Quantitative analysis of decoherence of entangled microwave signals in free space

Abstract

Entangled microwave signals are the new quantum information resource of microwave frequency. Based on the evolution model of two-mode squeezed vacuum state with Fokker–Planck equation, we consider the absorption of atmospheric gas and the attenuation of cloud, rain and fog in the typical environment, and then investigate decoherence of entangled microwave signals. To quantify the entanglement degree using logarithmic negativity, we quantitatively discuss the relation of the entanglement degree and propagation distance and estimate the effective operating distance of entangled microwave signals in free space. Results demonstrate that entangled microwave signals can still maintain a higher entanglement degree when the propagation distance is in the order of 10 km, which prove the availability.

Appendix

Here we provide the exact derivation and calculation process of logarithmic negativity given in Eq. (9).

Substituting the two evolutive formulas $\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} )$ and $\sinh (2r)e^{ - \gamma t}$ into Eq. (8),

$$ \begin{aligned} \lambda_{1} & = \frac{1}{2}\left[ {\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} } \right] \\ \lambda_{1} & = \frac{1}{2}\left[ {\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) - \sinh (2r)e^{ - \gamma t} } \right] \\ \end{aligned} $$

Then, substituting them into Eq. (7),

When $\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} \ge 1$,

$$ F(\lambda_{1} ) = 0 $$

When $\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} < 1$,

$$ F(\lambda_{1} ) = - \log_{2} \left[ {\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} } \right] $$

When $\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) - \sinh (2r)e^{ - \gamma t} \ge 1$,

$$ F(\lambda_{2} ) = 0 $$

When $\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} < 1$,

$$ F(\lambda_{2} ) = - \log_{2} \left[ {\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} } \right] $$

Due to $\sinh (2r)e^{ - \gamma t}$ is always greater than 0, we have

$$ \cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} > \cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) - \sinh (2r)e^{ - \gamma t} $$

Then, substituting $F(\lambda_{1} )$ and $F(\lambda_{2} )$ into Eq. (6),

• 1. When $\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) - \sinh (2r)e^{ - \gamma t} \ge 1$,

  $$ E(\rho ) = 0 $$

• 2. When $\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} \ge 1$, $\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) - \sinh (2r)e^{ - \gamma t} < 1$, and

  $$ E(\rho ) = - \log_{2} \left[ {\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) - \sinh (2r)e^{ - \gamma t} } \right] $$

• 3. When $\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} < 1$,

  $$ \begin{aligned} & E(\rho ) = - \log_{2} \left[ {\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) + \sinh (2r)e^{ - \gamma t} } \right] \\ & \quad \quad \quad - \,\log_{2} \left[ {\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} ) - \sinh (2r)e^{ - \gamma t} } \right] \\ & \quad \quad \quad = - \log_{2} \left\{ {\left[ {\cosh (2r)e^{ - \gamma t} + (2\bar{N} + 1)(1 - e^{ - \gamma t} )} \right]^{2} - \left[ {\sinh (2r)e^{ - \gamma t} } \right]^{2} } \right\} \\ \end{aligned} $$

Combine the above three situations, Eq. (9) is final logarithmic negativity expression.

As results of the given $r$, $\cosh (2r) \approx \sinh (2r)$. Therefore, calculation can be simplified.