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?

Question

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?

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

Answer 1

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.

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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: $$ \det(\tilde\Psi^\dagger ({\bf1}))=\det\big(K_1^\dagger \Psi^\dagger (K_2K_2^\dagger )K_1\big)=|\det(K_1)|^2\det\big( \Psi^\dagger (K_2K_2^\dagger ) \big)=0\tag*{$\square$} $$