# Given $\Psi$ completely positive when do there exist $K_1,K_2$ such that $K_2\Psi(K_1^\dagger(\cdot)K_1)K_2^\dagger$ is also trace preserving?

In quantum information it occasionally happens that one ends up with a completely positive but not yet trace-preserving map $$\Psi$$ which one wants to make trace preserving somehow; this often comes up in corresponding numerical considerations. A simple and well-known way of turning such maps into trace-preserving ones is to "skew" the input using the value of the adjoint map $$\Psi^\dagger$$ applied to the identity, i.e. $$\tilde\Psi:=\Psi((\Psi^\dagger ({\bf1}))^{-1/2}(\cdot)(\Psi^\dagger ({\bf1}))^{-1/2}) \tag1$$ is (still completely positive and) readily verified to be trace preserving, hence a channel. When doing numerics this is all one needs because $$\Psi^\dagger({\bf1})$$ is generically invertible so (1) always works. However, in other situations it may happen that the operator $$\Psi^\dagger({\bf1})$$ has non-trivial kernel in which case (1) is of course ill defined. The most obvious idea to fix this would be to replace the main operator $$(\Psi^\dagger ({\bf1}))^{-1}$$ (resp. its square root) by its Moore-Penrose inverse $$(\Psi^\dagger ({\bf1}))^+$$; however it is not clear whether the resulting map (1) is still trace preserving in this case.

But this procedure may be too restrictive altogether. Thus, given $$\Psi$$ completely positive one may ask whether there always exist completely positive maps $$\Phi_1,\Phi_2$$ with completely positive inverse (as we do not want to lose any information on $$\Psi$$, i.e. whatever we do to make $$\Psi$$ trace preserving we want to be able to undo in a "physical" manner) such that $$\Phi_1\circ\Psi\circ\Phi_2$$ is trace preserving? Recalling that a CP map has CP inverse only if its Kraus rank is 1 this is equivalent to the following

Question. Given $$\Psi$$ completely positive do there always exist $$K_1,K_2$$ such that $$\tilde\Psi:=K_2\Psi(K_1^\dagger(\cdot)K_1)K_2^\dagger$$ is also trace preserving?

(This is a Q&A style question meant as a contribution to the list of counterexamples in quantum information)

For a counterexample consider the qubit map $$\Psi\begin{pmatrix}x_{11}&x_{12}\\x_{21}&x_{22}\end{pmatrix}:=\begin{pmatrix}x_{11}&0\\0&0\end{pmatrix}.$$ This map is completely positive as it can be expressed using a single Kraus operator $$K=|0\rangle\langle 0|$$. In particular, $$K^\dagger=K$$ which implies $$\Psi^\dagger =\Psi$$. Now assume there existed $$K_1,K_2\in\mathbb C^{2\times 2}$$ such that $$\tilde\Psi(X):=K_2\Psi(K_1^\dagger XK_1)K_2^\dagger$$ were trace preserving. Then $$\tilde\Psi^\dagger ({\bf1})=K_1\Psi^\dagger (K_2^\dagger K_2)K_1^\dagger=K_1\Psi(K_2^\dagger K_2)K_1^\dagger =\|K_2|0\rangle\|^2K_1|0\rangle\langle 0|K_1^\dagger$$ is a rank-1 operator meaning it cannot be the identity (which is well known to be equivalent to trace preservation), regardless of how $$K_1,K_2$$ are chosen.
Seeing how $$\Psi^\dagger ({\bf1})=|0\rangle\langle 0|$$ is non-invertible in this counterexample this motivates the following characterization for when the original procedure (1) works:
Theorem. Let a positive, linear map $$\Psi:\mathbb C^{n\times n}\to\mathbb C^{k\times k}$$ be given. The following statements are equivalent.
1. $$\Psi^\dagger ({\bf1})>0$$
2. There exist $$K_1\in\mathbb C^{n\times n}$$, $$K_2\in\mathbb C^{k\times k}$$ such that $$\tilde\Psi(X):=K_2\Psi(K_1^\dagger XK_1)K_2^\dagger$$ is trace preserving.
Proof. Formula (1) from the original post shows (i) $$\Rightarrow$$ (ii) so we only have to prove the converse. "(ii) $$\Rightarrow$$ (i)": We argue via contraposition. Assume $$\Psi^\dagger ({\bf1})\not> 0$$. Because $$\Psi$$ is positive the same is true for $$\Psi^\dagger$$ meaning $$\Psi^\dagger ({\bf1})\not> 0$$ implies that $$\Psi^\dagger (YY^\dagger )$$ is never positive definite as it will always have non-trivial kernel, i.e. $$\det(\Psi^\dagger (YY^\dagger ))=0$$ for all $$Y\in\mathbb C^{k\times k}$$. But this forces $$\tilde\Psi^\dagger ({\bf1})$$ to be non-invertible (meaning it cannot be $${\bf1}$$ & $$\tilde\Psi$$ cannot be trace preserving) as the following computation shows: $$\det(\tilde\Psi^\dagger ({\bf1}))=\det\big(K_1^\dagger \Psi^\dagger (K_2K_2^\dagger )K_1\big)=|\det(K_1)|^2\det\big( \Psi^\dagger (K_2K_2^\dagger ) \big)=0\tag*{\square}$$