# What is the correct notation to denote operations conditional on a measurement outcome?

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.

• – glS
Jul 9, 2021 at 9:25

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 $$N_0\rho\otimes a|0\rangle\langle 0|+N_1\rho\otimes b|1\rangle\langle 1|$$.
• 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 $\text{Tr}_B\circ\sum_i N_i \otimes \vert i\rangle\langle i\vert$ Jul 9, 2021 at 9:28