figure uses $q=x/\ell$, $\kappa =k/k_0$ and $s=t/(\zeta /k_0)$, with arbitrary positive reference length $\ell$ and stiffness $k_0$. Thus $U/(k_0{\ell }^2)=\kappa q^2/2$, $F/(k_0\ell )=-\kappa q$ and, from $q(0)=1$, $q(s)=e^{-\kappa s}$. At fixed drag, greater stiffness gives both a steeper potential energy well and faster relaxation. Thermal Brownian motion is omitted from these deterministic curves.

A bead in a liquid keeps undergoing Brownian motion even when an optical trap holds its average position fixed. For a stationary harmonic optical trap in thermal equilibrium at absolute temperature $T$, the equipartition theorem assigns mean potential energy $k_{B}T/2$ to this one quadratic coordinate. With $k_B$ the Boltzmann constant and $x=0$ at equilibrium,At the same temperature, quadrupling the spring constant divides the position variance by four and the standard deviation by two. The Boltzmann factor gives the complete equilibrium probability density function, a Gaussian distribution:The prefactor makes ${\int }_{-\infty }^{\infty }p(x)dx=1$; its second moment is ${\int }_{-\infty }^{\infty }{x}^{2}p(x)dx=k_{B}T/k$. The probability density function has units of inverse length, and its area gives probability. This density is exact for the ideal quadratic model. In a finite optical trap it approximates the occupied well when thermal excursions remain within the harmonic region and escape is negligible during observation. Driven motion, appreciable nonconservative forces or laser heating need not give this equilibrium density.

Here $\ell =\sqrt{k_{B}T/k_0}$, $q=x/\ell$ and $\kappa =k/k_0$. The plotted density per unit $q$ is $p_q(q)=\ell p(\ell q)=\sqrt{\kappa /(2\pi )}{e}^{-\kappa q^2/2}$. Each curve has total area one; their dimensionless standard deviations are $1$, $1/\sqrt{2}$ and $1/2$. The curves are exact model densities, not noisy experimental data.

Brownian motion becomes a measuring tool: a calibrated position measurement supplies ${\sigma }^2=⟨(x-⟨x⟩{)}^2⟩$, and the known temperature supplies $k=k_{B}T/{\sigma }^2$. Once $k$ is known, a slowly applied constant external force $f$ shifts the mean by $⟨x⟩=f/k$, so $f=k⟨x⟩$ in the same harmonic range. Fluctuations therefore calibrate the conversion from displacement to force. Svoboda and Block used this equipartition theorem method when calibrating their bead traps. ^[ref].

For the example stiffness $0.01pN/nm$ and $T=300\mathrm{K}$, $k_{B}T\approx 4.142\mathrm{pN}\mathrm{nm}$ and $\sigma \approx 20.35\mathrm{nm}$. Using a variance measured in uncalibrated detector units would not determine that physical stiffness. Measurement noise, finite exposure time, drift and insufficient sampling of the correlated Brownian motion can bias the estimate; record positions over many optical trap relaxation times and account for the detector response. The deterministic relaxation curve above describes relaxation without thermal forcing; an individual thermal trajectory continues to fluctuate.