Recovering quantum information through partial access to the environment

Quantum noise is the main obstacle in realizing quantum information tasks. It results from the errors introduced on the system's state by the unavoidable interaction with the surrounding environment [1]. As a consequence, the quantum coherence features of the system's state are washed out. To restore them, one could think of measuring the system (gathering information about its state) and then applying a correction procedure. This is the idea underlying the quantum feedback control mechanism [2]. Actually, also quantum error correcting codes can be thought of as belonging to this kind of strategy [3].

Indeed, one could make a measurement on the final state of the environment and consider its classical result to recognize what kind of error has occurred on the system due to the interaction with the environment. Then, a proper correction should be performed on the system to reduce the effect of quantum noise [4]. In recent years, much attention has been devoted to this scheme from different aspects. In [5–7], the capacity for this scenario has been studied, and in [8], it has been shown that in certain cases repeated application of this scheme allows one to remove the effects of quantum noise completely. For a given measurement, the optimal recovery scheme (the recovery necessary for restoring the maximum value of quantum information) has been derived in [4], while in [9] it has been shown that the optimal measurement depends on the dimension of the system's Hilbert space.

In extending this quantum control strategy to multipartite systems, we must deal with a more intricate scenario. For instance, access to all subsystems' environments may not be available. Then we will address the problem of recovering quantum information by feedback partial control; that is, the measurement is only made on some of the subsystems' environments, while the error correction is performed on all subsystems. In this case, the feedback scheme will be effective if errors occurring on different subsystems are somehow correlated, so that gaining information on the measured subsystems also means to indirectly gain information about non-measured ones. This will help in designing the recovery operation on the whole system. We will consider quite a general kind of correlated error on qubits and determine the optimal recovery depending on the degree of the errors' correlation. We will also find the scaling of the performance versus the number of qubits (subsystems) while monitoring the error on just one of them.

This paper is organized as follows. In section 2, we briefly present the main conceptual and computational tools needed to recover quantum information by means of a quantum feedback control scheme. To get some insights, we apply, in section 3, this strategy to the correlated depolarizing channel for two qubits when only one is monitored. We then derive the main result for the system of n qubits in section 4 and present our conclusions in section 5.

The evolution of a system interacting with an environment can be described by a completely positive and trace-preserving map $T\!:\mathcal {L}(\mathcal {H}_{\mathrm {initial}})\rightarrow \mathcal {L}(\mathcal {H}_{\mathrm {final}})$ transforming the initial system's density operator in Hilbert space $\mathcal {H}_{\mathrm {initial}}$ to a final density operator in Hilbert space $\mathcal{H}_{\mathrm{final}}$ ($\mathcal {L}(\mathcal {H})$ is the space of linear operators on $\mathcal {H}$). At the same time, the initial state of the environment in Hilbert space $\mathcal {K}_{\mathrm {initial}}$ is mapped into a final one in $\mathcal {K}_{\mathrm {final}}$. The evolution of the system can be described as the unitary evolution of the system and the environment given by the unitary operator $U\!: \mathcal {H}_{\mathrm {initial}}\otimes \mathcal {K}_{\mathrm {initial}}\rightarrow \mathcal {H}_{\mathrm {final}}\otimes \mathcal {K}_{\mathrm {final}}$. By denoting by ρ and σ the initial state of the system and the environment, respectively, the map of the system evolution reads

$$\begin{eqnarray*} &&T(\rho)={\rm tr}_{_{{\mathcal{K}}_{{\rm final}}}}[U(\rho\otimes\sigma) U^{\dagger}], \end{eqnarray*}$$

where tr_• denotes the trace on the space •.

To acquire some information about the errors occurring on the system, one can carry out a measurement on the environment after the interaction with the system has taken place. In general, this is described by a positive operator-valued measure on $\mathcal {K}_{\mathrm {final}}$, namely a set of operators $M_{\alpha }\in \mathcal {L}(\mathcal {K}_{\mathrm {final}})$ satisfying

$$\begin{equation} \sum_{\alpha}M_{\alpha}=I,\quad~ M_{\alpha}\geqslant 0. \end{equation} \tag{ 1 }$$

The index α labels the classical measurement outcomes. Considering an arbitrary observable $A\in \mathcal {L}(\mathcal {H}_{\mathrm {final}})$, the expectation value of this observable is

$$\begin{equation} \langle A\rangle={\rm tr}_{_{\mathcal{H}_{{\rm final}}}}{\rm tr}_{_{\mathcal{K}_{{\rm final}}}}[ U(\rho\otimes\sigma )U^{\dagger} (A\otimes I)], \end{equation} \tag{ 2 }$$

where I is the identity on $\mathcal {L}(\mathcal {K}_{\mathrm {final}})$.

Definition 1. We define by $T_{\alpha}\!:\mathcal {L}(\mathcal {H}_{\mathrm {initial}})\rightarrow \mathcal {L}(\mathcal {H}_{\mathrm {final}})$,

$$\begin{eqnarray*} &&T_{\alpha}(\rho):={\rm tr}_{_{\mathcal{K}_{{\rm final}}}}[U(\rho\otimes\sigma )U^{\dagger} (I\otimes M_{\alpha} )], \end{eqnarray*}$$

the selected channel output corresponding to the outcome α.

Then, replacing I in (2) with the identity resolution (1), we obtain

$$\begin{eqnarray*} &&\langle A\rangle=\sum_{\alpha}{\rm tr}_{_{\mathcal{H}_{{\rm final}}}}(T_{\alpha}(\rho)A). \end{eqnarray*}$$

Rewriting the expectation value of A in the following way:

$$\begin{eqnarray*} &&\langle A\rangle=\sum_{\alpha}p_{\alpha}\,{\rm tr}\left[\frac{T_{\alpha}(\rho)}{p_{\alpha}}A\right], \end{eqnarray*}$$

we can conclude that p_α = tr(T_α(ρ)) is the probability of getting α as the result of the measurement and the density matrix $\frac {1}{p_{\alpha }}T_{\alpha }(\rho )$ as the selected state of the system after performing the measurement on the environment.

We can also define the most informative measurement [4] in terms of Kraus operators [10] composing the channel T_α.

Definition 2. Given a channel $T =\sum _{\alpha }T_{\alpha }$, the most informative measurement on the environment is such that we can describe the selected output of the channel T_α by a single Kraus operator T_α(ρ) = t_αρt^†_α.

Therefore

$$\begin{equation} T=\sum_{\alpha}t_{\alpha}\rho t_{\alpha}^{\dagger}, \quad \sum_{\alpha}t_{\alpha}^{\dagger}t_{\alpha}=I. \end{equation} \tag{ 3 }$$

In order to correct the errors due to the interaction with the environment, we have to introduce a recovery operation.

Definition 3. Let $R_{\alpha}\!: \mathcal {L}(\mathcal {H}_{\mathrm {final}})\rightarrow \mathcal {L}(\mathcal {H}_{\mathrm {initial}})$ be the recovery operator that acts on the selected output of the channel T_α(ρ) and depends on the classical outcome of the measurement α. Then, the overall corrected channel takes the form

$$\begin{equation} T_{{\rm corr}}:=\sum_{\alpha}R_{\alpha}\circ T_{\alpha}. \end{equation} \tag{ 4 }$$

Using (3) and a Kraus representation [10] for the recovery channel R_α,

$$\begin{eqnarray*} &&R_{\alpha}(\rho')=\sum_{\beta}r_{\beta}^{(\alpha)}\rho'r_{\beta}^{(\alpha)\dagger},\quad\sum_{\beta}r_{\beta}^{(\alpha\dagger)}r_{\beta}^{(\alpha)}=I, \end{eqnarray*}$$

we can decompose the corrected channel as

$$\begin{equation} T_{{\rm corr}}(\rho)=\sum_{\alpha,\beta}r_{\beta}^{(\alpha)}t_{_{\alpha}}\rho \,t_{_{\alpha}}^{\dagger}r_{\beta}^{(\alpha) \dagger}. \end{equation} \tag{ 5 }$$

To quantify the performance of the correction scheme, we use the entanglement fidelity [11, 12].

Definition 4. For a general map $\Phi\!: \mathcal {L}(\mathcal {H})\rightarrow \mathcal {L}(\mathcal {H})$ with Kraus operators A_k, the entanglement fidelity is defined as

$$\begin{equation} F(\Phi):=\langle\Psi| \Phi\otimes I(|\Psi\rangle\langle\Psi|)|\Psi\rangle=\frac{1}{d^2}\sum_k|{\rm tr} (A_k)|^2,\end{equation} \tag{ 6 }$$

where $d=\mathrm {dim}\,\mathcal {H}$ and $|\Psi \rangle \in \mathcal {H}\otimes \mathcal {H}$ is a maximally entangled state.

We are interested in F(T_corr), the entanglement fidelity of the corrected map (4). As a consequence of (5) and (4), we have

$$\begin{equation} F(T_{{\rm corr}})=\frac{1}{d^2}\sum_{\alpha,\beta}|{\rm tr}(r_{\beta}^{(\alpha)}t_{_{\alpha}})|^2. \end{equation} \tag{ 7 }$$

The entanglement fidelity reaches its maximum value if quantum information is completely recovered or in other words if the corrected channel becomes an identity map. In [4], it has been shown that there exists a family of operators that completely recover quantum information if and only if

$$\begin{equation} t_{\alpha}^{\dagger}t_{\alpha}=c_{\alpha}I,\quad \forall \alpha, \end{equation} \tag{ 8 }$$

with $c_{\alpha }\in \mathbb {R}_+$ and $\sum _{\alpha }c_{\alpha }=1$.

These results are obtained with the assumption that full access to the environment is available and it is possible to carry out a measurement on the whole environment after the interaction with the system. However, more generally we should assume that our access to the environment is partial. Here, we want to investigate how the performance of this correction scheme behaves in this case and to see if we can still completely retrieve quantum information. To shed light on this problem, we study a map for which the complete recovery of quantum information is possible, provided that we have complete access to the environment.

Specifically we are going to consider the depolarizing quantum channel. In the following we will consider

$$\begin{eqnarray*} &&\mathcal{H}:=\mathbb{C}^2,\quad \mathcal{K}:=\mathbb{C}^2\otimes \mathbb{C}^2. \end{eqnarray*}$$

Definition 5. The single-qubit depolarizing channel is defined by

$$\begin{eqnarray*} &&\mathcal{H}_{{\rm initial}}=\mathcal{H}_{{\rm final}}=\mathcal{H}, \end{eqnarray*}$$

$$\begin{eqnarray*} &&\mathcal{K}_{{\rm initial}}=\mathcal{K}_{{\rm final}}=\mathcal{K}, \end{eqnarray*}$$

and

$$\begin{eqnarray*} &&t_\alpha=\sqrt{p_\alpha}\sigma_\alpha, \end{eqnarray*}$$

where the operators σ_α, with α = 0,1,2,3 (the Greek indices go from 0 to 3 while Latin indices go from 1 to 3), denote the Pauli operators (including the identity operator), while p₀ = 1 − p and $p_1=p_2=p_3=\frac {p}{3}$.

Remark 1. Since the Pauli operators satisfy the condition (8), the quantum information in this case can be completely recovered. To achieve this it is enough to consider a recovery channel described by a single Kraus operator σ_α, where α is the classical outcome of the measurement. Hence from (7) we have

$$\begin{eqnarray*} &&F(T_{{\rm corr}})=\sum_{\alpha=0}^3 \left| \sqrt{p_\alpha} \right|^2=1. \end{eqnarray*}$$

However, the situation will be different when we enlarge the Hilbert spaces of the system and the environment while carrying out a measurement just on a subsystem of the environment. In the following sections, we show how the performance of this scheme behaves when our access to the environment is partial.

To study the feedback control scheme with partial access to the environment, we start by analyzing the depolarizing channel $T\!:\mathcal {L}(\mathcal {H}^{\otimes 2})\rightarrow \mathcal {L}(\mathcal {H}^{\otimes 2})$ acting on two qubits. In the following, we assume that we can carry out a measurement on $\mathcal {L}(\mathcal {K})$ while the state of the environment belongs to $\mathcal {L}(\mathcal {K}^{\otimes 2})$. Since the access to the environments is partial, the measurement cannot be the most informative measurement (see definition 2) and therefore the selected output of the channel is, in general, given by

$$\begin{eqnarray*} &&T_{\alpha}(\rho)=\sum_{\beta}t_{_{\alpha,\beta}}\rho t_{_{\alpha,\beta}}^{\dagger}. \end{eqnarray*}$$

The Kraus operators t_α,β will be

$$\begin{eqnarray*} &&t_{\alpha,\beta}=\sqrt{p_{\alpha}p_{\beta}}\sigma_{\alpha}\otimes \sigma_{\beta} \end{eqnarray*}$$

in the case of no correlations and

$$\begin{eqnarray*} &&t_{\alpha,\beta}=\sqrt{p_{\alpha}}\delta_{_{\alpha,\beta}}\sigma_{\alpha}\otimes \sigma_{\beta} \end{eqnarray*}$$

in the case of perfect correlations. In the first case, the outcome of the measurement does not give any information about the error occurring on the second qubit; therefore the selected output of the channel is

$$\begin{equation} T_{\alpha}^{{\rm uc}}=\sum_{\beta}p_{\alpha}p_{\beta}(\sigma_{\alpha}\otimes \sigma_{\beta})\rho(\sigma_{\alpha}\otimes \sigma_{\beta})^{\dagger}. \end{equation} \tag{ 9 }$$

In the second case, the measurement of the environment is the most informative one (according to definition 2); hence the selected output of the channel is

$$\begin{equation} T_{\alpha}^{{\rm cc}}(\rho)=p_{\alpha} (\sigma_{\alpha}\otimes \sigma_{\alpha})\rho(\sigma_{\alpha}\otimes \sigma_{\alpha})^{\dagger}. \end{equation} \tag{ 10 }$$

We will now consider a more general situation, which interpolates between the above two situations. More explicitly, we consider a correlated noise model that is a convex combination of the two cases described above, namely the uncorrelated noise and the completely correlated noise for two qubits [13].

Definition 6. The following convex combination of channels (9) and (10),

$$\begin{eqnarray*} &&T_{\alpha}:=(1-\mu)T_{\alpha}^{{\rm uc}}+\mu T_{\alpha}^{{\rm cc}}, \end{eqnarray*}$$

where μ∈[0,1] quantifies the amount of correlation in noise, defines the selected output.

Our aim now is to design the recovery channel in order to achieve the maximum value of the entanglement fidelity for the corrected channel.

Lemma 1. The recovery map

$$\begin{eqnarray*} &&R_{\alpha}(\rho'):=\sum_{\gamma}q_{_{\alpha,\gamma}}(\sigma_{\alpha}\otimes \sigma_{\gamma})\rho'(\sigma_{\alpha}\otimes \sigma_{\gamma}),\quad \sum_{\gamma}q_{_{\alpha,\gamma}}=1, \end{eqnarray*}$$

is optimal for the channel of definition 6.

Proof. The recovery map can be described, without loss of generality, as

$$\begin{eqnarray*} &&R_{\alpha}(\rho')=\sum_{\gamma}(\sigma_{\alpha}\otimes A_{\gamma}^\alpha)\rho'(\sigma_{\alpha}\otimes A_{\gamma}^\alpha)^{\dagger}, \end{eqnarray*}$$

where the single-qubit operators A^α_γ can be expressed in terms of the identity and the Pauli operators as

$$\begin{eqnarray*} &&A_{\gamma}^\alpha=\sum_{\delta} c_{\gamma,\delta}^{\alpha} \sigma_{\delta}. \end{eqnarray*}$$

The completeness condition $\sum _{\gamma } (\sigma _{\alpha }\otimes A_{\gamma }^\alpha )^{\dag }(\sigma _{\alpha }\otimes A_{\gamma }^\alpha )=I$ gives the following normalization condition for the coefficients c^α_γ,δ:

$$\begin{eqnarray*} &&\sum_{\gamma,\delta}|c_{\gamma,\delta}^{\alpha}|^2=1. \end{eqnarray*}$$

Note that this is the most general map we can use as recovery. Actually, on the first qubit the optimal action is to invert the action of σ_α by σ_α itself, while on the second one we consider a generic operator A^α_γ possibly correlated with that on the first qubit (this is the reason for the presence of the index α on A^α_γ). Then the entanglement fidelity of the corrected channel, using (7), takes the form

$$\begin{eqnarray*} &&F(T_{{\rm corr}})=(1-\mu)\sum_{\alpha,\beta,\gamma}p_\alpha p_{\beta}|c_{\gamma,\beta}^{\alpha}|^2 +\mu \sum_{\alpha,\gamma}p_\alpha |c_{\gamma,\alpha}^{\alpha}|^2. \end{eqnarray*}$$

Note that this can be rewritten as

$$\begin{equation} F(T_{{\rm corr}})=(1-\mu)\sum_{\alpha,\beta}p_\alpha p_{\beta}q_{\alpha,\beta} +\mu \sum_{\alpha}p_\alpha q_{\alpha,\alpha}, \end{equation} \tag{ 11 }$$

where we defined the probabilities

$$\begin{eqnarray*} &&q_{\alpha,\beta}=\sum_{\gamma}|c_{\gamma,\beta}^{\alpha}|^2. \end{eqnarray*}$$

Equation (11) is the same expression that we would obtain by assuming the recovery as a Pauli channel, namely with the Kraus operators

$$\begin{eqnarray*} &&\sigma_{\alpha}\otimes A_{\gamma}^\alpha=\sqrt{q_{\alpha,\gamma}} \sigma_{\alpha}\otimes\sigma_{\gamma} \end{eqnarray*}$$

with $\sum _{\gamma }q_{\alpha ,\gamma }=1$ for all γ.   □

By virtue of lemma 1, the corrected channel can be written as

$$\begin{eqnarray*} &&\fl T_{{\rm corr}}(\rho)=(1-\mu)\sum_{\alpha,\beta,\gamma}p_{\alpha}p_{_{\beta}}q_{\alpha,\gamma}( {I}\otimes \sigma_{\gamma}\sigma_{\beta})\rho( {I}\otimes \sigma_{\beta}\sigma_{\gamma}) +\mu\sum_{\alpha,\gamma} p_{\alpha}q_{\alpha,\gamma}( {I}\otimes \sigma_{\gamma}\sigma_{\alpha})\rho ( {I}\otimes \sigma_{\alpha}\sigma_{\gamma}), \end{eqnarray*}$$

and its entanglement fidelity becomes

$$\begin{eqnarray*} F(T_{{\rm corr}})&=&\frac{1-\mu}{16}\sum_{\alpha,\beta,\gamma}p_{\alpha}p_{{\beta}}q_{_{\alpha,\gamma}}|{\rm tr}( {I}\otimes \sigma_{\gamma}\sigma_{\beta})|^2+\frac{\mu}{16}\sum_{\alpha,\gamma} p_{\alpha}q_{{\alpha,\gamma}}|{\rm tr}( {I}\otimes \sigma_{\alpha}\sigma_{\gamma})|^2\\ &=&(1-\mu)\sum_{\alpha,\beta}p_{\alpha}p_{{\beta}}q_{{\alpha,\beta}}+\mu\sum_{\alpha} p_{\alpha}q_{{\alpha,\alpha}}. \end{eqnarray*}$$

Taking into account that $\sum _{\gamma }q_{_{\alpha ,\gamma }}=1$ for all values of α, the above equation is simplified as follows:

$$\begin{eqnarray}&&\fl F(T_{{\rm corr}})\kern-1pt=\kern-1pt(1\kern-1pt-\kern-1pt\mu)\frac{p}{3}\kern-1pt+\kern-1pt(1\kern-1pt-\kern-1pt p)\kern-2.5pt\left(\kern-2pt(1\kern-1pt-\kern-1pt\mu)\kern-2.5pt\left(\kern-2pt1\kern-1pt-\kern-1pt\frac{4p}{3}\kern-1pt\right)\kern-1pt+\kern-1pt\mu\kern-2pt\right)q_{0,0}\kern-1pt+\kern-1pt\frac{p}{3}\sum_{i=1}^3\kern-2pt\left(\kern-2pt(1\kern-1pt-\kern-1pt\mu)\kern-2pt\left(\kern-2pt 1\kern-1pt-\kern-1pt\frac{4p}{3}\right)q_{i,0}+\mu q_{i,i}\kern-2pt\right)\!.\nonumber\\ \end{eqnarray} \tag{ 12 }$$

Then, the following theorem holds.

Theorem 1. Upon recovery, the maximum achievable entanglement fidelity for the channel of definition 6 is:

• Region A
  $$\begin{eqnarray*} &&F_{{\rm max}}^{A}(T_{{\rm corr}})=1-p, \end{eqnarray*}$$
  for $0<\mu <\mu _{AB}=\frac {3-4p}{6-4p}$.
• Region B
  $$\begin{eqnarray*} &&F_{{\rm max}}^{B}(T_{{\rm corr}})=(1-\mu)\left(1-2p+\frac{4p^2}{3}\right)+\mu, \end{eqnarray*}$$
  for
  $$\begin{eqnarray*} &&\mu>\mu_{AB} \end{eqnarray*}$$
  and
  $$\begin{eqnarray*} &&\mu>\mu_{BC}=\frac{4p-3}{4p}. \end{eqnarray*}$$
• Region C
  $$\begin{eqnarray*} &&F_{{\rm max}}^{C}(T_{{\rm corr}})=(1-\mu)\frac{p}{3}+\mu p, \end{eqnarray*}$$
  for 0 < μ < μ_BC.

Proof. The optimal recovery channel is achieved by maximizing expression (12) over the parameters q_α,γ. Our strategy to maximize the entanglement fidelity is to optimize the correction performance for each channel component

$$\begin{eqnarray*} &&T^{(\alpha)}_{{\rm corr}}=R_{\alpha}\circ T_{\alpha}. \end{eqnarray*}$$

When the outcome of the measurement is α = 0, the entanglement fidelity of the corrected map T^(α=0)_corr is

$$\begin{eqnarray*} &&F^{(0)}_{{\rm corr}}=(1-\mu)(1-p)\frac{p}{3} +(1-p)\left[(1-\mu)\left(1-\frac{4p}{3}\right)+\mu\right]q_{_{00}}. \end{eqnarray*}$$

For $(1-\mu )(1-\frac {4p}{3})+\mu >0$ the coefficient of q₀₀ is positive; therefore the maximum of F⁽⁰⁾_corr is attained by choosing q_0,0 = 1. For $(1-\mu )(1-\frac {4p}{3})+\mu <0$ the maximum is achieved for q_0,0 = 0. This means that for $\mu < \frac {4p-3}{4p}$, if the outcome of the measurement is 0 (no error on the first qubit), the most appropriate recovery is to perform a Pauli channel on the second qubit and leave the first qubit unchanged. For $\mu > \frac {4p-3}{4p}$, if the outcome of the measurement is 0, the amount of correlation on noise is large enough to ensure that the second qubit has passed through the channel safely and no correction is required on either of them.

To find the optimum recovery for the other possible outcomes of the measurement, we have to maximize the following expressions:

$$\begin{equation} F^{(i)}_{{\rm corr}}=(1-\mu)\frac{p^2}{9}+(1-\mu)\frac{p}{3}\left(1-\frac{4p}{3}\right)q_{_{i,0}}+\mu \frac{p}{3} q_{_{i,i}}. \end{equation} \tag{ 13 }$$

Note that the probabilities q_i,j with j ≠ i do not appear in (13). Moreover, since F⁽ⁱ⁾_corr is linear in the parameters q_i,j and at least one of the coefficients is positive, remembering the normalization condition $\sum _{\gamma } q_{_{\alpha ,\gamma }}=1$, we set q_i,j = 0 for j ≠ i to achieve the maximum value for F⁽ⁱ⁾_corr. Hence we can write q_i,0 = 1 − q_i,i. Substituting it into equation (13), we obtain

$$\begin{eqnarray*} &&F^{(i)}_{{\rm corr}}=(1-\mu)\frac{p}{3}(1-p)+\frac{p}{3}\left(\mu-(1-\mu)\left(1-\frac{4p}{3}\right)\right)q_{_{i,i}}. \end{eqnarray*}$$

Therefore if the outcome of the measurement is i = 1,2,3, for $\mu >\frac {3-4p}{6-4p}$ the optimum correction can be performed by taking q_i,i = 1 and for $\mu <\frac {3-4p}{6-4p}$ the best performance of the recovery is attainable by taking q_i,0 = 1. Therefore, the optimum correction varies depending on the values of p and μ, and can be summarized as:

Region A: In this region, $0<\mu <\mu _{AB}=\frac {3-4p}{6-4p}$. The optimum correction is achieved by choosing q_α,0 = 1:

$$\begin{eqnarray*} &&R_{\alpha}(\rho')=(\sigma_{\alpha}\otimes {I})\rho'(\sigma_{\alpha}\otimes {I}). \end{eqnarray*}$$

Therefore, the maximum entanglement fidelity in this region is given by

$$\begin{eqnarray*} &&F_{{\rm max}}^{A}(T_{{\rm corr}})=1-p. \end{eqnarray*}\noindent$$

Region B: In this region, μ > μ_AB and $\mu >\mu _{BC}=\frac {4p-3}{4p}$. For the measurement outcome α, the optimum recovery is given by q_α,α = 1:

$$\begin{eqnarray*} &&R_{\alpha}(\rho')=(\sigma_{\alpha}\otimes \sigma_{\alpha})\rho'(\sigma_{\alpha}\otimes\sigma_{\alpha}).\end{eqnarray*}\noindent$$

Therefore, the maximum entanglement fidelity in this region is given by

$$\begin{eqnarray*} &&F_{{\rm max}}^{B}(T_{{\rm corr}})=(1-\mu)\left(1-2p+\frac{4p^2}{3}\right)+\mu.\end{eqnarray*}\noindent$$

Region C: In this region 0 < μ < μ_BC. If the outcome of the measurement is α = 0 the optimal recovery is given by q_0,0 = 0:

$$\begin{eqnarray*} &&R_{0}(\rho')=\sum_{i=1}^3q_i( {I}\otimes \sigma_{i})\rho'( {I}\otimes \sigma_{i}),\quad \sum_{i=1}^3q_i=1, \end{eqnarray*}\noindent$$

and for the measurement outcome i = 1,2,3, the optimal recovery is given by q_i,i = 1:

$$\begin{eqnarray*}&&R_{i}(\rho')=(\sigma_i\otimes \sigma_{i})\rho'(\sigma_i\otimes \sigma_{i}),\quad i=1,2,3.\end{eqnarray*} \noindent$$

The maximum entanglement fidelity takes the form

$$\begin{eqnarray*} &&F_{{\rm max}}^{C}(T_{{\rm corr}})=(1-\mu)\frac{p}{3}+\mu p. \end{eqnarray*}$$

   □

Based on theorem 1, we can identify three different regions for the optimum correction in the plane of p and μ, as shown in figure 1.

Zoom In Zoom Out Reset image size

Remark 2. It is interesting to note that the critical value μ_AB for the correlation parameter μ has the same form as the one characterizing the correlated depolarizing channel in terms of classical information transmission [13]. In that context, the critical value μ_AB gives a threshold value for the optimal input states: the mutual information along the channel is maximized with product states for μ ⩽ μ_AB, while it achieves its maximum value with maximally entangled states for μ ⩾ μ_AB [13].

In the previous section, we have seen how the performance of the quantum feedback control scheme behaves if we can carry out measurements over $\mathcal {L}(\mathcal {K})$ while the total Hilbert space of the environment is $\mathcal {K}\otimes \mathcal {K}$. If we had full access to the environment, we could completely retrieve quantum information. However, we have shown that having only partial access to the environment and exploiting the correlation in noise we are still capable of partially recovering quantum information. Now we want to see how the performance of the correction behaves if we keep increasing the Hilbert space of the environment without increasing our access to it. To do so, we consider a correlated depolarizing channel defined by $T\!:\mathcal {L}(\mathcal {H}^{\otimes n})\rightarrow \mathcal {L} (\mathcal {H}^{\otimes n})$, resembling the long-term memory channels introduced in [14], and we carry out measurement on $\mathcal {L}(\mathcal {K})$ while the state of the environment belongs to $\mathcal {L}(\mathcal {K}^{\otimes n})$.

Definition 7. Let us define the selected output of the channel corresponding to the classical outcome of the measurement, α, as

$$\begin{eqnarray*} &&T_{\alpha}(\rho):=(1-\mu)T_{\alpha}^{{\rm uc}}(\rho)+\mu T_{\alpha}^{{\rm cc}}(\rho), \end{eqnarray*}$$

with

$$\begin{eqnarray*} &&T_{\alpha}^{{\rm uc}}(\rho):=\sum_{\beta_1\ldots\beta_n}p_{\alpha}\left(\prod_{i=2}^np_{{\beta_i}}\right)\left[\sigma_a\bigotimes\left(\bigotimes_{i=2}^n\sigma_{{\beta_i}}\right)\right]\rho\left[\sigma_a\bigotimes\left(\bigotimes_{i=2}^n\sigma_{{\beta_i}}\right)\right]\end{eqnarray*}$$

and

$$\begin{eqnarray*} &&T_{\alpha}^{{\rm cc}}(\rho):=p_{\alpha}\sigma_{{\alpha}}^{\otimes n}\rho\sigma_{{\alpha}}^{\otimes n}.\end{eqnarray*}$$

Similarly to the previous section, we have the following:

Lemma 2. The recovery operator

$$\begin{eqnarray*} &&R_{\alpha}(\rho')=\sum_{\gamma_2,\ldots,\gamma_n}q_{{\alpha,\gamma_2,\ldots,\gamma_n}}\left[\sigma_a\bigotimes\left(\bigotimes_{i=2}^n\sigma_{{\gamma_i}}\right)\right]\rho\left[\sigma_a\bigotimes\left(\bigotimes_{i=2}^n\sigma_{{\gamma_i}}\right)\right], \end{eqnarray*}$$

with the constraint

$$\begin{equation} \sum_{\gamma_2,\ldots,\gamma_n}q_{\alpha,\gamma_2,\ldots,\gamma_n}=1,\quad \forall\alpha, \end{equation} \tag{ 14 }$$

is optimal for the channel of definition 7.

Proof. Having the classical outcome of the measurement α, we know that error σ_α has occurred on the first qubit and the effect of error can be completely removed by performing σ_α on the first qubit for correction. Therefore, the recovery map should have the following form:

$$\begin{equation} R_{\alpha}(\rho')=\sum_{\gamma_2,\ldots,\gamma_n}\left(\sigma_{\alpha}\otimes \mathcal{A}^{(\alpha)}_{\gamma_2,\ldots,\gamma_n}\right)\rho' \left(\sigma_{\alpha}\otimes \mathcal{A}^{(\alpha)}_{\gamma_2,\ldots,\gamma_n}\right)^{\dagger}, \end{equation} \tag{ 15 }$$

where $\mathcal {A}^{(\alpha)}_{\gamma_2,\ldots,\gamma _n}$ is an operator acting on n − 1 qubits. Expanding it in terms of products of Pauli matrices, we have

$$\begin{equation} \mathcal{A}^{(\alpha)}_{{\gamma_2,\ldots,\gamma_n}}=\sum_{{\delta_2,\ldots,\delta_n}}c^{{\gamma_2,\ldots,\gamma_n}}_{\alpha,\delta_2,\ldots,\delta_n}\sigma_{\delta_2}\otimes\sigma_{\delta_3}\otimes\cdots\otimes\sigma_{\delta_n}. \end{equation} \tag{ 16 }$$

The completeness condition

$$\begin{eqnarray*} &&\sum_{\gamma_2,\ldots,\gamma_n}\left(\sigma_{\alpha}\otimes \mathcal{A}^{(\alpha)}_{{\gamma_2,\ldots,\gamma_n}}\right)^{\dagger}\left(\sigma_{\alpha}\otimes \mathcal{A}^{(\alpha)}_{{\gamma_2,\ldots,\gamma_n}}\right)=I\end{eqnarray*}$$

imposes the following constraint on the coefficients cγ2,...,γn2:

$$\begin{equation} \sum_{\gamma_2,\ldots,\gamma_n}|c^{{\gamma_2,\ldots,\gamma_n}}_{{\alpha,\delta_2,\ldots,\delta_n}}|^2=1. \end{equation} \tag{ 17 }$$

Considering this general form of Kraus operators for the recovery map and using (7), the entanglement fidelity of the corrected channel takes the form

$$\begin{eqnarray} &&\fl F(T_{{\rm corr}})=(1-\mu)\sum_{{\alpha,\beta_2,\ldots\beta_n}}\sum_{{\gamma_2\ldots\gamma_n}}|c^{{\gamma_2,\ldots,\gamma_n}}_{{\alpha,\beta_2,\ldots,\beta_n}}|^2p_{\alpha}\prod_{i=2}^np_{{\beta_i}}+\mu \sum_{\alpha,\gamma_2,\ldots,\gamma_n}|c^{{\gamma_2,\ldots,\gamma_n}}_{{\alpha,\alpha,\ldots,\alpha}}|^2p_{\alpha}. \end{eqnarray}\noindent \tag{ 18 }$$

The above expression can be written as

$$\begin{eqnarray} &&\fl F(T_{{\rm corr}})=(1-\mu)\sum_{{\alpha,\beta_2,\ldots\beta_n}}p_{\alpha}q_{{\alpha,\beta_2,\ldots,\beta_n}}\prod_{i=2}^np_{{\beta_i}}+\mu \sum_{\alpha}p_{\alpha}q_{{\alpha,\alpha,\ldots,\alpha}}.\end{eqnarray} \tag{ 19 }$$

By defining

$$\begin{equation} q_{{\alpha,\beta_2,\ldots,\beta_n}}=\sum_{{\gamma_2\ldots\gamma_n}}|c^{{\gamma_2,\ldots,\gamma_n}}_{{\alpha,\beta_2,\ldots,\beta_n}}|^2 \end{equation} \tag{ 20 }$$

the same value of entanglement fidelity in equation (19) will be obtained by assuming the recovery map with following Kraus operators:

$$\begin{equation} \sqrt{q_{{\alpha,\gamma_2,\ldots,\gamma_n}}}\sigma_{\alpha}\otimes\sigma_{\gamma_2}\otimes\cdots\otimes\sigma_{\gamma_n}\end{equation}\noindent \tag{ 21 }$$

with the constraint that $\sum_{\gamma_2\ldots \gamma _n}q_{\alpha,\gamma _2,\ldots,\gamma _n}=1$ for all α.   □

The entanglement fidelity (7) of the channel in definition 7 corrected using theorem 2 results in

$$\begin{eqnarray} &&F(T_{{\rm corr}})=(1-\mu)\sum_{\alpha,\beta_2\dots\beta_n}p_{\alpha}p_{{\beta_2}}\ldots p_{{\beta_{n}}}q_{{\alpha,\beta_2,\ldots,\beta_n}}+\mu\sum_{\alpha}p_{\alpha}q_{{\alpha,\alpha,\ldots,\alpha}}.\end{eqnarray}\noindent \tag{ 22 }$$

Then we have the following theorem.

Theorem 2. Upon recovery, the maximum achievable entanglement fidelity for the channel of definition 7 is:

• Region A
  $$\begin{eqnarray*} &&F_{{\rm max}}^{A}(T_{{\rm corr}})=(1-\mu)(1-p)^{n-1}+\mu(1-p), \end{eqnarray*}$$
  for
  $$\begin{eqnarray*} &&0<\mu<\frac{(1-p)^{n-1}-(\frac{p}{3})^{n-1}}{(1-p)^{n-1}-(\frac{p}{3})^{n-1}+1}. \end{eqnarray*}$$
• Region B
  $$\begin{eqnarray*} &&F_{{\rm max}}^{B}(T_{{\rm corr}})=(1-\mu)\left((1-p)^{n}+3\left(\frac{p}{3}\right)^n\right)+\mu, \end{eqnarray*}$$
  for
  $$\begin{eqnarray*} &&\mu>\frac{(1-p)^{n-1}-(\frac{p}{3})^{n-1}}{(1-p)^{n-1}-(\frac{p}{3})^{n-1}+1} \end{eqnarray*}$$
  and
  $$\begin{eqnarray*} &&\mu>-\frac{(1-p)^{n-1}-(\frac{p}{3})^{n-1}}{(1-p)^{n-1}-(\frac{p}{3})^{n-1}-1}. \end{eqnarray*}$$
• Region C
  $$\begin{eqnarray*} &&F_{{\rm max}}^{C}(T_{{\rm corr}})=(1-\mu)(\frac{p}{3})^{n-1}+\mu p, \end{eqnarray*}$$
  for
  $$\begin{eqnarray*} &&0<\mu<-\frac{(1-p)^{n-1}-(\frac{p}{3})^{n-1}}{(1-p)^{n-1}-(\frac{p}{3})^{n-1}-1}. \end{eqnarray*}$$

Proof. To maximize the entanglement fidelity in (22) over its parameters, we maximize it for each value of the measurement outcome α. If the outcome of the measurement is zero, then the entanglement fidelity is given by

$$\begin{eqnarray}&&\fl F^{(0)}_{\rm corr}=[(1-\mu)(1-p)^n+\mu (1-p)]q_{0,\ldots,0}+(1-\mu)(1-p)^{n-1}\nonumber\\ &&\fl\hspace{6pc}\times\frac{p}{3}\sum_{j,{\rm perm}}q_{0,j,0\ldots,0}+(1-\mu)(1-p)^{n-2}\left(\frac{p}{3}\right)^{2}\nonumber\\ &&\fl\hspace{6pc}\times\sum_{j,k,{\rm perm}}q_{0,j,k,0,\ldots,0}+\cdots+(1-\mu)(1-p)\left(\frac{p}{3}\right)^{n-1}\sum_{i_2\ldots i_n}q_{0,i_2,\ldots,i_n},\end{eqnarray} \tag{ 23 }$$

where the notation perm in the summations above refers to all possible permutations of the last n − 1 indices. Note that for p < 3/4 the largest coefficient in the above expression is that in front of q_n and therefore in this case, for any value of μ, F⁽⁰⁾_{α,δ2,...,δn} is always maximized by q_0,0,0...,0 = 1. In the case of p > 3/4 the largest coefficient, excluding the first, is given by the last one in (23). Therefore, in this case the optimal solution can be searched by setting to zero all values of q that contain at least one value 0 among the last n − 1 indices. The expression for the entanglement fidelity that we are going to maximize is now simplified as

$$\begin{eqnarray} &&\fl F^{(0)}_{{\rm corr}}=[(1-\mu)(1-p)^n+\mu (1-p)]q_{0,\ldots,0}+(1-\mu)(1-p)\left(\frac{p}{3}\right)^{n-1}\sum_{i_2\ldots i_n}q_{0,i_2,\ldots,i_n}.\end{eqnarray} \tag{ 24 }$$

Using the condition in equation (14), we know that

$$\begin{eqnarray*} &&\sum_{i_2\ldots i_n}q_{0,i_2,\ldots,i_n}=1-q_{{0,\ldots,0}}. \end{eqnarray*}$$

Replacing it in equation (24), we find that

$$\begin{eqnarray*} &&\fl F^{(0)}_{{\rm corr}}=(1-\mu)(1-p)\left(\frac{p}{3}\right)^{n-1} +\left((1-\mu)\left[(1-p)^n-(1-p)\left(\frac{p}{3}\right)^{n-1}\right]+\mu (1-p)\right)q_{{0,\ldots,0}}. \end{eqnarray*}$$

It is easy to see that for

$$\begin{eqnarray*} &&0<\mu<-\frac{(1-p)^{n-1}-(\frac{p}{3})^{n-1}} {(1-p)^{n-1}+(\frac{p}{3})^{n-1}-1}, \end{eqnarray*}$$

the coefficient of q_corr is negative; therefore the maximum of F⁽⁰⁾_0,0,0,...,0 is attainable for q_0,...,0 = 0. For

$$\begin{eqnarray*} &&\mu>-\frac{(1-p)^{n-1}-(\frac{p}{3})^{n-1}}{(1-p)^{n-1}+ (\frac{p}{3})^{n-1}-1}, \end{eqnarray*}$$

the coefficient of q_corr is positive, so the maximum of F⁽⁰⁾_0,...,0 is achieved by taking q_0,...,0 = 1.

When the outcome of the measurement is i = 1,2,3, then the entanglement fidelity of the corrected channel is

$$\begin{eqnarray}&&\fl F^{(i)}_{{\rm corr}}=(1-\mu)(1-p)^{n-1}\frac{p}{3}q_{i,0,\ldots,0}+(1-\mu)(1-p)^{n-2}\left(\frac{p}{3}\right)^{2}\nonumber\\ &&\fl\hspace{6pc}\times\sum_{j,{\rm perm}}q_{i,j,0\ldots,0}+(1-\mu)(1-p)^{n-3}\left(\frac{p}{3}\right)^{3}\nonumber\\ &&\fl\hspace{6pc}\times\sum_{j,k,{\rm perm}}q_{i,j,k,0,\ldots,0}+\cdots+\frac{p}{3}\sum_{i_2\cdots i_n}q_{i,i_2,\ldots,i_n}\left[(1-\mu)\left(\frac{p}{3}\right)^{n-1}+\mu\prod_{i_k,_{k=2}}^n \delta_{i,i_k}\right].\end{eqnarray} \tag{ 25 }$$

Note that for p > 3/4 the largest coefficient in the above expression is that in front of q_corr; therefore in this case the optimal solution corresponds to q_0,...,0 = 1 for any value of μ. Note also that in the last line the coefficient in front of q_i,i,...,i is always larger than the other ones, so in order to look for the maximum we can always set q_i,i,...,i = 0 for all cases except q_i,i,...,i. Moreover, for p < 3/4, the largest coefficients in (25), with the exclusion of the last line, are the ones in front of q_i,i2,...,in. Therefore, by these considerations, we can restrict our search for the maximum values to the case of vanishing q except for q_i,i,...,i and q_i,0,...,0. We can then write

$$\begin{eqnarray*} &&F^{(i)}(T_{{\rm corr}})=(1-\mu)(1-p)^{n-1}\frac{p}{3}q_{i,0,\ldots,0}+\left[(1-\mu)\left(\frac{p}{3}\right)^n+\mu \frac{p}{3}\right] q_{{i,i,\ldots,i}}. \end{eqnarray*}\noindent$$

From equation (14), we know that

$$\begin{eqnarray*}&&q_{i,i,\ldots,i}=1-q_{{i,0,\ldots,0}}.\end{eqnarray*}\noindent$$

Therefore

$$\begin{eqnarray*}&&\fl F^{(i)}(T_{{\rm corr}})=(1-\mu)\left(\frac{p}{3}\right)^n+\mu\frac{p}{3}+\frac{p}{3}\left((1-\mu)\left[(1-p)^{n-1}-\left(\frac{p}{3}\right)^{n-1}\right]-\mu\right) q_{i,0,\ldots,0}.\end{eqnarray*}\noindent$$

The coefficient of q_i,0,...,0 is positive for

$$\begin{eqnarray*}&&0<\mu<\frac{(1-p)^{n-1}-(\frac{p}{3})^{n-1}}{(1-p)^{n-1}+(\frac{p}{3})^{n-1}+1}.\end{eqnarray*}\noindent$$

Therefore in this region we should take q_i,i,...,i = 1 and for

$$\begin{eqnarray*} &&\mu>\frac{(1-p)^{n-1}-(\frac{p}{3})^{n-1}}{(1-p)^{n-1}+(\frac{p}{3})^{n-1}+1},\end{eqnarray*}\noindent$$

the coefficient of q_i,0,...,0 is negative and therefore we should take q_i,0,...,0 = 1.   □

Figure 2 shows regions A, B and C of theorem 2 for different values of n (including the previously analyzed case of n = 2). We can see that by increasing n, regions A and C become smaller. This can be understood by noting that region A corresponds to the case when although the measurement outcome shows that an error has occurred on the first qubit, we expect that the other qubits have not experienced any error. However, the chance of this holding true is lowered by increasing n. The same reasoning can be applied to region C.

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Figure 3 shows the entanglement fidelity versus the number of qubits in the system for p = 0.4 and for different values of μ. It is interesting to note that for large n the entanglement fidelity does not go to zero, due to the role of noise correlation in performing the recovery operation.

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The main result presented in this paper is the possibility of recovering quantum information on a multipartite system by using limited access to the environment. In particular, we have addressed the important question of how well quantum information can be recovered on a multiple qubit system by carrying out a measurement on one environmental subsystem. We have considered quite a general kind of correlated error on qubits and we have determined the optimal recovery depending on the degree of error correlation. We have also found the scaling of the performance versus the number of qubits while monitoring the error on just one of them. Interestingly enough, for a finite degree of error correlation, the recovery ability is preserved by increasing the number of non-measured subsystems of the environment.

As a final remark, we point out that when considering partial control, one could exploit correlations residing on the system's state itself rather than on the errors. This would be more in the spirit of [15] and is left for future investigations.

We acknowledge financial support from the European Commission under the FET-Open grant agreement CORNER (number FP7-ICT-213681).