Multi-Output Nondestructive Controlled Quantum Teleportation Through a Partially Entangled Channel

Abstract

Through a detailed study of nondestructive controlled quantum teleportation (NCQT) with two outputs in three-dimensional and high-dimensional Hilbert spaces, we propose an NCQT scheme that synchronously and probabilistically teleports N arbitrary unknown d-dimensional single-particle states from one sender to N different receivers under the supervision of a controller, by using a d-dimensional partially entangled $(2N+1)$-particle state as the quantum channel. The protocol succeeds if and only if all receivers recover their respective target states, and the optimal success probability of the N-output NCQT is determined by the minimum superposition coefficient of each product two-qudit state in the partially entangled channel. In our scheme, each receiver introduces an auxiliary qubit to assist in the local recovery test. When the auxiliary-qubit measurement outcome is $|0⟩$, the receiver can restore the target state; when the outcome is $|1⟩$, the corresponding unknown original state is retained by the sender. Accordingly, the N-output controlled teleportation process can be repeated as many times as additional quantum channels are available after a failed attempt. The results show that weakly entangled channels can still realize N-output controlled teleportation through sufficiently many repetitions, whereas strongly entangled channels require only a small number of repetitions to achieve the same goal.

1. Introduction

Over the past several decades, quantum communication has developed into a broad research field that includes quantum teleportation and networked transmission [1,2,3,4,5,6,7], quantum cryptography and secure direct communication [8,9,10,11]. Among these directions, quantum teleportation (QT) is one of the most representative applications of quantum entanglement [12,13]. First proposed by Bennett et al. [1], QT enables the transfer of an unknown single-qubit state from a sender to a distant receiver by exploiting a pre-shared Einstein–Podolsky–Rosen channel together with classical communication and local operations. Since then, QT has developed rapidly in both theoretical [5,14,15,16,17,18,19] and experimental [2,20,21,22,23,24,25,26] directions. Most early QT protocols were deterministic and relied on maximally entangled channels [27,28,29,30].

In the standard formulation of QT, the information carrier is a qubit, i.e., a quantum system described in a two-dimensional Hilbert space. Its high-dimensional generalization is a qudit, which is defined on a d-dimensional Hilbert space; the qutrit considered in this paper corresponds to the special case $d=3$. Compared with qubit-based encoding, qudit-based encoding can offer larger information capacity per carrier, richer entanglement structures, and improved robustness in several high-dimensional quantum communication tasks. These features make qudit teleportation and controlled qudit teleportation particularly attractive for scalable quantum communication and high-dimensional quantum information processing [19,31,32,33,34,35].

In realistic teleportation scenarios, unavoidable coupling between the quantum system and its environment can degrade a maximally entangled channel into a partially entangled one. In this setting, Li et al. [36] first proposed a probabilistic protocol for teleporting an arbitrary unknown single-qubit state through a partially entangled channel, whose success probability is below 100%. Subsequently, a variety of probabilistic QT schemes were developed based on Bell states [37], GHZ states [38], cluster states [6], and mixed states [39]. For example, Chen et al. [40] proposed a probabilistic teleportation scheme for an unknown multi-particle GHZ state and employed a positive operator-valued measurement (POVM) [7,26,41,42] to improve operational feasibility. Choudhury et al. [43] later studied probabilistic teleportation of an arbitrary unknown two-qubit state by using two partially entangled GHZ-like states. More recently, Peng et al. [19] reported a probabilistic scheme for teleporting a shared three-dimensional quantum secret and generalized it to the shared d-dimensional case. Javed et al. [44] showed that repeated generalized Bell-state measurements can realize high-success standard QT, while Aisan et al. [45] presented conclusive bidirectional QT schemes for arbitrary unknown single-qubit and single-qutrit states in amplitude-damping channels. Since probabilistic teleportation through non-maximally entangled states may fail, an important question is how to preserve the original state when recovery is unsuccessful. To address this issue, researchers introduced auxiliary particles to assist in state transmission and to realize probabilistic teleportation without information loss [46,47,48,49,50,51]. For instance, Roa and Groiseau [47] proposed a protocol for probabilistically teleporting an unknown single-qubit state without destroying the original information, thereby allowing repeated attempts when additional channels are available. Subsequently, Fu et al. [50] proposed a multi-hop nondestructive teleportation scheme in which the receiver performs a suitable modulation unitary so that the information can be restored when teleportation fails.

As a variant of QT, controlled quantum teleportation (CQT) was proposed by Karlsson and Bourennane in 1998 [52], which allows the sender to teleport an arbitrary unknown single-qubit state to the receiver through a three-qubit entangled state as the quantum channel under a third-party controller. This CQT scheme not only enhances security by introducing a controller into QT, but also extends the QT protocol [1] to multi-party or multi-particle quantum states. In 2004, Yang et al. [53] suggested a multi-party controlled teleportation protocol to teleport multi-qubit quantum information. In 2005, Deng et al. [54] introduced a symmetric multi-party controlled teleportation scheme for an arbitrary two-particle entangled state. In 2008, Dai et al. [55] investigated a remote state preparation scheme of an entangled two-qubit state with three parties from a sender to either of two receivers, where, however, the receivers could not obtain the quantum state to be prepared simultaneously. Li et al. [56] presented a CQT scheme of a three-qubit state by using a five-qubit cluster state. In 2024, Peng et al. [57] proposed a controlled scheme for cyclic quantum teleportation of unknown single-qutrit states, and two years later, Maihemuti et al. [58] put forward a CQT scheme of double-direction cyclic quantum communication of arbitrary two-particle states. In 2024, Liu et al. [59] proposed a controlled bidirectional scheme for remote implementations of partially unknown quantum operations, and the same year, Maihemuti et al. [60] also presented a controlled scheme for assisted cloning of arbitrary unknown states via maximal slice states.

Quantum information processing in high-dimensional quantum systems is also an active research topic because it offers enhanced information capacity, improved eavesdropping security, and stronger noise resistance in many settings [31,32]. The manipulation of high-dimensional quantum states has become experimentally feasible on several platforms, such as orbital-angular-momentum states of photons [33], time-bin encoding [34], and multiple degrees of freedom of single particles [35], thereby paving the way for more sophisticated quantum communication protocols. Broadcast encryption involves a sender and multiple receivers [61]; the sender first encrypts the content and then broadcasts it to a dynamically changing privileged set of users, while unauthorized users cannot recover the protected information. In this broader context, Yu et al. [62] proposed theoretical schemes for quantum broadcast communication and multi-output quantum teleportation via a four-qubit cluster state, and later extended the multi-output idea to the symmetric transmission of arbitrary m- and $(m+1)$-qubit GHZ-class states [63]. Peng et al. [64] studied quantum multicast and multi-output protocols of quantum state sharing and quantum operation sharing via a six-particle cluster state. Aisan et al. [65] presented bidirectional multi-output quantum communication of single-qubit states, and Peng et al. [66] proposed a multi-type-output assisted cloning scheme for unknown single-qubit states. These developments further demonstrate the continuing expansion of secure quantum communication tasks.

Despite some significant progress, the extension of multi-output probabilistic controlled teleportation to high-dimensional quantum systems remains largely unexplored. High-dimensional quantum teleportation presents unique challenges [67,68,69,70], including the complexity of designing appropriate measurement bases for high-dimensional states [69], and the difficulty in maintaining high-dimensional entanglement in noisy channels [70]. Moreover, the implementation of controlled teleportation in high dimensions requires novel approaches to ensure that the controller’s supervision remains effective while maintaining the protocol’s efficiency and security. Recent semi-quantum private comparison protocols based on high-dimensional entangled states further highlight the broader demand for secure multi-party communication in high-dimensional settings [71,72].

This paper addresses these challenges by proposing a comprehensive scheme for multi-output probabilistic controlled teleportation of arbitrary unknown single-qudit states. First, by using a partially entangled five-qutrit state as the quantum channel, we construct a two-output NCQT protocol that allows repeated attempts without loss of the original quantum information. We then generalize this idea to the teleportation of two unknown single-qudit states through a partially entangled five-qudit channel. Finally, based on these two-output constructions, we present a general NCQT protocol in which one sender teleports N unknown single-qudit states to N receivers, respectively, via a partially entangled $(2N+1)$-qudit channel under the supervision of a controller. For convenience, Table 1 summarizes the main symbols used throughout the manuscript.

Table 1. Summary of the main symbols used in the manuscript.

Symbol	Meaning
$|{\chi }_k{⟩}_{t_k}$	Unknown single-qudit state (or single-qutrit state when $d=3$) to be teleported.
$t_k$	Sender-side particle carrying the kth unknown input state.
$A_k,B_k$	Channel particles held by the sender and the kth receiver, respectively.
C	Controller’s particle used to authorize the controlled teleportation process.
$e_k$	Auxiliary qubit introduced by the kth receiver to detect whether recovery is successful.
$a_{kl}$	Schmidt-type coefficients of the partially entangled channel.
d	Dimension of a single qudit; $d=3$ corresponds to a qutrit.
N	Number of unknown single-qudit states (equivalently, the number of receivers).
$\omega =exp\{2\pi i/3\}$	Third root of unity used in the qutrit case.
$\zeta =exp\{2\pi i/d\}$	Primitive dth root of unity used in the general qudit case.
${\mathcal{N}}^{\left(d\right)}$	d-dimensional controlled-NOT gate.
${\mathcal{H}}^{\left(d\right)}$	d-dimensional Hadamard-type transform.
$U_{B_k{e}_k}$	Local unitary transformation performed by the kth receiver on $B_k$ and $e_k$.

2. Two-Output Teleportation of Two Unknown Single-Qutrit States

Let us start with two-output teleportation of two qutrits (a three-level quantum system). Suppose that the sender Alice wants to simultaneously teleport two arbitrary unknown single-qutrit states $|{\chi }_1{⟩}_{t_1}$ and $|{\chi }_2{⟩}_{t_2}$ to two receivers Bob₁ and Bob₂, separately, under the control of the supervisor Charlie, where these two single-qutrit states can be written as

$$\begin{array}{c}\hfill |{\chi }_k{⟩}_{t_k}=(c_{k0}|0⟩+c_{k1}|1⟩+c_{k2}{|2⟩)}_{t_k},k=1,2,\end{array}$$

(1)

where $c_{kj}$ ($k=1,2$; $j=0,1,2$) are complex numbers and satisfy the normalization conditions $|c_{k0}{|}^2+|c_{k1}{|}^2+{|c_{k2}|}^2=1$ for all $k\in \{1,2\}$. The participants pre-share a five-qutrit entangled state as a quantum channel,

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill {|\mathcal{Q}⟩}_{A_1{B}_1{A}_2{B}_{2}C}& ={\displaystyle \frac{1}{\sqrt{3}}}\sum _{j=0}^2{|j⟩}_C(a_{10}|00⟩+a_{11}{\omega }^j|11⟩+a_{12}{\omega }^{2j}{|22⟩)}_{A_1{B}_1}\hfill & \\ & \otimes (a_{20}|00⟩+a_{21}{\omega }^j|11⟩+a_{22}{\omega }^{2j}{|22⟩)}_{A_2{B}_2},\hfill & \end{array}\end{array}$$

(2)

where $\omega =exp\{2\pi \mathrm{i}/3\}$ is the triple root, and $a_{kl}$ ($k=1,2$; $l=0,1,2$) are the channel parameters that satisfy the relationships $|a_{k0}{|}^2+|a_{k1}{|}^2+{|a_{k2}|}^2=1$ for any $k\in \{1,2\}$. Without loss of generality, we assume that the Schmidt coefficients are non-negative real numbers, $a_{k0}=min\{a_{k0},a_{k1},a_{k2}\}$ ($k=1,2$). Here, the five qutrits are identified by the subscripts $A_1,A_2,B_1,B_2$ and C; qutrits $A_1$ and $A_2$ are in the hands of Alice; qutrits $B_1$ and $B_2$ belong to Bob₁ and Bob₂, respectively; and qutrit C belongs to Charlie.

In the beginning, the receivers Bob₁ and Bob₂ introduce auxiliary qubits $e_1$ and $e_2$ in their initial states ${|0⟩}_{e_1}$ and ${|0⟩}_{e_2}$, respectively. The state of the entire initial system can be represented as

$$\begin{array}{cc}\hfill |T⟩& =|{\chi }_1{⟩}_{t_1}|{\chi }_2{⟩}_{t_2}{|\mathcal{Q}⟩}_{A_1{B}_1{A}_2{B}_{2}C}{|0⟩}_{e_1}{|0⟩}_{e_2}\hfill \\ & ={\displaystyle \frac{1}{\sqrt{3}}}\sum _{j=0}^2{|j⟩}_C[(c_{10}|00⟩+c_{11}|10⟩+c_{12}{|20⟩)}_{t_1{A}_1}\otimes a_{10}{|00⟩}_{B_1{e}_1}\hfill \\ & +(c_{10}|01⟩+c_{11}|11⟩+c_{12}{|21⟩)}_{t_1{A}_1}\otimes a_{11}{\omega }^j{|10⟩}_{B_1{e}_1}\hfill \\ & +(c_{10}|02⟩+c_{11}|12⟩+c_{12}{|22⟩)}_{t_1{A}_1}\otimes a_{12}{\omega }^{2j}{|20⟩}_{B_1{e}_1}]\hfill \\ & \otimes [(c_{20}|00⟩+c_{21}|10⟩+c_{22}{|20⟩)}_{t_2{A}_2}\otimes a_{20}{|00⟩}_{B_2{e}_2}\hfill \\ & +(c_{20}|01⟩+c_{21}|11⟩+c_{22}{|21⟩)}_{t_2{A}_2}\otimes a_{21}{\omega }^j{|10⟩}_{B_2{e}_2}\hfill \\ & +(c_{20}|02⟩+c_{21}|12⟩+c_{22}{|22⟩)}_{t_2{A}_2}\otimes a_{22}{\omega }^{2j}{|20⟩}_{B_2{e}_2}].\hfill \end{array}$$

(3)

In order to directly teleport two unknown single-qutrit states $|{\chi }_1{⟩}_{t_1}$ and $|{\chi }_2{⟩}_{t_2}$ from sender Alice to receivers Bob₁ and Bob₂, respectively, our scheme is divided into five steps as follows.

Step 1: Alice applies three-dimensional controlled-NOT gates on qutrit pairs $(t_1,A_1)$ and $(t_2,A_2)$ respectively, and then performs three-dimensional Hadamard gates on qutrits $t_1$ and $t_2$ respectively, where the three-dimensional controlled-NOT gate ${\mathcal{N}}^{\left(3\right)}$ and three-dimensional Hadamard gate ${\mathcal{H}}^{\left(3\right)}$ are respectively defined as follows:

$$\begin{array}{c}\hfill {\mathcal{N}}^{\left(3\right)}|mn⟩=|m⟩|(m+n)\mathrm{mod}3⟩,\end{array}$$

(4)

and

$$\begin{array}{c}\hfill {\mathcal{H}}^{\left(3\right)}={\displaystyle \frac{1}{\sqrt{3}}}\sum _{l,l^{\prime }=0}^2|l⟩⟨l^{\prime }|.\end{array}$$

(5)

After the ${\mathcal{N}}^{\left(3\right)}$ and ${\mathcal{H}}^{\left(3\right)}$ operations, the state shown in Equation (3) becomes

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |T^{\prime }⟩& ={\left({\mathcal{H}}^{\left(3\right)}\right)}_{t_1}{\left({\mathcal{H}}^{\left(3\right)}\right)}_{t_2}{\left({\mathcal{N}}^{\left(3\right)}\right)}_{t_1{A}_1}{\left({\mathcal{N}}^{\left(3\right)}\right)}_{t_2{A}_2}|T⟩\hfill & \\ & ={\displaystyle \frac{1}{3\sqrt{3}}}\sum _{j=0}^2{|j⟩}_C\{[c_{10}(|0⟩+|1⟩+|2⟩)|0⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|1⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|2⟩)]}_{t_1{A}_1}\otimes a_{10}{|00⟩}_{B_1{e}_1}\hfill & \\ & +[c_{10}(|0⟩+|1⟩+|2⟩)|1⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|2⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|0⟩)]}_{t_1{A}_1}\otimes a_{11}{\omega }^j{|10⟩}_{B_1{e}_1}\hfill & \\ & +[c_{10}(|0⟩+|1⟩+|2⟩)|2⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|0⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|1⟩)]}_{t_1{A}_1}\otimes a_{12}{\omega }^{2j}{|20⟩}_{B_1{e}_1}\}\hfill & \\ & \otimes \{[c_{20}(|0⟩+|1⟩+|2⟩)|0⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|1⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|2⟩)]}_{t_2{A}_2}\otimes a_{20}{|00⟩}_{B_2{e}_2}\hfill & \\ & +[c_{20}(|0⟩+|1⟩+|2⟩)|1⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|2⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|0⟩)]}_{t_2{A}_2}\otimes a_{21}{\omega }^j{|10⟩}_{B_2{e}_2}\hfill & \\ & +[c_{20}(|0⟩+|1⟩+|2⟩)|2⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|0⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|1⟩)]}_{t_2{A}_2}\otimes a_{22}{\omega }^{2j}{|20⟩}_{B_2{e}_2}\}.\hfill & \end{array}\end{array}$$

(6)

Step 2: Bob₁ executes a unitary transformation $U_{B_1{e}_1}$ on his two particles $B_1$ and $e_1$ given by

$$\begin{array}{cc}\hfill U_{B_1{e}_1}& ={|00⟩}_{B_1{e}_1}⟨00|+{|01⟩}_{B_1{e}_1}⟨01|\hfill \\ & +{\displaystyle \frac{a_{10}}{a_{11}}}{|10⟩}_{B_1{e}_1}⟨10|+\sqrt{1-{({\displaystyle \frac{a_{10}}{a_{11}}})}^2}{|10⟩}_{B_1{e}_1}⟨11|\hfill \\ & -\sqrt{1-{({\displaystyle \frac{a_{10}}{a_{11}}})}^2}{|11⟩}_{B_1{e}_1}⟨10|+{\displaystyle \frac{a_{10}}{a_{11}}}{|11⟩}_{B_1{e}_1}⟨11|\hfill \\ & +{\displaystyle \frac{a_{10}}{a_{12}}}{|20⟩}_{B_1{e}_1}⟨20|+\sqrt{1-{({\displaystyle \frac{a_{10}}{a_{12}}})}^2}{|20⟩}_{B_1{e}_1}⟨21|\hfill \\ & -\sqrt{1-{({\displaystyle \frac{a_{10}}{a_{12}}})}^2}{|21⟩}_{B_1{e}_1}⟨20|+{\displaystyle \frac{a_{10}}{a_{12}}}{|21⟩}_{B_1{e}_1}⟨21|,\hfill \end{array}$$

(7)

and at the same time, Bob₂ also performs a unitary transformation $U_{B_2{e}_2}$ on his two particles $B_2$ and $e_2$ given by

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill U_{B_2{e}_2}& ={|00⟩}_{B_2{e}_2}⟨00|+{|01⟩}_{B_2{e}_2}⟨01|\hfill & \\ & +{\displaystyle \frac{a_{20}}{a_{21}}}{|10⟩}_{B_2{e}_2}⟨10|+\sqrt{1-{({\displaystyle \frac{a_{20}}{a_{21}}})}^2}{|10⟩}_{B_2{e}_2}⟨11|\hfill & \\ & -\sqrt{1-{({\displaystyle \frac{a_{20}}{a_{21}}})}^2}{|11⟩}_{B_2{e}_2}⟨10|+{\displaystyle \frac{a_{20}}{a_{21}}}{|11⟩}_{B_2{e}_2}⟨11|\hfill & \\ & +{\displaystyle \frac{a_{20}}{a_{22}}}{|20⟩}_{B_2{e}_2}⟨20|+\sqrt{1-{({\displaystyle \frac{a_{20}}{a_{22}}})}^2}{|20⟩}_{B_2{e}_2}⟨21|\hfill & \\ & -\sqrt{1-{\left({\displaystyle \frac{a_{20}}{a_{22}}}\right)}^2}{|21⟩}_{B_2{e}_2}⟨20|+{\displaystyle \frac{a_{20}}{a_{22}}}{|21⟩}_{B_2{e}_2}⟨21|.\hfill & \end{array}\end{array}$$

(8)

After the operations of Bob₁ and Bob₂, the entire nine-particle system shown in Equation (6) is transformed into

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |T^{″}⟩& =(U_{B_1{e}_1}\otimes U_{B_2{e}_2})|T^{\prime }⟩\hfill & \\ & ={\displaystyle \frac{1}{3\sqrt{3}}}\sum _{j=0}^2{|j⟩}_C\{[c_{10}(|0⟩+|1⟩+|2⟩)|0⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|1⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|2⟩)]}_{t_1{A}_1}\otimes a_{10}{|00⟩}_{B_1{e}_1}\hfill & \\ & +[c_{10}(|0⟩+|1⟩+|2⟩)|1⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|2⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|0⟩)]}_{t_1{A}_1}\otimes {\omega }^j(a_{10}|10⟩-\sqrt{a_{11}^2-a_{10}^2}{|11⟩)}_{B_1{e}_1}\hfill & \\ & +[c_{10}(|0⟩+|1⟩+|2⟩)|2⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|0⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|1⟩)]}_{t_1{A}_1}\otimes {\omega }^{2j}(a_{10}|20⟩-\sqrt{a_{12}^2-a_{10}^2}{|21⟩)}_{B_1{e}_1}\}\hfill & \\ & \otimes \{[c_{20}(|0⟩+|1⟩+|2⟩)|0⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|1⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|2⟩)]}_{t_2{A}_2}\otimes a_{20}{|00⟩}_{B_2{e}_2}\hfill & \\ & +[c_{20}(|0⟩+|1⟩+|2⟩)|1⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|2⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|0⟩)]}_{t_2{A}_2}\otimes {\omega }^j(a_{20}|10⟩-\sqrt{a_{21}^2-a_{20}^2}{|11⟩)}_{B_2{e}_2}\hfill & \\ & +[c_{20}(|0⟩+|1⟩+|2⟩)|2⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|0⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|1⟩)]}_{t_2{A}_2}\otimes {\omega }^{2j}(a_{20}|20⟩-\sqrt{a_{22}^2-a_{20}^2}{|21⟩)}_{B_2{e}_2}\}.\hfill & \end{array}\end{array}$$

(9)

Bob₁ (Bob₂) performs a von Neumann measurement on his auxiliary qubit $e_1$ ($e_2$) with the basis $\{|0⟩,|1⟩\}$, and there are two possible outcomes: ${|0⟩}_{e_1}$ or ${|1⟩}_{e_1}$ (${|0⟩}_{e_2}$ or ${|1⟩}_{e_2}$). If the measurement results of Bob₁ and Bob₂ are ${|0⟩}_{e_1}$ and ${|0⟩}_{e_2}$, respectively, step 5 will be executed to restore the original unknown states $|{\chi }_1⟩$ and $|{\chi }_2⟩$ on qutrits $B_1$ and $B_2$ by receivers Bob₁ and Bob₂, respectively, after Charlie performs measurement on qudit C in step 3 and Alice performs measurement on qudit pairs $(t_1,A_1)$ and $(t_2,A_2)$ in step 4. Otherwise, the teleportation will fail, and step $5^{\prime }$ will be performed to preserve the original information on Alice’s location.

Obviously, step 1 and step 2 can be executed independently by sender Alice and receivers Bob₁ and Bob₂, and so they could be performed simultaneously.

Step 3: If supervisor Charlie is willing to collaborate with Alice, Bob₁ and Bob₂, he should measure his qutrit C in the computational basis $\{|0⟩,|1⟩,|2⟩\}$, and then announce his measurement result ${|j⟩}_C$ ($j=0,1,2$) to other participants through classical communication.

Step 4: If Charlie’s measurement result is ${|j⟩}_C$, and Bob₁ and Bob₂’s outcomes are ${|0⟩}_{e_1}$ and ${|0⟩}_{e_2}$, the state of qutrits $t_1$, $t_2$, $A_1$, $A_2$, $B_1$ and $B_2$ will collapse into

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |T^{‴}⟩& ={\displaystyle \frac{a_{10}{a}_{20}}{3}}\{[c_{10}(|0⟩+|1⟩+|2⟩)|0⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|1⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|2⟩)]}_{t_1{A}_1}\otimes {|0⟩}_{B_1}\hfill & \\ & +[c_{10}(|0⟩+|1⟩+|2⟩)|1⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|2⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|0⟩)]}_{t_1{A}_1}\otimes {\omega }^j{|1⟩}_{B_1}\hfill & \\ & +[c_{10}(|0⟩+|1⟩+|2⟩)|2⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|0⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|1⟩)]}_{t_1{A}_1}\otimes {\omega }^{2j}{|2⟩}_{B_1}\}\hfill & \\ & \otimes \{[c_{20}(|0⟩+|1⟩+|2⟩)|0⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|1⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|2⟩)]}_{t_2{A}_2}\otimes {|0⟩}_{B_2}\hfill & \\ & +[c_{20}(|0⟩+|1⟩+|2⟩)|1⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|2⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|0⟩)]}_{t_2{A}_2}\otimes {\omega }^j{|1⟩}_{B_2}\hfill & \\ & +[c_{20}(|0⟩+|1⟩+|2⟩)|2⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|0⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|1⟩)]}_{t_2{A}_2}\otimes {\omega }^{2j}{|2⟩}_{B_2}\}.\hfill & \end{array}\end{array}$$

(10)

From Equation (10), it can be seen that Alice’s qutrits $t_1$ and $A_1$ are still entangled with Bob₁’s qutrit $B_1$, and Alice’s particles $t_2$ and $A_2$ are also entangled with Bob₂’s qutrit $B_2$. In order for the recipients Bob₁ and Bob₂ to reconstruct their respective target states, Alice needs to measure her qutrits to disentangle qutrit $B_1$ from qutrits $t_1$ and $A_1$, as well as disentangle qutrit $B_2$ from qutrits $t_2$ and $A_2$. So, we rewrite the state $|T^{‴}⟩$ as

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |T^{‴}⟩& ={\displaystyle \frac{a_{10}{a}_{20}}{3}}{\{|0⟩}_{t_1}{|0⟩}_{A_1}(c_{10}|0⟩+c_{11}{\omega }^{2j}|2⟩+c_{12}{\omega }^j{|1⟩)}_{B_1}\hfill & \\ & +{|0⟩}_{t_1}{|1⟩}_{A_1}(c_{10}{\omega }^j|1⟩+c_{11}|0⟩+c_{12}{\omega }^{2j}{|2⟩)}_{B_1}\hfill & \\ & +{|0⟩}_{t_1}{|2⟩}_{A_1}(c_{10}{\omega }^{2j}|2⟩+c_{11}{\omega }^j|1⟩+c_{12}{|0⟩)}_{B_1}\hfill & \\ & +{|1⟩}_{t_1}{|0⟩}_{A_1}(c_{10}|0⟩+c_{11}{\omega }^{1+2j}|2⟩+c_{12}{\omega }^{2+j}{|1⟩)}_{B_1}\hfill & \\ & +{|1⟩}_{t_1}{|1⟩}_{A_1}(c_{10}{\omega }^j|1⟩+c_{11}\omega |0⟩+c_{12}{\omega }^{2+2j}{|2⟩)}_{B_1}\hfill & \\ & +{|1⟩}_{t_1}{|2⟩}_{A_1}(c_{10}{\omega }^{2j}|2⟩+c_{11}{\omega }^{1+j}|1⟩+c_{12}{\omega }^2{|0⟩)}_{B_1}\hfill & \\ & +{|2⟩}_{t_1}{|0⟩}_{A_1}(c_{10}|0⟩+c_{11}{\omega }^{2+2j}|2⟩+c_{12}{\omega }^{1+j}{|1⟩)}_{B_1}\hfill & \\ & +{|2⟩}_{t_1}{|1⟩}_{A_1}(c_{10}{\omega }^j|1⟩+c_{11}{\omega }^2|0⟩+c_{12}{\omega }^{1+2j}{|2⟩)}_{B_1}\hfill & \\ & +{|2⟩}_{t_1}{|2⟩}_{A_1}(c_{10}{\omega }^{2j}|2⟩+c_{11}{\omega }^{2+j}|1⟩+c_{12}{\omega |0⟩)}_{B_1}\}\hfill & \\ & \otimes {\{|0⟩}_{t_2}{|0⟩}_{A_2}(c_{20}|0⟩+c_{21}{\omega }^{2j}|2⟩+c_{22}{\omega }^j{|1⟩)}_{B_2}\hfill & \\ & +{|0⟩}_{t_2}{|1⟩}_{A_2}(c_{20}{\omega }^j|1⟩+c_{21}|0⟩+c_{22}{\omega }^{2j}{|2⟩)}_{B_2}\hfill & \\ & +{|0⟩}_{t_2}{|2⟩}_{A_2}(c_{20}{\omega }^{2j}|2⟩+c_{21}{\omega }^j|1⟩+c_{22}{|0⟩)}_{B_2}\hfill & \\ & +{|1⟩}_{t_2}{|0⟩}_{A_2}(c_{20}|0⟩+c_{21}{\omega }^{1+2j}|2⟩+c_{22}{\omega }^{2+j}{|1⟩)}_{B_2}\hfill & \\ & +{|1⟩}_{t_2}{|1⟩}_{A_2}(c_{20}{\omega }^j|1⟩+c_{21}\omega |0⟩+c_{22}{\omega }^{2+2j}{|2⟩)}_{B_2}\hfill & \\ & +{|1⟩}_{t_2}{|2⟩}_{A_2}(c_{20}{\omega }^{2j}|2⟩+c_{21}{\omega }^{1+j}|1⟩+c_{22}{\omega }^2{|0⟩)}_{B_2}\hfill & \\ & +{|2⟩}_{t_2}{|0⟩}_{A_2}(c_{20}|0⟩+c_{21}{\omega }^{2+2j}|2⟩+c_{22}{\omega }^{1+j}{|1⟩)}_{B_2}\hfill & \\ & +{|2⟩}_{t_2}{|1⟩}_{A_2}(c_{20}{\omega }^j|1⟩+c_{21}{\omega }^2|0⟩+c_{22}{\omega }^{1+2j}{|2⟩)}_{B_2}\hfill & \\ & +{|2⟩}_{t_2}{|2⟩}_{A_2}(c_{20}{\omega }^{2j}|2⟩+c_{21}{\omega }^{2+j}|1⟩+c_{22}{\omega |0⟩)}_{B_2}\}.\hfill & \end{array}\end{array}$$

(11)

Alice performs the projective measurements on qutrit pairs $(t_1,A_1)$ and $(t_2,A_2)$ separately with the orthogonal computational basis $\{|rs⟩:r,s=0,1,2\}$, and then informs Bob₁ and Bob₂ of the corresponding measurement results via classical channels, separately. Generally, if Alice’s measurement results are ${|rs⟩}_{t_1{A}_1}$ and ${|mn⟩}_{t_2{A}_2}$, then the states of qutrits $B_1$ and $B_2$ will collapse into

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill {|\phi ⟩}_{B_1{B}_2}& =\{\sum _{k=0}^2{c}_{1k}{\omega }^{rk+j\left[\right(s-k\left)\mathrm{mod}3\right]}|(s-k){\mathrm{mod}3⟩}_{B_1}\}\hfill & \\ & \otimes \{\sum _{l=0}^2{c}_{2l}{\omega }^{ml+j\left[\right(n-l\left)\mathrm{mod}3\right]}|(n-l){\mathrm{mod}3⟩}_{B_2}\}.\hfill & \end{array}\end{array}$$

(12)

Step 5: According to Alice’s measurement results ${|rs⟩}_{t_1{A}_1}$ and ${|mn⟩}_{t_2{A}_2}$ as well as Charlie’s outcome ${|j⟩}_C$, Bob₁ and Bob₂ respectively perform local unitary operations $U_{B_1}^{\left(jrs\right)}$ and $U_{B_2}^{\left(jmn\right)}$ on qutrits $B_1$ and $B_2$ to recover their respective target states, where $U_{B_1}^{\left(jrs\right)}$ and $U_{B_2}^{\left(jmn\right)}$ are given by

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill U_{B_1}^{\left(jrs\right)}& =\sum _{k=0}^2{\omega }^{-rk-j\left[\right(s-k\left)\mathrm{mod}3\right]}{|k⟩}_{B_1}⟨(s-k)\mathrm{mod}3|\hfill & \\ \hfill U_{B_2}^{\left(jmn\right)}& =\sum _{l=0}^2{\omega }^{-ml-j\left[\right(n-l\left)\mathrm{mod}3\right]}{|l⟩}_{B_2}⟨(n-l)\mathrm{mod}3⟩\hfill & \end{array}\end{array}$$

(13)

for any $r,s,n,m,j\in \{0,1,2\}$. After $U_{B_1}^{\left(jrs\right)}$ and $U_{B_2}^{\left(jmn\right)}$ are performed, the state ${|\phi ⟩}_{B_1{B}_2}$ shown in Equation (12) becomes

$$\begin{array}{c}\hfill |{\phi }^{\prime }{⟩}_{B_1{B}_2}=(U_{B_1}^{\left(jrs\right)}\otimes U_{B_2}^{\left(jmn\right)}){|\phi ⟩}_{B_1{B}_2}=(\sum _{k=0}^2{c_{1k}|k⟩}_{B_1})\otimes (\sum _{l=0}^2{c_{2l}|l⟩}_{B_2}).\end{array}$$

(14)

Comparing Equation (14) with the original states given by Equation (1), the initial unknown single-qutrit states $|{\chi }_1⟩$ and $|{\chi }_2⟩$ in Equation (1) are respectively reconstructed by the receivers Bob₁ and Bob₂.

Step 5’: If one of Bob₁ and Bob₂’s measurement results is ${|1⟩}_{e_1}$ or ${|1⟩}_{e_2}$, and Charlie’s outcome is ${|j⟩}_C$, then quantum teleportation fails, and so the state of qutrits $t_1,A_1$ and $B_1$ or the state of qutrits $t_2,A_2$ and $B_2$ becomes

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |\dot{T}⟩& ={\displaystyle \frac{1}{\sqrt{3}}}\{[c_{10}(|0⟩+|1⟩+|2⟩)|1⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|2⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|0⟩)]}_{t_1{A}_1}\otimes {\omega }^j\sqrt{a_{11}^2-a_{10}^2}{|1⟩}_{B_1}\hfill & \\ & +[c_{10}(|0⟩+|1⟩+|2⟩)|2⟩+c_{11}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|0⟩\hfill & \\ & +c_{12}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|1⟩)]}_{t_1{A}_1}\otimes {\omega }^{2j}\sqrt{a_{12}^2-a_{10}^2}{|2⟩}_{B_1}\}\hfill & \end{array}\end{array}$$

(15)

or

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |\ddot{T}⟩& ={\displaystyle \frac{1}{\sqrt{3}}}\{[c_{20}(|0⟩+|1⟩+|2⟩)|1⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|2⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|0⟩)]}_{t_2{A}_2}\otimes {\omega }^j\sqrt{a_{21}^2-a_{20}^2}{|1⟩}_{B_2}\hfill & \\ & +[c_{20}(|0⟩+|1⟩+|2⟩)|2⟩+c_{21}(|0⟩+\omega |1⟩+{\omega }^2|2⟩)|0⟩\hfill & \\ & +c_{22}\left(\right(|0⟩+{\omega }^2|1⟩+{\omega |2⟩)|1⟩)]}_{t_2{A}_2}\otimes {\omega }^{2j}\sqrt{a_{22}^2-a_{20}^2}{|2⟩}_{B_2}\}.\hfill & \end{array}\end{array}$$

(16)

Under the circumstances, Alice executes the inverse ${\mathcal{H}}^{\left(3\right)}$ gate operation ${\left({\mathcal{H}}^{\left(3\right)}\right)}^{-1}$ on her qutrit $t_1$ or $t_2$ and then carries out the inverse ${\mathcal{N}}^{\left(3\right)}$ gate operation ${\left({\mathcal{N}}^{\left(3\right)}\right)}^{-1}$ on her qutrit pair $(t_1,A_1)$ or $(t_2,A_2)$, where ${\left({\mathcal{H}}^{\left(3\right)}\right)}^{-1}$ is given by

$$\begin{array}{c}\hfill {\left({\mathcal{H}}^{\left(3\right)}\right)}^{-1}={\displaystyle \frac{1}{\sqrt{3}}}\sum _{k,k^{\prime }=0}^2{\omega }^{k{k}^{\prime }}|k⟩⟨k^{\prime }|\end{array}$$

(17)

and ${\left({\mathcal{N}}^{\left(3\right)}\right)}^{-1}$ is defined as

$$\begin{array}{c}\hfill {\left({\mathcal{N}}^{\left(3\right)}\right)}^{-1}|mn⟩=|m⟩|(n-m)\mathrm{mod}3⟩.\end{array}$$

(18)

After Alice performs ${\left({\mathcal{H}}^{\left(3\right)}\right)}^{-1}$ and ${\left({\mathcal{N}}^{\left(3\right)}\right)}^{-1}$, the state $|\dot{T}⟩$ or $|\ddot{T}⟩$ shown in Equations (15) and (16) becomes

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |{\dot{T}}^{\prime }⟩& ={\left({\mathcal{N}}^{\left(3\right)}\right)}^{-1}{\left({\mathcal{H}}^{\left(3\right)}\right)}^{-1}|\dot{T}⟩\hfill & \\ & =(\sum _{k=0}^2{c}_{1k}{|k⟩}_{t_1})\otimes ({\omega }^j\sqrt{a_{11}^2-a_{10}^2}|11⟩+{\omega }^{2j}\sqrt{a_{12}^2-a_{10}^2}{|22⟩)}_{A_1{B}_1}\hfill & \end{array}\end{array}$$

(19)

or

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |{\ddot{T}}^{\prime }⟩& ={\left({\mathcal{N}}^{\left(3\right)}\right)}^{-1}{\left({\mathcal{H}}^{\left(3\right)}\right)}^{-1}|\ddot{T}⟩\hfill & \\ & =(\sum _{k=0}^2{c}_{2k}{|k⟩}_{t_2})\otimes ({\omega }^j\sqrt{a_{21}^2-a_{20}^2}|11⟩+{\omega }^{2j}\sqrt{a_{22}^2-a_{20}^2}{|22⟩)}_{A_2{B}_2}.\hfill & \end{array}\end{array}$$

(20)

From Equations (19) and (20), it is obvious that the unknown state $|{\chi }_1{⟩}_{t_1}$ or $|{\chi }_2{⟩}_{t_2}$ is preserved. The teleportation fails if Bob₁ obtains outcome ${|1⟩}_{e_1}$ or Bob₂ obtains outcome ${|1⟩}_{e_2}$; Alice could recover the original state $|{\chi }_1{⟩}_{t_1}$ or $|{\chi }_2{⟩}_{t_2}$ via executing the ${\left({\mathcal{H}}^{\left(3\right)}\right)}^{-1}$ and ${\left({\mathcal{N}}^{\left(3\right)}\right)}^{-1}$ operations, thus maintaining the integrity of the unknown state $|{\chi }_1{⟩}_{t_1}$ or $|{\chi }_2{⟩}_{t_2}$.

All in all, we propose the scheme of two-output nondestructive and controlled teleportation, where with the control of supervisor Charlie, two unknown single-qutrit states can be recovered by the two receivers Bob₁ and Bob₂, respectively, or at least one can be preserved by Alice according to Bob₁ and Bob₂’s measurement outcomes of their auxiliary qubits.

3. Two-Output Teleportation of Two Unknown Single-Qudit States

Now, we turn to the general scheme for teleporting two qudits (d-level quantum systems) through a partially entangled five-qudit quantum channel. Since it is direct to extend the above scheme to d-level systems, we present the following results in compact general formulae in this part. Let two initial unknown d-dimensional single-qudit states be

$$\begin{array}{c}\hfill |{\xi }_k{⟩}_{t_k}=\sum _{j=0}^{d-1}{\beta }_{kj}{|j⟩}_{t_k},k=1,2,\end{array}$$

(21)

where ${\beta }_{kj}$ ($k=1,2;j=0,1,\cdots ,d-1$) are complex numbers satisfying the normalization conditions ${\sum }_{j=0}^{d-1}{|{\beta }_{kj}|}^2=1$ for all $k\in \{1,2\}$. The sender Alice intends to teleport the states $|{\xi }_1{⟩}_{t_1}$ and $|{\xi }_2{⟩}_{t_2}$ to the receivers Bob₁ and Bob₂, respectively, under the control of supervisor Charlie. In addition, Alice, Bob₁, Bob₂ and Charlie share a d-dimensional five-qudit partially entangled state as the quantum channel in advance, which can be generally written in the form of the Schmidt disposition as

$$\begin{array}{c}\hfill {|Q⟩}_{A_1{B}_1{A}_2{B}_{2}C}={\displaystyle \frac{1}{\sqrt{d}}}\sum _{j=0}^{d-1}{|j⟩}_C(\sum _{k=0}^{d-1}{b}_{1k}{\zeta }^{jk}{|kk⟩}_{A_1{B}_1}\sum _{l=0}^{d-1}{b}_{2l}{\zeta }^{jl}{|ll⟩}_{A_2{B}_2}),\end{array}$$

(22)

where $\zeta ={\mathrm{e}}^{2\pi \mathrm{i}/d}$, and real coefficients $b_{1k}$ and $b_{2l}$ ($k,l=0,1,\cdots ,d-1$) satisfy the conditions ${\sum }_{k=0}^{d-1}{b}_{1k}^2=1$ and ${\sum }_{l=0}^{d-1}{b}_{2l}^2=1$. Without loss of generality, we assume that $b_{1k}$ and $b_{2l}$ ($k,l=0,1,\cdots ,d-1$) are non-negative real numbers and $b_{k0}=min\{b_{kj}:j=0,1,\cdots ,d-1\}$ for all $k\in \{1,2\}$. Alice possesses qudits $A_1$ and $A_2$, while qudits $B_1,B_2$ and C belong to Bob₁, Bob₂ and Charlie, respectively.

In the beginning, Bob₁ and Bob₂ introduce two auxiliary qubits $e_1$ and $e_2$ with the initial states ${|0⟩}_{e_1}$ and ${|0⟩}_{e_2}$, respectively. The state of the entire initial system can be represented as

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |\mathcal{T}⟩& =|{\xi }_1{⟩}_{t_1}|{\xi }_2{⟩}_{t_2}{|Q⟩}_{A_1{B}_1{A}_2{B}_{2}C}{|0⟩}_{e_1}{|0⟩}_{e_2}\hfill & \\ & ={\displaystyle \frac{1}{\sqrt{d}}}\sum _{j=0}^{d-1}{|j⟩}_C[(\sum _{u=0}^{d-1}{\beta }_{1u}|u{⟩}_{t_1})\otimes (\sum _{k=0}^{d-1}{b}_{1k}{\zeta }^{jk}|kk{⟩}_{A_1{B}_1}{)\otimes |0⟩}_{e_1}]\hfill & \\ & \otimes [(\sum _{v=0}^{d-1}{\beta }_{2v}|v{⟩}_{t_2})\otimes (\sum _{l=0}^{d-1}{b}_{2l}{\zeta }^{jl}|ll{⟩}_{A_2{B}_2}{)\otimes |0⟩}_{e_2}]\hfill & \\ & ={\displaystyle \frac{1}{\sqrt{d}}}\sum _{j=0}^{d-1}{|j⟩}_C[\sum _{k=0}^{d-1}((\sum _{u=0}^{d-1}{\beta }_{1u}|uk{⟩}_{t_1{A}_1})\otimes b_{1k}{\zeta }^{jk}|k0{⟩}_{B_1{e}_1})]\hfill & \\ & \otimes [\sum _{l=0}^{d-1}((\sum _{v=0}^{d-1}{\beta }_{2v}|{vl⟩}_{t_2{A}_2})\otimes b_{2l}{\zeta }^{jl}|{l0⟩}_{B_2{e}_2})].\hfill & \end{array}\end{array}$$

(23)

To directly teleport two unknown single-qudit states $|{\xi }_1{⟩}_{t_1}$ and $|{\xi }_2{⟩}_{t_2}$ from sender Alice to receivers Bob₁ and Bob₂, respectively, our scheme is also divided into five steps as follows.

Step 1: Alice performs the d-dimensional controlled-NOT gate ${\mathcal{N}}^{\left(d\right)}$ operations on her own qudit pairs $(t_1,A_1)$ and $(t_2,A_2)$, respectively, and then carries out the d-dimensional Hadamard gate ${\mathcal{H}}^{\left(d\right)}$ operations on her qudits $t_1$ and $t_2$, respectively, where ${\mathcal{N}}^{\left(d\right)}$ can be defined as

$$\begin{array}{c}\hfill {\mathcal{N}}^{\left(d\right)}|mn⟩=|m⟩|(m+n)\mathrm{mod}d⟩\end{array}$$

(24)

and ${\mathcal{H}}^{\left(d\right)}$ can be expressed as

$$\begin{array}{c}\hfill {\mathcal{H}}^{\left(d\right)}={\displaystyle \frac{1}{\sqrt{d}}}\sum _{\lambda =0}^{d-1}{\zeta }^{\lambda {\lambda }^{\prime }}|\lambda ⟩⟨{\lambda }^{\prime }|.\end{array}$$

(25)

After the ${\mathcal{N}}^{\left(d\right)}$ and ${\mathcal{H}}^{\left(d\right)}$ operations, the state shown in Equation (23) becomes

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |{\mathcal{T}}^{\prime }⟩& =[{({\mathcal{H}}^{(d)})}_{t_1}\otimes {({\mathcal{H}}^{(d)})}_{t_2}][{({\mathcal{N}}^{\left(d\right)})}_{t_1{A}_1}\otimes {({\mathcal{N}}^{\left(d\right)})}_{t_2{A}_2}]|\mathcal{T}⟩\hfill & \\ & ={\displaystyle \frac{1}{d\sqrt{d}}}\sum _{j=0}^{d-1}{|j⟩}_C[\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{\beta }_{1r}(\sum _{u=0}^{d-1}{\zeta }^{ur}|u,(r+s)\mathrm{mod}d{⟩}_{t_1{A}_1})\otimes b_{1s}{\zeta }^{js}|s0{⟩}_{B_1{e}_1})]\hfill & \\ & \otimes [\sum _{m=0}^{d-1}(\sum _{n=0}^{d-1}{\beta }_{2n}(\sum _{v=0}^{d-1}{\zeta }^{vn}|v,(n+m)\mathrm{mod}d{⟩}_{t_2{A}_2})\otimes b_{2m}{\zeta }^{jm}|m0{⟩}_{B_2{e}_2})].\hfill & \end{array}\end{array}$$

(26)

Step 2: Bob₁ and Bob₂ carry out the unitary operations ${\tilde{U}}_{B_1{e}_1}$ and ${\tilde{U}}_{B_2{e}_2}$ on the qudit pairs $(B_1,e_1)$ and $(B_2,e_2)$, respectively, which are given by

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill {\tilde{U}}_{B_1{e}_1}& =\sum _{k=0}^{d-1}[(({\displaystyle \frac{b_{10}}{b_{1k}}})|k0{⟩}_{B_1{e}_1}⟨k0|+\sqrt{1-{({\displaystyle \frac{b_{10}}{b_{1k}}})}^2}{|k0⟩}_{B_1{e}_1}⟨k1|)\hfill & \\ & +(-\sqrt{1-{({\displaystyle \frac{b_{10}}{b_{1k}}})}^2}|k1{⟩}_{B_1{e}_1}⟨k0|+({\displaystyle \frac{b_{10}}{b_{1k}}}){|k1⟩}_{B_1{e}_1}⟨k1|)],\hfill & \\ \hfill {\tilde{U}}_{B_2{e}_2}& =\sum _{l=0}^{d-1}[(({\displaystyle \frac{b_{20}}{b_{2l}}})|l0{⟩}_{B_2{e}_2}⟨l0|+\sqrt{1-{({\displaystyle \frac{b_{20}}{b_{2l}}})}^2}{|l0⟩}_{B_2{e}_2}⟨l1|)\hfill & \\ & +(-\sqrt{1-{({\displaystyle \frac{b_{20}}{b_{2l}}})}^2}|l1{⟩}_{B_2{e}_2}⟨l0|+({\displaystyle \frac{b_{20}}{b_{2l}}}){|l1⟩}_{B_2{e}_2}⟨l1|)].\hfill & \end{array}\end{array}$$

(27)

After implementing ${\tilde{U}}_{B_1{e}_1}$ and ${\tilde{U}}_{B_2{e}_2}$ operations by Bob₁ and Bob₂, respectively, the state $|{\mathcal{T}}^{\prime }⟩$ shown in Equation (26) is transformed to

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |{\mathcal{T}}^{″}⟩& =({\tilde{U}}_{B_1{e}_1}\otimes {\tilde{U}}_{B_2{e}_2})|{\mathcal{T}}^{\prime }⟩\hfill & \\ & ={\displaystyle \frac{1}{d\sqrt{d}}}\sum _{j=0}^{d-1}{|j⟩}_C[\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{\beta }_{1r}(\sum _{u=0}^{d-1}{\zeta }^{ur}|u,(r+s)\mathrm{mod}d{⟩}_{t_1{A}_1})\hfill & \\ & \otimes {\zeta }^{js}(b_{10}|s0⟩-\sqrt{b_{1s}^2-b_{10}^2}{|s1⟩)}_{B_1{e}_1})]\hfill & \\ & \otimes [\sum _{m=0}^{d-1}(\sum _{n=0}^{d-1}{\beta }_{2n}(\sum _{v=0}^{d-1}{\zeta }^{vn}|v,(n+m)\mathrm{mod}d{⟩}_{t_2{A}_2})\hfill & \\ & \otimes {\zeta }^{jm}(b_{20}|m0⟩-\sqrt{b_{2m}^2-b_{20}^2}{|m1⟩)}_{B_2{e}_2})].\hfill & \end{array}\end{array}$$

(28)

Bob₁ and Bob₂ measure their respective auxiliary qubits $e_1$ and $e_2$ with the basis $\{|0⟩,|1⟩\}$. There are four possible measurement outcomes: ${|0⟩}_{e_1}{|0⟩}_{e_2}$ or ${|0⟩}_{e_1}{|1⟩}_{e_2}$ or ${|1⟩}_{e_1}{|0⟩}_{e_2}$ or ${|1⟩}_{e_1}{|1⟩}_{e_2}$. If Bob₁ and Bob₂’s measurement result is ${|0⟩}_{e_1}{|0⟩}_{e_2}$, step 5 will be executed to recover the original unknown single-qudit states by receivers Bob₁ and Bob₂ after supervisor Charlie’s local operation in step 3 and Alice’s measurements in step 4. Otherwise, step $5^{\prime }$ will be implemented to preserve the original information on Alice’s location.

It is worth mentioning that steps 1 and 2 can be independently executed by the sender and receivers, so they could be performed synchronously.

Step 3: If supervisor Charlie agrees to assist other participants, he measures his qudit C in the computational basis $\{|0⟩,|1⟩,\cdots ,|d-1⟩,\}$, and publishes his measurement result ${|j⟩}_C$ ($j=0,1,\cdots ,d-1$) to the receivers via classical communication. Then, the state of the left qudits may collapse into

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |{\mathcal{T}}^{‴}⟩& ={\displaystyle \frac{1}{d}}[\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{\beta }_{1r}(\sum _{u=0}^{d-1}{\zeta }^{ur}|u,(r+s)\mathrm{mod}d{⟩}_{t_1{A}_1})\hfill & \\ & \otimes {\zeta }^{js}(b_{10}|s0⟩-\sqrt{b_{1s}^2-b_{10}^2}{|s1⟩)}_{B_1{e}_1})]\hfill & \\ & \otimes [\sum _{m=0}^{d-1}(\sum _{n=0}^{d-1}{\beta }_{2n}(\sum _{v=0}^{d-1}{\zeta }^{vn}|v,(n+m)\mathrm{mod}d{⟩}_{t_2{A}_2})\hfill & \\ & \otimes {\zeta }^{jm}(b_{20}|m0⟩-\sqrt{b_{2m}^2-b_{20}^2}{|m1⟩)}_{B_2{e}_2})].\hfill & \end{array}\end{array}$$

(29)

Step 4: If Charlie’s outcome is ${|j⟩}_C$, and Bob₁ and Bob₂’s output is ${|0⟩}_{e_1}{|0⟩}_{e_2}$, then the state of qudits $t_1,A_1,B_1,t_2,A_2$ and $B_2$ will collapse into

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill {|\psi ⟩}_{t_1{A}_1{B}_1{t}_2{A}_2{B}_2}& ={\displaystyle \frac{b_{10}{b}_{20}}{d}}[\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{\beta }_{1r}(\sum _{u=0}^{d-1}{\zeta }^{ur}|u,(r+s)\mathrm{mod}d{⟩}_{t_1{A}_1})\otimes {\zeta }^{js}|s{⟩}_{B_1})]\hfill & \\ & \otimes [\sum _{m=0}^{d-1}(\sum _{n=0}^{d-1}{\beta }_{2n}(\sum _{v=0}^{d-1}{\zeta }^{vn}|v,(n+m)\mathrm{mod}d{⟩}_{t_2{A}_2})\otimes {\zeta }^{jm}|m{⟩}_{B_2})]\hfill & \\ & ={\displaystyle \frac{b_{10}{b}_{20}}{d}}[\sum _{r,s=0}^{d-1}{(|rs⟩}_{t_1{A}_1}(\sum _{u=0}^{d-1}{\beta }_{1u}{\zeta }^{ur+j\left[\right(s-u\left)\mathrm{mod}d\right]}|(s-u)\mathrm{mod}d{⟩}_{B_1}))]\hfill & \\ & \otimes [\sum _{n,m=0}^{d-1}{(|nm⟩}_{t_2{A}_2}(\sum _{v=0}^{d-1}{\beta }_{2v}{\zeta }^{vn+j\left[\right(m-v\left)\mathrm{mod}d\right]}|(m-v)\mathrm{mod}d{⟩}_{B_2}))].\hfill & \end{array}\end{array}$$

(30)

Alice executes the projective measurements on her qudit pairs $(t_1,A_1)$ and $(t_2,A_2)$ in the orthogonal basis of $\{|kl⟩:k,l\in \{0,1,\cdots ,d-1\}\}$, and then informs Bob₁ and Bob₂ of her measurement results ${|rs⟩}_{t_1{A}_1}$ and ${|nm⟩}_{t_2{A}_2}$, respectively, through classical communication. Then, the corresponding collapsed state of the remaining qudits $B_1$ and $B_2$ is

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill {|\psi ⟩}_{B_1{B}_2}& =(\sum _{u=0}^{d-1}{\beta }_{1u}{\zeta }^{ur+j\left[\right(s-u\left)\mathrm{mod}d\right]}|(s-u)\mathrm{mod}d{⟩}_{B_1})\hfill & \\ & \otimes (\sum _{v=0}^{d-1}{\beta }_{2v}{\zeta }^{vn+j\left[\right(m-v\left)\mathrm{mod}d\right]}|(m-v)\mathrm{mod}d{⟩}_{B_2}).\hfill & \end{array}\end{array}$$

(31)

Step 5: Based on the measurement information from Alice and Charlie, Bob₁ performs local operation

$$\begin{array}{c}\hfill {\mathcal{U}}_{B_1}^{\left(jrs\right)}=\sum _{u=0}^{d-1}{\zeta }^{-ur-j\left[\right(s-u\left)\mathrm{mod}d\right]}{|u⟩}_{B_1}⟨(s-u)\mathrm{mod}d|\end{array}$$

(32)

on his qudit $B_1$ to restore the initial unknown d-dimensional single-qudit state. At the same time, Bob₂ also carries out local operation

$$\begin{array}{c}\hfill {\mathcal{U}}_{B_2}^{\left(jnm\right)}=\sum _{v=0}^{d-1}{\zeta }^{-vn-j\left[\right(m-v\left)\mathrm{mod}d\right]}{|v⟩}_{B_2}⟨(m-v)\mathrm{mod}d|\end{array}$$

(33)

on his qudit $B_2$ to restore the initial unknown d-dimensional single-qudit state. As a result,

$$\begin{array}{c}\hfill ({\mathcal{U}}_{B_1}^{\left(jrs\right)}\otimes {\mathcal{U}}_{B_2}^{\left(jnm\right)}){|\psi ⟩}_{B_1{B}_2}=(\sum _{u=0}^{d-1}{\beta }_{1u}{|u⟩}_{B_1})\otimes (\sum _{v=0}^{d-1}{\beta }_{2v}|v{⟩}_{B_2}{)=|\xi ⟩}_{B_1}\otimes {|\xi ⟩}_{B_2},\end{array}$$

(34)

which means that Bob₁ and Bob₂ have successfully reconstructed their respective target states, thus achieving quantum teleportation in d-dimensional space.

Step 5’: If one of Bob₁ and Bob₂’s measurement outputs is ${|1⟩}_{e_1}$ or ${|1⟩}_{e_2}$ and Charlie’s outcome is ${|j⟩}_C$, then the teleportation fails, and so the state of qudits $t_1,A_1$ and B collapses into

$$\begin{array}{c}\hfill |{\psi }^{″}{⟩}_{B_1{B}_2}={\displaystyle \frac{1}{\sqrt{d}}}\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{\beta }_{1r}(\sum _{u=0}^{d-1}{\zeta }^{ur}|u,(r+s)\mathrm{mod}d{⟩}_{t_1{A}_1})\otimes {\zeta }^{js}\sqrt{b_{1s}^2-b_{10}^2}{|s⟩}_{B_1})\end{array}$$

(35)

or

$$\begin{array}{c}\hfill |{\psi }^{‴}{⟩}_{B_1{B}_2}={\displaystyle \frac{1}{\sqrt{d}}}\sum _{m=0}^{d-1}(\sum _{n=0}^{d-1}{\beta }_{2n}(\sum _{v=0}^{d-1}{\zeta }^{vn}|v,(n+m)\mathrm{mod}d{⟩}_{t_2{A}_2})\otimes {\zeta }^{jm}\sqrt{b_{2m}^2-b_{20}^2}{|m⟩}_{B_2}).\end{array}$$

(36)

In this situation, Alice carries out an inverse ${\mathcal{H}}^{\left(d\right)}$ gate operation ${\left({\mathcal{H}}^{\left(d\right)}\right)}^{-1}$ on her qudit $t_1$ or $t_2$, and then performs an inverse ${\mathcal{N}}^{\left(d\right)}$ gate operation ${\left({\mathcal{N}}^{\left(d\right)}\right)}^{-1}$ on her qudit pair $(t_1,A_1)$ and $(t_2,A_2)$. ${\left({\mathcal{H}}^{\left(d\right)}\right)}^{-1}$ can be written as

$$\begin{array}{c}\hfill {\left({\mathcal{H}}^{\left(d\right)}\right)}^{-1}={\displaystyle \frac{1}{\sqrt{d}}}\sum _{k,k^{\prime }=0}^{d-1}{\zeta }^{-k{k}^{\prime }}|k⟩⟨k^{\prime }|,\end{array}$$

(37)

and ${\left({\mathcal{N}}^{\left(d\right)}\right)}^{-1}$ is defined as

$$\begin{array}{c}\hfill {\left({\mathcal{N}}^{\left(d\right)}\right)}^{-1}|mn⟩=|m⟩|(n-m)\mathrm{mod}d⟩.\end{array}$$

(38)

After ${\left({\mathcal{H}}^{\left(d\right)}\right)}^{-1}$ and ${\left({\mathcal{N}}^{\left(d\right)}\right)}^{-1}$ are performed by Alice, the state shown in Equation (35) or (36) becomes

$$\begin{array}{c}\hfill {\left({\mathcal{N}}^{\left(d\right)}\right)}^{-1}{\left({\mathcal{H}}^{\left(d\right)}\right)}^{-1}|{\psi }^{″}{⟩}_{B_1{B}_2}=(\sum _{r=0}^{d-1}{\beta }_{1r}|r{⟩}_{t_1})\otimes (\sum _{s=0}^{d-1}{\zeta }^{js}\sqrt{b_{1s}^2-b_{10}^2}|ss{⟩}_{A_1{B}_1})\end{array}$$

(39)

or

$$\begin{array}{c}\hfill {\left({\mathcal{N}}^{\left(d\right)}\right)}^{-1}{\left({\mathcal{H}}^{\left(d\right)}\right)}^{-1}|{\psi }^{‴}{⟩}_{B_1{B}_2}=(\sum _{n=0}^{d-1}{\beta }_{2n}|n{⟩}_{t_2})\otimes (\sum _{m=0}^{d-1}{\zeta }^{jm}\sqrt{b_{2m}^2-b_{20}^2}|mm{⟩}_{A_2{B}_2}).\end{array}$$

(40)

From Equation (39) or (40), it is obvious that the unknown single-qudit state $|{\xi }_1⟩$ or $|{\xi }_2⟩$ is preserved in qubit $t_1$ or $t_2$. The controlled teleportation fails if one of Bob₁ and Bob₂’s measurement results is ${|1⟩}_{e_1}$ or ${|1⟩}_{e_2}$; Alice could restore the corresponding original state $|{\xi }_1{⟩}_{t_1}$ or $|{\xi }_2{⟩}_{t_2}$, via executing an ${\left({\mathcal{H}}^{\left(d\right)}\right)}^{-1}$ and ${\left({\mathcal{N}}^{\left(d\right)}\right)}^{-1}$ operation, thus maintaining the integrity of the unknown singe-qudit state $|{\xi }_1{⟩}_{t_1}$ or $|{\xi }_2{⟩}_{t_2}$, no matter what Charlie’s measurement result is.

In a word, we present the nondestructive and controlled scheme for two-output teleportation, in which with the permission of controller Charlie, two unknown single-qudit states are either restored by Bob₁ and Bob₂, respectively, or at least one of these unknown states is preserved by Alice according to Bob₁ and Bob₂’s measurement outcomes of their respective auxiliary qubits. In fact, for any measurement result ${|1⟩}_{e_j}$ of auxiliary qubit $e_j$ ($j=1,2$), its corresponding original unknown state is preserved by Alice, while for any measurement result ${|0⟩}_{e_j}$ of auxiliary qubit $e_j$ ($j=1,2$), Bob_j can successfully reconstruct the original unknown single-qudit state $|{\xi }_j⟩$ on his qudit $t_j$ by using an appropriate local operation on his qudit $B_j$.

4. Multi-Output Teleportation of Unknown $\mathit{d}$-Dimensional Single-Qudit States

This section further promotes the scheme in Section 2. Assume that Alice has the following N unknown d-dimensional single-qudit states:

$$\begin{array}{c}\hfill |{\chi }_n{⟩}_{t_n}=\sum _{j=0}^{d-1}{c}_{nj}{|j⟩}_{t_n},n=1,2,\cdots ,N,\end{array}$$

(41)

where N is a positive integer greater than 2, and complex coefficients $c_{n0},c_{n1},\cdots ,c_{n(d-1)}$ satisfy the normalization condition ${\sum }_{j=0}^{d-1}{|c_{nj}|}^2=1$ for all $n\in \{1,2,\cdots ,N\}$. Alice wants to teleport the k-th state $|{\chi }_k{⟩}_{t_k}$ to the k-th receiver Bob_k for any $k\in \{1,2,\cdots ,N\}$, under the control of supervisor Charlie. Sender Alice; receivers Bob₁, Bob₂, ⋯, Bob_N; and supervisor Charlie share a d-dimensional $(2N+1)$-qudit channel in advance, which can be represented as

$$\begin{array}{c}\hfill |\varphi ⟩={\displaystyle \frac{1}{\sqrt{d}}}\sum _{l=0}^{d-1}{|l⟩}_C(\prod _{n=1}^N(\sum _{k=0}^{d-1}{a}_{nk}{\zeta }^{lk}|kk{⟩}_{A_n{B}_n})),\end{array}$$

(42)

where real number $a_{nk}$ ($n=1,2,\cdots ,N$; $k=0,1,\cdots ,d-1$) is the channel parameters that satisfy the relationship ${\sum }_{k=0}^{d-1}{a}_{nk}^2=1$ for any $n\in \{1,2,\cdots ,N\}$. Without loss of generality, we assume that $a_{n0}=min\{a_{n0},a_{n1},\cdots ,a_{n(d-1)}\}$ for any $n\in \{1,2,\cdots ,N\}$. Alice possesses qudit groups $(t_1,t_2,\cdots ,t_N)$ and $(A_1,A_2,\cdots ,A_N)$, while Bob₁, Bob₂, ⋯, Bob_N and Charlie possess qudits $B_1,B_2,\cdots ,B_N$ and C, respectively.

In the beginning, each Bob_n ($n=1,2,\cdots ,N$) introduces an auxiliary qubit $e_n$ with the initial state ${|0⟩}_{e_n}$. The state of the entire initial system can be written as

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |\Theta ⟩& =\prod _{n=1}^N|{\chi }_n{⟩}_{t_n}\otimes |\varphi ⟩\otimes \prod _{n=1}^N{|0⟩}_{e_n}\hfill & \\ & =\prod _{n=1}^N(\sum _{j=0}^{d-1}{c}_{nj}|j{⟩}_{t_n})\otimes [{\displaystyle \frac{1}{\sqrt{d}}}\sum _{l=0}^{d-1}|l{⟩}_C(\prod _{n=1}^N(\sum _{k=0}^{d-1}{a}_{nk}{\zeta }^{lk}|kk{⟩}_{A_n{B}_n}))]\otimes \prod _{n=1}^N{|0⟩}_{e_n}\hfill & \\ & ={\displaystyle \frac{1}{\sqrt{d}}}\sum _{l=0}^{d-1}{|l⟩}_C\{\prod _{n=1}^N[(\sum _{j=0}^{d-1}{c}_{nj}|j{⟩}_{t_n})\otimes (\sum _{k=0}^{d-1}{a}_{nk}{\zeta }^{lk}|kk{⟩}_{A_n{B}_n}){\otimes |0⟩}_{e_n}]\}\hfill & \\ & ={\displaystyle \frac{1}{\sqrt{d}}}\sum _{l=0}^{d-1}{|l⟩}_C\{\prod _{n=1}^N[\sum _{k=0}^{d-1}((\sum _{j=0}^{d-1}{c}_{nj}|jk{⟩}_{t_n{A}_n})\otimes a_{nk}{\zeta }^{lk}|k0{⟩}_{B_n{e}_n})]\}.\hfill & \end{array}\end{array}$$

(43)

To complete the quantum task, one needs to go through the following five steps in sequence.

Step 1: Alice applies the d-dimensional controlled-NOT gate ${\mathcal{N}}^{\left(d\right)}$ operations shown in Equation (24) on her qudit pairs $(t_1,A_1)$, $(t_2,A_2)$, ⋯, $(t_N,A_N)$, separately, and then performs the d-dimensional Hadamard gate ${\mathcal{H}}^{\left(d\right)}$ operations shown in Equation (25) on her qudits $t_1,t_2,\cdots ,t_N$, respectively. Then, the state shown in Equation (43) becomes

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |{\Theta }^{\prime }⟩& =\prod _{n=1}^N[{\left({\mathcal{H}}^{\left(d\right)}\right)}_{t_n}{\left({\mathcal{N}}^{\left(d\right)}\right)}_{t_n{A}_n}]|\Theta ⟩\hfill & \\ & ={\displaystyle \frac{1}{\sqrt{d^{N+1}}}}\sum _{l=0}^{d-1}{|l⟩}_C\{\prod _{n=1}^N[\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{c}_{nr}(\sum _{j=0}^{d-1}{\zeta }^{jr}|j,(r+s)\mathrm{mod}d{⟩}_{t_n{A}_n})\otimes a_{ns}{\zeta }^{ls}|s0{⟩}_{B_n{e}_n})]\}.\hfill & \end{array}\end{array}$$

(44)

Step 2: Each Bob_n ($n=1,2,\cdots ,N$) executes a unitary operation $U_{B_n{e}_n}$ on his two qudits $B_n$ and $e_n$ given by

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill U_{B_n{e}_n}& =\sum _{k=0}^{d-1}(({\displaystyle \frac{a_{n0}}{a_{nk}}}|k0{⟩}_{B_n{e}_n}⟨k0|+\sqrt{1-{({\displaystyle \frac{a_{n0}}{a_{nk}}})}^2}{|k0⟩}_{B_n{e}_n}⟨k1|)\hfill & \\ & -(\sqrt{1-{\left({\displaystyle \frac{a_{n0}}{a_{nk}}}\right)}^2}|k1{⟩}_{B_n{e}_n}⟨k0|-{\displaystyle \frac{a_{n0}}{a_{nk}}}{|k1⟩}_{B_n{e}_n}⟨k1|)),\hfill & \end{array}\end{array}$$

(45)

where $n=1,2,\cdots ,N$. After all Bob operations, the entire $(4N+1)$-qudit state shown in Equation (44) is transformed to

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |{\Theta }^{″}⟩& =\prod _{n=1}^N{U}_{B_n{e}_n}|{\Theta }^{″}⟩\hfill & \\ & ={\displaystyle \frac{1}{\sqrt{d^{N+1}}}}\sum _{l=0}^{d-1}{|l⟩}_C\{\prod _{n=1}^N[\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{c}_{nr}(\sum _{j=0}^{d-1}{\zeta }^{jr}|j,(r+s)\mathrm{mod}d{⟩}_{t_n{A}_n})\hfill & \\ & \otimes {\zeta }^{ls}(a_{n0}|s0⟩-\sqrt{a_{ns}^2-a_{n0}^2}{|s1⟩)}_{B_n{e}_n})]\}.\hfill & \end{array}\end{array}$$

(46)

Each Bob_n ($n=1,2,\cdots ,N$) measures his auxiliary qubit $e_n$ in the basis of $\{|0⟩,|1⟩\}$. There are a total of $2^N$ possible measurement outputs. The teleportation is considered successful as a whole only when all $2^N$ measurement outputs are $vert1⟩$. Otherwise, the teleportation will be considered a failure. When the teleportation is successful, step 5 will be implemented for reconstructing N original unknown states by their respective corresponding receivers after Charlie’s operation in step 3 and Alice’s measurements in step 4. If the teleportation fails, among these $N^2$ measurement outputs, there may be q measurement outputs that are all $|1⟩$, and $(2^N-q)$ measurement outputs that are all $|0⟩$. Each receiver with a measurement result $|0⟩$ can reconstruct his desired original unknown state; Alice can preserve the original unknown states corresponding to the measurement results all being $|1⟩$ for the recipients.

Step 3: Charlie agrees to cooperate with the recipients; he measures his qudit C with the computational basis $\{|l⟩:l=0,1,\cdots ,d-1\}$ and publicly informs the recipients of his measurement result ${|l⟩}_C$ via classical communication. Then, the state of the left particles collapses into

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |\Xi ⟩& ={\displaystyle \frac{1}{\sqrt{d^N}}}\prod _{n=1}^N[\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{c}_{nr}(\sum _{j=0}^{d-1}{\zeta }^{jr}|j,(r+s)\mathrm{mod}d{⟩}_{t_n{A}_n})\hfill & \\ & \otimes {\zeta }^{ls}(a_{n0}|s0⟩-\sqrt{a_{ns}^2-a_{n0}^2}{|s1⟩)}_{B_n{e}_n}\left)\right].\hfill & \end{array}\end{array}$$

(47)

Step 4: If all Bob’s measurement results are $|0⟩$ and Charlie’s outcome is ${|l⟩}_C$, the state of the left qudits will become

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill |{\Xi }^{\prime }⟩& ={\displaystyle \frac{{\prod }_{n=1}^N{a}_{n0}}{\sqrt{d^N}}}\prod _{n=1}^N[\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{c}_{nr}(\sum _{j=0}^{d-1}{\zeta }^{jr}|j,(r+s)\mathrm{mod}d{⟩}_{t_n{A}_n})\otimes {\zeta }^{ls}|s⟩{)}_{B_n}]\hfill & \\ & ={\displaystyle \frac{{\prod }_{n=1}^N{a}_{n0}}{\sqrt{d^N}}}\prod _{n=1}^N[\sum _{r,s=0}^{d-1}{(|rs⟩}_{t_n{A}_n}\otimes (\sum _{j=0}^{d-1}{c}_{nj}{\zeta }^{jr+l\left[\right(s-j\left)\mathrm{mod}d\right]}|(s-j)\mathrm{mod}d{⟩}_{B_n}))].\hfill & \end{array}\end{array}$$

(48)

Alice performs the projective measurements on her qudit pairs $(t_1,A_1)$, $(t_2,A_2)$, ⋯, $(t_N,A_N)$, respectively, in the orthogonal basis $\{|rs⟩:r,s=0,1,2,\cdots ,d-1\}$, and informs Bob_n of her measurement result ${|rs⟩}_{t_n{A}_n}$ for any $n\in \{1,2,\cdots ,N\}$. Then, the state of qudits $B_1,B_2,\cdots ,d-i$ will collapse into

$$\begin{array}{c}\hfill |{\Xi }^{″}⟩=\prod _{n=1}^N(\sum _{j=0}^{d-1}{c}_{nj}{\zeta }^{jr+l\left[\right(s-j\left)\mathrm{mod}d\right]}|(s-j)\mathrm{mod}d{⟩}_{B_n}).\end{array}$$

(49)

Step 5: According to Alice’s measurement outcome ${\sum }_{n=0}^N{|rs⟩}_{t_n{A}_n}$ and Charlie’s outcome ${|l⟩}_C$, for any $n\in \{1,2,\cdots ,N\}$, each Bob_n performs local unitary operation ${\mathcal{U}}_{B_n}^{lrs}$ on her qudit $B_n$ to recover his target state, where ${\mathcal{U}}_{B_n}^{lrs}$ is given by

$$\begin{array}{c}\hfill {\mathcal{U}}_{B_n}^{lrs}=\sum _{j=0}^{d-1}{\zeta }^{-jr-l\left[\right(s-j\left)\mathrm{mod}d\right]}{|j⟩}_{B_n}⟨(s-j)\mathrm{mod}d|.\end{array}$$

(50)

After all ${\mathcal{U}}_{B_n}^{jrs}$ ($n=1,2,\cdots ,N$) are executed, the state of qubits $B_1$, $B_2$, ⋯, $B_N$ becomes

$$\begin{array}{c}\hfill (\prod _{n=1}^N{\mathcal{U}}_{B_n}^{lrs})|{\Xi }^{″}⟩=\prod _{n=1}^N(\sum _{j=0}^{d-1}{c}_{nj}|j{⟩}_{B_n}).\end{array}$$

(51)

Comparing Equation (51) with the original states given by Equation (41), the initial unknown states $|{\chi }_1⟩$, $|{\chi }_1⟩$, ⋯, $|{\chi }_N⟩$ in Equation (41) are recovered by receivers Bob₁, Bob₂, ⋯, Bob_N, respectively.

Step 5’: If there is a receiver, say Bob_n, whose measurement result on qubit $e_n$ is ${|1⟩}_{e_n}$, then the state of qudits $t_n$, $A_n$ and $B_n$ is

$$\begin{array}{c}\hfill |{\Omega }_n⟩={\displaystyle \frac{1}{\sqrt{d}}}\sum _{s=0}^{d-1}(\sum _{r=0}^{d-1}{c}_{nr}(\sum _{j=0}^{d-1}{\zeta }^{jr}|j,(r+s)\mathrm{mod}d{⟩}_{t_n{A}_n})\otimes {\zeta }^{ls}\sqrt{a_{ns}^2-a_{n0}^2}{|s⟩}_{B_n}).\end{array}$$

(52)

Now, Alice performs a unitary operation ${\left({\mathcal{H}}^{\left(d\right)}\right)}^{-1}$ shown in Equation (37) on her qudit $t_n$, and then applies a unitary operation ${\left({\mathcal{N}}^{\left(d\right)}\right)}^{-1}$ shown in Equation (38) on her two qudits $t_n$ and $A_n$. Then, the state shown in Equation (52) becomes

$$\begin{array}{c}\hfill {({\mathcal{N}}^{\left(d\right)})}^{-1}{({\mathcal{H}}^{\left(d\right)})}^{-1}|{\Omega }_n⟩=\sum _{r=0}^{d-1}{(c_{nr}|r⟩}_{t_n})\otimes (\sum _{s=0}^{d-1}{\zeta }^{ls}\sqrt{a_{ns}^2-a_{n0}^2}|ss{⟩}_{A_n{B}_n}).\end{array}$$

(53)

From Equation (53), it is obvious that the unknown d-dimensional single-qudit state $|{\chi }_n⟩$ is preserved in qudit $t_n$. If there are other Bobs whose measurement outputs on their respective auxiliary qubits are $|1⟩$, then the original unknown states corresponding to these Bobs can also be preserved by Alice through the above analysis method. Any receiver whose measurement result for auxiliary qubit is $|0⟩$ among the receivers can reconstruct the target state by performing appropriate local operation on his own qudit.

5. Discussion and Conclusions

A key component of deterministic controlled quantum teleportation is the quantum channel connecting the sender, receiver, and controller, which should be a maximally entangled pure state. In the real world, however, quantum channels are always inevitably affected by mutual coupling with the surrounding environment, e.g., different types of decoherence, dephasing, and dissipation mechanisms can reduce the purity and entanglement of the channel. Therefore, the sender, receiver, and controller often share the partially entangled state instead of sharing the maximum entangled state. For this reason, probabilistic controlled QT is more general and practical in the laboratory.

From the security viewpoint, the proposed NCQT protocol is inherently controlled and distributed. No receiver can reconstruct a target state from local channel particles alone, because successful recovery requires Alice’s joint-measurement outcomes, Charlie’s control result, and the receiver’s own auxiliary-qubit outcome. If any part of this classical or quantum side information is missing, the appropriate correction operation cannot be completed and the target state cannot be recovered faithfully. In addition, when a recovery attempt fails, the corresponding unknown state is retained at the sender’s side rather than leaked to an unauthorized party. Therefore, the controller authorization mechanism and the distributed dependence on measurement information together provide the basic security foundation of the scheme.

Compared with conventional destructive QT [1,36], standard controlled QT [52,53,54], and recent competitive protocols such as double-direction cyclic controlled QT [17], high-dimensional teleportation via arbitrary entangled states [18], shared high-dimensional secret teleportation [19], and bidirectional multi-output quantum communication [65], the present scheme has four distinctive features. It simultaneously combines high-dimensional transmission, multiple outputs, controller authorization, and partially entangled channels within a unified $(2N+1)$-qudit framework. Each receiver only needs one auxiliary qubit and a local unitary correction device to determine whether the corresponding target state can be recovered locally. Most importantly, a failed attempt does not irreversibly destroy all of the sender’s original unknown states; instead, the unrecovered states remain available at the sender’s side, which makes repeated attempts possible when additional partially entangled channels are available. For clarity, Table 2 summarizes the main differences between representative destructive and controlled teleportation settings and the protocol proposed in this work, while Table 3 further positions our method relative to recent related schemes.

Table 2. Systematic comparison between representative teleportation settings and the protocol proposed in this work.

Scheme	Channel/Control Requirement	Auxiliary Particles	Failure Consequence	Repeatability/Success Characteristic
Standard destructive QT	Typically a bipartite maximally entangled channel; no controller.	Usually none	The sender’s unknown state is consumed once the teleportation measurement is completed.	No repeatability after failure; deterministic only with an ideal maximally entangled channel.
Standard controlled QT	Multipartite entangled channel with an additional controller.	Usually none	Controller authorization is required, but an unsuccessful attempt still destroys the original state in the usual destructive setting.	Controlled transmission is achieved, but failed attempts are generally not resumable.
Existing probabilistic multi-output QT/CQT	Multipartite channel; may involve multiple outputs and, in some cases, partial entanglement.	Often receiver-side auxiliaries are needed in nondeterministic implementations.	At least part of the transmitted quantum information is lost once the probabilistic attempt fails.	Single-shot success probability is below 1 and the protocol is generally not fully repeatable.
Protocol in this work	One partially entangled $(2N+1)$-qudit channel with a single controller supervising all outputs.	N auxiliary qubits, one for each receiver	If some receiver-side recovery measurements fail, the corresponding original states remain at the sender’s side.	Per-trial success probability is $d^N{\prod }_{n=1}^N{a}_{n0}^2$; repeated attempts can make the total success probability arbitrarily close to 1.

Table 3. Comparison with recent related high-dimensional, controlled, or multi-output protocols.

Representative Scheme	High-Dimensional	Controlled	Multi-Output	Nondestructive/ Repeatable	Main Distinguishing Feature
Double-direction cyclic controlled QT [17]	Yes	Yes	No	No	Cyclic bidirectional controlled transmission of single-qutrit states.
High-dimensional QT via arbitrary entangled states [18]	Yes	No	No	No	Uses high-dimensional entangled resources for teleporting a two-qubit state.
Shared high- dimensional quantum secret QT [19]	Yes	No	No	No	Probabilistic teleportation of shared high-dimensional quantum secrets.
Bidirectional multi- output quantum communication [65]	No	No	Yes	No	Realizes bidirectional multi-output transmission of single-qubit states.
Protocol in this work	Yes	Yes	Yes	Yes	Unifies high-dimensionality, controller authorization, multiple outputs, and nondestructive repeated recovery within one partially entangled $(2N+1)$-qudit channel.

To further clarify the position of the present work, Table 3 compares the proposed protocol with several recent related schemes.

The protocol can in principle be simulated or partially implemented on existing quantum platforms for reduced-size instances. For qubit-encoded proof-of-principle circuits, the sequence of entangling gates, conditional unitary corrections, projective measurements, and classical feed-forward is compatible with existing superconducting or trapped-ion architectures; related multi-output teleportation demonstrations on IBM Quantum have already been reported [63]. For the genuinely high-dimensional versions considered here, photonic orbital-angular-momentum systems, time-bin encodings, and other high-dimensional carriers provide natural experimental candidates [31,33,34]. Nevertheless, a full platform-level implementation of the general $(2N+1)$-qudit protocol remains experimentally demanding because it requires high-dimensional entangled-state preparation, synchronized multi-party control, and low-noise feed-forward. We therefore regard detailed hardware simulation and implementation as important directions for future work.

In addition, our controlled N-output scheme in Section 4 differs from controlled single-output QT via partially entangled $(N+1)$-particle channels. The former performs synchronous transmission of N unknown single-qudit states under a single controller and requires one partially entangled $(2N+1)$-qudit channel, whereas the latter concerns asynchronous single-output transmission and typically requires repeated control and multiple channel uses.

The existing probabilistic N-output QT is destructive and usually requires N auxiliary particles together with local unitary operations after the sender’s Bell-state measurements. Its success probability cannot reach 1 in a single attempt, and once teleportation fails, at least one transmitted state is destroyed, so the process cannot be repeated directly, because of the quantum no-cloning principle. In contrast, in our scheme the teleported single-qudit states that are not successfully recovered by the receivers can be retained by the sender when the controlled N-output QT process fails. Therefore, although the scheme does not increase the single-trial success probability, it provides the possibility of repeating the process until success, provided that additional quantum channels are available. Equation (48) in Section 4 shows that, under the condition that all receivers obtain the auxiliary-qubit outcome $|0⟩$, the probability that Alice obtains the outcome ${|rs⟩}_{t_n{A}_n}$ is $a_{n0}^2/\sqrt{d}$, and thus the probability associated with a full set of qudit-pair outcomes is ${({\prod }_{n=1}^N{a}_{n0}/\sqrt{d^N})}^2$. Because Alice’s measurement has ${(d\times d)}^N$ possible outcomes, the success probability of the QT is $d^N{\prod }_{n=1}^N{a}_{n0}^2$. Hence, the N-output controlled QT can be regarded as a Bernoulli experiment. After M attempts, the probability of observing k successful events follows the binomial distribution $P_{M,k}=C_M^k{\left[d^N{\prod }_{n=1}^N{a}_{n0}^2\right]}^k{[1-d^N{\prod }_{n=1}^N{a}_{n0}^2]}^{M-k}$, where $C_M^k$ is the number of combinations of choosing k elements from M elements. Therefore, the total success probability is given by

$$\begin{array}{c}\hfill P=\sum _{k=1}^M{P}_{M,k}=1-{[1-d^N\prod _{n=1}^N{a}_{n0}^2]}^M.\end{array}$$

(54)

Figure 1 illustrates the behavior of P as a function of ${\prod }_{n=1}^N{a}_{n0}$ for different values of M. As noted in the caption, this plot is drawn for the representative case $d^N=4$ (equivalently, $d=4^{1/N}$), so that the maximally entangled channel corresponds to ${\prod }_{n=1}^N{a}_{n0}=1/2$. More generally, because a maximally entangled pair satisfies $a_{n0}=1/\sqrt{d}$, the admissible range in the general d-dimensional case is $0\le {\prod }_{n=1}^N{a}_{n0}\le d^{-N/2}$. Therefore, Figure 1 is intended to provide an illustrative visualization of the monotonic dependence of P on ${\prod }_{n=1}^N{a}_{n0}$ and M, while the analytical expression in Equation (50) remains valid for arbitrary d. The factor $a_{n0}$ in Equation (42) is the minimum Schmidt coefficient of the partially entangled pair ${\sum }_{k=0}^{d-1}{a}_{nk}{\zeta }^{lk}{|kk⟩}_{A_n{B}_n}$ and therefore quantifies the weakest entanglement resource relevant to receiver-side recovery. Equation (50) shows that the overall success probability after M attempts is completely determined by the single-trial success probability $d^N{\prod }_{n=1}^N{a}_{n0}^2$ and the number of repetitions M. Hence, for fixed M, stronger entanglement yields larger P, whereas for fixed channel parameters, repeated nondestructive attempts monotonically increase P. This provides a direct theoretical explanation of the trends observed in Figure 1. Accordingly, we draw the following conclusions:

Figure 1. The total success probability P as a function of ${\prod }_{n=1}^N{a}_{n0}$ for different values of M, plotted for the representative case $d^N=4$ (equivalently, $d=4^{1/N}$ in this parameterization). Under this condition, the admissible range is $0\le {\prod }_{n=1}^N{a}_{n0}\le 1/2$.

(a)

When $a_{nj}=1/\sqrt{d}$ for all $j\in \{0,1,2,\cdots ,d-1\}$ and $n\in \{1,2,\cdots ,N\}$, the quantum channel in Equation (42) becomes maximally entangled. In this case, the success probability of the N-output controlled QT is $d^N{\prod }_{n=1}^N{a}_{n0}^2=1$, and the scheme becomes deterministic. Therefore, our protocol reduces to the corresponding deterministic controlled multi-output QT scheme.

(b)

For any fixed $a_{n0}\ne 0$ and $n\in \{1,2,\cdots ,N\}$, there always exists a positive integer M such that the total success probability becomes arbitrarily close to one. Hence, our scheme provides a way to simultaneously teleport N different unknown single-qudit states successfully even through weakly entangled channels, which demonstrates its resumable nature.

(c)

The success probability of the N-output controlled QT process increases significantly when repeated attempts are allowed ($M>1$).

(d)

When the entanglement is weak, more repetitions are required to make the success probability close to one. Consequently, the efficiency is relatively low under these circumstances, and improving the practical efficiency remains an interesting direction for future work.

On the other hand, we note that there are $d^{2N+1}$ possible outcomes in the joint measurement process of Alice and Charlie. Therefore, transmitting the measurement results to the N receivers Bob₁, Bob₂, ⋯, Bob_N requires $(2N+1){log}_{2}d$ classical bits in one-way classical communication. It is interesting that when N-output controlled teleportation fails in a trial, there may be k ($k<N$) receivers whose measurement results for each auxiliary qubit are all $|0⟩$, while the measurement results of the remaining $N-k$ receivers regarding their respective auxiliary qubits must all be $|1⟩$. In the former case, the receivers can recover their respective target states, while in the latter case, sender Alice can retain these corresponding initial states.

In summary, following the approach from special to general, this article first considers the two-output QT in three-dimensional Hilbert space, then extends it to the two-output case in high-dimensional Hilbert space, and finally generalizes it to the multi-output case in high-dimensional Hilbert space. The specific contributions are as follows:

(i)

By using a three-dimensional partially entangled five-qutrit state as the quantum channel, we present a nondestructive scheme to teleport two arbitrary unknown single-qutrit states simultaneously and separately to distant receivers under the permission of the supervisor. In our scheme, each of the two receivers needs to introduce an auxiliary qubit, perform a special unitary transformation on the qutrit in his/her hand and the auxiliary qubit, and measure the auxiliary qubit with the Z-basis. After the sender’s series of operations and the controller’s measurement, both original unknown states are reconstructed by the receivers, or one is reconstructed by the receiver and the other is retained by the sender, or both are retained by the sender. The former achieves successful teleportation, while the latter two lead to the failure of quantum teleportation, providing the opportunity to repeat the teleportation process until successful teleportation.

(ii)

By replacing the quantum channel with a high-dimensional partially entangled five-qudit state, we extend the two-output NCQT to the case of teleporting two arbitrary unknown single-qudit states in a d-dimensional Hilbert space and obtain analogous conclusions.

(iii)

We further extend the two-output NCQT scheme in (ii) to the general case of N outputs and analyze the corresponding success probability.