Scrambling and quantum feedback in a nanomechanical system

Abstract

The question of how swiftly entanglement spreads over a system has attracted vital interest. In this regard, the out-of-time-ordered correlator (OTOC) is a quantitative measure of the entanglement spreading process. Particular interest concerns the propagation of quantum correlations in the lattice systems, e.g., spin chains. In a seminal paper of Roberts et al. (J. High Energy Phys. 03:051, 2015), the concept of the OTOC’s radius was introduced. The radius of the OTOC defines the front line reached by the spread of entanglement. Beyond this radius operators commute. In the present work, we propose a model of two nanomechanical systems coupled with two nitrogen-vacancy (NV) center spins. Oscillators are coupled to each other directly, while NV spins are not. Therefore, the correlation between the NV spins may arise only through the quantum feedback exerted from the first NV spin to the first oscillator and transferred from the first oscillator to the second oscillator via the direct coupling. Thus, nonzero OTOC between NV spins quantifies the strength of the quantum feedback. We show that NV spins cannot exert quantum feedback on classical nonlinear oscillators. We also discuss inherently quantum case with a linear quantum harmonic oscillator indirectly coupling the two spins and verify that in the classical limit of the oscillator, the OTOC vanishes.

Appendix: Calculation of thermally averaged OTOC \(C_\rho \)

From Eq. (16) after re-scaling, the total Hamiltonian will be written as

$$\begin{aligned} H_{tot}= & {} (\omega _{0R}+\varOmega _0)(\sigma _{1}^z+ \sigma _{2}^z)\nonumber \\&+\varOmega _n(\sigma _{1}^+ \sigma _{2}^- + \sigma _{1}^- \sigma _{2}^+) \end{aligned}$$

(24)

or in the matrix form, in the standard basis,

$$\begin{aligned} \begin{pmatrix} 2(\varOmega _0+\omega _{0R}) &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad \varOmega _{n} &{}\quad 0\\ 0 &{}\quad \varOmega _{n}&{}\quad 0&{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad -2(\varOmega _0+\omega _{0R}) \end{pmatrix},\quad \end{aligned}$$

(25)

where $n=\langle a^{\dagger }a\rangle$, $\omega _{0R}=\frac{\omega _{0}}{2n+1}$, and $\varOmega _0=\frac{g^2 }{\omega _{0}-\omega }$, and $\varOmega _{n}=\frac{g^2 }{(\omega _{0}-\omega )(2n+1)}$. The eigenstates of the above Hamiltonian are

$$\begin{aligned} \vert \phi _{1}\rangle= & {} \vert 0,0\rangle , \end{aligned}$$

(26)

$$\begin{aligned} \vert \phi _{2}\rangle= & {} \frac{1}{\sqrt{2}}(\vert 1,0\rangle +\vert 0,1\rangle ), \end{aligned}$$

(27)

$$\begin{aligned} \vert \phi _{3}\rangle= & {} \frac{1}{\sqrt{2}}(\vert 1,0\rangle -\vert 0,1\rangle ), \end{aligned}$$

(28)

$$\begin{aligned} \vert \phi _{4}\rangle= & {} \vert 1,1\rangle , \end{aligned}$$

(29)

with corresponding eigenvalues:

$$\begin{aligned} E_{1}= & {} 2(\varOmega _0+\omega _{0R}) \end{aligned}$$

(30)

$$\begin{aligned} E_{2}= & {} \varOmega _{n} \end{aligned}$$

(31)

$$\begin{aligned} E_{3}= & {} -\varOmega _{n} \end{aligned}$$

(32)

$$\begin{aligned} E_{4}= & {} -2(\varOmega _0+\omega _{0R}). \end{aligned}$$

(33)

At a finite temperature, in the equilibrium state the density matrix $\hat{\varrho }=Z^{-1}e^{-\beta H_{tot}}$ of the system in the diagonal basis of the Hamiltonian is

$$\begin{aligned} \hat{\varrho }= & {} Z^{-1}\big (e^{-\beta E_{1}}\vert \phi _{1}\rangle \langle \phi _{1}\vert \nonumber \\&+e^{-\beta E_{2}}\vert \phi _{2}\rangle \langle \phi _{2}\vert +e^{-\beta E_{3}}\vert \phi _{3}\rangle \langle \phi _{3}\vert \nonumber \\&+ e^{-\beta E_{4}}\vert \phi _{4}\rangle \langle \phi _{4}\vert \big ) \end{aligned}$$

(34)

$$\begin{aligned}&\hat{\varrho }=Z^{-1}\begin{pmatrix} e^{-2\beta (\varOmega _0+\omega _{0R})}&{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad e^{- \beta \varOmega _{n}} &{}\quad 0&{}\quad 0\\ 0 &{}\quad 0&{}\quad e^{ \beta \varOmega _{n}}&{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad e^{2\beta (\varOmega _0+\omega _{0R})} \end{pmatrix},\nonumber \\&Z=2\cosh {\beta 2(\varOmega _0+\omega _{0R})}+2\cosh {\beta \varOmega _{n}}. \end{aligned}$$

(35)

Pauli operators $\sigma _{1}^{z}$ and $\sigma _{2}^{z}$ in the diagonal basis of Hamiltonian are written as

$$\begin{aligned} \sigma _{1}^{z}= & {} \vert \phi _{1}\rangle \langle \phi _{1}\vert +\vert \phi _{3}\rangle \langle \phi _{2}\vert +\vert \phi _{2}\rangle \langle \phi _{3}\vert -\vert \phi _{4}\rangle \langle \phi _{4}\vert \end{aligned}$$

(36)

$$\begin{aligned} \sigma _{2}^{z}= & {} \vert \phi _{1}\rangle \langle \phi _{1}\vert -\vert \phi _{3}\rangle \langle \phi _{2}\vert -\vert \phi _{2}\rangle \langle \phi _{3}\vert -\vert \phi _{4}\rangle \langle \phi _{4}\vert \end{aligned}$$

(37)

Also, the time evolution operators $\exp {(-i H_{tot} t)}$ in the diagonal basis can be given as

$$\begin{aligned} \exp {(-i H_{tot} t)}= & {} e^{- i E_{1}t}\vert \phi _{1}\rangle \langle \phi _{1}\vert +e^{- i E_{2}t}\vert \phi _{2}\rangle \langle \phi _{2}\vert \nonumber \\&+ e^{- i E_{3}t}\vert \phi _{3}\rangle \langle \phi _{3}\vert +e^{= i E_{4}t}\vert \phi _{4}\rangle \langle \phi _{4}\vert , \end{aligned}$$

(38)

We calculate $\sigma _{1}^{z}(t)$ as

$$\begin{aligned} \sigma _{1}^{z}(t)= & {} e^{iH_{tot}t}\sigma _{1}^{z}e^{-iH_{tot}t}=\vert \phi _{1}\rangle \langle \phi _{1}\vert +e^{-2i\varOmega _{n}t}\vert \phi _{3}\rangle \langle \phi _{2}\vert \nonumber \\&+ e^{2i\varOmega _{n}t}\vert \phi _{2}\rangle \langle \phi _{3}\vert -\vert \phi _{4}\rangle \langle \phi _{4}\vert . \end{aligned}$$

(39)

By successive application of operators in the sequence $\sigma _{1}^{z}(t)\sigma _{2}^{z}\sigma _{1}^{z}(t)\sigma _{2}^{z}$, we get

$$\begin{aligned} \sigma _{1}^{z}(t)\sigma _{2}^{z}\sigma _{1}^{z}(t)\sigma _{2}^{z}= & {} \big (\vert \phi _{1}\rangle \langle \phi _{1}\vert -e^{4i\varOmega _{n}t}\vert \phi _{2}\rangle \langle \phi _{2}\vert \nonumber \\&- e^{-4i\varOmega _{n}t}\vert \phi _{3}\rangle \langle \phi _{3}\vert +\vert \phi _{4}\rangle \langle \phi _{4}\vert \big ). \end{aligned}$$

(40)

We can calculate $\rho \sigma _{1}^{z}(t)\sigma _{2}^{z}\sigma _{1}^{z}(t)\sigma _{2}^z$ as

$$\begin{aligned} \rho \sigma _{1}^{z}(t)\sigma _{2}^{z}\sigma _{1}^{z}(t)\sigma _{2}^{z}= & {} Z^{-1}\big (e^{-2\beta (\varOmega _0+\omega _{0R})}\vert \phi _{1}\rangle \langle \phi _{1}\vert \nonumber \\&-e^{-\beta \varOmega _n}e^{4i\varOmega _{n}t}\vert \phi _{2}\rangle \langle \phi _{2}\vert \nonumber \\&- e^{\beta \varOmega _n}e^{-4i\varOmega _{n}t}\vert \phi _{3}\rangle \langle \phi _{3}\vert \nonumber \\&+ e^{2\beta (\varOmega _0+\omega _{0R})}\vert \phi _{4}\rangle \langle \phi _{4}\vert \big ). \end{aligned}$$

(41)

Further, we calculate the thermally averaged OTOC $C_\rho (t)=1-Re(Tr \{\rho \sigma _{1}^{z}(t)\sigma _{2}^{z}\sigma _{1}^{z}(t)\sigma _{2}^z\})$ as

$$\begin{aligned}&C_\rho (t)=1\nonumber \\&\quad -\frac{2\cosh 2\beta (\varOmega _0+\omega _{0R})+2\cos 4\varOmega _n t\cosh {\beta \varOmega _n}}{Z},\nonumber \\ \end{aligned}$$

(42)