Influence of Magnetic Field on Qutrit Teleportation under Intrinsic Decoherence

Abstract

We study qutrit teleportation through a qutrit xyz chain, in the presence of intrinsic decoherence and a non-homogeneous magnetic field. We study the effects of intrinsic phase change, magnetic field and entanglement of the initial state of the channel. It is observed that while the intrinsic phase change and the non-homogeneity of the magnetic field have adverse effects on the teleportation fidelity, the entanglement of the initial state of the channeled enhances the latter. Moreover, the intrinsic decoherence may remove the ripples from the time curve that is delivered by the Schrödinger channel.

1 Introduction

Quantum entanglement, a non-classical property of the physical systems, has been considered a resource to perform several information processing tasks which are not possible in the classical realm [1]. There has been a concerted effort to study the entanglement properties of the spin chains in the last decade; they have been used in several quantum information processes including quantum computations [2, 3], quantum communications [4] and have been considered as relatively realistic models to study quantum dots [5, 6], superconductivity [7] and phase transitions [8]. Spin chains, specifically the spin-$\frac{1}{2}$ones, have been also considered as appropriate media to perform quantum teleportation and their properties in the absence [9,10,11,12] or in the presence of the environment decoherence [13,14,15,16]. A few papers also consider the effect of the intrinsic decoherence, as was formulated by Milburn [17], on teleportation via the$\frac{1}{2}$-spin chains [13, 18]. It has been also reported that application of a homogeneous magnetic field may reduce the adverse effect of the intrinsic decoherence on the teleportation [13], while another research reports a similar effect, in a xxz chain, if a non-homogeneous magnetic field is applied [18].

Reports regarding qutrit teleportation are few and far between [19,20,21]. A reference reports qutrit state transfer between two cavities [19]. Probabilistic qutrit teleportation has been also discussed [20]. Teleportation of one – and two-qutrit systems, through a maximally entangled quantum channel of three-qutrit channel, has been also reported [21]. We have also considered the teleportation of a qutrit state through an xyz chain model, and have studied the effect of the entanglement of the initial state of the channel on that, previously [22]. Our aim in this work is to consider qutrit teleportation through a qutrit xyz chain, in the presence of the intrinsic decoherence; while a non-homogeneous magnetic field is also applied, to study its possible improving effects onthe teleportation, as has been reported in the case of spin-$\frac{1}{2}$ models [18].

2 Theoretical Model

We intend to study qutrit teleportation through a qutritxyzHeisenberg chain under non-homogeneous magnetic field. Considering a two-qutrit chain, its Hamiltonian is given by

$$ H={J}_X{S}_1^X\otimes {S}_2^X+{J}_Y{S}_1^Y\otimes {S}_2^Y+{J}_Z{S}_1^Z\otimes {S}_2^Z+\left(B+b\right)\left({S}_1^Z\otimes I\right)+\left(B-b\right)\left(I\otimes {S}_2^Z\right), $$

(1)

Where, J_i‘s are the interaction parameters, B is homogeneous magnetic field, b is the inhomogeneous magnetic field and the qutrit operators are given by [23].

$$ {S}^x=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}0& 1& 0\\ {}1& 0& 1\\ {}0& 1& 0\end{array}\right), $$

(2-a)

$$ {S}^y=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}0& -i& 0\\ {}i& 0& -i\\ {}0& i& 0\end{array}\right), $$

(2-b)

$$ {S}^z=\left(\begin{array}{ccc}1& 0& 0\\ {}0& 0& 0\\ {}0& 0& -1\end{array}\right). $$

(2-c)

The matrix representation of the Hamiltonian in the computational basis is given by

$$ H=\left(\begin{array}{ccccccccc}2+2B& 0& 0& 0& -1& 0& 0& 0& 0\\ {}0& B+b& 0& 2& 0& -1& 0& 0& 0\\ {}0& 0& -2+2b& 0& 2& 0& 0& 0& 0\\ {}0& 2& 0& -b+B& 0& 0& 0& -1& 0\\ {}-1& 0& 2& 0& 0& 0& 2& 0& -1\\ {}0& -1& 0& 0& 0& b-B& 0& 2& 0\\ {}0& 0& 0& 0& 2& 0& -2-2b& 0& 0\\ {}0& 0& 0& -1& 0& 2& 0& -b-B& 0\\ {}0& 0& 0& 0& -1& 0& 0& 0& 2-2B\end{array}\right), $$

(3)

whose eigenvalues are expressed by

$$ {E}_1=-{\left(5+{b}^2+{B}^2-2{A}_1\right)}^{\frac{1}{2}}, $$

(4-a)

$$ {E}_2=-{E}_1, $$

(4-b)

$$ {E}_3=-{\left(5+{b}^2+{B}^2+2{A}_1\right)}^{\frac{1}{2}}, $$

(4-c)

$$ {E}_4=-{E}_3, $$

(4-d)

$$ {\displaystyle \begin{array}{l}{E}_5= Roo{t}_1\Big[-48-16{b}^2+64{B}^2+\left(56-8{b}^2+16{B}^2+16{b}^2{B}^2\right)X\\ {}+\left(12+16{b}^2-16{B}^2\right){X}^2+\left(-18-4{b}^2-4{B}^2\right){X}^3+{X}^5=0\Big],\end{array}} $$

(4-e)

$$ {\displaystyle \begin{array}{l}{E}_6= Roo{t}_2\Big[-48-16{b}^2+64{B}^2+\left(56-8{b}^2+16{B}^2+16{b}^2{B}^2\right)X\\ {}+\left(12+16{b}^2-16{B}^2\right){X}^2+\left(-18-4{b}^2-4{B}^2\right){X}^3+{X}^5=0\Big],\end{array}} $$

(4-f)

$$ {\displaystyle \begin{array}{l}{E}_7= Roo{t}_3\Big[-48-16{b}^2+64{B}^2+\left(56-8{b}^2+16{B}^2+16{b}^2{B}^2\right)X\\ {}+\left(12+16{b}^2-16{B}^2\right){X}^2+\left(-18-4{b}^2-4{B}^2\right){X}^3+{X}^5=0\Big],\end{array}} $$

(4-g)

$$ {\displaystyle \begin{array}{l}{E}_8= Roo{t}_4\Big[-48-16{b}^2+64{B}^2+\left(56-8{b}^2+16{B}^2+16{b}^2{B}^2\right)X\\ {}+\left(12+16{b}^2-16{B}^2\right){X}^2+\left(-18-4{b}^2-4{B}^2\right){X}^3+{X}^5=0\Big],\end{array}} $$

(4-h)

$$ {\displaystyle \begin{array}{l}{E}_9= Roo{t}_5\Big[-48-16{b}^2+64{B}^2+\left(56-8{b}^2+16{B}^2+16{b}^2{B}^2\right)X\\ {}+\left(12+16{b}^2-16{B}^2\right){X}^2+\left(-18-4{b}^2-4{B}^2\right){X}^3+{X}^5=0\Big].\end{array}} $$

(4-i)

And the corresponding eigenfunctions are

$$ {\displaystyle \begin{array}{l}\mid {\psi}_1\left\rangle =\right\{\left(1-{A}_1^{\ast }+{B}^{\ast}\left({B}^{\ast }-{E}_2^{\ast}\right)\right)\mid {1}_{\widehat{1}},{0}_{\widehat{2}}\left\rangle +{E}_2^{\ast}\right(2+{A}_2^{\ast}\left(-{B}^{\ast }+{\left(b+{E}_2\right)}^{\ast}\right)\mid {0}_{\widehat{1}},{1}_{\widehat{2}}\Big\rangle \\ {}+{A}_2^{\ast }{E}_2^{\ast}\mid {1}_{\widehat{1}},{2}_{\widehat{2}}\left\rangle +{E}_2^{\ast}\mid {2}_{\widehat{1}},{1}_{\widehat{2}}\right\rangle \left\}\right(1+{\left|{A}_2\right|}^2+{\left|\left(1-{A}_1+B\left(B-{E}_2\right)\right)\right|}^2{\left({\left|{E}_2\right|}^2\right)}^{-1}\\ {}+{\left|\left(2+{A}_2\left(b-B+{E}_2\right)\right)\right|}^2\Big){}^{-\frac{1}{2}},\end{array}} $$

(5-a)

$$ {\displaystyle \begin{array}{l}\mid {\psi}_2\left\rangle =\right\{\left(-1+{\left({A}_1-B\left(B+{E}_2\right)\right)}^{\ast}\right)\mid {1}_{\widehat{1}},{0}_{\widehat{2}}\left\rangle +{E}_2^{\ast}\left(2+{A}_3^{\ast}\left({b}^{\ast }-{\left(B+{E}_2\right)}^{\ast}\right)\right)\mid {0}_{\widehat{1}},{1}_{\widehat{2}}\right\rangle \\ {}+{A}_3^{\ast }{E}_2^{\ast}\mid {1}_{\widehat{1}},{2}_{\widehat{2}}\left\rangle +{E}_2^{\ast}\mid {2}_{\widehat{1}},{1}_{\widehat{2}}\right\rangle \left\}\right(1+{\left|{A}_3\right|}^2+{\left|\left(-1+{A}_1-B\left(B+{E}_2\right)\right)\right|}^2{\left({\left|{E}_2\right|}^2\right)}^{-1}\\ {}+{\left|\left(2+{A}_3\left(b-B+{E}_2\right)\right)\right|}^2\Big){}^{-\frac{1}{2}},\end{array}} $$

(5-b)

$$ {\displaystyle \begin{array}{l}\mid {\psi}_3\left\rangle =\right\{\left(1+{\left({A}_1+B\left(B-{E}_4\right)\right)}^{\ast}\right)\mid {1}_{\widehat{1}},{0}_{\widehat{2}}\left\rangle +{E}_4^{\ast}\left(2+{A}_4^{\ast}\left(-{B}^{\ast }+{\left(b+{E}_4\right)}^{\ast}\right)\right)\mid {0}_{\widehat{1}},{1}_{\widehat{2}}\right\rangle \\ {}+{A}_4^{\ast }{E}_4^{\ast}\mid {1}_{\widehat{1}},{2}_{\widehat{2}}\left\rangle +{E}_4^{\ast}\mid {2}_{\widehat{1}},{1}_{\widehat{2}}\right\rangle \left\}\right(1+{\left|{A}_4\right|}^2+{\left|\left(1+{A}_1+B\left(B-{E}_4\right)\right)\right|}^2{\left({\left|{E}_4\right|}^2\right)}^{-1}\\ {}+{\left|\left(2+{A}_4\left(b-B+{E}_4\right)\right)\right|}^2\Big){}^{-\frac{1}{2}},\end{array}} $$

(5-c)

$$ {\displaystyle \begin{array}{l}\mid {\psi}_4\left\rangle =\right\{-\left(1+{A}_1^{\ast }+{\left(B\left(B+{E}_4\right)\right)}^{\ast}\right)\mid {1}_{\widehat{1}},{0}_{\widehat{2}}\left\rangle +{E}_4^{\ast}\left(2+{A}_5^{\ast}\left({b}^{\ast }-{\left(B+{E}_4\right)}^{\ast}\right)\right)\mid {0}_{\widehat{1}},{1}_{\widehat{2}}\right\rangle \\ {}+{A}_5^{\ast }{E}_4^{\ast}\mid {1}_{\widehat{1}},{2}_{\widehat{2}}\left\rangle +{E}_4^{\ast}\mid {2}_{\widehat{1}},{1}_{\widehat{2}}\right\rangle \left\}\right(1+{\left|{A}_5\right|}^2+{\left|\left(1+{A}_1+B\left(B+{E}_4\right)\right)\right|}^2{\left({\left|{E}_4\right|}^2\right)}^{-1}\\ {}+{\left|\left(2-{A}_5\left(-b+B+{E}_4\right)\right)\right|}^2\Big){}^{-\frac{1}{2}},\end{array}} $$

(5-d)

$$ {\displaystyle \begin{array}{l}\mid {\psi}_5\left\rangle =\right\{\left(-9-2{A}_6^{\ast }+2{A}_7^{\ast }+2\left({B}^{\ast }-1\right){E}_5^{\ast }+{\left({E}_5^{\ast}\right)}^2\right)\mid {0}_{\widehat{1}},{0}_{\widehat{2}}\left\rangle +\left(-2-{A}_6^{\ast}\right)\mid {0}_{\widehat{1}},{2}_{\widehat{2}}\right\rangle \\ {}+\left(2-2{B}^{\ast }-{E}_5^{\ast}\right)\mid {1}_{\widehat{1}},{1}_{\widehat{2}}\left\rangle +\left(-2+{A}_7^{\ast}\right)\mid {2}_{\widehat{1}},{0}_{\widehat{2}}\right\rangle +\mid {2}_{\widehat{1}},{2}_{\widehat{2}}\left\rangle \right\}\\ {}{\left(1+{\left|-2-{A}_6\right|}^2+{\left|-2+{A}_7\right|}^2+{\left|2-2B-{E}_5\right|}^2+{\left|-9-2{A}_6+2{A}_7+2\left(B-1\right){E}_5+{\left({E}_5\right)}^2\right|}^2\right)}^{-\frac{1}{2}},\end{array}} $$

(5-e)

$$ {\displaystyle \begin{array}{l}\mid {\psi}_6\left\rangle =\right\{\left(-9-2{A}_8^{\ast }+2{A}_9^{\ast }+2\left({B}^{\ast }-1\right){E}_6^{\ast }+{\left({E}_6^{\ast}\right)}^2\right)\mid {0}_{\widehat{1}},{0}_{\widehat{2}}\left\rangle +\left(-2-{A}_8^{\ast}\right)\mid {0}_{\widehat{1}},{2}_{\widehat{2}}\right\rangle \\ {}+\left(2-2{B}^{\ast }-{E}_6^{\ast}\right)\mid {1}_{\widehat{1}},{1}_{\widehat{2}}\left\rangle +\left(-2+{A}_9^{\ast}\right)\mid {2}_{\widehat{1}},{0}_{\widehat{2}}\right\rangle +\mid {2}_{\widehat{1}},{2}_{\widehat{2}}\left\rangle \right\}\\ {}{\left(1+{\left|-2-{A}_8\right|}^2+{\left|-2+{A}_9\right|}^2+{\left|2-2B-{E}_6\right|}^2+{\left|-9-2{A}_8+2{A}_9+2\left(B-1\right){E}_6+{\left({E}_6\right)}^2\right|}^2\right)}^{-\frac{1}{2}},\end{array}} $$

(5-f)

$$ {\displaystyle \begin{array}{l}\mid {\psi}_7\left\rangle =\right\{\left(-9-2{A}_{10}^{\ast }+2{A}_{11}^{\ast }+2\left({B}^{\ast }-1\right){E}_7^{\ast }+{\left({E}_7^{\ast}\right)}^2\right)\mid {0}_{\widehat{1}},{0}_{\widehat{2}}\left\rangle +\left(-2-{A}_{10}^{\ast}\right)\mid {0}_{\widehat{1}},{2}_{\widehat{2}}\right\rangle \\ {}+\left(2-2{B}^{\ast }-{E}_7^{\ast}\right)\mid {1}_{\widehat{1}},{1}_{\widehat{2}}\left\rangle +\left(-2+{A}_{11}^{\ast}\right)\mid {2}_{\widehat{1}},{0}_{\widehat{2}}\right\rangle +\mid {2}_{\widehat{1}},{2}_{\widehat{2}}\left\rangle \right\}\\ {}{\left(1+{\left|-2-{A}_{10}\right|}^2+{\left|-2+{A}_{11}\right|}^2+{\left|2-2B-{E}_7\right|}^2+{\left|-9-2{A}_{10}+2{A}_{11}+2\left(B-1\right){E}_7+{\left({E}_7\right)}^2\right|}^2\right)}^{-\frac{1}{2}},\end{array}} $$

(5-g)

$$ {\displaystyle \begin{array}{l}\mid {\psi}_8\left\rangle =\right\{\left(-9-2{A}_{12}^{\ast }+2{A}_{13}^{\ast }+2\left({B}^{\ast }-1\right){E}_8^{\ast }+{\left({E}_8^{\ast}\right)}^2\right)\mid {0}_{\widehat{1}},{0}_{\widehat{2}}\left\rangle +\left(-2-{A}_{12}^{\ast}\right)\mid {0}_{\widehat{1}},{2}_{\widehat{2}}\right\rangle \\ {}+\left(2-2{B}^{\ast }-{E}_8^{\ast}\right)\mid {1}_{\widehat{1}},{1}_{\widehat{2}}\left\rangle +\left(-2+{A}_{13}^{\ast}\right)\mid {2}_{\widehat{1}},{0}_{\widehat{2}}\right\rangle +\mid {2}_{\widehat{1}},{2}_{\widehat{2}}\left\rangle \right\}\\ {}{\left(1+{\left|-2-{A}_{12}\right|}^2+{\left|-2+{A}_{13}\right|}^2+{\left|2-2B-{E}_8\right|}^2+{\left|-9-2{A}_{12}+2{A}_{13}+2\left(B-1\right){E}_8+{\left({E}_8\right)}^2\right|}^2\right)}^{-\frac{1}{2}},\end{array}} $$

(5-h)

$$ {\displaystyle \begin{array}{l}\mid {\psi}_9\left\rangle =\right\{\left(-9-2{A}_{14}^{\ast }+2{A}_{15}^{\ast }+2\left({B}^{\ast }-1\right){E}_9^{\ast }+{\left({E}_9^{\ast}\right)}^2\right)\mid {0}_{\widehat{1}},{0}_{\widehat{2}}\left\rangle +\left(-2-{A}_{14}^{\ast}\right)\mid {0}_{\widehat{1}},{2}_{\widehat{2}}\right\rangle \\ {}+\left(2-2{B}^{\ast }-{E}_9^{\ast}\right)\mid {1}_{\widehat{1}},{1}_{\widehat{2}}\left\rangle +\left(-2+{A}_{15}^{\ast}\right)\mid {2}_{\widehat{1}},{0}_{\widehat{2}}\right\rangle +\mid {2}_{\widehat{1}},{2}_{\widehat{2}}\left\rangle \right\}\\ {}{\left(1+{\left|-2-{A}_{14}\right|}^2+{\left|-2+{A}_{15}\right|}^2+{\left|2-2B-{E}_9\right|}^2+{\left|-9-2{A}_{14}+2{A}_{15}+2\left(B-1\right){E}_9+{\left({E}_9\right)}^2\right|}^2\right)}^{-\frac{1}{2}},\end{array}} $$

(5-i)

Where, we have defined

$$ {A}_1={\left(\left(4+{b}^2\right)\left(1+{B}^2\right)\right)}^{\frac{1}{2}}. $$

(6-a)

$$ {A}_2=\left({A}_1-4+b\left({E}_2-b\right)\right){\left(2{E}_2\right)}^{-1}, $$

(6-b)

$$ {A}_3=\left(-{A}_1+4+b\left({E}_2+b\right)\right){\left(2{E}_2\right)}^{-1}, $$

(6-c)

$$ {A}_4=\left(-{A}_1-4+b\left({E}_4-b\right)\right){\left(2{E}_4\right)}^{-1}, $$

(6-d)

$$ {A}_5=\left({A}_1+4+b\left({E}_4+b\right)\right){\left(2{E}_4\right)}^{-1}, $$

(6-f)

$$ {A}_6=4\left(-2+b+B\right){\left(2-2b+{E}_5\right)}^{-1}, $$

(6-f)

$$ {A}_7=4\left(2+b-B\right){\left(2+2b+{E}_5\right)}^{-1}, $$

(6-g)

$$ {A}_8=4\left(-2+b+B\right){\left(2-2b+{E}_6\right)}^{-1}, $$

(6-h)

$$ {A}_9=4\left(2+b-B\right){\left(2+2b+{E}_6\right)}^{-1}, $$

(6-i)

$$ {A}_{10}=4\left(-2+b+B\right){\left(2-2b+{E}_7\right)}^{-1}, $$

(6-j)

$$ {A}_{11}=4\left(2+b-B\right){\left(2+2b+{E}_7\right)}^{-1}, $$

(6-k)

$$ {A}_{12}=4\left(-2+b+B\right){\left(2-2b+{E}_8\right)}^{-1}, $$

(6-l)

$$ {A}_{13}=4\left(2+b-B\right){\left(2+2b+{E}_8\right)}^{-1}, $$

(6-m)

$$ {A}_{14}=4\left(-2+b+B\right){\left(2-2b+{E}_9\right)}^{-1}, $$

(6-n)

$$ {A}_{15}=4\left(2+b-B\right){\left(2+2b+{E}_9\right)}^{-1}, $$

(6-o)

3 Schrödinger and Milburn Evolution of an Initial Product State

The evolution under Schrödinger equation is given by

$$ {\rho}_{Sh}(t)=U(t)\rho (0){U}^{\dagger }(t) $$

(7)

where,

$$ U(t)={e}^{-i\frac{H\left(t-{t}_0\right)}{\hslash }}. $$

(8)

The evolution of the initial state under Millburn model is given by

$$ \frac{d\widehat{\rho}(t)}{dt}=\frac{1}{\eta}\left\{\left({e}^{\frac{- i\eta \widehat{H}}{\hslash }}\widehat{\rho}(0){e}^{\frac{i\eta \widehat{H}}{\hslash }}\right)-\widehat{\rho}(0)\right\}, $$

(9)

where, η aris the minimum phase change and ρ(0) the initial density matrix of the system [17]. Solving this equation we find [24].

$$ \widehat{\rho}{(t)}_M=\sum \limits_{mn}\exp \left[\frac{- t\eta}{2}{\left({E}_m-{E}_n\right)}^2- it\left({E}_m-{E}_n\right)\right]\left\langle {\psi}_m|\widehat{\rho}(0)|{\psi}_n\right\rangle \mid {\psi}_m\left\rangle \right\langle {\psi}_n\mid . $$

(10)

where, ∣ψ_n〉 and E_nare given in eqs. (5-a) to (5-i) and (4-a) to (4-i) respectively. Now considering the initial product state

$$ \mid {\psi}_0\left\rangle =\mid {0}_1,{1}_2\right\rangle . $$

(11)

The matrix representation is given by

$$ {\rho}_{Sh-P}(t)=\left(\begin{array}{ccccccccc}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& {\left|{I}_1\right|}^2& 0& {I}_2{I}_1^{\ast }& 0& {I}_3{I}_1^{\ast }& 0& {I}_4{I}_1^{\ast }& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& {I}_1{I}_2^{\ast }& 0& {\left|{I}_2\right|}^2& 0& {I}_3{I}_2^{\ast }& 0& {I}_4{I}_2^{\ast }& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& {I}_1{I}_3^{\ast }& 0& {I}_2{I}_3^{\ast }& 0& {\left|{I}_3\right|}^2& 0& {I}_4{I}_3^{\ast }& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& {I}_1{I}_4^{\ast }& 0& {I}_2{I}_4^{\ast }& 0& {I}_3{I}_4^{\ast }& 0& {\left|{I}_4\right|}^2& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right), $$

(12)

where, theI_i parameters are too complicated to write it down and they are saved in our computer program. Similarly, we find

$$ {\rho}_{M-P}=\left(\begin{array}{ccccccccc}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& {H}_1& 0& {H}_2& 0& {H}_3& 0& {H}_4& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& {H}_5& 0& {H}_6& 0& {H}_7& 0& {H}_8& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& {H}_9& 0& {H}_{10}& 0& {H}_{11}& 0& {H}_{12}& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& {H}_{13}& 0& {H}_{14}& 0& {H}_{15}& 0& {H}_{16}& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right), $$

(13)

where, the H_i parameters are too complicated to write it down and they are saved in our computer program.

4 Teleportation using Schrödinger and Milburn Channel Whose Initial State is a Product One

We consider the following incoming state to be teleported

$$ \mid {\varLambda}_{in}\Big\rangle =\frac{1}{\sqrt{3}}\left(|0\Big\rangle +|1\Big\rangle +|2\Big\rangle \right), $$

(14)

whose corresponding density matrix is expressed by

$$ {\rho}_{in}=\frac{1}{3}\left(\begin{array}{ccc}1& 1& 1\\ {}1& 1& 1\\ {}1& 1& 1\end{array}\right). $$

(15)

The outgoing state is given by [25].

$$ {\rho}_{out}=\sum \limits_{j=0}^8 Tr\left({E}^j\rho (t)\right)\left\{{\varGamma}^j{\rho}_{in}{\varGamma}^j\right\}, $$

(16)

where, Γ^j(j = 1, ..., 8) are Gell-Mann matrices [26] and the density matrices E^j_‘s represent the two-qutrit maximally entangled states (ME) [27] which are expressed in appendix 1 and 2, respectively.

The fidelity of the teleportation may be defined by [28].

$$ F\left({\rho}_{in},{\rho}_{out}\right)={\left[ Tr{\left({\rho_{in}}^{\frac{1}{2}}{\rho}_{out}{\rho_{in}}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\right]}^2. $$

(17)

Now using eqs. (11), (14) and (15) we find

$$ {F}_{Sh-P}=\frac{2}{27}\left[\left(4{I}_1+\left(1-{3}^{\frac{1}{2}}i\right){I}_2\right){I}_1^{\ast }+\left(\left(1+{3}^{\frac{1}{2}}i\right){I}_1+4{I}_2\right){I}_2^{\ast}\right] $$

(18)

and using eqs. (12), (14) and (15) also we find

$$ {F}_{M-P}=\frac{2}{27}\left({3}^{\frac{1}{2}}\left(\mathit{\operatorname{Im}}\left[{H}_2\right]-\mathit{\operatorname{Im}}\left[{H}_5\right]\right)+\mathit{\operatorname{Re}}\left[4{H}_1+{H}_2+{H}_5+4{H}_6\right]\right). $$

(19)

In Fig. 1, fidelities of Schrödinger and Milburn non-entangled channels as a function of time are plotted: It is observed that both start at the same value, but F_M − P decreases monotonically while F_Sh − P fluctuates in time. Moreover, it is observed that at specific periods of time, F_Sh − P gains larges values than that of Milburn fidelity and it is vice versa in specific periods of time.

Fig. 1

F_Sh − P (solid line) and F_M − P (dotted line) as a function oft; J_x = 1, J_y = 3, J_z = 2, b = 1, η = 1, B = 1

Figure 2, depicts the fidelity of Schrödinger non-entangled channel as a function of time in the presence of an inhomogeneous magnetic field (b). By increasing an inhomogeneous magnetic field, the time periods of the fidelity become shorter and the latter decreases. Also, in Fig. 3, the fidelity of the Schrödinger non-entangled channel as a function of time for three different values of homogeneous magnetic field (B) is plotted. Increasing the homogeneous magnetic field, does not have any appreciable effect on the fidelity; only the latter increases slightly.

Fig. 2

F_Sh − P as a function of t and b; J_x = 1, J_y = 3, J_z = 2, B = 10

Fig. 3

F_Sh − P (B = 3: thick solid line, B = 6: dotted line, B = 10: thin solid line) as a function of t; J_x = 1,J_y = 3, J_z = 2, b = 1, η = 1

In Fig. 4 we have plotted fidelity of the non-entangled Milburn channel as a function of time and the inhomogeneous magnetic field (b). Also Fig. 5, depicts the fidelity of the non-entangled Milburn channel as a function of time, for three different values of homogeneous magnetic field (B). It is deduced that fidelity increases as the homogenous or the inhomogeneous magnetic field are increased.

Fig. 4

F_M − P as a function of t and b; J_x = 1, J_y = 3, J_z = 2, η = 1, B = 10

Fig. 5

F_M − P (B = 3: thick solid line, B = 6: dotted line, B = 10: thin solid line) as a function of t; J_x = 1, J_y = 3, J_z = 2, b = 1, η = 1

In Fig. 6, the fidelity of non-entangled Milburn channel is shown as a function of time for three different values of the decoherence phase (η). It is observed that by increasing (η), the fidelity first decreases slightly but finally follows a plateau in time.

Fig. 6

F_M − P (η = 1: thick solid line, η = 2: dotted line, η = 3: thin solid line) as a function of t; J_x = 1, J_y = 3, J_z = 2, b = 1, B = 10

5 Teleportation using Schrödinger and Milburn Channel Whose Initial State is Entangled

Now we consider the entangled state

$$ \mid {\phi}^0\Big\rangle =\frac{1}{\sqrt{3}}\left(|0,0\Big\rangle +|1,1\Big\rangle +|2,2\Big\rangle \right), $$

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Which following the same procedure as the previous section leads to the following density matrices for the channel

$$ {\rho}_{Sh-E}(t)=\left(\begin{array}{ccccccccc}{C}_1{C}_1^{\ast }& 0& {C}_2{C}_1^{\ast }& 0& {C}_3{C}_1^{\ast }& 0& {C}_4{C}_1^{\ast }& 0& {C}_5{C}_1^{\ast}\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}{C}_1{C}_2^{\ast }& 0& {C}_2{C}_2^{\ast }& 0& {C}_3{C}_2^{\ast }& 0& {C}_4{C}_2^{\ast }& 0& {C}_5{C}_2^{\ast}\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}{C}_1{C}_3^{\ast }& 0& {C}_2{C}_3^{\ast }& 0& {C}_3{C}_3^{\ast }& 0& {C}_4{C}_3^{\ast }& 0& {C}_5{C}_3^{\ast}\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}{C}_1{C}_4^{\ast }& 0& {C}_2{C}_4^{\ast }& 0& {C}_3{C}_4^{\ast }& 0& {C}_4{C}_4^{\ast }& 0& {C}_5{C}_4^{\ast}\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}{C}_1{C}_5^{\ast }& 0& {C}_2{C}_5^{\ast }& 0& {C}_3{C}_5^{\ast }& 0& {C}_4{C}_5^{\ast }& 0& {C}_5{C}_5^{\ast}\end{array}\right), $$

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where, the C_i parameters are too complicated to write it down and they are saved in our computer program, and

$$ {\rho}_{M-E}=\left(\begin{array}{ccccccccc}{G}_1& 0& {G}_2& 0& {G}_3& 0& {G}_4& 0& {G}_5\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}{G}_6& 0& {G}_7& 0& {G}_8& 0& {G}_9& 0& {G}_{10}\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}{G}_{11}& 0& {G}_{12}& 0& {G}_{13}& 0& {G}_{14}& 0& {G}_{15}\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}{G}_{16}& 0& {G}_{17}& 0& {G}_{18}& 0& {G}_{19}& 0& {G}_{20}\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}{G}_{21}& 0& {G}_{22}& 0& {G}_{23}& 0& {G}_{24}& 0& {G}_{25}\end{array}\right), $$

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where, the G_i parameters are too complicated to write it down and they are saved in our computer program. The above two density matrices also lead to the following fidelities, following the same procedure as the previous section.

$$ {\displaystyle \begin{array}{l}{F}_{Sh-E}=\frac{1}{27}\left(2\ast {3}^{\frac{1}{2}}\mathit{\operatorname{Im}}\left[\left({C}_3-{C}_2\right){C}_4^{\ast}\right]+\mathit{\operatorname{Re}}\right[\left(13{C}_2+\left(7+2\ast {3}^{\frac{1}{2}}i\right){C}_3+\left(7-2\ast {3}^{\frac{1}{2}}i\right){C}_4\right){C}_2^{\ast}\\ {}+\left(\left(7-2\ast {3}^{\frac{1}{2}}i\right){C}_2+13{C}_3+\left(7+2\ast {3}^{\frac{1}{2}}i\right){C}_4\right){C}_3^{\ast }+\left(7\left({C}_2+{C}_3\right)+13{C}_4\right){C}_4^{\ast }+8{\left|{C}_5\right|}^2\left]\right).\end{array}} $$

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And

$$ {\displaystyle \begin{array}{l}{F}_{M-E}=\frac{1}{27}\Big(2\ast {3}^{\frac{1}{2}}\mathit{\operatorname{Im}}\left[{G}_{12}-{G}_{14}-{G}_{17}+{G}_{18}-{G}_8+{G}_9\right]\\ {}+\mathit{\operatorname{Re}}\left[13\left({G}_7+{G}_{13}+{G}_{19}\right)+7\left({G}_8+{G}_9+{G}_{12}+{G}_{14}+{G}_{17}+{G}_{18}\right)+8{G}_{25}\right]\Big).\end{array}} $$

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In Fig. 7, fidelity of the entangled Schrödinger and Milburn channel as a function of time are plotted. It is noted that the Schrödinger fidelity is larger than the Milburn one most of the time.

Fig. 7

F_Sh − E (solid line) and F_M − E (dotted line) as a function oft; J_x = 1, J_y = 3, J_z = 2, b = 1, η = 1, B = 1

In Figs. 8 and 9, fidelities of entangled Schrödinger channel as a function of time for three different values of inhomogeneous (b) and homogeneous (B) magnetic fields are plotted. It is observed that as the inhomogeneous (b) and homogeneous (B) magnetic fields are increased, fidelities on the average decrease.

Fig. 8

F_Sh − E (b = 0: thick solid line, b = 1: dotted line, b = 3: thin solid line) as a function of t; J_x = 1, J_y = 3, J_z = 2, B = 10

Fig. 9

F_Sh − E (B = 3: thick solid line, B = 6: dotted line, B = 10: thin solid line) as a function of t; J_x = 1, J_y = 3, J_z = 2, b = 1

In Figs. 10 and 11, fidelities of the entangled Milburn channel as a function of time for three different values of the inhomogeneous (b) and the homogeneous (B) magnetic fields are shown. As the inhomogeneous magnetic field increases, the fidelity decreases; while, as the homogeneous magnetic field increases the fidelity is increased. Moreover, for large values of the homogeneous field, fidelity follows a plateau.

Fig. 10

F_M − E (b = 0.1: thick solid line, b = 1: dotted line, b = 3: thin solid line) as a function of t; J_x = 1,J_y = 3, J_z = 2, η = 1, B = 10

Fig. 11

F_M − E (B = 3: thick solid line, B = 6: dotted line, B = 10: thin solid line) as a function of t; J_x = 1, J_y = 3, J_z = 2, b = 1, η = 1

In Fig. 12, fidelity of entangled Milburn channel as a function of time for three different values of decoherence phase (η) are shown. It is observed that, when the channel is entangled, an increase in (η), initially increases the fidelity slightly for small periods of time; however, does not have an appreciable effect for large times.

Fig. 12

F_M − E (η = 1: thick solid line, η = 2: dotted line, η = 3: thin solid line) as a function oft; J_x = 1, J_y = 3, J_z = 2, b = 1, B = 10

6 Comparison of the Fidelity for Entangled and Non-Entangled Channels Under the Schrödinger and Milburn Evolution

In Fig. 13, F_Sh − P and F_Sh − E and in Fig. 14F_M − P and F_M − E as a function of time are shown. In both of Schrödinger and Milburn evolution, fidelity of entangled channel is superior to non-entangled channel in most of the times. However, for small times, the situation is vice versa.

Fig. 13

F_Sh − P (solid line) andF_Sh − E (dotted line) as a function of t: J_x = 1, J_y = 3, J_z = 2, b = 1, η = 1, B = 1

Fig. 14

F_M − P (solid line) and F_M − E (dotted line) as a function of t: J_x = 1, J_y = 3, J_z = 2, b = 1, η = 1, B = 1

In Fig. 15F_M − P and F_M − E as a function of decoherence phase η are plotted. It is observed that as η increases, both entangled and non-entangled Milburn fidelity increase, but finally they follow their respective plateaus.

Fig. 15

F_M − P (solid line) and F_M − E (dotted line) as a function of η: J_x = 1, J_y = 3, J_z = 2, b = 1, t = 1, B = 1

7 Conclusions and Discussion

For non-entangled Schrödinger channel, in the absence of decoherence, an increase in the inhomogeneous magnetic field, increases fidelity but its fluctuation periods decrease. An increase of the homogeneous magnetic field for this channel does not have any appreciable effect on the fidelity, only increases slightly. For entangled Schrödinger channel, an increase in the inhomogeneous or homogeneous magnetic field increases the fidelity on the average.

For the non-entangled Milburn channel, an increase of the homogeneous or inhomogeneous magnetic field increases the fidelity. For entangled Milburn channel, an increase of the inhomogeneous magnetic field decreases the fidelity, but an increase of the homogeneous magnetic field increases the latter.

We also note an interesting observation regarding the Milburn channel: The fidelity of the entangled Milburn channel increases due to decoherence, while that of the nonentangled channel decreases, due to small decoherence effects; however, for the larger values of the decoherence, the fidelity is not appreciably changed for both entangled and non-entangled channels.

Finally we consider some results obtained in the case of two-qubit teleportation in the presence of the intrinsic decoherence for comparison. In reference (13) it is found that both entangled and non-entangled initial states are appropriate for teleportation and the tuning of the magnetic field can improve the fidelity. We have also observed a similar result regarding our initial-qutrit entangled and non-entangled states in the presence of the magnetic fields. In reference (15) the Dzyaloshinskii–Moriya interaction has also been considered; it is observed that the Dm interaction and both homogeneous and inhomogeneous part of the field can influence the fidelity, depending on the entangled initial state. We observed in our work that for the qutrit-entangled Milburn channel, an increase of the inhomogeneous magnetic field decreases the fidelity, but an increase of the homogeneous magnetic field increases the latter.

Appendices

Appendix 1

$$ {\varGamma}^0=\left(\begin{array}{ccc}1& 0& 0\\ {}0& 1& 0\\ {}0& 0& 1\end{array}\right), $$

(25)

$$ {\varGamma}^1=\left(\begin{array}{ccc}0& 1& 0\\ {}1& 0& 0\\ {}0& 0& 0\end{array}\right), $$

(26)

$$ {\varGamma}^2=\left(\begin{array}{ccc}0& -i& 0\\ {}i& 0& 0\\ {}0& 0& 0\end{array}\right), $$

(27)

$$ {\varGamma}^3=\left(\begin{array}{ccc}1& 0& 0\\ {}0& -1& 0\\ {}0& 0& 0\end{array}\right), $$

(28)

$$ {\varGamma}^4=\left(\begin{array}{ccc}0& 0& 1\\ {}0& 0& 0\\ {}1& 0& 0\end{array}\right), $$

(29)

$$ {\varGamma}^5=\left(\begin{array}{ccc}0& 0& -i\\ {}0& 0& 0\\ {}i& 0& 0\end{array}\right), $$

(30)

$$ {\varGamma}^6=\left(\begin{array}{ccc}0& 0& 0\\ {}0& 0& 1\\ {}0& 1& 0\end{array}\right), $$

(31)

$$ {\varGamma}^7=\left(\begin{array}{ccc}0& 0& 0\\ {}0& 0& -i\\ {}0& i& 0\end{array}\right), $$

(32)

$$ {\varGamma}^8=\frac{1}{\sqrt{3}}\left(\begin{array}{ccc}1& 0& 0\\ {}0& 1& 0\\ {}0& 0& -2\end{array}\right). $$

(33)

Appendix 2

$$ \mid {\phi}^0\Big\rangle =\frac{1}{\sqrt{3}}\left(|2,0\Big\rangle +|1,1\Big\rangle +|0,2\Big\rangle \right) $$

(34)

$$ \mid {\phi}^1\Big\rangle =\frac{1}{\sqrt{3}}\left(|1,0\Big\rangle +|0,1\Big\rangle +|2,2\Big\rangle \right) $$

(35)

$$ \mid {\phi}^2\Big\rangle =\frac{1}{\sqrt{3}}\left(|0,0\Big\rangle +|2,1\Big\rangle +|1,2\Big\rangle \right) $$

(36)

$$ \mid {\phi}^3\Big\rangle =\frac{1}{\sqrt{3}}\left(|2,0\Big\rangle +{e}^{\frac{2\pi i}{3}}|1,1\Big\rangle +{e}^{-\frac{2\pi i}{3}}|0,2\Big\rangle \right) $$

(37)

$$ \mid {\phi}^4\Big\rangle =\frac{1}{\sqrt{3}}\left(|1,0\Big\rangle +{e}^{\frac{2\pi i}{3}}|0,1\Big\rangle +{e}^{-\frac{2\pi i}{3}}|2,2\Big\rangle \right) $$

(38)

$$ \mid {\phi}^5\Big\rangle =\frac{1}{\sqrt{3}}\left(|0,0\Big\rangle +{e}^{\frac{2\pi i}{3}}|2,1\Big\rangle +{e}^{-\frac{2\pi i}{3}}|1,2\Big\rangle \right) $$

(39)

$$ \mid {\phi}^6\Big\rangle =\frac{1}{\sqrt{3}}\left(|2,0\Big\rangle +{e}^{-\frac{2\pi i}{3}}|1,1\Big\rangle +{e}^{\frac{2\pi i}{3}}|0,2\Big\rangle \right) $$

(40)

$$ \mid {\phi}^7\Big\rangle =\frac{1}{\sqrt{3}}\left(|1,0\Big\rangle +{e}^{-\frac{2\pi i}{3}}|0,1\Big\rangle +{e}^{\frac{2\pi i}{3}}|2,2\Big\rangle \right) $$

(41)

$$ \mid {\phi}^8\Big\rangle =\frac{1}{\sqrt{3}}\left(|0,0\Big\rangle +{e}^{-\frac{2\pi i}{3}}|2,1\Big\rangle +{e}^{\frac{2\pi i}{3}}|1,2\Big\rangle \right) $$

(42)