---
title: "How much work can be extracted by measuring a trapped bead and moving the trap"
description: "Derive the information in a noisy reading of a trapped bead, the mean work a trap shift extracts from it, and why a shift alone cannot use all of it."
canonical_url: "https://sandboxphysics.com/stories/stochastic-thermodynamics/feedback-and-information/"
language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/stochastic-thermodynamics/feedback-and-information/"
last_modified: "2026-10-29"
---

[Stochastic thermodynamics](https://sandboxphysics.com/stories/stochastic-thermodynamics/)

L3 · Chapter 5

# How much work can be extracted by measuring a trapped bead and moving the trap

Sandbox Physics · 2026-10-29

In the earlier chapters the trap followed a fixed plan, and the mean work never fell below the free-energy change. Now we play differently: first look at where the bead has wandered, then decide where to move the trap. If the reading is useful, the bead drops into the new trap and the heat of the water does work for us. This chapter computes that work and its limit, the thermal energy times the information the measurement carries.

## Before you begin

The Boltzmann distribution and relaxation time of Chapter 1, the rule from Chapter 2 that a sudden trap move costs the change in potential energy, and the Jarzynski equality of Chapter 4. You also need to average products of two Gaussian variables.

- Write the work and heat of each step of a measure–shift–wait cycle, and derive the mean extracted work in the steady state.
- Compute the mutual information of a Gaussian reading, and use the feedback fluctuation relation to show why the mean extracted work cannot exceed the thermal energy times that information.
- Find the best gain and its efficiency, and explain why shifting the trap alone keeps the efficiency below one.

## Work and heat in one cycle

Let $z$ be the bead offset from the trap centre, in units of the thermal width $\sqrt{k_BT/k}$, with energies in units of $k_BT$. At the start of a cycle $z$ is Gaussian with variance $S$. A camera reads $y=z+\varepsilon$, where the error $\varepsilon$ has variance $R$. The trap then jumps by $gy$; $g$ is called the gain.

The bead has no time to move during the jump, so the work is just the change in potential energy, given below. Then the trap stays put while the bead relaxes for a time $t_w$. No work is done during the wait, and every drop in potential energy leaves as heat into the water. With $a=e^{-2t_w/\tau_r}$, the offset variance must return to $S$ after the wait, which fixes the steady variance.

Averaging uses $\langle zy\rangle=S$ and $\langle y^2\rangle=S+R$. A positive extracted work means the reading really did let the bead fall lower; the second term is the cost of the error, and it grows with the gain.

$$
\beta W=\tfrac12\left[(z-gy)^2-z^2\right],\qquad S=\frac{a g^2R+1-a}{1-a(1-g)^2}
$$

The first expression is the work done on the bead by one jump; the second is the condition that the variance is unchanged after one round of jump and relaxation.

$$
-\beta\langle W\rangle=\tfrac12\left[(2g-g^2)S-g^2R\right]
$$

Mean work extracted per cycle. It vanishes at zero gain: no measurement, no work.

## How much information one reading carries

How much the reading $y$ tells us about the bead is measured by the stochastic information $I=\ln[p(y\mid z)/p(y)]$, Eq. (151) of the Seifert review. It is large when the reading lands unexpectedly close to the true position, and it can be negative when the reading lands far off.

Its average is the mutual information. For a Gaussian reading $p(y)$ has variance $S+R$ and $p(y\mid z)$ has variance $R$. Averaging the difference of their logarithms gives the formula below. It depends only on the signal-to-noise ratio $S/R$ and is measured in nats.

$$
\langle I\rangle=\tfrac12\ln\!\left(1+\frac{S}{R}\right)
$$

An error-free reading carries unlimited information; a reading whose error far exceeds the bead jitter carries almost none.

Source: U. Seifert

## The bound set by information

Sagawa and Ueda proved that with feedback $\langle W\rangle\ge\Delta F-k_BT\langle I\rangle$. After a full cycle the trap has the same shape, so $\Delta F=0$ and the mean extracted work cannot exceed $k_BT\langle I\rangle$. Our cycle starts from the steady state of the loop, and the matching exact equality is Eq. (156) of the Seifert review, where the total entropy production $\Delta s_{\mathrm{tot}}$ is the system entropy change plus $\beta Q$.

Averaging Eq. (156), and using the fact that in the steady state the mean system entropy change equals $-I_{\mathrm{left}}$, gives a tighter bound. $I_{\mathrm{left}}$ is the information about the bead that the reading still holds at the end of the cycle. Nobody used it, and the next measurement makes it obsolete. After a long wait it is almost zero.

The bound limits only the average. A single cycle can extract more than $k_BT\,I$, just as a single pull in Chapter 4 could cost less than the free-energy change. The sample average of the equality behaves like the Jarzynski average too: a few cycles whose reading points badly the wrong way dominate it, so small samples fall short of one.

$$
\left\langle e^{-(\Delta s_{\mathrm{tot}}+I)}\right\rangle=1\quad\Rightarrow\quad -\beta\langle W\rangle\le\langle I\rangle-I_{\mathrm{left}}\le\langle I\rangle
$$

The left side averages over the random quantities of each cycle; Jensen’s inequality turns it into the bound on the right.

Source: T. Sagawa and M. Ueda · U. Seifert

## The best gain and the efficiency

After a long wait $a\to0$ and $S\to1$. Setting the derivative with respect to the gain to zero gives the best gain $g^\ast=1/(1+R)$, which extracts $\tfrac12\,\mathrm{SNR}/(1+\mathrm{SNR})$ thermal energies per cycle, with $\mathrm{SNR}=1/R$. The sharper the reading, the closer the best gain gets to one, which means moving the trap all the way to the reading.

The extracted work never exceeds half a thermal energy, while the information grows with the logarithm of the signal-to-noise ratio. Their ratio is the efficiency below. It approaches one as the signal-to-noise ratio goes to zero and is only about 0.21 at a ratio of 100. Section 7.3.5 of the Seifert review states that this bound cannot be reached by controlling the trap centre alone; using all the information also requires changing the trap stiffness.

$$
\eta^\ast=\frac{-\langle W\rangle}{k_BT\langle I\rangle}=\frac{\mathrm{SNR}}{(1+\mathrm{SNR})\ln(1+\mathrm{SNR})}
$$

Efficiency at the best gain after a long wait. It is always below one and approaches one only for blurry readings.

Source: U. Seifert

## Where the second law survives

The engine seems to draw work steadily from water at a single temperature. The missing piece is the controller’s memory: each reading must be recorded and then erased before the next one. Sagawa and Ueda showed in 2009 that measuring and erasing together cost at least $k_BT\,I$ of work. Include that cost, and the whole machine gains no net work per cycle; the second law holds.

Information therefore acts like a fuel. It can be turned into work, at most one thermal energy per nat, and making and clearing it costs at least as much.

Source: T. Sagawa and M. Ueda

## Worked example: a blurry reading and fast cycles

Take a trap stiffness of $10\,\mathrm{pN/\mu m}$, so the thermal width is $20.4\,\mathrm{nm}$ and the relaxation time $\tau_r=0.942\,\mathrm{ms}$. A measurement error of $10\,\mathrm{nm}$ gives $R=(10/20.35)^2=0.241$, a signal-to-noise ratio of 4.14 and a best gain of 0.806. Use $g=0.8$ and wait $3\,\mathrm{ms}$, so $a=0.0017$ and $S=0.999$.

The mean work formula gives $-\beta\langle W\rangle=\tfrac12[(1.6-0.64)(0.999)-0.64(0.241)]=0.402$. The information is $\tfrac12\ln(1+0.999/0.241)=0.818\,\mathrm{nat}$, so the efficiency is 0.49. In physical units each cycle extracts $1.67\,\mathrm{pN\,nm}$, an average power of about $0.56\,\mathrm{aW}$.

Now cycle fast: error $2\,\mathrm{nm}$, $g=1$, and a wait of only $0.25\,\mathrm{ms}$. Then $a=0.588$, $R=0.0097$ and $S=0.417$: the bead has no time to spread back to its thermal width, so the feedback has cooled it. Each cycle extracts only $0.204\,k_BT$, less than the 0.494 of a full $3\,\mathrm{ms}$ wait, but cycles are twelve times shorter and the power rises from $0.165$ to $0.815\,k_BT/\mathrm{ms}$.

## Try it in the Lab

### [Turn information into work](https://sandboxphysics.com/labs/feedback-information-engine/)

1. Predict first: raise the gain from 0.8 to 1.8. How do the information and the extracted work change? Then select Overshoot costs work to check.
2. Select Sharp eye, wasted bits and then Blurry eye, frugal use. Compare the information and the efficiency, and test both efficiencies against the $\eta^\ast$ formula.
3. Look at the middle scatter plot: cycles above the line beat the bound. Note what share they make up, and which side of the line the mean point sits on.
4. Drag the handle of the large trap and move the trap under the bead when it wanders. Compare your extracted work with the automatic feedback, and think about how accurate and how fast your own reading is.

## Exercises with explained answers

1. After a long wait, with zero measurement error and a gain of one, what mean work is extracted per cycle, and what is the efficiency?

   Show the answer

   Here $S=1$ and $R=0$, so each cycle extracts $\tfrac12k_BT$, the whole mean potential energy of the bead. The information $\tfrac12\ln(1+1/R)$ diverges, so the efficiency goes to zero: almost all the information in a perfect reading goes unused.

2. With a signal-to-noise ratio of one and a long wait, find the best gain, the work extracted per cycle and the efficiency.

   Show the answer

   The best gain is $g^\ast=1/2$, the work is $\tfrac12\cdot\tfrac12=0.25\,k_BT$, the information is $\tfrac12\ln2=0.347\,\mathrm{nat}$ and the efficiency is $1/(2\ln2)=0.72$.

3. Why does the engine pay work at gains above two after a long wait, however sharp the reading?

   Show the answer

   After a long wait $-\beta\langle W\rangle=\tfrac12[(2g-g^2)-g^2R]$. For $g>2$ the term $2g-g^2$ is negative, so both terms are negative. The trap overshoots so far that the bead sits higher in the new trap than in the old one, and every jump adds energy to it.

[Previous: How the Jarzynski equality and the Crooks relation recover free energy from irreversible pulls](https://sandboxphysics.com/stories/stochastic-thermodynamics/jarzynski-and-crooks/) [Next: How much entropy a fuel-driven enzyme produces each second](https://sandboxphysics.com/stories/stochastic-thermodynamics/entropy-production-in-networks/)

## Sources and what we recomputed

1. [U. Seifert, Stochastic thermodynamics, fluctuation theorems and molecular machines, arXiv:1205.4176v1; Reports on Progress in Physics 75, 126001 (2012)](https://arxiv.org/abs/1205.4176v1)

   Version 1, 18 May 2012. Section 2: overdamped Langevin dynamics, Eq. (14) for the heat released into the medium and Eqs. (15)–(19) for the first law along one trajectory. Section 4: Eqs. (58) and (61), the Jarzynski and Crooks relations. Section 6.2: entropy production for master equations, Eqs. (120)–(129). Section 7.2: relative-entropy bounds on dissipation, Eqs. (146)–(150). Section 7.3: measurement and feedback, Eqs. (151)–(163).

2. [T. Sagawa and M. Ueda, Generalized Jarzynski equality under nonequilibrium feedback control, arXiv:0907.4914v3; Physical Review Letters 104, 090602 (2010)](https://arxiv.org/abs/0907.4914v3)

   Version 3, 20 January 2010. Eq. (3): with feedback, the mean work can fall below the free-energy change by at most the thermal energy times the mutual information. Eq. (4): the corresponding equality for a process that starts in equilibrium. The trapped-bead engine of this course and its numbers are our own.

3. [T. Sagawa and M. Ueda, Minimal energy cost for thermodynamic information processing: measurement and information erasure, arXiv:0809.4098v3; Physical Review Letters 102, 250602 (2009)](https://arxiv.org/abs/0809.4098v3)

   Version 3, 30 May 2009. Eq. (4): the work to measure and then erase the record together is at least the thermal energy times the mutual information obtained.