7.11 Correctable channels

Another question that we might ask ourselves is if we can reverse the action of a quantum channel, recovering the input state. Let us first make precise what we mean by this.

We say that a quantum channel \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}') is correctable if there exists a recovery channel \mathcal{R}\colon\mathcal{B}(\mathcal{H}')\to\mathcal{B}(\mathcal{H}) such that the composition \mathcal{R}\circ\mathcal{E} is the identity channel.

The action of any unitary operation U can, of course, be reversed by simply applying the inverse operation, U^\dagger; the same holds for any isometry V, because V^\dagger V=\mathbf{1}. For example, the process of first adding an auxiliary system in a fixed state and then applying a unitary U to the composite system can be reversed by first applying U^\dagger to the composed system and then discarding the auxiliary system. So how about a statistical mixture of unitaries? Or, in general, a statistical mixture of isometries?

If all we know is that an isometry V_i is chosen randomly according to some distribution \{p_i\}_{i\in I}, then the best we can do in an attempt to reverse the random process is to apply V_k^\dagger, where k\in I is such that p_k is the largest of all the p_i. Clearly, this approach succeeds in reversing the action of the channel with probability p_k. However, if there were a way to determine, post factum, which particular isometry was chosen, then, of course, we could perfectly reverse the action of the channel.

Consider a channel \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}') given by \mathcal{E}\colon\rho \longmapsto \rho' =\sum_i p_i V_i\rho V^\dagger_i in which the isometries V_i are mutually orthogonal (i.e. V_i^\dagger V_j =\delta_{ij}\mathbf{1}). Let \mathcal{H}'_i be the image of \mathcal{H} under V_i; these images are subspaces in \mathcal{H}', and they are mutually orthogonal. Indeed, for any vector |v\rangle in \mathcal{H}, the vectors V_i|v\rangle\in \mathcal{H}'_i and V_j|v\rangle\in \mathcal{H}'_j are orthogonal: \langle v|V_i^\dagger V_j|v\rangle = \delta_{ij} \langle v|\mathbf{1}|v\rangle = \delta_{ij}. We can now perform a measurement defined by the projections onto these mutually orthogonal subspaces \mathcal{H}'_i, and find out which particular isometry was chosen (and then perfectly reverse its action by applying the inverse isometry).

Note that, for a measurement on \mathcal{H}' to be well defined, we need to decompose \mathcal{H}' into mutually orthogonal subspaces. That is, the direct sum \bigoplus_i\mathcal{H}'_i might not “fill up” the space \mathcal{H}', so we might need to pad out with whatever is “left over” in order to obtain the decomposition of the whole \mathcal{H}' into mutually orthogonal subspaces. That extra subspace with which we pad out the direct sum is not in the image of the channel: we will never see the result of the measurement corresponding to the projection onto that subspace. However, we need to add it in order to obtain a complete decomposition of \mathcal{H}'. Here the Kraus operators E_i=\sqrt{p_i}V_i satisfy E^\dagger_i E_j = \sqrt{p_i p_j}\delta_{ij}\mathbf{1} but remember that the operator-sum decomposition is not unique, and so the same channel can be described by another, unitarily related, set of Kraus operators, K_i = \sum_k U_{ik}E_k, which then satisfy \begin{aligned} K_i^\dagger K_j &= \sum_{kl} U^\star_{ik} E^\dagger_k E_l U_{jl} \\&= \sum_{kl} U^\star_{ik} (\sqrt{p_k p_l}\delta_{kl}\mathbf{1}) U_{jl} \\&= \sum_{k} U^\star_{ik} p_k U_{jk}\mathbf{1} \\&= \sigma_{ij}\mathbf{1} \end{aligned} where \sigma_{ij} are elements of a density matrix with eigenvalues p_i. If this condition is satisfied, then the action of the channel can be reversed, and the original state \rho can be recovered.

Conversely, if the channel \mathcal{E} is correctable, then the above condition above holds. Indeed, in terms of Kraus representations for \mathcal{E} and \mathcal{R}, we require that \mathcal{R}\circ\mathcal{E}(\rho) = \sum_{lj} R_l E_j \rho R^\dagger_l E^\dagger_j = \rho for any state \rho. This means that identity channel \mathcal{R}\circ\mathcal{E}=\mathbf{1} must have all the Kraus operators proportional to the identity: R_lE_j=\lambda_{lj}\mathbf{1} for some complex numbers \lambda_{lj} such that120 \sum_{lj} |\lambda_{lj}|^2=1. Then we can write \begin{aligned} \sum_l E_i^\dagger R_l^\dagger R_l E_j &= E_i^\dagger E_j \\&=\sum_l \lambda^*_{il}\lambda_{jl}\mathbf{1} \\&=\sigma_{ij}\mathbf{1} \end{aligned} where \sigma_{ij} = \sum_l \lambda^*_{il}\lambda_{jl}. Clearly, \sigma_{ij} is a positive matrix such that \operatorname{tr}(\sigma_{ij})=1 and \sum_i\sigma_{ii}=\sum_{il}|\lambda_{il}|^2=1. So the condition E_i^\dagger E_j = \sigma_{ij}\mathbf{1} is both necessary and sufficient in order for the channel \mathcal{E} to be correctable. In summary:

Let \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}') be a quantum channel with Kraus decomposition \mathcal{E}(\rho)=\sum_i E_i\rho E^\dagger_i. Then the following statements are equivalent:

  1. \mathcal{E} is correctable;
  2. E_i^\dagger E_j = \sigma_{ij}\mathbf{1} for some density matrix \sigma_{ij};
  3. there exists a set of orthogonal isometries \{V_i\} and a probability distribution \{p_i\} such that \mathcal{E} (\rho) = \sum_i p_i V_i\rho V^\dagger_i for every state \rho.

  1. This is the normalisation condition for the Kraus operators R_lE_j↩︎