## 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 that \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:

- \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.