# Microcantilever Thermomechanical Fluctuations¶

Overview. In this tutorial we examine the steady-state response of a microcantilever to both a coherent and an incoherent driving force. The cantilever is modeled classically as a damped harmonic oscillator.

When the cantilever is driven by a coherent sinusoidal force, the steady-state solution of the equation of motion is greatly simplified by recasting the equation in terms of a complex variable whose real and imaginary parts track the in-phase and out-of-phase response, respectively, of the cantilever. Using this approach we derive the steady-state in-phase and out-of-phase amplitude of the driven cantilever as a function of frequency.

Even when no external driving force is applied, the cantilever is acted upon by its surroundings. The action of the surroundings can be treated as an incoherent driving force that perturbs the cantilever. This incoherent driving force casuses fluctuations in the cantilever position which are desribed in terms of correlation functions. We establish a link between the time-domain correlation function and the frequency-domain power spectrum of the fluctuating cantilever position. For a cantilever in thermal contact with surroundings having a well-defined temperature, the form of the incoherent environmental force can be derived using the equipartition theorem of statistical mechanics. We derive an expression of the force’s power spectrum in terms of the cantilever’s measured friction coefficient. We show that if the cantilever’s temperature, resonance frequency, and ringdown time are known, then the cantilever spring constant can be determined from the area under the measured power spectrum of cantilever position fluctuations.

Equation of Motion. The equation of motion for a damped harmonic oscillator is

(1)$m \: \ddot{x} + \Gamma \: \dot{x} + k \: x = F$

The variables are

$\begin{split}\begin{array}{lll} x & \text{oscillator position} & [\text{m}] \\ m & \text{mass} & [\text{kg}] \\ \Gamma & \text{friction coefficient} & [\text{kg} \: {\text{s}}^{-1} = \text{N} \: \text{s} \: {\text{m}}^{-1}] \\ k & \text{spring constant} & [\text{N} \: {\text{m}}^{-1}] \\ F & \text{applied force} & [\text{N}] \end{array}\end{split}$

It is useful to rewrite this equation in a more canonical form. Divide equation (1) by $$m$$, and define new variables according to the following equations:

$\frac{\Gamma}{m} = \frac{\omega_0}{Q}, \: \: \frac{k}{m} = \omega_0^2, \: \mbox{and} \: \frac{F}{m} = \frac{\omega_0^2 F}{k} = A$

The new variables are

$\begin{split}\begin{array}{lll} \omega_0 &\text{resonance frequency} & [\text{rad} \: {\text{s}}^{-1}] \\ Q & \text{quality factor} & [\text{unitless}] \\ A & \text{acceleration} & [\text{m} \: {\text{s}}^{-2}] \end{array}\end{split}$

The canonical equation of motion for a classical harmonic oscillator is thus

(2)$\ddot{x} + \frac{\omega_0}{Q} \dot{x} + \omega_0^2 \: x = A = \frac{\omega_0^2 \: F}{k}$

We now wish to calculate the response of the oscillator to a resonant force,

$F(t) \propto \cos{\omega t}$

Here $$\omega$$ is the driving frequency, close to but not necessarily equal to $$\omega_0$$. At steady state, the cantilever response must also be periodic, of the general form

$x(t) = x_c \cos{\omega t} + x_s \sin{\omega t}$

We wish to solve for $$x_c$$ and $$x_s$$ as a function of driving frequency. It is convenient to introduce a complex number $$z$$ that tracks cantilever displacement, $$x = \mathrm{Re}(z)$$. If we make the ansatz that $$z = z_0 \exp{(\imath \omega t)}$$ then

$\begin{split}\begin{split} x(t) & = \mathrm{Re} \{ z \} \\ & = \mathrm{Re} \{ z_0 \: e^{\, \imath \omega t} \} \\ & = \underbrace{\mathrm{Re} \{ z_0 \} }_{x_c} \cos{\omega t} - \underbrace{\mathrm{Im} \{ z_0 \} }_{x_s} \sin{\omega t} \end{split}\end{split}$

If we can recast equation (2) in terms of the complex variable $$z$$ then we can reduce the problem of solving for two real variables, $$x_c$$ and $$x_s$$, to solving for one complex variable, $$z_0$$. With this goal in mind, let us introduce another complex variable $$F_c$$ that tracks the applied force. If the force is a sinusoidal function of time, then $$F_c = F_0 \exp{(\imath \omega t)}$$ where $$F_0 = | F_0 | \: \exp{(\imath \phi)}$$ is a complex number that describes the magnitude and phase of the harmonic driving force:

$\begin{split}\begin{split} F(t) & = \mathrm{Re}\{ F_c \} \\ & = \mathrm{Re} \{ F_0 \: e^{\, \imath \omega t} \} \\ & = \mathrm{Re} \{ F_0 \} \cos{\omega t} - \mathrm{Im} \{ F_0 \} \sin{\omega t} \\ & = | F_0 | \cos{(\omega t + \phi)} \end{split}\end{split}$

The equation of motion for $$z$$ is

(3)$\ddot{z} + \frac{\omega_0}{Q} \dot{z} + \omega_0^2 \, z = \frac{\omega_0^2 \, F_c}{k}$

Steady State Response: It is convenient to work with the frequency expressed in experimental units of $$[\mathrm{cyc}/{\mathrm{s}}] = [{\mathrm{Hz}}]$$ instead of $$[\mathrm{rad}/{\mathrm{s}}]$$. Let us therefore define

$f_0 = \frac{\omega_0}{2 \pi} \: \sim \: [\frac{\mathrm{cyc}}{{\mathrm{s}}}] = [{\mathrm{Hz}}]$

and work from now on with frequencies in $${\mathrm{Hz}}$$.

We are now ready to explore the response of the cantilever to a oherent sinusoidal driving force. Substitute $$F_c = F_0 \exp{(2 \pi \imath f t)}$$ into equation (3) and assume that the response $$z$$ is of the form $$z_0 \exp{(2 \pi \imath f t)}$$. This will be true at steady state.

$(-f^2 + \imath f \frac{f_0}{Q} + f_0^2 ) \: z_0 \: e^{\, 2 \pi \imath f t} = \frac{f_0^2}{k} F_0 \: e^{\, 2 \pi \imath f t}$

where we have canceled a factor of $$4 \pi^2$$ from every term. We infer that

$z_0 = \frac{F_0}{k} \: \frac{f_0^2}{f_0^2 - f^2 + \imath f \: f_0 / Q}$

$\begin{split}\begin{split} z(f) & = z_0 \: e^{\, 2 \pi \imath f t} \\ & = \frac{F_0 \: e^{\, 2 \pi \imath f t}}{k} \: \frac{f_0^2}{f_0^2 - f^2 + \imath f \: f_0 / Q} \end{split}\end{split}$

It is useful to write $$z$$ as follows:

$z(f) = \frac{| F_0 |}{k} \left( \frac{f_0^2 (f_0^2 - f^2)}{(f_0^2 - f^2)^2 + f^2 \: f_0^2 / Q^2} - \imath \frac{f \: f_0^3 / Q}{(f_0^2 - f^2)^2 + f^2 \: f_0^2 / Q^2} \right) \: e^{\, \imath ( 2 \pi f t + \phi)}$

Using $$x = \mathrm{Re} \{ z \}$$ we can infer that $$x(t)$$ is of the form

$x(t) = x_c \: \cos{(2 \pi f t + \phi)} + x_s \: \sin{(2 \pi f t + \phi)}$

where

$x_c(f) = \frac{| F_0 |}{k} \frac{f_0^2 ( f_0^2 - f^2)} {(f_0^2 - f^2)^2 + f^2 \: f_0^2 / Q^2}$
$x_s(f) = \frac{| F_0 |}{k} \frac{f \: f_0^3 / Q} {(f_0^2 - f^2)^2 + f^2 \: f_0^2 / Q^2}$

The signal $$x_c$$ is the part of the response detected with a lock-in as in phase with the driving force. The signal $$x_s$$ is the out of phase part of the response. We can see that when the applied force drives the oscillator right on resonance, $$\omega = \omega_0$$ and

$\begin{split}\begin{split} x_s(\omega_0) & =0 \\ x_s(\omega_0) & =\frac{Q \: | F_0 |}{k} \end{split}\end{split}$

This is to be compared to the steady-state response to a non-oscillating (DC) force

$\begin{split}\begin{split} x_c(0) & = \frac{| F_0 |}{k} \\ x_s(0) & = 0 \end{split}\end{split}$

We conclude that the response to a resonant force is $$Q$$ times larger than the response to a static DC force. The response at resonance is also ninety degrees out of phase with the applied oscillating force. These two results are captured in the single equation

$z_0(\omega_0) = - \imath \frac{Q \: F_0}{k}$

The response on resonance is purely imaginary and therefore ninety degrees out of phase with the applied force.

Correlation Functions. The section explores a connection between a function’s correlation function and its power spectrum. Correlation functions are usually applied to fluctuating quantities having zero mean. In our case, we wish to apply correlation functions to understand fluctuations in cantilever position:

$\delta x(t) = x(t) - \mathrm{mean}(x(t))$

The correlation function of $$\delta x(t)$$ is defined as

(4)$C_{\delta x}(\tau) = \int_{-\infty}^{\infty} dt \: \delta x(t) \: \delta x(t+\tau) \: \sim \: [\frac{{\mathrm{m}}^2}{{\mathrm{Hz}}}]$

The Fourier and inverse Fourier transforms of $$x(t)$$ are:

$\widehat{\delta x}(f) = \int_{-\infty}^{\infty} dt \: \delta x(t) \: e^{-2 \pi \imath f t}$
$\delta x(t) = \int_{-\infty}^{\infty} df \: \widehat{\delta x}(f) \: e^{\, 2 \pi \imath f t}$

Substitute for $$\delta x(t)$$ and $$\delta x(t+\tau)$$ the appropriate Fourier transform relation

$C_{\delta x}(\tau) = \int df \int df^{\prime} \: \widehat{\delta x}(f^{\prime}) \: \widehat{\delta x}(f) \: e^{\, 2 \pi \imath f \tau} \underbrace{\int dt \: e^{-2 \pi \imath f t} e^{-2 \pi \imath f^{\prime} t}}_{\delta(-f-f^{\prime}) \Longrightarrow f^{\prime} = -f}$

The integral over time involving exponentials reduces to a Dirac delta function. Only frequencies $$f^{\prime} = -f$$ contribute to the final double integral, so that

$C_{\delta x}(\tau) = \int_{-\infty}^{\infty} df \: \: \widehat{\delta x}(-f) \: \widehat{\delta x}(f) \: e^{\, 2 \pi \imath f \tau}$

If $$\delta x(t)$$ is a real function of time, then it can be shown that

$\widehat{\delta x}(-f) = {\widehat{\delta x}}^{\: *}(f)$

where the star indicates the complex conjugate. We have finally

$\begin{split}\begin{split} C_{\delta x}(\tau) & = \int_{-\infty}^{\infty} df \: {\widehat{\delta x}}^{*}(f) \: \widehat{\delta x}(f) \: e^{\, 2 \pi \imath f \tau} \\ & = \int_{-\infty}^{\infty} df \: | \widehat{\delta x}(f) |^2 \: e^{\, 2 \pi \imath f \tau} \end{split}\end{split}$

If we define the one-sided power spectral density as

$P_{\delta x}(f) = | \widehat{\delta x}(f) |^2 + | \widehat{\delta x}(-f) |^2 \: \sim \: [\frac{{\mathrm{m}}^2}{{\mathrm{Hz}}^2}]$

then

(5)$C_{\delta x}(\tau) = \int_{0}^{\infty} df \: P_{\delta x}(f) \: e^{\, 2 \pi \imath f \tau}$

This is an important result: The correlation function and the power spectrum are Fourier transform pairs.

While equations (4) and (5) can in principle be used to analyze thermomechanical fluctuations in the position of a microcantilever, in practice we need to introduce a modified correlation function to analyze the fluctuations. The reason for this can be seen by considering the correlation function of equation (4) at $$\tau = 0$$:

$C_{\delta x}(0) = \int_{-\infty}^{\infty} dt \: \delta x(t)^2 \longrightarrow \infty$

As indicated, this integral will diverge if applied to a real-world laboratory signal such as a cantilever oscillation. Following Weissbluth [1], let’s define a more physically-relevant correlation function as follows.

$G(\tau) \equiv \langle \delta x(t) \: \delta x(t+\tau) \rangle$
(6)$G(\tau) \equiv \lim_{T \rightarrow \infty} \: \frac{1}{T} \int_{0}^{T} \delta x(t) \: \delta x(t+\tau) \: dt \: \sim \: [{\mathrm{m}}^2]$

The units of this correlation function are $$[{\mathrm{m}}^2]$$, if the units of x are $$[{\mathrm{m}}]$$. This correlation function is quite different from the mathematically-defined correlation function $$C(\tau)$$ of equation (4) whose units are $$[{\mathrm{m}}^2/{\mathrm{Hz}}]$$. The correlation function at $$\tau=0$$ (zero delay) has special significance:

$\begin{split}\begin{split} G(0) & = \lim_{T \rightarrow \infty} \: \frac{1}{T} \int_{0}^{T} \delta x^2(t) \: dt & = x_{\mathrm{rms}}^2 \end{split}\end{split}$

We see that $$G(0)$$ is the mean square value of $$\delta x(t)$$ and therefore the root-mean-square is $${\delta x}_{\mathrm{rms}} = \sqrt{G(0)}$$.

We will now reproduce Weissbluth’s treatment [1] relating the (physically-relevant) correlation function $$G(\tau)$$ to an analogous power spectrum. Following Weissbluth, let us define the function $${\delta x}_{T}(t)$$ which is equal to $$\delta x(t)$$ on the time interval $$(0,T)$$ and is zero at all other times:

$\begin{split}{\delta x}_{T}(t) = \left\{ \begin{array}{cc} \delta x(t) & 0 \leq t \leq T \\ 0 & \mathrm{otherwise} \end{array} \right.\end{split}$

Let us define correlation function for $${\delta x}_T$$ as follows:

$\begin{split}\begin{split} G_{T}(\tau) & = \frac{1}{T} \int_{0}^{T} {\delta x}_T(t) \: {\delta x}_T(t+\tau) \: dt \\ & = \frac{1}{T} \int_{-\infty}^{+\infty} {\delta x}_T(t) \: {\delta x}_T(t+\tau) \: dt \end{split}\end{split}$

Since we’ve confined $${\delta x}_T$$ to the time interval $$(0,T)$$ we can extend the limits in integration out to infinity. Now take the Fourier transform of $$G_{T}(\tau)$$:

$\begin{split} \begin{multline} \int_{-\infty}^{+\infty} G_{T}(\tau) \: e^{-2 \pi \imath f \tau} \: d\tau = \frac{1}{T} \int_{-\infty}^{+\infty} d\tau \: e^{-2 \pi \imath f \tau} \int_{-\infty}^{+\infty} dt \: {\delta x}_{T}(t) \: {\delta x}_{T}(t+\tau) \\ = \frac{1}{T} \int_{-\infty}^{+\infty} dt \: {\delta x}_{T}(t) \: e^{\, 2 \pi \imath f t} \int_{-\infty}^{+\infty} d\tau \: {\delta x}_{T}(t+\tau) \: e^{-2 \pi \imath f (t+\tau)} \end{multline}\end{split}$

where we have inserted $$1 = \exp{(-\imath 2 \pi f t)} \exp{(+\imath 2 \pi f t)}$$. In the second integral, change the variable of integration to $$t^{\prime} = t+\tau$$. This lets us write

$\int_{-\infty}^{+\infty} G_{T}(\tau) \: e^{-2 \pi \imath f \tau} \: d\tau = \frac{1}{T} \underbrace{\int_{-\infty}^{+\infty} dt \: {\delta x}_{T}(t) \: e^{\, 2 \pi \imath f t}}_{{\widehat{\delta x}}_T(-f) = {\widehat{\delta x}}^{\: *}_{T}(f)} \underbrace{\int_{-\infty}^{+\infty} dt^{\prime} \: {\delta x}_{T}(t^{\prime}) \: e^{-2 \pi \imath f t^{\prime}}}_{{\widehat{\delta x}}_T(f)}$

Since $$x(t)$$ is a real function, it follows that $${\widehat{\delta x}}_{T}(-f) = {\widehat{\delta x}}^{\: *}_{T}(f)$$. We can thus write

(7)$\int_{-\infty}^{+\infty} G_{T}(\tau) \: e^{-2 \pi \imath f \tau} \: d\tau = \frac{1}{T} \: | {\widehat{\delta x}}_{T}(f) |^{2}$

We recover the “real” correlation function by a limiting procedure.

$G(\tau) = \lim_{T \rightarrow \infty} \: G_{T}(\tau)$

Take the limit on each side of equation (7) as $$T \rightarrow \infty$$. On the left-hand side, $$G_T$$ becomes $$G$$; the terms on the right-hand side motivate us to define

(8)$J(f) \equiv \lim_{T \rightarrow \infty} \: \frac{1}{T} \: | {\widehat{\delta x}}_{T}(f) |^{2} \: \sim \: [\frac{{\mathrm{m}}^2}{{\mathrm{Hz}}}]$

as the physically relevant spectral density. It still holds that

$J(f) = \int_{-\infty}^{+\infty} G(\tau) \: e^{-2 \pi \imath f \tau} \: d\tau$

and

(9)$\begin{split}\begin{split} G(\tau) & = \int_{-\infty}^{+\infty} J(f) \: e^{\, 2 \pi \imath f \tau} \: df \\ & = \int_{0}^{+\infty} P_{\delta x}(f) \: e^{\, 2 \pi \imath f \tau} \: df. \end{split}\end{split}$

We have defined the one-sided power spectral density as

(10)$\begin{split}\begin{split} P_{\delta z}(f) & = J(f) + J(-f) \\ & = \lim_{T \rightarrow \infty} \frac{1}{T} \: ( | {\widehat{\delta x}}_{T}(f) |^{2} + | {\widehat{\delta x}}_{T}(-f) |^{2}) \end{split}\end{split}$

With these definitions of correlation function (equation (6)) and spectral density (equation (8)), we still have that the correlation function $$G(\tau)$$ and the power spectrum $$J(f)$$ of $$\delta x(t)$$ are Fourier transform pairs. Finally, equation (9) can be used to calculate the root-mean-square of $$x(t)$$ given a measured one-sided power spectral density:

(11)${\delta x}_{\mathrm{rms}}^2 = \langle {\delta x}^2(t) \rangle = G(0) = \int_{0}^{+\infty} P_{\delta x}(f) \: df$

We conclude that the area under the one-sided spectrum is the mean-square displacement. We note that this connection is not valid for the mathematically-defined power-spectrum of the last section.

Steady-State Response Revisited. In this section we explore the response of the cantilever to an incoherent driving force. We assume that the driving force averages to zero over long times:

$\langle F(t) \rangle = \lim_{T \rightarrow \infty} \: \frac{1}{T} \int_{0}^{T} F(t) \: dt \longrightarrow 0$

The change in cantilever position resulting from such a force will likewise average to zero at long times. At short times, however, the cantilever will experience force fluctuations $$\delta F(t)$$ and these force fluctuations will stimulate fluctuations $$\delta z(t)$$ in the cantilever’s position. Let us define correlation functions for both $$\delta z$$ and $$\delta F$$ as above,

$G_{\delta z}(\tau) \equiv \lim_{T \rightarrow \infty} \: \frac{1}{T} \int_{0}^{T} \delta z(t) \: \delta z(t+\tau) \: dt \: \sim \: [{\mathrm{m}}^2]$
$G_{\delta F}(\tau) \equiv \lim_{T \rightarrow \infty} \: \frac{1}{T} \int_{0}^{T} \delta F(t) \: \delta F(t+\tau) \: dt \: \sim \: [{\mathrm{N}}^2]$

With each of these correlation functions is associated a power spectrum:

\begin{split}\begin{aligned} G_{\delta z}(\tau) \overset{\: \mathrm{\small FT}}{\iff} J_{\delta z}(f) \: \text{or} \: P_{\delta z}(f) \\ G_{\delta F}(\tau) \overset{\: \mathrm{\small FT}}{\iff} J_{\delta F}(f) \: \text{or} \: P_{\delta F}(f) \end{aligned}\end{split}

Because $$z$$ and $$F$$ are connected by an equation of motion, we can write $$J_{\delta z}$$ in terms of $$J_{\delta F}$$, as we will now show. Let us use a Fourier expansion to write the fluctuating quantities as follows

(12)$\delta F(t) = \int_{-\infty}^{\infty} df \: \widehat{\delta F}(f) \: e^{\, 2 \pi \imath f t}$
(13)$\delta z(t) = \int_{-\infty}^{\infty} df \: \widehat{\delta z}(f) \: e^{\, 2 \pi \imath f t}$

Substitute equations (12) and (13) into the equation of motion connecting $$F$$ and $$z$$, equation (3).

$\int_{-\infty}^{+\infty} (-f^2 + \imath f \: \frac{f_0}{Q} + f_0^2 ) \: \widehat{\delta z}(f) \: e^{\, 2 \pi \imath f t} \: df = \int_{-\infty}^{+\infty} \frac{f_0^2}{k} \widehat{\delta F}(f) \: e^{\, 2 \pi \imath f t} \: df$

For both sides to be equal, we must have that at each frequency

$\widehat{\delta z}(f) = \frac{\widehat{\delta F}(f)}{k} \frac{f_0^2}{f_0^2 - f^2 + \imath f \: f_0 / Q}$

Taking the magnitude of each side, we infer that the power spectra are related by

$| \widehat{\delta z}(f) |^2 = \frac{| \widehat{\delta F}(f) |^2}{k^2} \frac{f_0^4}{(f_0^2 - f^2)^2 + f^2 f_0^2 / Q^2}$

This equation relates “mathematical” correlation functions. It is a straightforward matter to introduce the time-averaging and limiting procedure employed above to obtain this result in terms of “physically-relevant” correlation functions:

$P_{\delta z}(f) = \lim_{T \rightarrow \infty} \frac{1}{T} \: ( | {\widehat{\delta z}}_{T}(f) |^{2} + | {\widehat{\delta z}}_{T}(-f) |^{2}) \: \sim \: [\frac{{\text{m}}^2}{{\text{Hz}}}]$
(14)$P_{\delta F}(f) = \lim_{T \rightarrow \infty} \frac{1}{T} \: ( | {\widehat{\delta F}}_{T}(f) |^{2} + | {\widehat{\delta F}}_{T}(-f) |^{2}) \: \sim \: [\frac{{\text{N}}^2}{{\text{Hz}}}]$

The result, which we write in terms of one-sided power spectral densities is:

(15)$P_{\delta z}(f) = \frac{P_{\delta F}(f)}{k^2} \frac{f_0^4}{(f_0^2 - f^2)^2 + f^2 f_0^2 / Q^2}$

Given a fluctuating force $$\delta F(t)$$, we can form a one-sided power spectrum $$P_{\delta F}(f)$$ by Fourier transforming the time-domain spectrum of $$\delta F$$ and averaging (equation (14)). We can then predict the resulting one-sided power spectrum $$P_{\delta z}(f)$$ of the response $$\delta z(t)$$ using equation (15). Finally, if we wish, we could determine the time-correlation function $$G_{\delta z}(\tau)$$ of $$\delta z(t)$$ by inverse Fourier-transforming $$P_{\delta z}(f)$$.

We can proceed no further in discussing the response of the harmonic oscillator to an incoherent driving force unless we specify a form for either $$\delta F(t)$$, $$G_{\delta F}(\tau)$$, $$J_{\delta F}(f)$$, or the power spectrum $$P_{\delta F}(f)$$. The simplest approximation is to assume that the environmental force fluctuation driving the cantilever is well-described as being white noise, e.g., a randomly-fluctuating with a power spectrum that is flat up to some very high frequency cutoff:

(16)$\begin{split}P_{\delta F}(f) = \left\{ \begin{array}{cc} P_{\delta F} & 0 \leq f \leq f_m \\ 0 & f_m \leq f \end{array} \right.\end{split}$

The cutoff frequency’s numerical value is determined by the physical process giving rise to the force fluctuation. Atomic force microscope cantilevers experience force fluctuations due to random collisions with gas molecules and fluctuating cantilever phonon populations, for example. Both of these processes have characteristic timescales on the order of nanoseconds, which implies, by Fourier transforming the associated correlation function, that $$f_m \sim 1 / {\mathrm{ns}} = \mathrm{GHz}$$. The resonance frequencies of atomic-force microscope cantilevers are in the range of $$f_0 \sim 1 \: \text{to} \: 500 \: \text{kHz}$$; consequently, $$f_0 << f_m$$, and thus when considering a cantilever’s response to the above-mentioned force fluctuations the approximation of equation (16) is a good one. An example of a case where the white-force-noise approximation of equation (16) would not be valid is the cantilever being driven by acoustic room vibrations. The power spectrum of doors closing, mechanical vibrations from transformers, and people walking by the cantilever is generally not flat near the cantilever resonance frequency.

We conclude that a cantilever driven by white-noise force fluctuations exhibits a power spectrum of position fluctuations given by

(17)$P_{\delta z}(f) = \underbrace{\frac{P_{\delta F}}{k^2}}_{\mathrm{\small freq. independent}} \underbrace{\frac{f_0^4}{(f_0^2 - f^2)^2 + f^2 f_0^2 / Q^2}}_{\mathrm{\small freq. dependent}}$

Equipartition Theorem. As may be derived using statistical mechanics, a harmonic oscillator in equilibrium with a bath of temperature $$T$$ has a energy expectation value for each mode equal to $$k_b T/2$$. Thus

(18)$\frac{1}{2} \: k \: \langle \delta x(t)^2 \rangle = \frac{1}{2} \: k_b T$

where $$k_b = 1.3806 \times {10}^{-23} \: {\text{J}} \: {{\text{K}}}^{-1}$$ is Boltzmann’s constant and $$T \: \sim \: [{\text{K}}]$$ is the absolute temperature. Here $$\langle \delta x(t)^2 \rangle = \delta x_{\mathrm{rms}}^2$$ is the mean-square cantilever displacement. The mean-square displacement can be calculated directly from observations of the cantilever displacement versus time. According to (18), the cantilever’s spring constant can then be calculated using

(19)$k = \frac{k_b T}{{\delta x}_{\mathrm{rms}}^2} \: \sim \: [\frac{{\mathrm{N}}}{{\mathrm{m}}}]$

The only assumption in this procedure is that the cantilever is in thermal equilibrium with an environment of known temperature.

An alternative and often more accurate way to determine $$x_{\text{rms}}$$ is to measure the power spectrum of cantilever position fluctuations instead. According to equation (11), $${\delta x}_{\text{rms}}^2$$ is the area under this power spectrum. The relevant integral is

${\delta x}_{\text{rms}}^2 = \int_{0}^{\infty} P_{\delta z}(f) \: df = \frac{P_{\delta F} f_0^4}{k^2} \int_{0}^{\infty} \frac{df}{(f_0^2-f^2)^2 + f^2 f_0^2 / Q^2}$

This integral may be rewritten in terms of a unitless variable $$F \equiv f/f_0$$ to give

(20)${\delta x}_{\text{rms}}^2 = \frac{P_{\delta F}(0) Q f_0}{k^2} \int_{0}^{\infty} \frac{Q dF}{Q^2 (F^2 - 1)^2 + F^2}$

The integral in equation (20) is unitless and is equal to $$\pi/2$$ (see this section’s Appendix). Thus

${\delta x}_{\text{rms}}^2 = \frac{P_{\delta F}(0) Q f_0}{k^2} \frac{\pi}{2},$

Recognizing

$\Gamma = \frac{k}{\omega_0 Q} = \frac{k}{2 \pi f_0 Q}$

and solving for $$P_{\delta F}$$, we find

(21)$\boxed{ P_{\delta F} = 4 \: \Gamma \: k_b T \: \sim \: [\dfrac{\text{N}^2}{\text{Hz}}] }$

Substituting this result into equation (17) gives

(22)$P_{\delta z}(f) = \frac{ 4 \: \Gamma \: k_b T}{k^2} \frac{f_0^4}{(f_0^2 - f^2)^2 + f^2 f_0^2 / Q^2}$

Finally, let us recast this result into a form that is somewhat more practical for curve fitting. We do this by writing the cantilever quality factor $$Q$$ in terms of the cantilever ringdown time $$\tau_0$$ uaing

(23)$Q = \pi f_0 \tau_0$

and by writing the spring constant in terms of the dissipation constant, resonance frequency, and ringdown time using

(24)$k = 2 \pi^2 f_0^2 \tau_0 \Gamma.$

Substituting equations (23) and (24) into equation (22) and simplifying the result we find

(25)$\boxed{ P_{\delta z}^{\text{therm}}(f) = \dfrac{k_b T \tau_0^2}{\Gamma} \dfrac{1}{(\pi \tau_0)^4(f_0^2 - f^2)^2 + (\pi \tau_0)^2 f^2} \: \sim \: [\dfrac{\text{m}^2}{\text{Hz}}] }$

Equations (21) and (25) are the central results of this section.

References

 [1] (1, 2) Weissbluth, M. Photon-Atom Interactions. Academic Press, New York (1989). We modify Weiessbluth’s treatment a little. He considers a signal that extends in time from $$t = -T$$ to $$t = +T$$ while we consider instead a signal that extends in time from $$t = 0$$ to $$t = T$$.

Appendix. We wish to compute the following integral

$I = \int_{0}^{\infty} \frac{Q dF}{Q^2 (F^2 - 1)^2 + F^2}$

Mathematica will carry out this integral, but the result is not easily rewritten in a simple form. We will carry out the integral using a trigonometric substitution to convert the integral into a contour integral in the complex plane; the resulting contour integral can be computed using the residue theorem.

Let $$F = \tan{\theta}$$. Substituting, $$dF = \cos^{-2}{\theta} \: d\theta$$ and consequently

$I = \frac{Q}{4} \int_{0}^{2 \pi} \frac{d\theta} {\cos^2{\theta} \: (Q^2 (\tan^2{\theta} - 1)^2 + \tan^2{\theta})}$

In carrying out the substitution we have used

$\int_{0}^{\pi/2} \cdots \: d\theta \rightarrow \frac{1}{4} \int_{0}^{2 \pi} \cdots \: d\theta,$

which is valid given the periodicity of the integrand. We can convert this integral into an integral over the $$|z| = 1$$ contour in the complex plane. Along this contour, $$z = e^{\imath \theta}$$, $$\theta \in (0,2 \pi)$$, and

$\cos{\theta} = \frac{1}{2} (e^{\, \imath \theta} + e^{-\imath \theta}) = \frac{1}{2} (z + \frac{1}{z})$

likewise

$\sin{\theta} = \frac{1}{2 \imath} (z + \frac{1}{z}), \quad \tan{\theta} = -\imath \frac{z^2 - 1}{z^2 + 1}, \quad d\theta = - \imath \frac{dz}{z}$

Substituting, after considerable simplification,

$I = - \imath \int_{|z| = 1} \frac{Q \: (z^2 + 1)^2} {4 Q^2 (z^4 + 1)^2 - (z^4 - 1)^2} \: dz$

Setting

$4 Q^2 (z^4 + 1)^2 - (z^4 - 1)^2 = 0$

and solving for $$z$$ we find that the denominator in the integrand has eight poles. Four of them outsize the $$|z| = 1$$ contour and four of them are inside. The four poles inside the contour are given by

$z_j = \left( \frac{2 Q - 1}{2 Q + 1} \right)^{1/4} \: ( e^{\, \imath \pi/4}, \: e^{\, \imath 3 \pi/4}, \: e^{\, \imath 5 \pi/4}, \: e^{\, \imath 7 \pi/4})$

According to the residue theorem of complex analysis

$I = 2 \pi \imath \sum_{J = 1}^{4} \text{Res}\left( - \frac{\imath Q (z^2 + 1)^2} {4 Q^2 (z^4 + 1)^2 - (z^4 - 1)^2}; \: z_j \right)$

Computing the residues with Mathematica’s help and carrying out the sum we find, remarkably, that

$I = \frac{\pi}{2}.$