# Erratum: Virtual double-well potential for an underdamped oscillator created by a feedback loop

Salambô Dago, Jorge Pereda, Sergio Ciliberto, Ludovic Bellon

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Salambô Dago, Jorge Pereda, Sergio Ciliberto, Ludovic Bellon. Erratum: Virtual double-well potential for an underdamped oscillator created by a feedback loop. Journal of Statistical Mechanics: Theory and Experiment, 2023, 2023 (5), pp.059901. 10.1088/1742-5468/acd697. hal-04298666

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Submitted on 22 Nov 2023

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# Erratum: Virtual double-well potential for an underdamped oscillator created by a feedback loop.

Salambò Dago, Jorge Pereda, Sergio Ciliberto and Ludovic Bellon*

Univ Lyon, ENS de Lyon, CNRS, Laboratoire de Physique, F-69342 Lyon, France

In this note, we correct the expression of the switching rate $\Gamma(\mathcal{B}, T)$ of the potential barrier $\mathcal{B}$ for a system at temperature $T$ expressed in Eq. 8 and derived in appendix B. We also update below all the equations using the switching rate expression. The error is nevertheless small enough to not change any of the other results or conclusions of the paper.

The error came from the assumption that the total energy in the double well potential $E$ was simply distributed according to the Boltzmann probability distribution $P(E) = \frac{1}{k_B T} \exp\{-E/(k_B T)\}$, which would be valid only in an harmonic potential. Indeed, in the double well potential, the motion period $T$ depends on the system energy. Therefore the probability distribution of the energy $P(E)$ is obtained by integrating the Boltzmann canonical distribution on the time range $[-T(E)/2 + T(E)/2]$ (time range required to explore the full phase space). Contrary to the harmonic case where $T(E) = T_0$ is constant, here the energy dependence modifies the expression of $P(E)$.

## I. ERRATUM: EQ. 8

In this section we present the erratum regarding the switching rate expression in the case of a bi-quadratic potential $U(x) = \frac{1}{2}k(|x| - x_1)^2$, with an energetic barrier $\mathcal{B} = \frac{1}{2}kx_1^2$, for a system temperature $T$.

### A. Replacement to Eq. 8

Eq. 8 of the paper has to be replaced by Eq. (2):

$$ \Gamma = \omega_0 \frac{e^{-\beta \mathcal{B}}}{\beta \int_0^\infty dE e^{-\beta E} \left[ \pi + 2 \sin^{-1} \left( \sqrt{\frac{\mathcal{B}}{E}} \right) \right]} \quad (1) $$

$$ = \omega_0 \frac{k_B T}{\mathcal{B}} \frac{e^{-\mathcal{B}/(k_B T)}}{\int_0^\infty d\epsilon e^{-\epsilon \mathcal{B}/(k_B T)} \left[ \pi + 2 \sin^{-1} \left( \epsilon^{-\frac{1}{2}} \right) \right]}, \quad (2) $$

with $\beta = \frac{1}{k_B T}$, and $\omega_0 = \sqrt{k/m}$ the angular resonance frequency in a single well of the bi-quadratic potential. Note that in this expression, we extend the definition of the $\sin^{-1}$ function to arguments greater than one, with

$$ \sin^{-1}(x) = \pi/2 \text{ for } x > 1. \quad (3) $$

### B. Replacement to Eq. 14

As the $\Gamma$ expression was corrected, Eq. 14 also has to be updated. The correct expression of the function $g(z)$ is:

$$ g(z) = \frac{4}{\sqrt{\pi}} \frac{1}{z} \frac{e^{-z^2}}{\int_0^\infty d\epsilon e^{-\epsilon z^2} \left[ \pi + 2 \sin^{-1} \left( \epsilon^{-\frac{1}{2}} \right) \right]} \quad (4) $$

It presents a global maximum $g^* = 0.19$ in $z^* = 0.59$.

## II. ERRATUM TO APPENDIX B. SWITCHING RATE

We study a one degree of freedom mechanical system of mass $m$ and position $x$ evolving in a double well potential of potential energy $U(x) = \frac{1}{2}k(|x| - x_1)^2$. The barrier between the two wells is placed in $x = 0$ and its energy is $\mathcal{B} = \frac{1}{2}kx_1^2$. The system kinetic energy is $K = p^2/(2m)$, with $p = m \frac{dx}{dt}$ the momentum, so that the system Hamiltonian is:

$$ H(x, p) = \frac{1}{2m} p^2 + U(x). \quad (5) $$

In the limit of weak damping, the total energy $E = H(x, y)$ of the cantilever is conserved in time. Therefore, the Hamiltonian dynamics can be solved for a given total energy $E$ by integrating the following equation:

$$ \frac{dx}{dt} = \pm \sqrt{\frac{2}{m} [E - U(x)]}. \quad (6) $$

The system motion is periodic: the time $T(E)$ to explore the full phase space available for a given $E$, depends on the value of $E$ with respect to the barrier height $\mathcal{B}$.

* If $E > \mathcal{B}$ the system explores both wells every period, so there are 2 barrier crossing every period, with the period being:

$$ \mathcal{T}(E > \mathcal{B}) = 2 \int_{-x_M}^{x_M} \frac{dx}{\sqrt{2[E - U(x)]/m}} \quad (7) $$

where $x_M = x_1 + \sqrt{2E/k}$ is the maximum excursion of the cantilever allowed by the criterion $U = E - K \le E$.

* If $E < \mathcal{B}$, then the motion is confined to a single well, there are no switches, and the period is $T_0 = 2\pi/\omega_0$. As there are 2 wells, the time to explore the full phase space available for an energy $E$ is

$$ \mathcal{T}(E < \mathcal{B}) = 2T_0. \quad (8) $$

<footnote>* ludovic.bellon@ens-lyon.fr</footnote>
Using the the convention of Eq. (3), we can compute for all values of E:

$$ \mathcal{T}(E) = \frac{2}{\omega_0} \left[ \pi + 2 \sin^{-1} \left( \sqrt{\frac{\mathcal{B}}{E}} \right) \right]. \quad (9) $$

Eq. (9) is the same as Eq. B3 of the paper, simply extended to $E < \mathcal{B}$. This time is twice $\mathcal{T}_0$ when $E \le \mathcal{B}$ or $E \gtrsim \mathcal{B}$, and tends to $\mathcal{T}_0$ for $E \gg \mathcal{B}$.

It remains to compute the energy probability distribution $P(E)$. The first simplistic (and wrong) approach that we originally applied consists in using the Boltzmann distribution $P^B(E) = \exp(-E/k_B T)/k_B T$. The latter would be valid in a single well where the time to explore the full phase space available doesn't depend on the system's energy. But in the double well potential case, the computation of $P(E)$ has to take into account the energy dependency of the motion period $[\mathcal{T}(E)]$. In the next paragraph we derive $P(E)$.

The first step consists in going back to the canonical motion variables of position and momentum, that are distributed at all time from a Boltzmann distribution:

$$ P(x,p) = \frac{1}{Z} e^{-\beta H(x,p)}, \quad (10) $$

$$ \text{with } Z = \iint dx dp e^{-\beta H(x,p)}. \quad (11) $$

Since time is a natural variable for this problem, we operate a change of variables from $(x, p)$ to $(t, E)$: for any given position and momentum, we can define a unique time and energy, and vice-versa. We consider here that as the system is periodic, the time range is restricted from $-\frac{1}{2}\mathcal{T}(E)$ to $+\frac{1}{2}\mathcal{T}(E)$ for a given total energy $E$. The above change is a canonical transformation for the Hamiltonian since the Jacobian determinant is 1:

$$ dx dp = \left| \frac{\partial x}{\partial E} \frac{\partial p}{\partial t} - \frac{\partial x}{\partial t} \frac{\partial p}{\partial E} \right| dt dE = dt dE, \quad (12) $$

The change of variable is therefore straightforward in the expression of the probability:

$$ P(t, E) = \frac{1}{Z} e^{-\beta E}, \quad (13) $$

where $P(t, E)$ is defined only for $-\frac{1}{2}\mathcal{T}(E) < t < \frac{1}{2}\mathcal{T}(E)$ (0 otherwise), as the motion is periodic. The probability of having an energy $E$ is deduced by integrating on time,

leading to:

$$ P(E) = \frac{1}{Z} e^{-\beta E} \mathcal{T}(E), \quad (14) $$

$$ \text{with } Z = \int_{0}^{\infty} dE e^{-\beta E} \mathcal{T}(E). \quad (15) $$

Now we have all we need to compute the crossing rate of the barrier. For a given energy $E$, the escape rate is 0 if $E < \mathcal{B}$. On the contrary, for $E > \mathcal{B}$ the crossing occurs twice every $\mathcal{T}(E)$. The mean transition rate in the double potential, that is the rate at which the system crosses the barrier, regardless of the direction is therefore:

$$ \Gamma = \int_{\mathcal{B}}^{\infty} dE \frac{2}{\mathcal{T}(E)} P(E) \quad (16) $$

$$ = \frac{2e^{-\beta \mathcal{B}}}{\beta \int_{0}^{\infty} dE e^{-\beta E} \mathcal{T}(E)}, \quad (17) $$

which taking into account Eq. (9) is equivalent to Eq. (2).

If the barrier $\mathcal{B}$ is high (with respect to $k_B T = 1/\beta$), then the denominator to Eq. (17) can be seen as the average time to explore the bottom of the wells, $\mathcal{T}(0) = 2\mathcal{T}_0$. We get in this case a very simple expression matching Kramers' expression:

$$ \Gamma(\mathcal{B} \gg k_B T) = \frac{1}{\mathcal{T}_0} e^{-\beta \mathcal{B}}. \quad (18) $$

<figure><img src="image_2.png" /><figcaption>FIG. 1. switching rate on a bi-quadratic well $U(x) = \frac{1}{2}k(|x|-x_1)^2$. The barrier height is $\mathcal{B} = \frac{1}{2}kx_1^2$. The period in a single harmonic well is $\mathcal{T}_0$. At low barrier height, Kramers' simple expression is a factor 2 below the result from Eq. (2). The latter perfectly matches the switching rate extracted from a numerical simulation of a Langevin dynamics in the potential $U(x)$, using a quality factor $Q = 10$.</figcaption></figure>

In Fig. 1, we apply Eq. (2) and compare the predicted escape rate to the one measured on a numerical simulation of a Langevin dynamics in such a potential: the agreement is excellent. The Kramers approximation holds only in the limit of large barriers.