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 $$K(\Phi) = \sum_a A_a\otimes \bar B_a.$$

  • $\begingroup$ I was initially surprised that a partially depolarizing channel can be inverted, but I guess the crucial piece is that you need to know $p$, so then you can always just subtract an appropriate multiple of the identity from your state. $\endgroup$ Aug 20 at 15:08
  • $\begingroup$ Change suggestion: I think this question should assume that $\mathcal{X}=\mathcal{Y}$. Otherwise $K(\Phi)$ is a rectangular matrix and neither its inverse nor its determinant are defined. We can of course work around this by looking for pseudo-inverse, so rewording to this effect is also ok, but TBH this seems hard enough as it is without such generalization. $\endgroup$ Aug 28 at 19:53
  • $\begingroup$ I suspect there may not be a simple general condition here since the problem appears to be related to finding the inverse of a sum of matrices as is evident from your last equation. A few related sources of inspiration and ideas are the Matrix determinant lemma, Woodbury identity and Weinstein-Aronszajn identity. $\endgroup$ Aug 28 at 19:56
  • $\begingroup$ It may also be possible to say something interesting when $\mathcal{X}$ has low dimension using results such as those in this answer. $\endgroup$ Aug 28 at 19:59
  • $\begingroup$ @AdamZalcman I think I added something about the dimensions in a previous version but then ended up not including it in the end for some reason. I agree that it unnecessarily complicates the issue. About the determinant of a sum, yes I was also not very hopeful for such a nice characterisation in the general case, but figured that something might be doable assuming that the Kraus ops are orthogonal (which they can always be chosen to be) (also, thanks for the bounty!) $\endgroup$
    – glS
    Aug 28 at 20:33

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