# Is the extension of a pure operation also a pure operation?

Given some quantum channel—or, more generally, some positive linear map—$$\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$$ one usually calls $$\Phi$$ a pure operation if for all $$\rho\geq 0$$ pure (rank 1), $$\Phi(\rho)$$ is pure, as well. For example, every unitary channel or, more generally, every completely positive map of Kraus rank 1 is pure. Based on this let us ask the following question:

If $$\Phi$$ is a pure operation, is $${\rm id}_k\otimes\Phi$$ a pure operation for all $$k\in\mathbb N$$, as well?

To return to the example, if $$\Phi$$ has Kraus rank 1, then the same is true for $${\rm id}\otimes\Phi$$ so in this case the statement in question holds.

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

The form of all possible pure operations is well known, cf. also Ch. 2, Theorem 3.1 ff. in Davies' book "Quantum Theory of Open Systems" (1976). In the special case where $$\Phi$$ is a channel (i.e. completely positive and trace-preserving) there are exactly two possibilities:

1. $$\Phi=U(\cdot)U^\dagger$$ for some $$U\in\mathbb C^{n\times n}$$ unitary
2. $$\Phi={\rm tr}(\cdot)|\psi\rangle\langle\psi|$$ for some $$\psi\in\mathbb C^n$$ with $$\|\psi\|=1$$

This second case is what gives rise to a counter-example. More precisely, consider $$\Phi={\rm tr}(\cdot)|0\rangle\langle 0|$$ in which case \begin{align*} ({\rm id}_2\otimes\Phi)(X)&=\begin{pmatrix} \Phi\begin{pmatrix} x_{11}&x_{12}\\ x_{21}&x_{22} \end{pmatrix}&\Phi\begin{pmatrix} x_{13}&x_{14}\\ x_{23}&x_{24} \end{pmatrix}\\ \Phi\begin{pmatrix} x_{31}&x_{32}\\ x_{41}&x_{42} \end{pmatrix}&\Phi\begin{pmatrix} x_{33}&x_{34}\\ x_{43}&x_{44} \end{pmatrix} \end{pmatrix}\\ &=\begin{pmatrix} x_{11}+x_{22}&0&x_{13}+x_{24}&0\\ 0&0&0&0\\ x_{31}+x_{42}&0&x_{33}+x_{44}&0\\ 0&0&0&0 \end{pmatrix}\,. \end{align*} This allows us to find a 2-qubit pure state which is mapped to a mixed state: choosing $$|\psi\rangle:=\frac1{\sqrt2}(1,0,0,1)^\top$$ to be the maximally entangled state, $$({\rm id}_2\otimes\Phi)(|\psi\rangle\langle\psi|)= \frac12({\rm id}_2\otimes\Phi)\begin{pmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix}=\frac12\begin{pmatrix}1&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&0\end{pmatrix}$$ is obviously rank 2 and hence not pure.

This also shows that the extension of rank non-increasing channels—i.e. $${\rm rank}(\Phi(\rho))\leq{\rm rank}(\rho)$$ for all $$\rho\geq 0$$—need not be rank non-increasing, because every rank non-increasing channel is a pure operation (apply the definition to $$\rho$$ pure). This is interesting because

• it is true in the classical case, i.e. for all non-negative matrices $$A\in\mathbb R_+^{n\times n}$$ such that $${\bf e}^\top A\,{\rm sgn}(x)\leq {\bf e}^\top{\rm sgn}(x)$$ one readily verifies that the same is true for $${\bf1}\otimes A$$.
• when one replaces rank non-increasing to rank non-decreasing, then then statement—at least in the unital case—is true, cf. Appendix C in this paper.
• Why not simply feed in a maximally entangled state (or just any entangled state)? Commented Aug 2 at 13:20
• @NorbertSchuch Good call, that's of course much easier! Edited my answer accordingly. Commented Aug 2 at 13:50
• My point was also (sorry for not elaborating, typed on the phone) that writing a specific state makes it look overly technical and obfuscates the conceptual insight -- namely, since the channel works by tracing one subsystem, inserting an entanged state will output a mixed state at the other system (in fact, by inserting a purification you can get any mixed state you want). Commented Aug 2 at 14:49
• With regard to your very last point: Isn't a unital pure channel just a unitary channel, in which case it should be rank conserving? Commented Aug 2 at 14:51
• @NorbertSchuch While you are correct that unital pure channels are necessarily unitary, the last paragraph of my answer is not about pure channels anymore but rather about general rank non-decreasing, resp. rank non-increasing channels. (to be fair, looking back at my answer maybe this distinction/generalization wasn't as clear as it could've been) Commented Aug 2 at 16:20