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, 2021 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, 2021 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, 2021 at 19:56
  • 2
    $\begingroup$ The existence of a left inverse for a quantum channel is given by the Knill-Laflamme conditions, which is an explicit condition based on the Kraus operators i.e. there needs to be a projection in the commutant of the span of $\{A_a^\dagger A_b\}$. However, if your quantum maps are just linear maps between linear operators (not necessarily completely positive) then they don't actually all necessarily have Kraus representations, or at least it's not obvious to me that they do. $\endgroup$
    – Condo
    Aug 30, 2021 at 14:20
  • 1
    $\begingroup$ @glS ah yes I see, I couldn't remember all the details from Watrous. But anyways I think the same machinery from OAQEC could be used in this case. Following arxiv.org/pdf/quant-ph/0504189.pdf with the caveat that the maps aren't CP, I would imagine it stills goes through, except now you want to look at the algebra spanned by $\{A_a^\dagger B_b\}$. $\endgroup$
    – Condo
    Aug 30, 2021 at 15:39

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.