Exploring the influence of intrinsic decoherence on residual entanglement and tripartite uncertainty bound in XXZ Heisenberg chain model

Abstract

The uncertainty principle restricts our capacity to precisely predict the measurement outcome of two incompatible observables. By taking into account two more particles as the quantum memories that correlate with the observed particle, the tripartite entropic uncertainty bound can be promoted. This paper discusses the impact of intrinsic decoherence on the dynamics of tripartite uncertainty bound (TUB) and residual entanglement in a three-qubit XXZ Heisenberg chain coupled to Dzyaloshinskii–Moriya interaction (DMI) and exposed to a uniform magnetic field; we consider the case where both interactions are working in the z-direction. To identify residual entanglement, we employ negativity (N) as an entanglement measure. The impact of different system parameters on the dynamics of residual entanglement, bipartite negativity, and TUB is discussed. We highlight the relationship between these entanglement measures (residual entanglement and bipartite entanglement) and the TUB. Two cases of initial state for the nearest-neighbor particle (A and B) are considered in this work: the maximally entangled state (Bell-state) and the separable state. Regarding the maximal entangled state, it has been observed that the negative effect of magnetic field is significantly stronger than the DMI in the presence of decoherence, and that it has the ability to completely destroy the entanglements and maximize TUB value. Moreover, it has been ascertained that we can improve the residual entanglement and TUB by controlling the decoherence rate. As for the separable initial state, it has been found that the DMI in this case stimulates the effect of intrinsic decoherence, which can highlight the entanglement sudden death phenomenon and reduce the accuracy of the prediction.

Graphical abstract

Appendix

(Milburn’s Solution):

Firstly, the matrix representation of the initial state density matrix (Eq. 14) is

$$\begin{aligned}{} & {} {\hat{\rho }}_{ABC}(0)\nonumber \\{} & {} =\left( \begin{array}{cccccccc} p &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \sqrt{p(1-p)} &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ \sqrt{p(1-p)} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1-p &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ \end{array} \right) .\nonumber \\ \end{aligned}$$

(47)

Also, the matrix form of the eigenvectors is:

$$\begin{aligned} \begin{aligned}&|\psi _{1}\rangle =\frac{1}{\sqrt{2}}\left( \begin{array}{c} 0 \\ \frac{D_z-i}{D_z+i} \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{array} \right) \quad \quad |\psi _{2}\rangle =\frac{1}{\sqrt{2}}\left( \begin{array}{c} 0 \\ \frac{D_z-i}{D_z+i} \\ 0 \\ 0 \\ 1\\ 0 \\ 0 \\ 0 \\ \end{array} \right) \\&|\psi _{3}\rangle = \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ \end{array} \right) \quad \quad |\psi _{4}\rangle = \left( \begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{array} \right) \\ {}&|\psi _{5}\rangle = \mu _+\left( \begin{array}{c} 0 \\ 0 \\ 0 \\ -\frac{D_z-i}{D_z+i} \\ 0 \\ -\frac{i (J+Q)}{2(D_z+i)} \\ 1 \\ 0 \\ \end{array} \right) \quad \quad |\psi _{6}\rangle = \mu _+\left( \begin{array}{c} 0 \\ \frac{-(D_z-i) }{D_z+i} \\ -\frac{i (J+Q) }{2 (D_z+i)} \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{array} \right) \\&|\psi _{7}\rangle = \mu _-\left( \begin{array}{c} 0 \\ 0 \\ 0 \\ -\frac{D_z-i}{D_z+i} \\ 0 \\ -\frac{i (J-Q)}{2(D_z+i)} \\ 1 \\ 0 \\ \end{array} \right) \quad \quad |\psi _{8}\rangle = \mu _-\left( \begin{array}{c} 0 \\ \frac{-(D_z-i) }{D_z+i} \\ -\frac{i (J-Q) }{2 (D_z+i)} \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{array} \right) . \end{aligned}\nonumber \\ \end{aligned}$$

(48)

Using eigenvectors (Eq. 48) of the Hamiltonian and the initial state (Eq. 47) together with the definition of dynamical density matrix (Eq. 12), the components of $\delta _{mn}$ ($\delta _{mn}=\langle \psi _m| \rho _{ABC}(0) |\psi _n\rangle$) can be written as:

$$\begin{aligned} \begin{aligned} \delta _{11}&=\frac{1-p}{2}\\ \delta _{14}&=\delta _{41}=\frac{\sqrt{p(1-p)}}{2}\\ \delta _{15}&= \delta _{51}=\frac{1-p}{\sqrt{2}}\mu _+ \\ \delta _{44}&=p \delta _{17}=\delta _{71}=\frac{1-p}{\sqrt{2}}\mu _- \\ \delta _{45}&=\delta _{54}=\sqrt{p(1-p)}\mu _+ \\ \delta _{55}&= (1-p)\mu _+^{2} \delta _{47}=\delta _{74}=\sqrt{p(1-p)}\mu _- \\ \delta _{57}&=\delta _{75}=(1-p)\mu _+\mu _- \\ \delta _{77}&= (1-p)\mu _-^{2}. \end{aligned} \end{aligned}$$

(49)

Furthermore, we can obtain the matrix representation for all $\psi _{m,n}$, by using these eigenvectors and $\psi _{m,n}=|\psi _m\rangle \langle \psi _n|$ directly.

On the other hand, we can deduce the corresponding components of $\beta _{mn}$ by using the eigenvalues of H (Eq. 6) such that:

$$\begin{aligned} \begin{aligned}&\beta _{11}= =\beta _{44}=\beta _{55}=\beta _{77}=1 \\&\beta _{14}=\beta _{41}^*=e^{(i\eta -\frac{\gamma }{2}\eta ^2)t} \\&\beta _{15}= \beta _{51}^*=e^{-(i\xi _++\frac{\gamma }{2}\xi _+^2)t} \\&\beta _{17}=\beta _{71}^*=e^{-(i\xi _-+\frac{\gamma }{2}\xi _-^2)t} \\&\beta _{45}=\beta _{54}^*=e^{-\left( i(\eta +\xi _+)+\frac{\gamma }{2}(\eta +\xi _+)^2\right) t} \\&\beta _{47}= \beta _{74}^*=e^{-\left( i(\eta +\xi _-)+\frac{\gamma }{2}(\eta +\xi _-)^2\right) t} \\&\beta _{57}=\beta _{75}^*=e^{\left( i(\xi _+-\xi _-)-\frac{\gamma }{2}(\xi _+-\xi _-)^2\right) t}. \end{aligned} \end{aligned}$$

(50)

Finally, the general solution is

$$\begin{aligned} {\hat{\rho }}_{ABC}(t)=\sum _{i,j=1,4,5,7}\beta _{ij}\delta _{ij}\psi _{ij}. \end{aligned}$$

(51)