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Quantum communication with SU(2) invariant separable \(2\times N\) level systems

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Abstract

Information is encoded in a qubit in the form of its Bloch vector. In this paper, we propose protocols for remote transfers of information in a known and an unknown qubit to qudits using SU(2)-invariant \(2 \times N\)-level separable discordant states as quantum channels. These states have been identified as separable equivalents of the two-qubit entangled Werner states in Bharath and Ravishankar (Phys Rev A 89:062110, 2014). Due to \(SU(2) \times SU(2)\) invariance of these states, the remote qudit can be changed by performing appropriate measurements on the qubit. We also propose a protocol for transferring information of a family of unknown qudits to remote qudits using \(2 \times N\)-level states as channels. Finally, we propose a protocol for swapping of quantum discord from \(2\times N\)-level systems to \(N \times N\)-level systems. All the protocols proposed in this paper involve separable states as quantum channels.

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Notes

  1. For \(S=20\), the equivalent state \(\frac{1}{2S+1}(\mathbb {1}+\frac{S_z}{S})\) (of a pure single-qubit state \(\frac{1}{2}(\mathbb {1}+\sigma _z)\)) has fidelity 0.876 with the completely mixed state \(\frac{1}{2S+1}\mathbb {1}\).

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Acknowledgements

It is a pleasure to thank Soumik Adhikary for discussions. We thank the anonymous referees, whose comments have really helped us improve the quality of the presentation and for bringing several relevant references to our attention. Sooryansh and Rajni thank CSIR (Grant no.: 09/086 (1278)/ 2017-EMR-I) and UGC for funding their research, respectively.

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  1. Department of Physics, Indian Institute of Technology Delhi, New Delhi, 110016, India

    Sooryansh Asthana, Rajni Bala & V. Ravishankar

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  1. Sooryansh Asthana
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  2. Rajni Bala
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  3. V. Ravishankar
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Correspondence to Sooryansh Asthana.

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Appendices

Transfer of information from an unknown qubit to a remote qudit

Detailed calculations leading to Eqs. (22), (23), (24) and (25) are as follows. They essentially employ three properties, viz., tracelessness of Pauli matrices, \(\mathrm{Tr}(\varvec{\sigma }\cdot \hat{m}\varvec{\sigma }\cdot \hat{n}) = 2\hat{m}\cdot \hat{n}\), and, \(\mathrm{Tr}(A\otimes B)=\mathrm{Tr}(A)\mathrm{Tr}(B).\) It is sufficient to illustrate it for one case. The same method works for the other three equations as well.

We start with the case when Alice measures a singlet state out of the four Bell states, i.e. \(\rho _1^{AA}=\frac{1}{4}(\mathbb {1}-\varvec{{\sigma }_1} \cdot \varvec{\sigma }_2)\), which leads to equation (22), as follows:

$$\begin{aligned}&\dfrac{1}{(2S+1)2^4} \mathrm{Tr}_{12}\Big \{\Big (\mathbb {1}-\varvec{{\sigma }_1} \cdot \varvec{\sigma }_2\Big )(\mathbb {1}+\varvec{{\sigma }_1} \cdot \hat{p})\otimes (1-\alpha \varvec{\sigma }_2\cdot \hat{S}_3)\Big \}\nonumber \\&\quad =\dfrac{1}{(2S+1)2^4} \mathrm{Tr}_{12}\Big \{\Big (\mathbb {1}-\varvec{{\sigma }_1} \cdot \varvec{\sigma }_2\Big )\Big (\mathbb {1}+{\varvec{\sigma }_1} \cdot \hat{p}-\alpha \varvec{\sigma }_2\cdot \hat{S}_3-\alpha \varvec{\sigma }_{\mathbf{1}}\cdot \hat{p}\otimes \varvec{\sigma }_{\mathbf{2}}\cdot \hat{S}_3\Big )\Big \}\nonumber \\&\quad =\dfrac{1}{(2S+1)2^4} \mathrm{Tr}_{12}\Big \{\mathbb {1}+\varvec{\sigma }_{\mathbf{1}}\cdot \hat{p}-\alpha \varvec{\sigma }_2\cdot \hat{S}_3-\alpha {\varvec{\sigma }_1} \cdot \hat{p}\otimes \varvec{\sigma }_2\cdot \hat{S}_3\nonumber \\&\qquad -\sigma _{1x}\sigma _{2x}(\mathbb {1}+{\varvec{\sigma }_1}\cdot \hat{p}-\alpha \varvec{\sigma }_2\cdot \hat{S}_3-\alpha {\varvec{\sigma }_1}\cdot \hat{p}\otimes \varvec{\sigma }_2\cdot \hat{S}_3)-\sigma _{1y}\sigma _{2y}(\mathbb {1}+{\varvec{\sigma }_1}\cdot \hat{p}\nonumber \\&\qquad -\alpha \mathbb {1}\otimes \varvec{\sigma }_2\cdot \hat{S}_3-\alpha {\varvec{\sigma }_1}\cdot \hat{p}\otimes \varvec{\sigma }_2\cdot \hat{S}_3)-\sigma _{1z}\sigma _{2z}(\mathbb {1}+{\varvec{\sigma }_1}\cdot \hat{p}-\alpha \varvec{\sigma }_2\cdot \hat{S}_3-\alpha {\varvec{\sigma }_1}\cdot \hat{p}\nonumber \\&\qquad \otimes \varvec{\sigma }_2\cdot \hat{S}_3)\Big \}\nonumber \\&\quad =\dfrac{1}{(2S+1)2^4}\Big \{4\mathbb {1}+4p_x\hat{S}_{3x}+4p_y\hat{S}_{3y}+4p_z\hat{S}_{3z}\Big \}\nonumber \\&\quad =\dfrac{1}{4}\times \dfrac{1}{2S+1}(\mathbb {1}+\alpha \hat{S}_3\cdot \hat{p}), \end{aligned}$$
(39)

which is the same as the equation (22).

Transfer of information from an unknown qudit to a remote qudit

When Alice measures \(\hat{S}_1\cdot \varvec{\sigma }_2\), she gets the states \(\dfrac{1}{2S+1}\Big (\mathbb {1}-\dfrac{\alpha }{3}\hat{S}_3\cdot \mathbf {p}\Big )\) and \(\dfrac{1}{2S+1}\Big (\mathbb {1}+\dfrac{\alpha }{3}\dfrac{S+1}{S}\hat{S}_3\cdot \mathbf {p}\Big )\) with respective probabilities of \(\frac{S+1}{2S+1}\) and \(\frac{S}{2S+1}\). The detailed calculations are as follows, which essentially employ the following two properties:

  1. 1.

    The spin operators \(S_x, S_y, S_z\) are traceless, i.e. \(\mathrm{Tr}~ S_i=0\) and \(\mathrm{Tr}~ \sigma _j=0\), \(i, j \in {x, y, z}\).

  2. 2.

    Trace of tensor product of two operators is equal to the product of the trace of operators, i.e. \(\mathrm{Tr}(A\otimes B)=\mathrm{Tr}(A)\mathrm{Tr}(B).\)

Case I When Alice’s state is projected into the eigenstate corresponding to the eigenvalue +1, which occurs with a probability \(\frac{S+1}{2S+1}\), then

$$\begin{aligned}&\dfrac{1}{2(2S+1)^2} \mathrm{Tr}_{12}\Big \{{\Big (\mathbb {1}+\dfrac{S}{S+1}\hat{S}_1\cdot \varvec{\sigma }_{2}\Big )}(\mathbb {1}+\hat{S}_1\cdot \mathbf {p})\otimes (\mathbb {1}-\alpha \varvec{\sigma }_2\cdot \hat{S}_3)\Big \}\nonumber \\&\quad =\dfrac{1}{2(2S+1)^2}\mathrm{Tr}_{12}\Big \{{\Big (\mathbb {1}+\dfrac{S}{S+1}\hat{S}_1\cdot \varvec{\sigma }_{2}\Big )}\nonumber \\&\quad \quad \times \Big (\mathbb {1}+\hat{S}_1\cdot \mathbf {p}-\alpha \varvec{\sigma }_2\cdot \hat{S}_3-\alpha \hat{S}_1\cdot \mathbf {p}\otimes \varvec{\sigma }_2\cdot \hat{S}_3 \Big )\Big \}\nonumber \\&\quad =\dfrac{1}{2(2S+1)^2}\mathrm{Tr}_{12}\Big \{\mathbb {1}+\hat{S}_1\cdot \mathbf {p}-\alpha \varvec{\sigma }_2\cdot \hat{S}_3-\alpha \hat{S}_1\cdot \mathbf {p}\nonumber \\&\qquad \otimes \varvec{\sigma }_2\cdot \hat{S}_3+\dfrac{S}{S+1}\hat{S}_1\cdot \varvec{\sigma }_{2}\nonumber \\&\qquad +\dfrac{S}{S+1}(\hat{S}_1\cdot \varvec{\sigma }_{2})(\hat{S}_1\cdot \mathbf {p})-\alpha \dfrac{S}{S+1}(\hat{S}_1\cdot \varvec{\sigma }_{2})(\varvec{\sigma }_2\cdot \hat{S}_3)\nonumber \\&\qquad -\alpha \dfrac{S}{S+1}(\hat{S}_1\cdot \varvec{\sigma }_{2})(\hat{S}_1\cdot \mathbf {p}\otimes \varvec{\sigma }_2\cdot \hat{S}_3) \Big \}\nonumber \\&\quad =\dfrac{1}{2(2S+1)^2}\Big \{\mathbb {1}+\hat{S}_1\cdot \mathbf {p}-\alpha { \varvec{\sigma }_2\cdot \hat{S}_3}-\alpha (\hat{S}_1\cdot \mathbf {p})\otimes \varvec{\sigma }_2\cdot \hat{S}_3\nonumber \\&\qquad +\dfrac{S}{S+1}\hat{S}_1\cdot \varvec{\sigma }_{2}\nonumber \\&\qquad +\dfrac{S}{S+1}(\hat{S}_1\cdot \varvec{\sigma }_{2})(\hat{S}_1\cdot \mathbf {p}\nonumber \\&\qquad -\alpha \dfrac{S}{S+1}(\hat{S}_1\cdot \varvec{\sigma }_{2})(\varvec{\sigma }_2\cdot \hat{S}_3) -\alpha \dfrac{S}{S+1}(\hat{S}_1\cdot \varvec{\sigma }_{2})(\hat{S}_1\cdot \mathbf {p}\otimes \varvec{\sigma }_2\cdot \hat{S}_3) \Big \}\nonumber \\&\quad =\dfrac{1}{2S+1}\Big (\mathbb {1}-\dfrac{\alpha }{3}\hat{S}_3\cdot \mathbf {p}\Big ). \end{aligned}$$
(40)

The formula that we have used \(\mathrm{Tr}(\hat{S}_i\hat{S}_j) = \frac{(S+1)(2S+1)}{3S}\delta _{ij}\) can be proved as follows. We have

$$\begin{aligned} S_x^2+S_y^2+S_z^2 = S(S+1)\mathbb {1}. \end{aligned}$$
(41)

Taking trace of both sides and employing the property that \(\mathrm{Tr}S_x^2 = \mathrm{Tr}S_y^2 = \mathrm{Tr}S_z^2\), we obtain

$$\begin{aligned} 3\mathrm{Tr}S_x^2&= S(S+1)\mathrm{Tr}\mathbb {1} = S(S+1)(2S+1)\nonumber \\ \implies \mathrm{Tr}S_x^2&= \frac{S(S+1)(2S+1)}{3}. \end{aligned}$$
(42)

Similarly, \(\mathrm{Tr}S_y^2 = \mathrm{Tr}S_z^2 = \frac{S(S+1)(2S+1)}{3}\). Furthermore, we have the following expression of invariance of trace under a unitary transformation,

$$\begin{aligned} \mathrm{Tr}(S_xS_y) = \mathrm{Tr}(US_xS_yU^{\dagger }), \end{aligned}$$
(43)

where U is a unitary transformation. Consider a rotation about the X axis such that \(\hat{y} \rightarrow -\hat{y}\) and \(\hat{z}\rightarrow -\hat{z}\). Thus, under the effect of this transformation

$$\begin{aligned} \mathrm{Tr}(S_xS_y) = \mathrm{Tr}(-S_xS_y)\implies \mathrm{Tr}(S_xS_y) =0. \end{aligned}$$
(44)

Both of these results can be combined to the following equation:

$$\begin{aligned} \mathrm{Tr}(\hat{S}_i\hat{S}_j) = \frac{(S+1)(2S+1)}{3S}\delta _{ij}. \end{aligned}$$
(45)

Case II When Alice’s state is projected to the eigenstate corresponding to the eigenvalue \(-\frac{S+1}{S}\), which occurs with a probability \(\frac{S}{2S+1}\), then

$$\begin{aligned}&\dfrac{1}{2(2S+1)^2}\mathrm{Tr}_{12}\Big \{(\mathbb {1}-\hat{S}_1\cdot \varvec{\sigma }_{2})(\mathbb {1}+\hat{S}_1\cdot \mathbf {p})\otimes (1-\alpha \varvec{\sigma }_2\cdot \hat{S}_3)\Big \}\nonumber \\&\quad =\dfrac{1}{2S+1}\Big (\mathbb {1}+\alpha \dfrac{(S+1)}{3S}\hat{S}_3\cdot \mathbf {p}\Big ).\nonumber \\ \end{aligned}$$
(46)

This result is calculated following the same methodology as done for the Case I.

Swapping of quantum discord

In this appendix, we present those cases of swapping of quantum discord, proposed in Sect. (3.4), in which the remaining three Bell states are measured. The calculations for rest of the three projection operators of Bell basis are as follows. They essentially employ three properties, viz., tracelessness of Pauli matrices, \(\mathrm{Tr}(\varvec{\sigma }\cdot \hat{m}\varvec{\sigma }\cdot \hat{n}) = 2\hat{m}\cdot \hat{n}\), and, \(\mathrm{Tr}(A\otimes B)=\mathrm{Tr}(A)\mathrm{Tr}(B).\)

  1. 1.

    When the measurement of \(\dfrac{1}{4}\Big (\mathbb {1}-({\sigma }_{1x}{\sigma }_{4x}-{\sigma }_{1y}{\sigma }_{4y}-{\sigma }_{1z}{\sigma }_{4z})\Big )\) is made, the post-measurement state is

    $$\begin{aligned}&\mathrm{Tr}_{14}\dfrac{1}{2^4(2S+1)^2}\Big \{\Big (\mathbb {1}-({\sigma }_{1x}{\sigma }_{4x}-{\sigma }_{1y}{\sigma }_{4y}-{\sigma }_{1z}{\sigma }_{4z})\Big )(\mathbb {1}-\alpha {\varvec{\sigma }_1}\cdot \hat{S}_2)\nonumber \\&\qquad (\mathbb {1}-\beta \hat{S}_3\cdot \varvec{\sigma }_4)\Big \}\nonumber \\&\quad =\dfrac{1}{2^4(2S+1)^2}\mathrm{Tr}_{14}\Big (\mathbb {1}-\alpha \beta \varvec{\sigma }_{1}\cdot \hat{S}_2({\sigma }_{1x}{\sigma }_{4x}-{\sigma }_{1y}{\sigma }_{4y}-{\sigma }_{1z}{\sigma }_{4z})\hat{S}_3\cdot \varvec{\sigma }_4\Big )\nonumber \\&\quad =\dfrac{1}{2^4(2S+1)^2}\mathrm{Tr}_{14}\Big (\mathbb {1} -\alpha \beta \varvec{\sigma }_{1}\cdot \hat{S}_2({\sigma }_{1x}{\sigma }_{4x}\hat{S_{3x}} \sigma _{4x}\nonumber \\&\qquad +\,\sigma _{1x}{\sigma }_{4x}\hat{S}_{3y}\sigma _{4y} +\sigma _{1x}{\sigma }_{4x}\hat{S}_{3z}\sigma _{4z}\nonumber \\&\qquad - {\sigma }_{1y}{\sigma }_{4y}\hat{S}_{3x}\sigma _{4x}-{\sigma }_{1y}{\sigma }_{4y}\hat{S}_{3y}\sigma _{4y}-{\sigma }_{1y}{\sigma }_{4y}\hat{S}_{3z}\sigma _{4z}-{\sigma }_{1z}{\sigma }_{4z}\hat{S}_{3x}\sigma _{4x}\nonumber \\&\qquad - {\sigma }_{1z}{\sigma }_{4z}\hat{S}_{3y}\sigma _{4y}-{\sigma }_{1z}{\sigma }_{4z}\hat{S}_{3z}\sigma _{4z}\Big )\nonumber \\&\quad =\dfrac{1}{2^4(2S+1)^2}\mathrm{Tr}_{1}\Big (2\mathbb {1}-\alpha \beta \varvec{\sigma }_{1}\cdot \hat{S}_2({2\sigma }_{1x}\hat{S}_{3x}-2{\sigma }_{1y}\hat{S}_{3y}-2{\sigma }_{1z}\hat{S}_{3z}\Big )\nonumber \\&\quad =\dfrac{1}{2^3(2S+1)^2}\mathrm{Tr}_{1}\Big (\mathbb {1}-\alpha \beta \varvec{\sigma }_{1}\cdot \hat{S}_2({\sigma }_{1x}\hat{S}_{3x}-{\sigma }_{1y}\hat{S}_{3y}-{\sigma }_{1z}\hat{S}_{3z}\Big )\nonumber \\&\quad =\dfrac{1}{2^3(2S+1)^2}\mathrm{Tr}_{1}\Big (\mathbb {1}-\alpha \beta (\sigma _{1x}\hat{S}_{2x}({\sigma }_{1x}\hat{S}_{3x} -{\sigma }_{1y}\hat{S}_{3y}-{\sigma }_{1z}\hat{S}_{3z})\nonumber \\&\qquad +\,\sigma _{1y}\hat{S}_{2y}({\sigma }_{1x}\hat{S}_{3x}-{\sigma }_{1y}\hat{S}_{3y} -{\sigma }_{1z}\hat{S}_{3z})\nonumber \\&\qquad +\,\sigma _{1z}\hat{S}_{2z}({\sigma }_{1x}\hat{S}_{3x}-{\sigma }_{1y}\hat{S}_{3y}-{\sigma }_{1z}\hat{S}_{3z})\Big )\nonumber \\&\quad =\dfrac{1}{2^3(2S+1)^2}\mathrm{Tr}_{1}\Big (\mathbb {1}-\alpha \beta (\sigma _{1x}\hat{S}_{2x}({\sigma }_{1x}\hat{S}_{3x}-{\sigma }_{1y}\hat{S}_{3y}-{\sigma }_{1z}\hat{S}_{3z})\nonumber \\&\qquad +\,\sigma _{1y}\hat{S}_{2y}({\sigma }_{1x}\hat{S}_{3x}-{\sigma }_{1y}\hat{S}_{3y}-{\sigma }_{1z}\hat{S}_{3z})+\sigma _{1z}\hat{S}_{2z}({\sigma }_{1x}\hat{S}_{3x}-{\sigma }_{1y}\hat{S}_{3y}-{\sigma }_{1z}\hat{S}_{3z})\Big )\nonumber \\&\quad =\dfrac{1}{2^3(2S+1)^2}\Big (2\mathbb {1}-\alpha \beta (2\hat{S}_{2x}\hat{S}_{3x}-2\hat{S}_{2y}\hat{S}_{3y}-2\hat{S}_{3z}\hat{S}_{3z})\Big )\nonumber \\&\quad =\dfrac{1}{4}\times \dfrac{1}{(2S+1)^2}\Big (\mathbb {1}-\frac{\alpha \beta }{S^2}(S_{2x}S_{3x}-S_{2y}S_{3y}-S_{2z}S_{3z})\Big ). \end{aligned}$$
    (47)

    The factor of \(\frac{1}{4}\) represents the probability with which this state is prepared. Other calculations are performed using the same approach.

  2. 2.

    When the measurement of \(\dfrac{1}{4}\Big (\mathbb {1}-(-{\sigma }_{1x}{\sigma }_{4x}-{\sigma }_{1y}{\sigma }_{4y}+{\sigma }_{1z}{\sigma }_{4z})\Big )\) is made

    $$\begin{aligned}&\mathrm{Tr}_{14}\dfrac{1}{2^4(2S+1)^2}\Big \{\Big (\mathbb {1}-(-{\sigma }_{1x}{\sigma }_{4x}-{\sigma }_{1y}{\sigma }_{4y}+{\sigma }_{1z}{\sigma }_{4z})\Big )\nonumber \\&\qquad \times (\mathbb {1}-\alpha {\varvec{\sigma }_1} \cdot \hat{S}_2)(\mathbb {1}-\beta \hat{S}_3\cdot \varvec{\sigma }_4)\Big \}\nonumber \\&\quad =\dfrac{1}{2^4(2S+1)^2}\mathrm{Tr}_{14}\Big (\mathbb {1}-\alpha \beta \varvec{\sigma }_{1}\cdot \hat{S}_2(-{\sigma }_{1x}{\sigma }_{4x}-{\sigma }_{1y}{\sigma }_{4y}+{\sigma }_{1z}{\sigma }_{4z})\hat{S}_3\cdot \varvec{\sigma }_4\Big )\nonumber \\&\quad =\dfrac{1}{4}\times \dfrac{1}{(2S+1)^2}\Big (\mathbb {1}-\frac{\alpha \beta }{S^2}(-S_{2x}S_{3x}-S_{2y}S_{3y}+S_{2z}S_{3z})\Big ). \end{aligned}$$
    (48)

    The factor of \(\frac{1}{4}\) represents the probability with which this state is prepared.

  3. 3.

    When the measurement of \(\dfrac{1}{4}\Big (\mathbb {1}-(-{\sigma }_{1x}{\sigma }_{4x}+{\sigma }_{1y}{\sigma }_{4y}-{\sigma }_{1z}{\sigma }_{4z})\Big )\) is made

    $$\begin{aligned}&\mathrm{Tr}_{14}\dfrac{1}{2^4(2S+1)^2} \Big \{\Big (\mathbb {1}-(-{\sigma }_{1x}{\sigma }_{4x}+{\sigma }_{1y}{\sigma }_{4y}-{\sigma }_{1z}{\sigma }_{4z})\Big )(\mathbb {1}-\alpha {\varvec{\sigma }_1}\cdot \hat{S}_2)\nonumber \\&\qquad (\mathbb {1}-\beta \hat{S}_3\cdot \varvec{\sigma }_4)\Big \}\nonumber \\&\quad =\dfrac{1}{2^4(2S+1)^2}\mathrm{Tr}_{14}\Big (\mathbb {1}-\alpha \beta \varvec{\sigma }_{1}\cdot \hat{S}_2(-{\sigma }_{1x}{\sigma }_{4x}+{\sigma }_{1y}{\sigma }_{4y}-{\sigma }_{1z}{\sigma }_{4z})\hat{S}_3\cdot \varvec{\sigma }_4\Big )\nonumber \\&\quad =\dfrac{1}{4}\times \dfrac{1}{(2S+1)^2}\Big (\mathbb {1}-\frac{\alpha \beta }{S^2}(-S_{2x}S_{3x}+S_{2y}S_{3y}-S_{2z}S_{3z})\Big ). \end{aligned}$$
    (49)

    The factor of \(\frac{1}{4}\) represents the probability with which this state is prepared.

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Asthana, S., Bala, R. & Ravishankar, V. Quantum communication with SU(2) invariant separable \(2\times N\) level systems. Quantum Inf Process 21, 35 (2022). https://doi.org/10.1007/s11128-021-03358-9

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