---
abstract: 'The impossibility of perfect cloning and state estimation are two fundamental results in Quantum Mechanics. It has been conjectured that quantum cloning becomes equivalent to state estimation in the asymptotic regime where the number of clones tends to infinity. We prove this conjecture using two known results of Quantum Information Theory: the monogamy of quantum correlations and the properties of entanglement breaking channels.'
author:
- Joonwoo Bae and Antonio Acín
title:  Asymptotic quantum cloning is state estimation 
---

The impossibility of perfect state estimation is a major consequence of the nonorthogonality of quantum states: *the state of a single quantum system cannot be perfectly measured*. In other words, a measurement on a system in order to acquire information on its quantum state perturbs the system itself. The full reconstruction of the state is only possible by computing statistical averages of different observables on a large number of identically prepared systems. Thus, any measurement at the single-copy level only provides partial information.

The fact that state estimation is in general imperfect leads in a natural way to the problem of building *optimal measurements*. Being a perfect reconstruction impossible, it is relevant to find the measurement strategy that maximizes the gain of information about the unknown state. A standard approach to this problem in Quantum Information Theory (QIT) is to quantify the quality of a measurement by means of the so-called *fidelity* [@MP]. This quantity is defined as follows. Consider the situation in which a quantum state $\ket{\psi}$ is chosen from the ensemble $\{p_i,\ket{\psi_i}\}$, i.e. $\ket{\psi}$ can be equal to $\ket{\psi_i}$ with probability $p_i$. A measurement, defined by $N_M$ positive operators, $M_j\geq 0$, summing up to the identity, $\sum_j M_j=\one$, is applied on this unknown state. For each obtained outcome $j$, a guess $\ket{\phi_j}$ for the input state is made. The overlap between the guessed state and the input state, $|\braket{\psi_i}{\phi_j}|^2$, quantifies the quality of the estimation process. The averaged fidelity of the measurement then reads $$\label{avfidmeas}
    \bar F_M=\sum_{i,j}
    p_i\, \tr(M_j\proj{\psi_i})\,|\braket{\psi_i}{\phi_j}|^2 .$$ A measurement is optimal according to the fidelity criterion when it provides the largest possible value of $\bar F_M$, denoted in what follows by $F_M$.

The No-cloning theorem [@WZ], one of the cornerstones of QIT [@reviews], represents another known consequence of the nonorthogonality of quantum states. It proves that *given a quantum system in an unknown state $\ket{\psi}$, it is impossible to design a device producing two identical copies, $\ket{\psi}\ket{\psi}$*. Indeed, two nonorthogonal quantum states suffice to prove the no-cloning theorem.

As it happens for state estimation, the impossibility of perfect cloning leads to the characterization of *optimal cloning machines* [@BH]. In this case, one looks for the quantum map $\L$ that, given a state $\ket{\psi}$ chosen from an ensemble $\{p_i,\ket{\psi_i}\}$ in $\compl^d$, produces a state $\L(\psi)=\rho_{C_1\ldots C_N}$ in $(\compl^d)^{\otimes N}$, such that each individual clone $\rho_{C_k}=\tr_{\bar
k}(\rho_{C_1\ldots C_N})$ resembles as much as possible the input state, where $\tr_{\bar k}$ denotes the trace with respect to all the systems $C_1,\ldots,C_N$ but $C_k$. The average fidelity of the cloning process is then $$\label{avfidcl}
    \bar F_C(N)=\sum_{i,k}
    p_i\, \frac{1}{N}\bra{\psi_i}\tr_{\bar k}\L(\psi_i)\ket{\psi_i} .$$ The goal of the optimal machine is to maximize this quantity, this optimal value being denoted by $F_C(N)$.

One can easily realize that the no-cloning theorem and the impossibility of perfect state estimation are closely related. On the one hand, if perfect state estimation was possible, one could use it to prepare any number of clones of a given state, just by measurement and preparation. On the other hand, if perfect cloning was possible, one could perfectly estimate the unknown state of a quantum system by preparing infinite clones of it and then measuring them. Beyond these qualitative arguments, the connection between state estimation and cloning was strengthened in [@GM; @BEM]. The results of these works suggested that asymptotic cloning, i.e. the optimal cloning process when $N\to\infty$, is equivalent to state estimation, in the sense that $$\label{conj}
    F_C=F_C(N\to\infty)=F_M .$$ Later, this equality was rigorously shown to hold in the cases of (i) universal cloning [@KW], where the initial ensemble consists of an arbitrary pure state in $\compl^d$, chosen with uniform probability, and (ii) phase covariant qubit cloning [@BCDM], where the initial ensemble corresponds to a state in $\compl^2$ lying on one of the equators of the Bloch sphere. Since then, the validity of this equality for any ensemble has been conjectured and, indeed, has been identified as one of the open problems in QIT [@web].

In this work, we show that the fidelities of optimal asymptotic cloning and of state estimation are indeed equal for any initial ensemble of pure states. Actually, we prove that *asymptotic cloning does effectively correspond to state estimation*, from which the equality of the two fidelities trivially follows. The proof of this equivalence is based on two known results of QIT: the monogamy of quantum correlations and the properties of the so-called entanglement breaking channels (EBC).

It is easy to prove that $F_M\leq F_C$. Indeed, given the initial state $\ket{\psi}$, a possible asymptotic cloning map, not necessarily optimal, consists of first applying state estimation and then preparing infinite copies of the guessed state. It is sometimes said that the opposite has to be true since “asymptotic cloning cannot represent a way of circumventing optimal state estimation". As already mentioned in [@web], this reasoning is too naive, since it neglects the role correlations play in state estimation. For instance, take the simplest case of universal cloning of a qubit, i.e. a state in $\compl^2$ isotropically distributed over the Bloch sphere. The optimal cloning machines produces $N$ approximate clones pointing in the same direction in the Bloch sphere as the input state, but with a shrunk Bloch vector [@KW]. If the output of the asymptotic cloning machine was in a product form, it would be possible to perfectly estimate the direction of the local Bloch vector, whatever the shrinking was. Then, a perfect estimation of the initial state would be possible. And of course, after the perfect estimation one could prepare an infinite number of perfect clones! This simple reasoning shows that the correlations between the clones play an important role in the discussion. Actually, it has recently been shown that the correlations present in the output of the universal cloning machine are the worst for the estimation of the reduced density matrix [@rafael].

As announced, the proof of the conjecture is based on two known results of QIT: the monogamy of entanglement and the properties of EBC. For the sake of completeness, we state here these results, without proof.

Quantum correlations, or entanglement, represent a monogamous resource, in the sense that they cannot be arbitrarily shared. One of the strongest results in this direction was obtained by Werner in 1989 [@Werner]. There, it was shown that the only states that can be arbitrarily shared are the separable ones. Recall that a bipartite quantum state $\rho_{AC}$ in $\compl^d\otimes\compl^d$ is said to be $N$-shareable when it is possible to find a quantum state $\rho_{AC_1\ldots C_N}$ in $\compl^d\otimes(\compl^d)^{\otimes N}$ such that $\rho_{AC_k}=\tr_{\bar k}\rho_{AC_1\ldots C_N}=\rho_{AC},\,\forall
k$. The state $\rho_{AC_1\ldots C_N}$ is then said to be an $N$-extension of $\rho_{AC}$. The initial correlations between subsystems $A$ and $C$ are now shared between $A$ and each of the $N$ subsystems $C_i$, see Fig. \[entshar\]. It is straightforward to see that $$\rho_{AC_1\ldots C_N}=\sum_i q_i\proj{\alpha_i}\otimes
    \proj{\gamma_i}^{\otimes N}$$ gives a valid $N$-extension of a separable state $\rho_{AC}^s=\sum_i q_i\proj{\alpha_i}\otimes\proj{\gamma_i}$ for all $N$. As proven by Werner, if the state is entangled, there exists a finite $N$ where no valid extension can be found.

![The state $\rho_{AC}$ is said to be $N$-shareable when there exists a global state $\rho_{AC_1\ldots C_N}$ such that the local state shared between $A$ and $C_i$ is equal to $\rho_{AC}$, for all $i$.[]{data-label="entshar"}](rhoAcccc "fig:"){width="8cm"}\

The second ingredient needed in what follows are the properties of EBC. A channel $\Upsilon$ is said to be entanglement breaking when it cannot be used to distribute entanglement. In Ref. [@HSR] it was proven that the following three statements are equivalent: (1) $\Upsilon$ is entanglement breaking, (2) $\Upsilon$ can be written in the form $\Upsilon(\rho)=\sum_j\tr(M_j\rho)\rho_j$, where $\rho_j$ are quantum states and $\{M_j\}$ defines a measurement and (3) $(\one\otimes\Upsilon)\ket{\Phi^+}$ is a separable state, where $\ket{\Phi^+}=\sum_i \ket{ii}/\sqrt d$ is a maximally entangled state in $\compl^d\otimes\compl^d$. The equivalence of (1) and (2) simply means that any EBC can be understood as the measurement of the input state, $\rho$, followed by the preparation of a new state $\rho_j$ depending on the obtained outcome. The equivalence of (1) and (3) reflects that the intuitive strategy for entanglement distribution where half of a maximally entangled state is sent through the channel is enough to detect if $\Upsilon$ is entanglement breaking.

After collecting all these results, we are now ready to prove the following

[**Theorem:**]{} Asymptotic cloning corresponds to state estimation. Thus, $F_M=F_C$ for any ensemble of states.

[*Proof:*]{} The idea of the proof is to characterize the quantum maps $\L$ associated to asymptotic cloning machines. First of all, note that, for any number of clones, we can restrict our considerations to symmetric cloning machines, $\L^s$, where the clones are all in the same state. Indeed, given a machine where this is not the case, one can construct a symmetric machine achieving the same fidelity $F_C(N)$, just by making a convex combination of all the permutations of the $N$ clones [@note]. Now, denote by $\L^c$ the effective cloning map consisting of, first, the application of a symmetric machine $\L^s$ and then tracing all but one of the clones, say the first one. The cloning problem can be rephrased as, see Eq. (\[avfidcl\]), $$\label{asclon}
    \max_{\L^c}\sum_ip_i\bra{\psi_i}\L^c(\psi_i)\ket{\psi_i} .$$ Note that this maximization runs over all channels that can be written as $\L^c=\tr_{\bar 1}\L^s$. For instance, the identity map, where $\psi\to\psi,\,\forall\psi$, does not satisfy this constraint.

If the $N$-cloning map is applied to half of a maximally entangled state, the resulting state, $$\rho_{AC_1\ldots C_N}=(\one_A\otimes\L^s_B)\ket{\Phi^+}_{AB},$$ is such that, for all $i$, $$\label{rhoac}
    \rho_{AC_i}=(\one\otimes\L^c)\ket{\Phi^+}=\rho_{AC}.$$ That is, the output of the $N$-cloning machine acting on half of a maximally entangled state is a valid $N$-extension of $\rho_{AC}$. When taking the limit of an infinite number of clones, and because of the monogamy of entanglement, this implies that $\rho_{AC}$ has to be separable and, thus, $\L^c$ is entanglement breaking (\[rhoac\]). Since any EBC can be seen as measurement followed by state preparation, asymptotic cloning (\[asclon\]) can be written as [@note2] $$\max_{\{M_j,\phi_j\}}\sum_{i,j}
    p_i\, \tr(M_j\proj{\psi_i})\,|\braket{\psi_i}{\phi_j}|^2 ,$$ which defines the optimal state estimation problem. Therefore, $F_M=F_C$ for any ensemble of states. $\Box$

The same argument applies to the case in which $L$ copies of the initial state $\ket{\psi}$ are given. The measurement and cloning fidelities now read, see Eqs. (\[avfidmeas\]) and (\[avfidcl\]), $$\begin{aligned}
% \nonumber to remove numbering (before each equation)
  \bar F_M(L) &=& \sum_{i,j}
    p_i\, \tr(M_j\proj{\psi_i}^{\otimes L})\,|
    \braket{\psi_i}{\phi_j}|^2  \nonumber\\
  \bar F_C(N,L) &=& \sum_{i,k} p_i\, \frac{1}{N}\bra{\psi_i}
  \tr_{\bar k}\L(\psi_i^{\otimes L})\ket{\psi_i} .\end{aligned}$$ Using the same ideas as in the previous Theorem, it is straightforward to prove that $$\label{eqfidL}
    F_M(L)=F_C(N\to\infty,L),$$ where $F_M(L)$ and $F_C(N,L)$ denote the optimal values of $\bar
F_M(L)$ and $\bar F_C(N,L)$, as above.

One can also extend this result to asymmetric scenarios. An asymmetric cloning machine [@asclon], given an initial input state $\ket{\psi}$, produces $N_A$ clones of fidelity $F_C(N_A)$ and $N_B$ clones of fidelity $F_C(N_B)$. The machine is optimal when it gives the largest $F_C(N_B)$ for fixed $F_C(N_A)$. In the case of measurement, we are thinking of measurement strategies where the goal is to obtain information on an unknown state introducing the minimal disturbance. As above, we consider that a guess for the input state is done depending on the measurement outcome. The information vs disturbance trade-off can again be expressed in terms of fidelities [@banaszek]: the information gain is given by the overlap, $G$, between the initial and the guessed state, while the disturbance is quantified by the overlap, $F$, between the state after the measurement and the initial state. A measurement is optimal when for fixed disturbance, $F$, it provides the largest value of $G$. The optimal trade-off between $F$ and $G$ has been derived in [@banaszek] for the case in which the input ensemble consists of any pure state in $\compl^d$ with uniform probability.

As it happens for the symmetric case, a connection between this state estimation problem and asymmetric cloning machines can be expected when $N_A=1$ and $N_B\to\infty$. Indeed, the previous measurement strategy gives a possible realization of this asymptotic and asymmetric cloning machine, not necessarily optimal. Actually, when the input state is any pure state, with uniform probability, the optimal measurement strategy of [@banaszek] turns out to saturate the optimal cloning $1\to
N_A+N_B$ fidelities of [@IACFFG], with $N_A=1$ and $N_B\to\infty$. Now, the equality between the measurement and asymptotic cloning fidelities in the asymmetric scenario for any ensemble of input states can be proven using the same arguments as above: one has to symmetrize the $N_B$ clones and then use the connection with entanglement shareability and EBC when $N_B\to\infty$.

From a more speculative point of view, there exist several works relating the impossibility of perfect cloning to the no-signaling principle, namely the impossibility of having faster-than-light communication (see for instance [@gisin]). Actually, a no-cloning theorem can be derived just from the no-signaling principle, without invoking any additional quantum feature [@ncsign]. In view of the strong connection between cloning and state estimation, it would be interesting to study whether a similar link between the no-signaling principle and the impossibility of perfect state estimation could also be established.

To conclude, this work proves the long-standing conjecture on the equivalence between asymptotic cloning and state estimation. It represents the strongest link between two fundamental no-go theorems of Quantum Mechanics, namely the impossibilities of perfect cloning and state estimation.

We thank Emili Bagan, John Calsamiglia, Sofyan Iblisdir and Ramon Muñoz-Tapia for discussion. This work is supported by the Spanish MCyT, under “Ramón y Cajal" grant, and the Generalitat de Catalunya, 2006FIR-000082 grant.

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