How does the invertibility of a quantum map reflect on its Kraus operators?

Question

Consider a quantum map $\Phi\in\mathrm T(\mathcal X)$, that is, a linear operator $\Phi:\mathrm{Lin}(\mathcal X)\to\mathrm{Lin}(\mathcal X)$ for some finite-dimensional complex vector spaces $\mathcal X$.

In the specific case in which $\Phi$ is also completely positive and trace-preserving, we call it a quantum channel. One can show that a quantum channel is reversible, in the sense of there being another quantum channel $\Psi\in\mathrm T(\mathcal X)$ such that $\Phi\circ\Psi=\Psi\circ\Phi=\mathrm{Id}_{\cal X}$, iff it is a unitary/isometric channel, meaning $\Phi(X)=UXU^\dagger$ for some isometry $U:\mathcal X\to\mathcal X$.

More generally, I'll call a quantum map invertible if there is some other map $\Psi$ such that $\Phi\circ\Psi=\Psi\circ\Phi=\mathrm{Id}_{\cal X}$. The difference compared to the previous definition is that we don't impose further constraints on the inverse. In particular, $\Phi$ might be invertible, but its inverse not be a channel. Consider for example channels of the form $$\Phi_p(X)=p X + (1-p) \operatorname{Tr}(X)\frac{I_{\cal X}}{\dim\mathcal X}.$$ These can be written, with respect to an arbitrary orthogonal basis $\{\sigma_j\}_j\subset\mathrm{Herm}(\mathcal X)$ with $\operatorname{Tr}(\sigma_i\sigma_j)=\dim(\mathcal X)\delta_{ij}$, $$\langle\sigma_i,\Phi_p(\sigma_j)\rangle = p \delta_{ij} + (1-p).$$ One can thus see that $\Phi_p$, as a linear map, has $\det(\Phi_p)=p^{d-1}(d-(d-1)p)$, with $d\equiv \dim(\mathcal X)$, and is thus invertible for all $p\neq0,\frac{d}{d-1}$ (although it is obviously not a channel unless $0\le p\le 1$). Its inverse is $$\Phi_p^{-1}(Y) = \left(\frac{p-1}{p}\right)\operatorname{Tr}(Y) \frac{I_{\mathcal X}}{\dim(\mathcal X)} + \frac{Y}{p}.$$ However, $\Phi_p^{-1}$ is not a channel for $p\in(0,1]$.

Is there a general way to predict the invertibility of a quantum map, for example based on its Kraus or Choi representations? Clearly, a naive way is to just write the linear map as a matrix in some basis and compute its determinant, but does this somehow translate nicely into some property in Choi or Kraus (or other) representations? For a general (not CPTP) map, by "Kraus representation" I mean a decomposition of the form $\Phi(X)=\sum_a A_a X B_a^\dagger$ for some linear operators $A_a:\mathcal X\to\mathcal X$ and $B_a:\mathcal X\to\mathcal X$.

I suppose the question thus boils down to the following: given $\Phi(X)=\sum_a A_a X B_a^\dagger$, is there a nice enough way to write $\det(K(\Phi))$? Here $K(\Phi):\mathcal X\otimes\mathcal X\to\mathcal X\otimes\mathcal X$ is the natural representation of the channel, which can be seen to be writable as $$K(\Phi) = \sum_a A_a\otimes \bar B_a.$$

Answer 1

This is too long for a comment but I wanted to contribute an example which hopefully argues in favor of Danylo's remark that there is "[probably no] simpler way than using the representation $\sum_aA_a\otimes\overline{B_a}$ directly". Starting from the qubit channel $\Phi$ defined via $$ \Phi(\rho):=\frac13\begin{pmatrix} \rho_{11}+2\rho_{22}&-\rho_{12}\\-\rho_{21}&2\rho_{11}+\rho_{22} \end{pmatrix} $$ consider the mixture $\Psi_\lambda:=\lambda\Phi+(1-\lambda){\rm Id}$, $\lambda\in[0,1]$. From the Choi matrix of $\Phi$ one readily finds Kraus operators $\{ \sqrt2|0\rangle\langle 1|,\sqrt2|1\rangle\langle 0|,\sigma_z \}/\sqrt3$ meaning a set of Kraus operators of $\Psi_\lambda$ is given by $$ \Big\{\sqrt{1-\lambda}\,I,\sqrt{\frac{2\lambda}3}|0\rangle\langle 1|,\sqrt{\frac{2\lambda}3}|1\rangle\langle 0|,\sqrt{\frac{\lambda}3}\sigma_z\Big\}\,.\tag{1} $$ Now I would argue that from this set it is not at all obvious that $\Psi_\lambda$ fails to be bijective for $\lambda=\frac34$, as then $\Psi_{\frac34}(X)={\rm tr}(X)\frac{I}2$ is the completely depolarizing channel. The problem & the reason that there is no obvious connection between the Kraus operators and the determinant $\det(K(\Phi))=(1-\frac43\lambda)^3$ in this case is of course the non-trivial interplay between the summands $A_a\otimes\overline{A_a}$, i.e. between $(1-\lambda)I_4$, $\frac23\lambda|00\rangle\langle 11|$, $\frac23\lambda|11\rangle\langle 00|$, and $\frac\lambda3\sigma_z\otimes\sigma_z$ in terms of the resulting eigenvalues.

To be fair, in this particular example things could have been simplified slightly by looking at the equivalent set of Kraus operators $\{\sigma_x,\sigma_y,\sigma_z\}/\sqrt3$ so the counterpart of (1) for $\lambda=\frac34$ becomes the (weighted) orthonormal basis $\{{\bf1},\sigma_x,\sigma_y,\sigma_z\}/2$ which gives rise to the unital reset channel (as alredy observed, e.g., in the original GKS-paper, Eq.(2.5)). However, we can easily modify this example such that bijecitivity fails at an arbitrary reset channel $X\mapsto{\rm tr}(X)\omega$ (with $\omega$ full rank) in which case it becomes even harder to connect lack of bijectivity for one specific value of $\lambda$ to the Kraus operators.