Exploration of entropic uncertainty relation for two accelerating atoms immersed in a bath of electromagnetic field


The uncertainty principle as an elementary theory of quantum mechanics plays an important role in quantum information science. In this paper, we study the dynamics of quantum-memory-assisted entropic uncertainty relation for two accelerating atoms coupled with a bath of fluctuating electromagnetic field. The master equation that the system evolution obeys is firstly derived. We find that the mixedness is bound up with the entropic uncertainty. For equilibrium state, the tightness of uncertainty vanishes. For the initial maximum entangled state, the tightness of uncertainty experiences a slight increase and then declines to zero with evolution time. It is found that the greater acceleration makes the uncertainty faster reach a maximum value and stable value. For a fixed acceleration, the uncertainty with different two-atom separations converges to a fixed value. Furthermore, we utilize weak measurement reversal to manipulate the entropic uncertainty. Our explorations may suggest a method of probing entanglement, acceleration effect and vacuum fluctuation effect with entropic uncertainty.

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The work is supported by the National Natural Science Foundation of China (61871205), the Innovation Project of Department of Education of Guangdong Province (2017KTSCX180), the Jiangmen Science and Technology Plan Project for Basic and Theoretical Research (2018JC01010), the Young Science and Technology Talent Growth Fund Project of Education Department of Guizhou Province of China (Qian Jiao He KY Zi[2018]426), the Major Special Fund Project of Research and Innovation for Qiannan Normal university for Nationalities of China (QNSY2018BS015), the Industrial Technology Foundation of Qiannan State of China (Qiannan Ke He Gong Zi (2017) 9 Hao) and the Scientific Research Foundation for High-level Talents of Qiannan Normal University for Nationalities (qnsyrc201716).

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Correspondence to Zhiming Huang.


Appendix A: Kossakowski–Lindblad master equation

The Kossakowski–Lindblad master equation that governs the atom evolution is [43,44,45]

$$\begin{aligned} \frac{\partial \rho (\tau )}{\partial \tau }=-i[H_{\text {eff}},\rho (\tau )]+{\mathcal {L}}[\rho (\tau )], \end{aligned}$$


$$\begin{aligned} H_{\text {eff}}=H_{S}-\frac{i}{2}\sum _{\alpha ,\beta =1}^{2}\sum _{i,j=1}^{3}H_{ij}^{(\alpha \beta )}\sigma _{i}^{(\alpha )}\sigma _{j}^{(\beta )}, \end{aligned}$$


$$\begin{aligned} {\mathcal {L}}[\rho ]=\frac{1}{2}\sum _{\alpha ,\beta =1}^{2}\sum _{i,j=1}^{3}C^{(\alpha \beta )}_{ij}\left[ 2\sigma _{j}^{(\beta )}\rho \sigma _{i}^{(\alpha )}-\sigma _{i}^{(\alpha )}\sigma _{j}^{(\beta )}\rho -\rho \sigma _{i}^{(\alpha )}\sigma _{j}^{(\beta )}\right] . \end{aligned}$$

\(C_{ij}^{(\alpha \beta )}\) and \(H_{ij}^{(\alpha \beta )}\) are determined by the electromagnetic field correlation functions:

$$\begin{aligned} G^{(\alpha \beta )}_{mn}(\tau -\tau ') =\langle E_{m}(\tau ,{\mathbf {x}}_{\alpha }) E_{n}(\tau ',{\mathbf {x}}_\beta ) \rangle \;. \end{aligned}$$

\({\mathcal {G}}^{(\alpha \beta )}_{mn}(\omega )\) and \({\mathcal {K}}^{(\alpha \beta )}_{mn}(\omega )\) denote Fourier and Hilbert transforms, respectively, defined as

$$\begin{aligned}&{\mathcal {G}}^{(\alpha \beta )}_{mn}(\lambda )=\int _{-\infty }^{\infty }\mathrm{{d}}{\varDelta }\tau e^{i \lambda {\varDelta }\tau }G^{(\alpha \beta )}_{mn}({\varDelta }\tau ), \end{aligned}$$
$$\begin{aligned}&{\mathcal {K}}^{(\alpha \beta )}_{mn}(\lambda )=\frac{P}{\pi i}\int _{-\infty }^{\infty }\mathrm{{d}}\omega \frac{{\mathcal {G}}^{(\alpha \beta )}_{mn}(\omega )}{\omega -\lambda }, \end{aligned}$$

where P is the principal value. The Kossakowski matrix \(C_{ij}^{(\alpha \beta )}\) can be written explicitly as

$$\begin{aligned} C_{ij}^{(\alpha \beta )} = A^{(\alpha \beta )}\delta _{ij} -iB^{(\alpha \beta )}\epsilon _{ijk}\,\delta _{3k} -A^{(\alpha \beta )}\delta _{3i}\,\delta _{3j} , \end{aligned}$$


$$\begin{aligned} A^{(\alpha \beta )} ={\frac{1}{4}}\,[\,\mathcal{G}^{(\alpha \beta )}(\omega ) +{{{\mathcal {G}}}}^{(\alpha \beta )}(-\omega )] ,\nonumber \\ B^{(\alpha \beta )} ={\frac{1}{4}}\,[\,\mathcal{G}^{(\alpha \beta )}(\omega ) -{{{\mathcal {G}}}}^{(\alpha \beta )}(-\omega )] , \end{aligned}$$


$$\begin{aligned} {{{\mathcal {G}}}}^{(\alpha \beta )}(\omega ) =\sum _{m,n=1}^3\, d^{(\alpha )}_m d^{(\beta )*}_n\, {{\mathcal {G}}}^{(\alpha \beta )}_{mn}(\omega ) . \end{aligned}$$

Similarly, \(H^{(\alpha \beta )}_{ij}\) can be obtained by replacing \({{\mathcal {G}}}^{(\alpha \beta )}_{mn}\) with \({{\mathcal {K}}}^{(\alpha \beta )}_{mn}\) in the above expressions.

Appendix B: The relative calculation of correlation function

Substituting Eq. (8) into Eq. (9), the correlation functions of accelerating two-atom system can be computed [32]:

$$\begin{aligned} G^{(11)}_{mn}(x,x')=G^{(22)}_{mn}(x,x') =\frac{a^4}{16\pi ^2} \frac{1}{\sinh ^4\left( \frac{a{\varDelta }\tau }{2}-i\epsilon \right) }\delta _{mn} , \end{aligned}$$


$$\begin{aligned} G^{(\alpha \beta )}_{mn}(x,x') =&\frac{a^4}{16\pi ^2}\frac{1}{ \left[ \sinh ^2 \left( \frac{a{\varDelta }\tau }{2}-i\epsilon \right) -{\frac{a^2l^2}{4}} \right] ^3} \left\{ [\delta _{mn}+al{\varepsilon _{\alpha \beta 3}}(\textit{q}_{m} \textit{k}_{n}-\textit{q}_{n}{} \textit{k}_{m})]{\sinh ^2 \frac{a{\varDelta }\tau }{2}} \right. \end{aligned}$$
$$\begin{aligned}&+\,\frac{a^2l^2}{4}\left[ (\delta _{mn}-2\textit{q}_{m} \textit{q}_{n}){\cosh ^2 \frac{a{\varDelta }\tau }{2}} \left. +(\delta _{mn}-2\textit{q}_{m}{} \textit{q}_{n}- 2\textit{k}_{m}{} \textit{k}_{n}){\sinh ^2 \frac{a{\varDelta }\tau }{2}}\right] \right\} , \end{aligned}$$

for \(\alpha \ne \beta \), where \({\varDelta }\tau =\tau -\tau '\), \({\mathbf {k}}=(1, 0, 0)\) and \({\mathbf {q}}=(0, 0, 1)\). According to Eq. (23), the Fourier transforms of the above field correlation function are [32]

$$\begin{aligned} {{\mathcal {G}}}^{(11)}_{mn}(\lambda )=\mathcal{G}^{(22)}_{mn}(\lambda )=\frac{1}{6\pi }{\lambda }^3 \left( 1+\coth \frac{\pi \lambda }{a} \right) f^{(11)}(\lambda ,a) \delta _{mn}, \end{aligned}$$


$$\begin{aligned} f^{(11)}(\lambda ,a)=1+\frac{a^2}{\lambda ^2}, \end{aligned}$$


$$\begin{aligned} {{\mathcal {G}}}^{(\alpha \beta )}_{mn}(\lambda )=\frac{1}{6\pi } {\lambda }^3 \left( 1+\coth \frac{\pi \lambda }{a} \right) f^{(\alpha \beta )}_{mn}(\lambda ,a,l), \end{aligned}$$

for \(\alpha \ne \beta \),

$$\begin{aligned} f^{(12)}_{11}(\lambda ,a,l)=&f^{(21)}_{11}(\lambda ,a,l)=\frac{12}{{{\lambda ^3}{l^3}{{(4 + {a^2}{l^2})}^{5/2}}}}\nonumber \\&\times \left\{ 2\lambda l(1 + {a^2}{l^2}){(4 + {a^2}{l^2})^{1/2}}\cos \left( \frac{{2\lambda }}{a}{\sinh ^{ - 1}}\frac{{al}}{2} \right) \right. \nonumber \\&\left. -\,[4-4{\lambda ^2}{l^2} + {a^2}{l^2}(2-{\lambda ^2}{l^2}+{a^2}{l^2})]\sin \left( \frac{{2\lambda }}{a}{\sinh ^{-1}}\frac{{al}}{2} \right) \right\} , \end{aligned}$$
$$\begin{aligned} f^{(12)}_{22}(\lambda ,a,l) =&f^{(21)}_{22}(\lambda ,a,l)=\frac{3}{{{\lambda ^3}{l^3}{{(4 + {a^2}{l^2})}^{3/2}}}}\nonumber \\&\times \left[ \lambda l(2 + {a^2}{l^2}){(4 + {a^2}{l^2})^{1/2}}\cos \left( \frac{{2\lambda }}{a}{\sinh ^{ - 1}}\frac{{al}}{2} \right) \right. \nonumber \\&\left. +\,(-4+4{\lambda ^2}{l^2} + {a^2}{\lambda ^2}{l^4})\sin \left( \frac{{2\lambda }}{a}{\sinh ^{-1}}\frac{{al}}{2} \right) \right] , \end{aligned}$$
$$\begin{aligned} f^{(12)}_{33}(\lambda ,a,l) =&f^{(21)}_{33}(\lambda ,a,l)=-\frac{3}{{{\lambda ^3}{l^3}{{(4 + {a^2}{l^2})}^{5/2}}}}\nonumber \\&\times \left\{ \lambda l(16 + 2{a^2}{l^2}+{a^4}{l^4}){(4 + {a^2}{l^2})^{1/2}}\cos \left( \frac{{2\lambda }}{a}{\sinh ^{ - 1}}\frac{{al}}{2} \right) \right. \nonumber \\&\left. +\,[-32+{a^4}{\lambda ^2}{l^6} + 4{a^2}{l^2}(-5+{\lambda ^2}{l^2})]\sin \left( \frac{{2\lambda }}{a}{\sinh ^{-1}}\frac{{al}}{2} \right) \right\} , \end{aligned}$$
$$\begin{aligned} f^{(12)}_{13}(\lambda ,a,l) =&-f^{(12)}_{31}(\lambda ,a,l)=-f^{(21)}_{13}(\lambda ,a,l)\nonumber \\ =&f^{(21)}_{31}(\lambda ,a,l)=-\frac{6a}{{{\lambda ^3}{l^2}{{(4 + {a^2}{l^2})}^{5/2}}}}\nonumber \\&\times \left\{ \lambda l(-2 + {a^2}{l^2}){(4 + {a^2}{l^2})^{1/2}}\cos \left( \frac{{2\lambda }}{a}{\sinh ^{ - 1}}\frac{{al}}{2} \right) \right. \nonumber \\&\left. +\,[4+4{\lambda ^2}{l^2} + {a^2}{l^2}(4+{\lambda ^2}{l^2})]\sin \left( \frac{{2\lambda }}{a}{\sinh ^{-1}}\frac{{al}}{2} \right) \right\} , \end{aligned}$$

with other components being zero. In this article, assume that the magnitudes of the electric dipoles of the atoms are the same, i.e., \(|{\mathbf {d}}^{(1)}|=|{\mathbf {d}}^{(2)}|=|{\mathbf {d}}|\), but the orientations may be different. In the interaction representation, the first item of the evolution master equation (19) vanishes. According to Eqs. (26) and (27), we can obtain the coefficients \(C_{ij}^{(\alpha \beta )}\) (25):

$$\begin{aligned}&C_{ij}^{(11)}=C_{ij}^{(22)}=A_1\,\delta _{ij}-iB_1\epsilon _{ijk}\,\delta _{3k} -A_1\delta _{3i}\,\delta _{3j} , \end{aligned}$$
$$\begin{aligned}&C_{ij}^{(12)}=C_{ij}^{(21)}=A_2\,\delta _{ij}-iB_2\epsilon _{ijk}\,\delta _{3k} -A_2\delta _{3i}\,\delta _{3j}, \end{aligned}$$


$$\begin{aligned}&A_1=\frac{{\varGamma }}{4} f^{(11)}(\omega ,a)\coth {\frac{\pi \omega }{a}} ,\quad A_2=\frac{{\varGamma }}{4} \sum _{i,j=1}^3 f^{(12)}_{ij}(\omega ,a,l){\hat{d}}^{(1)}_i{\hat{d}}^{(2)*}_j\coth {\frac{\pi \omega }{a}} ,\nonumber \\&B_1=\frac{{\varGamma }}{4} f^{(11)}(\omega ,a) ,\quad B_2=\frac{{\varGamma }}{4} \sum _{i,j=1}^3 f^{(12)}_{ij}(\omega ,a,l){\hat{d}}^{(1)}_i{\hat{d}}^{(2)*}_j , \end{aligned}$$

with \({\varGamma }= \omega ^3 |{\mathbf {d}}|^2 /3\pi \) being the spontaneous emission rate and \({\hat{d}}^{(\alpha )}_{i}\) being a unit vector defined as \({\hat{d}}^{(\alpha )}_{i}=d^{(\alpha )}_{i}/|{\mathbf {d}}|\).

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Huang, Z., Situ, H. Exploration of entropic uncertainty relation for two accelerating atoms immersed in a bath of electromagnetic field. Quantum Inf Process 18, 38 (2019). https://doi.org/10.1007/s11128-018-2151-z

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  • Entropic uncertainty
  • Mixedness
  • Dynamics
  • Fluctuating electromagnetic field
  • Weak measurement reversal