How to write the Kraus operators of the classical-quantum channel ${\cal E}(\rho_A)=\sum_i\langle i|\rho_A|i\rangle\tau_i$?

Consider a channel $$\mathcal{E}: A\rightarrow B$$ that maps quantum states $$\rho_A\in A$$ to

$$\mathcal{E}(\rho_A) = \sum_i \langle i\vert \rho_A\vert i\rangle\tau_i\ ,\tag{1}$$

where $$\tau_i\in B$$ is a quantum state and $$\langle i\vert \rho_A\vert i\rangle$$ is the probability of outcome $$i$$ when $$\rho_A$$ is measured in the standard basis. I have two questions

1) According to this answer on another question I asked, the quantity $$\langle i\vert \rho_A\vert i\rangle\tau_i$$ is not well-defined due to dimensional inconsistency. What is the mathematically correct way to express the action of this channel?

2) How do I write the Kraus operators of this channel?

• Can you elaborate how those Kraus operators map $A\to B$ (the Hilbert spaces)? Oct 19 '19 at 16:32
• @NorbertSchuch, perhaps I am using incorrect/confusing notation - sorry for this but maybe it is clearer after the edit? The idea is that the channel measures the state in $A$ and obtains some outcome $i$ with probability $p_i$ (at this point the register $A$ ceases to exist). Then the channel outputs the state $\tau_i\in B$ with this probability. So the Kraus operators must take us from $A$ to $B$. Oct 19 '19 at 17:45
• Yes. But the new channel you write takes A+B as an input and maps it onto B. And on the joint A+B system (with input $\rho_A\otimes I$) the map is not trace preserving. On the other hand, if the $I_B$ is not part of the input to the channel, then the form you write is not the Kraus form. Oct 19 '19 at 21:02
• @NorbertSchuch, I see your point - thank you for this comment. I realize I do not know how to correctly express the action of this channel. I have edited the question to reflect this as well along with my original goal to write our the Kraus operators. Oct 19 '19 at 22:25

2. As always, there are many ways of writing the Kraus operators. E.g., you could decompose $$\tau_i = \sum_j \vert a_{ij}\rangle\langle a_{ij}\vert$$ (e.g., an eigenvalue decomposition with the eigenvalues absorbed in the $$\vert a_{ij}\rangle$$, but any such decomposition will be fine), and choose Kraus operators $$K_{ij} = \vert a_{ij}\rangle\langle i|\ .$$ That they satisfy the completeness relation is left as an exercise to the reader ;) (We know they have to satisfy them, since they are Kraus operators of a trace preserving map.)