Energetics of Feedback: Application to Memory Erasure

A sample of polystyrene particles suspended in an aqueous solution (heat bath) is placed in an Optical Tweezer. A particle is trapped by a laser (see [13]) near the focus of the objective lens with high numerical aperture; thus creating a single harmonic potential $U(x)$ around the trap center, where $x$ is the particle position. The model for the single well potential is,

$\displaystyle U(x)=\begin{cases}\frac{1}{2}kx^{2}+U_{r},&\mbox{if }|x|\leq w,\\ \frac{1}{2}kw^{2}+U_{r},&\mbox{otherwise, }\end{cases}$

(1)

where $U_{r}$ is a constant reference potential energy, $k$ is the stiffness and $w$ is the width. The particle position is measured using a photodiode of $1nm$ accuracy at a bandwidth of $20kHz.$ The position data $x$ is collected for sufficient time for accurate estimation of probability distribution $P(x)$. With a particle in thermal equilibrium with the heat bath, the particle position is given by Boltzmann distribution; solving for $U(x)$ gives $-k_{B}T\ln(\frac{P(x)}{C})$. Experimentally computed values of $w$ and $k$ for a single potential well are $175nm$ and $0.0045\ pN/nm,$ respectively. Refer to [13, 10] for more details on experimental setup.

For our memory model, we require a double well potential with two stable equilibrium locations at $L$ and $-L$. This is achieved by alternately focusing a trapping laser at two trap locations using time multiplexing enabled by an Acousto Optical Deflector. The time spent by the laser at a trap location ($\sim 10\mu s$) is much smaller than the time constant of the particle ($1\ \sim ms$), which helps in creating a bistable potential well. Duty ratio ($d$) is defined as the fraction of the total time spent by the laser at left location. Duty ratio provides a control on the asymmetry of the bistable potential. For $d=0.5$, the two wells are symmetric; for a duty ratio of $0.7,$ the wells are asymmetric, as shown in FIG. 1(c).

For creating a stable memory bit (where the time of exit from one state to another is considerably longer than the experiment time), $L$ is chosen to be $550nm$. A particle is initialized either in the left or right well with equal probability and allowed to thermalize with the surrounding heat bath. The mathematical model of the bistable potential $U(x,d)$ is characterized by the model of the single well potential in Eq. (1) as follows:

$\displaystyle U(x,d)=\begin{cases}\frac{1}{2}k(x-L)^{2}+U_{r},&\mbox{if }|x-L|\leq w,\ r(t)=1,\\ \frac{1}{2}k(x+L)^{2}+U_{r},&\mbox{if }|x+L|\leq w,\ r(t)=0,\\ \frac{1}{2}kw^{2}+U_{r},&\mbox{otherwise, }\end{cases}$

(2)

where, $r(t)=1$ when the laser is present at right trap, otherwise $r(t)=0$; $w,\ k$ are the parameters that we computed for the single potential well earlier. The particle dynamics under the influence of time multiplexed potential is modeled by overdamped Langevin equation,

$\displaystyle-\gamma\frac{dx}{dt}+\xi(t)-\frac{\partial U(x,d)}{\partial x}=0.$

(3)

Here, $\xi$ is a zero-mean uncorrelated Gaussian noise with $\langle\xi(t),\xi(t^{\prime})\rangle=2D\delta(t-t^{\prime}),$ where $D=\gamma k_{B}T$ is the diffusion constant and $\gamma$ is the coefficient of viscosity. See [13] for more details. Monte Carlo simulations are performed using Eq. (3) in conjunction with Eq. (2) for a duty ratio of $0.5$. From the ensembles of $\{x(t)\}$ simulated, the potential wells are reconstructed using the canonical distribution. The bistable potential reconstructed using simulated data matched closely with the potential wells constructed from the position data collected from photodiode. The measurement device consists of sensor and a time sampler as shown in FIG. 1(b). The sensor is composed of a photodiode signal corrupted with a measurement noise $\eta.$ The measurement signal $M_{t_{m}}$ is modeled as $X_{t_{m}}+\eta$, with $\eta\sim\mathcal{N}(0,\sigma_{n})$ is the measurement noise.

Using the memory model and the measurement device, the feedback erasure protocol is implemented. Experimentally it was ascertained that the protocol achieves erasure with a probability $>95\%$. As alluded earlier in the Feedback step of the protocol, the potential wells are made asymmetric for successful transfer of particle to the left well. Duty ratio $d$ controls the asymmetric nature of the wells during the feedback step. For each $d$ in $\{0.65,0.7,0.75,0.8,0.85\},$ we instantiated $300$ independent runs of erasure based on feedback. For $d=0.65,$ the probability of the successful erasure under open-loop protocol is less than $95\%$; the results for $d=\{0.7,0.75,0.8,0.85\},$ where the open-loop protocol results in erasure with a probability higher than $0.95$ are reported.

The photodiode measurement $M_{t_{m}}$ is a continuous random variable, modeled as $X_{t_{m}}+\eta,$ where, $\eta\sim\mathcal{N}(0,\sigma_{n})$ is measurement noise. The conditional density $f_{M_{t}/X_{t}}(m,x)=f_{\eta}(m-x),$ where $f_{\eta}$ is the density function of a zero-mean Gaussian with variance $\sigma_{n}^{2}$. In an open-loop protocol [10], the probability of successful erasure depends on the duty ratio. A duty ratio greater than $0.7$ results in $p>0.95.$ However, in a feedback erasure protocol the value of $p$ depends on both $d$ and $\sigma_{n}.$ Because, the measurement may wrongly indicate that the particle is in the well corresponding to the reset state. Such a measurement will lead to no action or reset which will lead to an unsuccessful erasure. Moreover, the success of erasure has considerable impact on the minimal amount to work needed for an erasure.

An estimate on the work needed to erase a single bit memory with a success probability $p$ is provided by the Generalized Landauer bound (GLB), $k_{B}T[\ln 2+p\ln p+(1-p)\ln(1-p)]$ [15]. Here, the erasure probability $p>0.95$ ensures that GLB is within $29\%$ of the Landauer limit. As discussed earlier in the feedback erasure protocol, the erasure probability $p$ depends on the specifics of how the wells are altered to execute transfer of the state as well as the quality of the sensor that impacts whether the state transfer protocol is executed or not. We now relate the open-loop erasure probability to the closed-loop erasure probability. Let $p_{ol}(d)$ denote the probability of successful erasure when the open-loop protocol is performed at a duty ratio $d$. Let $p(d,\sigma_{n})$ be the probability of successful erasure under feedback erasure protocol. Then,

$\displaystyle p(d,\sigma_{n})=p_{ol}(d)-0.5\mathbb{P}(z<-\frac{L}{\sqrt{\sigma_{T}^{2}+\sigma_{n}^{2}}}),$

(4)

where, $z=\frac{M_{t_{m}}-L}{\sqrt{\sigma_{T}^{2}+\sigma_{n}^{2}}}\sim\mathcal{N}(0,1),$ and $\mathbb{P}(z<-\frac{L}{\sqrt{\sigma_{T}^{2}+\sigma_{n}^{2}}})$ represents the probability of error when a particle is initialized in the right well and with an incorrect decision that the particle is in the left well corresponding to the reset state. To ensure $p$ to be larger than $0.95$ for $d\geq 0.7,$ for the values of $\sigma_{T}$ and $p_{ol}(d)$ (as determined in [10]), it is determined that $\sigma_{n}\leq 300nm$. FIG. 1(d) confirms that $\sigma_{n}$ should not be more than $300nm$. In this article, $\sigma_{n}$ is taken to be $300nm.$ Such a choice allows for the assumption that the Landauer lower bound is a good approximation to the GLB. The mutual information between $X_{t_{m}}$ and $M_{t_{m}}$ is given by $I(X_{t_{m}},M_{t_{m}})\approx 0.6.$

Stochastic framework for Langevin systems [13] is utilized to compute the external work by the laser in the feedback process. This quantity depends on the value of $\sigma_{n}$, as wrong inference about the particle location is a possibility. For an erasure process, the work done on the particle $W_{fb}^{d}$ is,

$\displaystyle W_{fb}^{d}=\begin{cases}0&\mbox{if }M_{t_{m}}\leq 0,\\ \sum_{j=1}^{2}[U(x(t_{j}),d(t_{j}^{+}))-U(x(t_{j}),d(t_{j}^{-}))]&\mbox{if }M_{t_{m}}>0,\end{cases}$

(5)

where, $t_{1}=t_{m},\ t_{2}=t_{f}$ are the time instants when the duty ratio is switched from $0.5$ to $d$ and when switched back from $d$ to $0.5,$ respectively. The duration of erasure is $\tau:=t_{1}-t_{2}.$ The choice of $\tau$ is dependent on the value of $d$ for successful erasure (see [10]). For example, $\tau=30s$ for $d=0.7.$

For each memory sample, we computed $W_{fb}^{d}.$ FIG. 3 shows the average work done by the laser $\langle W_{fb}^{d}\rangle,$ computed for $d=\{0.7,0.75,0.8,0.85\}$. We fit the data to the model $\langle W_{fb}^{d}\rangle_{fit}=A+B\frac{exp(-\frac{0.99}{d-0.5})}{\sqrt{d-0.5}}$ [10] using Weighted Least Squares. $A,$ represents the minimal average external work required. The value of $A$ is $(0.133\pm 0.029)k_{B}T$ and $B$ is $(36.143\pm 0.958)k_{B}T.$ $A$ is less than the Landauer limit $k_{B}T\ln 2$ by a difference of $0.560k_{B}T,$ termed as energy deficit, and is approximately equal to the mutual information ($\approx 0.6k_{B}T$). $d\rightarrow 0.5$ would make the feedback action to approach quasistatic limit. However, there is a practical limitation. Lower $d$ ($<0.7$) would yield in low erasure performance in practice, as evident from FIG. 1(d). Moreover, $\tau$, which is the time chosen for successful erasure is proportional to $\frac{exp[0.99(d-0.5)^{-1}]}{(d-0.5)^{-0.5}}$ [10] and thus, $\tau$ grows unbounded as $d\rightarrow 0.5$. Hence, we extrapolate $\langle W_{fb}^{d}\rangle$ with the help of the model ($A$) to find the thermodynamic limit ($d\rightarrow 0.5$).

FIG. 2 shows the distribution of $W_{fb}^{d},$ where the y-axis represents the probability $P(W_{fb}^{d}).$ The spike at origin ($P(W_{fb}^{d}=0)=0.5$) is almost common for both $d=0.7$ and $d=0.75.$ Because, in the feedback step (second step), the potential wells are not tilted when a particle is initialized in the left well and thus the external work done is zero. The lobe present on the right side of the spike contributes significantly to $\langle W_{fb}^{d}\rangle.$ The right lobe represents the work done in tilting the potentials when a particle is initialized in the right well. By comparing the FIG. $10$ from [10] with FIG. 2, the difference is a spike present at the origin in FIG. 2.

We present the thermodynamics of feedback erasure protocol, which would explain the energy deficit observed in the experiments. An average external work $\langle W\rangle$ required in a process is bounded by the information free energy difference $\Delta\mathcal{F}$ (see [14]). We now apply second law of thermodynamics to each phase of the protocol.

(a) Measurement: Initial free energy of the system before measurement is $\mathcal{F}(X_{t_{m}}^{-},M_{t_{m}}^{-})=\mathcal{F}(X_{t_{m}}^{-})+\mathcal{F}(M_{t_{m}}^{-}),$ while the final free energy is $\mathcal{F}(X_{t_{m}},M_{t_{m}})=\mathcal{F}(X_{t_{m}})+\mathcal{F}(M_{t_{m}})+k_{B}TI(X_{t_{m}},M_{t_{m}})$. The measurement does not perturb the particle position and thus, $\mathcal{F}(X_{t_{m}}^{-})=\mathcal{F}(X_{t_{m}})$. From second law, $\langle W_{meas}\rangle\geq\mathcal{F}(M_{t_{m}})-\mathcal{F}(M_{t_{m}}^{-})+k_{B}TI(X_{t_{m}},M_{t_{m}}).$

(b) Feedback: After the measurement, the free energy of the system is $\mathcal{F}(X_{t_{m}},M_{t_{m}})$. The feedback action starts at $t_{m}$ and ends at $t_{f},$ and then the system relaxes until $t_{e}$. The final free energy of the system is $\mathcal{F}(X_{t_{e}},M_{t_{e}})=\mathcal{F}(X_{t_{e}})+\mathcal{F}(M_{t_{e}}).$ $\mathcal{F}(M_{t_{e}})=\mathcal{F}(M_{t_{m}}),$ as the measurement device is unaffected by the feedback action. From second law, $\langle W_{fb}^{d}\rangle\geq\mathcal{F}(X_{t_{e}})-\mathcal{F}(X_{t_{m}})-k_{B}TI(X_{t_{m}},M_{t_{m}}).$

(c) Reset: The initial free energy of the system is $\mathcal{F}(X_{t_{e}})+\mathcal{F}(M_{t_{e}})$ and the final free energy of the system is $\mathcal{F}(X_{t_{e}}^{+})+\mathcal{F}(M_{t_{e}}^{+}).$ Reset operation of the measurement device doesn’t affect the particle position and hence $\mathcal{F}(X_{t_{e}}^{+})=\mathcal{F}(X_{t_{e}})$. Note that $\mathcal{F}(M_{t_{e}}^{+})=\mathcal{F}(M_{t_{m}}^{-})$ since the state is reset to its initial value. Replacing $\mathcal{F}(M_{t_{e}}^{+})$ with $\mathcal{F}(M_{t_{m}}^{-}),$ $\mathcal{F}(M_{t_{e}})$ with $\mathcal{F}(M_{t_{m}}),$ and using second law, we get $\langle W_{res}\rangle\geq\mathcal{F}(M_{t_{m}}^{-})-\mathcal{F}(M_{t_{m}}).$

The average feedback work $\langle W_{fb}^{d}\rangle$ consumed by the system is at least $\mathcal{F}(X_{t_{e}})-\mathcal{F}(X_{t_{m}})-k_{B}TI(X_{t_{m}},M_{t_{m}});$ and the free energy change associated with system $1$ alone is $k_{B}T\ln 2$. Thus, the average feedback work is at least $k_{B}T[\ln 2-I(X_{t_{m}},M_{t_{m}})]$. The difference between the minimal average feedback work and the Landauer limit is $k_{B}TI(X_{t_{m}},M_{t_{m}})$ (mutual information). Experimentally we computed $\langle W_{fb}^{d}\rangle$ and observed to be quasistatically approaching $k_{B}T[\ln 2-I(X_{t_{m}},M_{t_{m}})]$ (See FIG. 3). The energy deficit is given by the mutual information, which was experimentally verified in Section II.C as $0.6k_{B}T$. The reduction in energy expenditure is not a violation of second law, as we spend an additional amount of $k_{B}TI(X_{t_{m}},M_{t_{m}})$ in the Measurement.