## Correctable channels

Another question that we might ask ourselves is when we can *reverse* the action of a quantum channel, recovering the input state.
This is something that we have already mentioned once or twice, but let us now 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 on \mathcal{B}(\mathcal{H}).

The action of any individual 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.
But the important question is if this is still true for a *statistical mixture* of unitaries, or, more generally, 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 pick the largest of the p_i, say p_k, and then apply V_k^\dagger.
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 we could, of course, 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 also mutually orthogonal: 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, since
\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 we have Kraus operators E_i=\sqrt{p_i}V_i that 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, say K_i = \sum_k U_{ik}E_k.
These other operators satisfy
\begin{aligned}
K_i^\dagger K_j
&= \sum_{k,l} U^\star_{ik} E^\dagger_k E_l U_{jl}
\\&= \sum_{k,l} 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.
So the existence of such a density matrix implies that 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 such a density matrix exists.
Indeed, in terms of Kraus representations for \mathcal{E} and \mathcal{R}, we require that
\rho
= (\mathcal{R}\circ\mathcal{E})(\rho)
\equiv \sum_{l,j} R_l E_j \rho R^\dagger_l E^\dagger_j
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 that \sum_{l,j} |\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_{i,l}|\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.

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:

\mathcal{E} is correctable;

E_i^\dagger E_j = \sigma_{ij}\mathbf{1} for some density matrix \sigma_{ij};

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.