Harmonic optical trap ($U(x)=k{x}^2/2,F(x)=-kx$)

A harmonic optical trap approximates one coordinate near a stable equilibrium by a quadratic potential energy. Measure displacement $x$ from that equilibrium, set its potential energy to zero there, and take a positive effective spring constant $k$, also called the trap stiffness:The restoring force points left when $x>0$ and right when $x<0$. This is a local model: a Taylor series of a smooth potential energy about a nondegenerate local minimum gives $k=U^{\prime \prime }(0)>0$ as its leading curvature. A real optical trap has a finite capture region; the endlessly rising parabola does not describe its escape boundary. A three-dimensional optical trap can have different stiffnesses along different axes, and nonconservative optical forces can limit the potential energy description.

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.

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 an overdamped harmonic optical trap, impose a known sinusoidal local fluid velocity of amplitude $u_0$ and angular frequency $\Omega$. The coherent bead response has amplitude $A_x=u_0/\sqrt{(k/\zeta {)}^2+{\Omega }^2}$. Measuring the thermal corner frequency $f_c=k/(2\pi \zeta )$ and detector-voltage amplitude $A_s=A_x/\beta$ therefore determines the position sensitivity in the displayed formula without a drag coefficient as input. Use a thermal record with the drive line removed to obtain stiffness from the equipartition theorem. Fluid-following, detector response and hydrodynamic corrections must remain valid. Combined thermal and stage measurements have experimentally calibrated an optical trap without drag as an input.^[ref]

Translate the liquid past a bead held in a stationary harmonic optical trap. At steady mean position, take $u$ as the local fluid velocity relative to the fixed trap and $x$ positive downstream. Averaging the overdamped force balance givesFor the invented optical trap Langevin calibration example, $u=50.0\mu m/s$ predicts $⟨x⟩=23.562\mathrm{nm}$, $V-V_0=0.23562\mathrm{V}$, and $\zeta u=k⟨x⟩=0.47124\mathrm{pN}$. Reversing $u$ reverses the mean shift but leaves the ideal thermal variance unchanged. A symmetric speed sweep tests the offset and the linear force range; the slope determines $\zeta /k$ if position is calibrated.

Near a wall, $\zeta =6\pi \eta a$ is not the appropriate drag without a boundary correction; record bead height and direction of motion. Stage velocity must correspond to the local fluid velocity after transients have settled. Escape under increasing flow estimates a maximum holding force in that geometry, whereas small-displacement drag calibration estimates the local spring constant. They are different quantities. The atomic force microscopy similarly needs a calibrated spring constant to convert a measured deflection to force, but its cantilever is a physical spring rather than this optical field.

Write $s_V^2$ for the detector's position-related voltage variance, after characterizing independent detector noise and the acquisition response. The equipartition theorem estimate is $k=k_{B}T/({\beta }^2{s}_V^2)$. For small independent input standard uncertainties, first-order propagation givesIn this expression $u_{s_V^2}$ is the uncertainty of the variance itself. If the assumed independent relative uncertainties are $1\%$, $2\%$, and $3\%$, respectively, the resulting relative standard uncertainty is $\sqrt{26}\%\simeq 5.10\%$. This invented budget excludes model bias. Shared calibration inputs introduce covariance terms: a force estimate $F=k\beta \overline{V}$ using this same thermal estimate has $F=k_{B}T\overline{V}/(\beta s_V^2)$, so treating $k$ and $\beta$ as independent would count their sensitivity incorrectly.

Correlated Brownian motion also makes sample count a poor proxy for independent information. For a long continuous stationary Gaussian record of duration $\mathcal{T}$, estimating ${\sigma }^2$ from ${\mathcal{T}}^{-1}\int x(t{)}^{2}dt$ gives, to leading order,The Gaussian distribution fourth-moment identity gives $\mathrm{Cov}(x(t{)}^2,x(0{)}^2)=2C_x(t{)}^2$; integrating this covariance over the observation interval gives the result. Subtracting a fitted mean changes higher-order terms when $\mathcal{T}\gg \tau$. The ideal $10\mathrm{s}$ example gives about $0.97\%$ relative uncertainty in the variance, even though $200000$ samples were recorded. This ideal statistical estimate does not include drift, detector noise or incorrect drag.