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What is the correct mathematical notation to describe the following setup? I have classical state in register $A$ which I can think of as $\sum_i p_i \vert i\rangle\langle i \vert_A$. I measure this to obtain outcome $i$ and then perform operation $N_i$ on a different register $A'$.

How does one denote this type of conditional channel? The channel has to have input system $AA'$ and output $B$ (which is the output of $N_i$ for any $i$). I currently say

$$\sum_i N_i \otimes \langle i\vert \cdot \vert i\rangle $$

but this looks awkward.

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I think $\sum_iN_i\otimes|i\rangle\langle i|$ will be enough.

For example, if the state to be measured is $a|0\rangle\langle 0|+b|1\rangle\langle 1|$ where $a$ and $b$ satisfy the condition of normalization. And the state to be operated by $N_i$ defined as $\rho$. Then the total state is $\rho\otimes(a|0\rangle\langle 0|+b|1\rangle\langle 1|)$, now $\sum_iN_i\otimes|i\rangle\langle i|$ becomes $N_0\otimes|0\rangle\langle 0|+N_1\otimes|1\rangle\langle 1|$. Easy to see after the operation, the total state becomes $N_0\rho\otimes a|0\rangle\langle 0|+N_1\rho\otimes b|1\rangle\langle 1|$.

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    $\begingroup$ I think you still need to partial trace out the second register. The final channel should not have the classical register anymore (it gets measured). But then you have $\text{Tr}_B\circ\sum_i N_i \otimes \vert i\rangle\langle i\vert$ $\endgroup$
    – koolaid
    Jul 9 at 9:28
  • $\begingroup$ @koolaid No. The result is the same as his 'awkward form'. $\endgroup$
    – narip
    Jul 9 at 11:12

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