IPhO 2004 · Theory problem 3

Atomic Probe Microscope

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Atomic probe microscopes (APMs) are powerful tools in the field of nano-science. The motion of a cantilever in APM can be detected by a photo-detector monitoring the reflected laser beam, as shown in Fig. 3.1. The cantilever can move only in the vertical direction and its displacement \(z\) as a function of time \(t\) can be described by the equation

\[ m\frac{d^2z}{dt^2}+b\frac{dz}{dt}+kz=F, \tag{3.1} \]

where \(m\) is the cantilever mass, \(k=m\omega_0^2\) is the spring constant of the cantilever, \(b\) is a small damping coefficient satisfying \(\omega_0\gg(b/m)>0\), and finally \(F\) is an external driving force of the piezoelectric tube.

Figure 1

Figure 3.1 A schematic diagram for a scanning probe microscope (SPM). The inset in the lower right corner represents a simplified mechanical model to describe the coupling of the piezotube with the cantilever.

[Part A]

(a) [1.5 points] When \(F=F_0\sin\omega t\), \(z(t)\) satisfying Eq. (3.1) can be written as \(z(t)=A\sin(\omega t-\phi)\), where \(A>0\) and \(0\leq\phi\leq\pi\). Find the expression of the amplitude \(A\) and \(\tan\phi\) in terms of \(F_0\), \(m\), \(\omega\), \(\omega_0\), and \(b\). Obtain \(A\) and the phase \(\phi\) at the resonance frequency \(\omega=\omega_0\).

(b) [1 point] A lock-in amplifier shown in Fig.3.1 multiplies an input signal by the lock-in reference signal, \(V_R=V_{R0}\sin\omega t\), and then passes only the dc (direct current) component of the multiplied signal. Assume that the input signal is given by \(V_i=V_{i0}\sin(\omega_i t-\phi_i)\). Here \(V_{R0}\), \(V_{i0}\), \(\omega_i\), and \(\phi_i\) are all positive given constants. Find the condition on \(\omega\) (>0) for a non-vanishing output signal. What is the expression for the magnitude of the non-vanishing dc output signal at this frequency?

(c) [1.5 points] Passing through the phase shifter, the lock-in reference voltage \(V_R=V_{R0}\sin\omega t\) changes to \(V'_R=V_{R0}\sin(\omega t+\pi/2)\). \(V'_R\), applied to the piezoelectric tube, drives the cantilever with a force \(F=c_1V'_R\). Then, the photo-detector converts the displacement of the cantilever, \(z\), into a voltage \(V_i=c_2z\). Here \(c_1\) and \(c_2\) are constants. Find the expression for the magnitude of the dc output signal at \(\omega=\omega_0\).

(d) [2 points] The small change \(\Delta m\) of the cantilever mass shifts the resonance frequency by \(\Delta\omega_0\). As a result, the phase \(\phi\) at the original resonance frequency \(\omega_0\) shifts by \(\Delta\phi\). Find the mass change \(\Delta m\) corresponding to the phase shift \(\Delta\phi=\pi/1800\), which is a typical resolution in phase measurements. The physical parameters of the cantilever are given by \(m=1.0\times10^{-12}\ \mathrm{kg}\), \(k=1.0\ \mathrm{N/m}\), and \((b/m)=1.0\times10^3\ \mathrm{s}^{-1}\). Use the approximations \((1+x)^a\approx1+ax\) and \(\tan(\pi/2+x)\approx-1/x\) when \(|x|\ll1\).

[Part B]

From now on let us consider the situation that some forces, besides the driving force discussed in Part A, act on the cantilever due to the sample as shown in Fig.3.1.

(e) [1.5 points] Assuming that the additional force \(f(h)\) depends only on the distance \(h\) between the cantilever and the sample surface, one can find a new equilibrium position \(h_0\). Near \(h=h_0\), we can write \(f(h)\approx f(h_0)+c_3(h-h_0)\), where \(c_3\) is a constant in \(h\). Find the new resonance frequency \(\omega'_0\) in terms of \(\omega_0\), \(m\), and \(c_3\).

(f) [2.5 points] While scanning the surface by moving the sample horizontally, the tip of the cantilever charged with \(Q=6e\) encounters an electron of charge \(q=e\) trapped (localized in space) at some distance below the surface. During the scanning around the electron, the maximum shift of the resonance frequency \(\Delta\omega_0\) \((=\omega'_0-\omega_0)\) is observed to be much smaller than \(\omega_0\). Express the distance \(d_0\) from the cantilever to the trapped electron at the maximum shift in terms of \(m\), \(q\), \(Q\), \(\omega_0\), \(\Delta\omega_0\), and the Coulomb constant \(k_e\). Evaluate \(d_0\) in nm (\(1\ \mathrm{nm}=1\times10^{-9}\ \mathrm{m}\)) for \(\Delta\omega_0=20\ \mathrm{s}^{-1}\). The physical parameters of the cantilever are \(m=1.0\times10^{-12}\ \mathrm{kg}\) and \(k=1.0\ \mathrm{N/m}\). Disregard any polarization effect in both the cantilever tip and the surface. Note that \(k_e=1/(4\pi\varepsilon_0)=9.0\times10^9\ \mathrm{N\cdot m^2/C^2}\) and \(e=-1.6\times10^{-19}\ \mathrm{C}\).

Official English text of the problem.

Answers

Write \(S_{\mathrm{dc}}=\langle V_RV_i\rangle\) for the dc output under the stated multiplication rule.

  • (a) \[ \boxed{A=\frac{F_0}{\sqrt{m^2(\omega_0^2-\omega^2)^2+b^2\omega^2}}}, \qquad \boxed{\tan\phi=\frac{b\omega}{m(\omega_0^2-\omega^2)}}. \] At \(\omega=\omega_0\), \[ \boxed{A=\frac{F_0}{b\omega_0},\qquad \phi=\frac{\pi}{2}}. \]

  • (b) A nonzero dc output requires \[ \boxed{\omega=\omega_i,\qquad \cos\phi_i\ne0}. \] Its magnitude is \[ \boxed{|S_{\mathrm{dc}}|=\frac{V_{R0}V_{i0}}{2}|\cos\phi_i|}. \]

  • (c) \[ \boxed{|S_{\mathrm{dc}}|=\frac{|c_1c_2|V_{R0}^{\,2}}{2b\omega_0}}. \]

  • (d) \[ \boxed{\Delta m\simeq\frac{b}{\omega_0}\Delta\phi \simeq1.7\times10^{-18}\ \mathrm{kg}}. \] The specified positive phase shift corresponds to an increase in mass.

  • (e) \[ \boxed{\omega'_0=\sqrt{\omega_0^2-\frac{c_3}{m}}}, \] provided \(m\omega_0^2-c_3>0\).

  • (f) The maximum upward frequency shift occurs directly above the electron, and \[ \boxed{d_0\simeq \left(\frac{k_eqQ}{m\omega_0\Delta\omega_0}\right)^{1/3} \simeq41\ \mathrm{nm}}. \]

Solution

We consider the steady periodic motion after the damped transient has decayed. Throughout, we use the problem’s weak-damping convention that the resonance frequency is \(\omega_0=\sqrt{k/m}\).

(a)

Let \(\theta=\omega t-\phi\). Substitution of \(z=A\sin\theta\) into the equation of motion gives

\[ A(k-m\omega^2)\sin\theta+Ab\omega\cos\theta =F_0\sin(\theta+\phi). \]

Expanding the right-hand side and equating the coefficients of \(\sin\theta\) and \(\cos\theta\),

\[ A(k-m\omega^2)=F_0\cos\phi, \qquad Ab\omega=F_0\sin\phi. \]

Squaring and adding these equations yields

\[ A^2\left[(k-m\omega^2)^2+b^2\omega^2\right]=F_0^2. \]

Using \(k=m\omega_0^2\),

\[ \boxed{A=\frac{F_0} {\sqrt{m^2(\omega_0^2-\omega^2)^2+b^2\omega^2}}}. \]

Dividing the coefficient equations gives

\[ \boxed{\tan\phi=\frac{b\omega}{m(\omega_0^2-\omega^2)}}. \]

The quadrant follows from the coefficient equations: for positive driving frequency, \(\sin\phi>0\), while \(\cos\phi\) has the sign of \(\omega_0^2-\omega^2\). Thus the lag is below \(\pi/2\) below resonance and above \(\pi/2\) above resonance.

At \(\omega=\omega_0\), the elastic and inertial terms cancel. Consequently,

\[ \boxed{\phi=\frac{\pi}{2},\qquad A=\frac{F_0}{b\omega_0}}. \]

(b)

Using the product-to-sum identity,

\[ \begin{aligned} V_RV_i &=V_{R0}V_{i0}\sin\omega t\, \sin(\omega_i t-\phi_i)\\ &=\frac{V_{R0}V_{i0}}{2} \left[ \cos\bigl((\omega-\omega_i)t+\phi_i\bigr) -\cos\bigl((\omega+\omega_i)t-\phi_i\bigr) \right]. \end{aligned} \]

The ideal dc filter retains only terms independent of time. Since both frequencies are positive, the sum-frequency term always averages to zero. The difference-frequency term is constant only when

\[ \boxed{\omega=\omega_i}. \]

At this frequency,

\[ S_{\mathrm{dc}}=\frac{V_{R0}V_{i0}}{2}\cos\phi_i, \]

so the magnitude is

\[ \boxed{|S_{\mathrm{dc}}| =\frac{V_{R0}V_{i0}}{2}|\cos\phi_i|}. \]

Frequency matching is necessary but does not ensure a nonzero result: if \(\phi_i=\pi/2+n\pi\), the input is in quadrature with the reference and the dc output vanishes. Hence non-vanishing output also requires \(\cos\phi_i\ne0\).

(c)

At the specified frequency, the shifted reference drives the cantilever with

\[ F=c_1V_{R0}\sin(\omega_0t+\pi/2) =c_1V_{R0}\cos\omega_0t. \]

The cantilever displacement lags the force by \(\pi/2\). The phase shifter therefore makes the displacement in phase with the original reference:

\[ z(t)=\frac{c_1V_{R0}}{b\omega_0}\sin\omega_0t. \]

This expression also follows directly by substitution into the equation of motion. The detector voltage is

\[ V_i(t)=\frac{c_1c_2V_{R0}}{b\omega_0}\sin\omega_0t. \]

Multiplication by the original reference gives

\[ V_RV_i= \frac{c_1c_2V_{R0}^{\,2}}{b\omega_0}\sin^2\omega_0t. \]

Since \(\langle\sin^2\omega_0t\rangle=1/2\),

\[ S_{\mathrm{dc}}=\frac{c_1c_2V_{R0}^{\,2}}{2b\omega_0}, \qquad \boxed{|S_{\mathrm{dc}}| =\frac{|c_1c_2|V_{R0}^{\,2}}{2b\omega_0}}. \]

The absolute value allows for either sign of the conversion constants.

(d)

The spring constant is unchanged when the mass changes. The new natural frequency is therefore

\[ \omega_0+\Delta\omega_0 =\sqrt{\frac{k}{m+\Delta m}} =\omega_0\left(1+\frac{\Delta m}{m}\right)^{-1/2}. \]

For \(|\Delta m|\ll m\), the supplied binomial approximation gives

\[ \boxed{\Delta\omega_0\simeq -\frac{\omega_0}{2}\frac{\Delta m}{m}}. \]

The measurement frequency remains at the original value \(\omega_0\). Taking \(b\) unchanged, the new phase satisfies

\[ \tan\left(\frac{\pi}{2}+\Delta\phi\right) =\frac{b\omega_0}{k-(m+\Delta m)\omega_0^2}. \]

Because \(k=m\omega_0^2\),

\[ \tan\left(\frac{\pi}{2}+\Delta\phi\right) =-\frac{b}{\omega_0\Delta m}. \]

Applying \(\tan(\pi/2+x)\simeq-1/x\),

\[ -\frac{1}{\Delta\phi}\simeq-\frac{b}{\omega_0\Delta m}, \]

and hence

\[ \boxed{\Delta m\simeq\frac{b}{\omega_0}\Delta\phi}. \]

The given parameters yield

\[ \omega_0=\sqrt{\frac{1.0}{1.0\times10^{-12}}} =1.0\times10^6\ \mathrm{s}^{-1}, \]

\[ b=m\frac bm =(1.0\times10^{-12})(1.0\times10^3) =1.0\times10^{-9}\ \mathrm{kg\,s^{-1}}. \]

Thus

\[ \Delta m\simeq \frac{1.0\times10^{-9}}{1.0\times10^6} \frac{\pi}{1800}\ \mathrm{kg} =1.7453\times10^{-18}\ \mathrm{kg}, \]

or, to two significant figures,

\[ \boxed{\Delta m\simeq1.7\times10^{-18}\ \mathrm{kg}}. \]

Here \(\Delta m/m\simeq1.7\times10^{-6}\) and \(\Delta\phi\simeq1.7\times10^{-3}\), consistent with both small-change approximations. The associated frequency change is approximately \(-0.87\ \mathrm{s}^{-1}\).

(e)

Take positive displacement and positive sample force to point away from the sample. Let \(z_{\mathrm{eq}}\) be the new static displacement, corresponding to the gap \(h_0\). Static force balance gives

\[ kz_{\mathrm{eq}}=f(h_0). \]

For a small displacement from this equilibrium, define

\[ x=z-z_{\mathrm{eq}}=h-h_0. \]

Including the sample force, the equation of motion is

\[ m\ddot z+b\dot z+kz=F+f(h). \]

Insert the linearized force \(f(h)\simeq f(h_0)+c_3x\), then subtract the static equilibrium equation:

\[ m\ddot x+b\dot x+kx=F+c_3x. \]

Therefore,

\[ m\ddot x+b\dot x+(k-c_3)x=F. \]

The force gradient changes the effective spring constant to \(k-c_3\), whereas the constant force merely shifts the equilibrium. Thus

\[ \boxed{\omega'_0 =\sqrt{\frac{k-c_3}{m}} =\sqrt{\omega_0^2-\frac{c_3}{m}}}. \]

A stable oscillatory equilibrium requires \(k-c_3>0\); the resonance interpretation also retains the weak-damping assumption.

(f)

Since \(q=e\) and \(Q=6e\),

\[ qQ=6e^2>0. \]

The interaction is repulsive. To identify where the frequency shift is greatest, let \(\rho\) be the horizontal separation between the tip and electron, and let \(d\) be their vertical separation. Coulomb’s law gives the upward force component

\[ f_z(\rho,d)= \frac{k_eqQ\,d}{(\rho^2+d^2)^{3/2}}. \]

A vertical cantilever displacement changes \(d\) by the same amount. The relevant force gradient is therefore

\[ c_3=\frac{\partial f_z}{\partial d} =k_eqQ\,\frac{\rho^2-2d^2}{(\rho^2+d^2)^{5/2}}. \]

For a small frequency shift, part (e) gives

\[ \omega'_0 =\omega_0\sqrt{1-\frac{c_3}{m\omega_0^2}} \simeq\omega_0-\frac{c_3}{2m\omega_0}, \]

so

\[ \Delta\omega_0(\rho)\simeq \frac{k_eqQ}{2m\omega_0} \frac{2d^2-\rho^2}{(\rho^2+d^2)^{5/2}}. \]

To leading order in the weak interaction, the scan height can be treated as fixed. This expression has its positive maximum at \(\rho=0\): while the numerator is positive, it decreases with \(\rho\) and the denominator increases; beyond \(\rho=\sqrt2d\), the shift is nonpositive.

Thus the maximum occurs directly above the electron, where the tip–electron distance is \(d_0\). At that point,

\[ f_z=\frac{k_eqQ}{d_0^2}, \qquad c_3=-\frac{2k_eqQ}{d_0^3}. \]

Substitution into part (e) yields

\[ (\omega_0+\Delta\omega_0)^2 =\omega_0^2+\frac{2k_eqQ}{md_0^3}. \]

Consequently,

\[ 2\omega_0\Delta\omega_0+(\Delta\omega_0)^2 =\frac{2k_eqQ}{md_0^3}. \]

Since \(\Delta\omega_0\ll\omega_0\), we neglect the squared shift:

\[ \boxed{d_0\simeq \left(\frac{k_eqQ}{m\omega_0\Delta\omega_0}\right)^{1/3}}. \]

Numerically,

\[ \omega_0=1.0\times10^6\ \mathrm{s}^{-1}, \]

and

\[ k_eqQ =(9.0\times10^9)\,6(1.6\times10^{-19})^2 =1.3824\times10^{-27}\ \mathrm{N\,m^2}. \]

Therefore,

\[ d_0^3\simeq \frac{1.3824\times10^{-27}} {(1.0\times10^{-12})(1.0\times10^6)(20)} \ \mathrm{m^3} =6.912\times10^{-23}\ \mathrm{m^3}, \]

giving

\[ d_0\simeq4.1039\times10^{-8}\ \mathrm{m} \quad\Longrightarrow\quad \boxed{d_0\simeq41\ \mathrm{nm}}. \]

The relative frequency shift is \(20/10^6=2.0\times10^{-5}\), confirming the validity of the small-shift approximation.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.