Comparing the quantum memory-assisted entropic uncertainties of spin-qubit in presence of spin-qutrit and spin-qubit quantum memories in Heisenberg spin chains

Abstract

Quantum memory-assisted entropic uncertainty relation (QMA-EUR) addresses that the entropic uncertainty of measured particle can be reduced by the aid of another particle acting as quantum memory. Especially, the measurement precision for two incompatible observables can be improved. In this paper, we have studied QMA-EUR in the qubit–qubit and qubit–qubit-spin chain models and examined quantum-memory-assisted entropic uncertainties (QMA-EUs) and their lower bounds (LBs) of the spin-qubit measured subsystem in the case of spin-qutrit and spin-qubit acting as the quantum memories, respectively. The numerical results show that spin-qutrit memory subsystems can effectively suppress the amounts of QMA-EUs and LBs of the spin-qubit measurement subsystem, which indicates that it can broke the limits of entropic uncertainty relationship without quantum memory subsystem. But, the suppression effects are very different in both antiferromagnetic and ferromagnetic spin coupling cases. In general, one can get lower values of QMA-EUs and LBs by adjusting the thermal equilibrium temperature and relevant coupling parameters in the case of antiferromagnetic spin coupling case. Furthermore, we compared the regulation effects of reducing the values of QMA-EUs and LBs when the quantum memory subsystem is severed by spin-qutrit and spin-qubit, respectively, in the same model parameters condition. It is found that effects of spin-qutrit memory subsystem on reducing QMA-EUs and LBs are inferior to that of spin-qubit memory one. This comparative analysis result indicates that the dimensions of quantum measurement subsystem and quantum memory subsystem have a significant effect on the reducing the QMA-EUs and LBs in the entropic uncertainty game.

Appendices

Appendix A: The nonzero elements in Eq.(6)

The nonzero elements of the density matrix of hybrid qubit–qutrit spin chain model in a thermal equilibrium at temperature T in Eq.(6) are given by

$$\begin{aligned} \rho _{11}^{(1)}= & {} \exp {\left( -\frac{E_1^{(1)}}{T}\right) }\nonumber \\ \rho _{22}^{(1)}= & {} |C_{2}|^{2}A_{1}^{2}\exp {\left( -\frac{E_{2}^{(1)}}{T}\right) }+|C_{3}|^{2}B_{1}^{2}\exp {\left( -\frac{E_{3}^{(1)}}{T}\right) }\nonumber \\ \rho _{33}^{(1)}= & {} |C_{4}|^{2}A_{2}^{2}\exp {\left( -\frac{E_{4}^{(1)}}{T}\right) }+|C_{5}|^{2}B_{2}^{2}\exp {\left( -\frac{E_{5}^{(1)}}{T}\right) }\nonumber \\ \rho _{44}^{(1)}= & {} |C_{2}|^{2}\exp {\left( -\frac{E_{2}^{(1)}}{T}\right) }+|C_{3}|^{2}\exp {\left( -\frac{E_{3}^{(1)}}{T}\right) }\nonumber \\ \rho _{55}^{(1)}= & {} |C_{4}|^{2}\exp {\left( -\frac{E_{4}^{(1)}}{T}\right) }+|C_{5}|^{2}\exp {\left( -\frac{E_{5}^{(1)}}{T}\right) }\nonumber \\ \rho _{66}^{(1)}= & {} \exp {\left( -\frac{E_6}{T}\right) }\nonumber \\ \rho _{24}^{(1)}= & {} |C_{2}|^{2}A_{1}\exp {\left( i\theta -\frac{E_{2}^{(1)}}{T}\right) }+|C_{3}|^{2}B_{1}\exp {\left( i\theta -\frac{E_{3}^{(1)}}{T}\right) }\nonumber \\ \rho _{35}^{(1)}= & {} |C_{4}|^{2}A_{2}\exp {\left( i\theta -\frac{E_{4}^{(1)}}{T}\right) }+|C_{5}|^{2}B_{2}\exp {\left( i\theta -\frac{E_{5}^{(1)}}{T}\right) }\nonumber \\ \rho _{42}^{(1)}= & {} \rho _{24}^{*},~~~\rho _{53}=\rho _{35}^{*} \end{aligned}$$

(20)

The partition function Z is obtained by summing the diagonal elements as $Z=\rho _{11}^{(1)}+\rho _{22}^{(1)}+\rho _{33}^{(1)}+\rho _{44}^{(1)}+\rho _{55}^{(1)}+\rho _{66}^{(1)}$, and some symbols are given by

$$\begin{aligned} A_{1}= & {} \frac{\sin \theta _{1}+1}{\cos \theta _{1}},~~~B_{1}=\frac{\sin \theta _{1}-1}{\cos \theta _{1}}\nonumber \\ A_{2}= & {} \frac{\sin \theta _{2}+1}{\cos \theta _{2}},~~~B_{2}=\frac{\sin \theta _{2}-1}{\cos \theta _{2}} \end{aligned}$$

(21)

$C_{2}$, $C_{3}$, $C_{4}$, $C_{5}$ are the normalization coefficients for quantum states of $|\psi _{2}\rangle$, $|\psi _{3}\rangle$, $|\psi _{4}\rangle$, $|\psi _{5}\rangle$ (the normalization coefficients for $|\psi _{1}\rangle$, $|\psi _{6}\rangle$ are equal to 1), which are given by

$$\begin{aligned} C_{2}= & {} \frac{2\sqrt{MN}}{\sqrt{(B-3b+\kappa {J}+\varLambda ^{-})^2+4MN}}\nonumber \\ C_{3}= & {} \frac{2\sqrt{MN}}{\sqrt{(B-3b+\kappa {J}-\varLambda ^{-})^2+4MN}}\nonumber \\ C_{4}= & {} \frac{2\sqrt{MN}}{\sqrt{(B-3b-\kappa {J}+\varLambda ^{+})^2+4MN}}\nonumber \\ C_{5}= & {} \frac{2\sqrt{MN}}{\sqrt{(B-3b-\kappa {J}-\varLambda ^{+})^2+4MN}} \end{aligned}$$

(22)

and

$$\begin{aligned} \theta _{1}= & {} \arctan \frac{B-3b+\kappa {J}}{2\sqrt{MN}},~~~\theta _{2}=\arctan \frac{B-3b-\kappa {J}}{2\sqrt{MN}}\nonumber \\ \theta= & {} \arctan \frac{D_{z}}{J} \end{aligned}$$

(23)

$E_{i}^{(1)}$ $(i=1,2,3,4,5,6)$ are eigenenergies of $H_{AB}^{(1)}$ in Eq.(2), which are given by

$$\begin{aligned} E_{1}= & {} \kappa {J}+2B,\nonumber \\ E_{2}= & {} \frac{B+b-\kappa {J}+\varLambda ^{+}}{2},~~~~~E_{3}=\frac{B+b-\kappa {J}-\varLambda ^{+}}{2}\nonumber \\ E_{4}= & {} \frac{-B-b-\kappa {J}+\varLambda ^{-}}{2},~~~E_{5}=\frac{-B-b-\kappa {J}-\varLambda ^{-}}{2}\nonumber \\ E_{6}= & {} \kappa {J}-2B \end{aligned}$$

(24)

with

$$\begin{aligned} \varLambda ^{\pm }=\sqrt{\left( B-3b\pm \kappa {J}\right) ^2+4MN} \end{aligned}$$

(25)

and $M=\sqrt{2}(J+iD_{z})$ and $N=M^*=\sqrt{2}(J-iD_{z})$.

Appendix B: The nonzero elements in Eq.(8)

The nonzero elements of the density matrix of qubit–qubit-spin chain model in a thermal equilibrium at temperature T in Eq.(8) are given by

$$\begin{aligned} \rho _{11}^{(2)}= & {} \exp {\left( -\frac{E_1^{(2)}}{T}\right) },~~~~~\rho _{44}=\exp {\left( -\frac{E_4^{(2)}}{T}\right) }\nonumber \\ \rho _{22}^{(2)}= & {} C_{0}^2A_{0}^{2}\exp {\left( -\frac{E_{2}^{(2)}}{T}\right) }+C_{0}^2B_{0}^{2}\exp {\left( -\frac{E_{3}^{(2)}}{T}\right) }\nonumber \\ \rho _{33}^{(2)}= & {} C_{0}^2A_{0}^{2}\exp {\left( -\frac{E_{3}^{(2)}}{T}\right) }+C_{0}^2B_{0}^{2}\exp {\left( -\frac{E_{2}^{(2)}}{T}\right) }\nonumber \\ \rho _{23}^{(2)}= & {} C_{0}\exp {\left( i\theta -\frac{E_{3}^{(2)}}{T}\right) }-C_{0}\exp {\left( i\theta -\frac{E_{2}^{(2)}}{T}\right) }\nonumber \\ \rho _{23}^{(2)}= & {} C_{0}\exp {\left( -i\theta -\frac{E_{3}^{(2)}}{T}\right) }-C_{0}\exp {\left( -i\theta -\frac{E_{2}^{(2)}}{T}\right) } \end{aligned}$$

(26)

with

$$\begin{aligned} C_{0}= & {} \frac{1}{2}\left( {\sqrt{J^2+D_{z}^{2}}}/{\sqrt{b^2+J^2+D_{z}^{2}}}\right) \end{aligned}$$

(27)

$$\begin{aligned} A_{0}= & {} \frac{\sin \theta _{1}+1}{\cos \theta _{1}},~~~B_{0}=\frac{\sin \theta _{0}-1}{\cos \theta _{1}} \end{aligned}$$

(28)

$$\begin{aligned} \theta _{0}= & {} \arctan \frac{b}{\sqrt{(J+iD_{z})(J-iD_{z})}},~~~~~\theta =\arctan \frac{D_{z}}{J} \end{aligned}$$

(29)

$E_{i}^{(2)}$ $(i=1,2,3,4)$ are eigenenergies of $H_{AB}^{(2)}$ in Eq.(5), which are given by

$$\begin{aligned} E_{1}^{(2)}= & {} +\kappa {J}+2B,~~~E_{2}^{(2)}=-\kappa {J}-2G\nonumber \\ E_{3}^{(2)}= & {} -\kappa {J}+2G,~~~E_{4}^{(2)}=+\kappa {J}-2B \end{aligned}$$

(30)

with

$$\begin{aligned} G=\sqrt{b^2+D_{z}^{z}+J^{2}} \end{aligned}$$

(31)