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Equation \eqref{eq:sp1} represents the reduced state of the system after tracing over environment.(Page number 358)

$$\mathcal{E}(\rho) = \mathrm{tr}_{env}(\lbrack U(\rho \otimes \rho_{env} )U^{\dagger}\rbrack). \tag{8.6} \label{eq:sp1}$$

And then they say in page 359 that initially $\rho_{env} = |0\rangle\langle0|$ and then we apply $U$ to the combined state.(here $U$ is CNOT). The equation \eqref{eq:sp1} becomes (after plugging these values)

$$ \mathcal{E}(\rho) = P_{0}\rho P_{0} + P_{1}\rho P_{1} \tag{8.7} \label{eq:sp2}$$ where $P_{m}=|m\rangle\langle m|$.

How are they arriving at \eqref{eq:sp2}?

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Just plug in all of the relevant stuff you state in the question, i.e. $$ U = |0\rangle \langle 0 | \otimes I + |1 \rangle \langle 1 | \otimes X $$ and $$ \rho_{\mathrm{env}} = |0\rangle \langle 0 |. $$ Then expand and simplify $$ \begin{aligned} \mathcal{E}(\rho) &= \mathrm{Tr}_{\mathrm{env}}[(P_0 \otimes I + P_1 \otimes X)(\rho \otimes |0\rangle \langle 0 |) (P_0 \otimes I + P_1 \otimes X)] \\ &= \mathrm{Tr}_{\mathrm{env}}[(P_0 \otimes I + P_1 \otimes X)(\rho P_0 \otimes |0\rangle \langle 0| + \rho P_1 \otimes |1\rangle \langle 0|)] \\ &= \mathrm{Tr}_{\mathrm{env}}[(P_0\rho P_0 \otimes |0\rangle \langle 0| + P_0\rho P_1 \otimes |1\rangle \langle 0| + P_1\rho P_0 \otimes |1\rangle\langle 0| + P_1 \rho P_1 \otimes |1\rangle\langle1|] \\ &= P_0 \rho P_0 + P_1 \rho P_1. \end{aligned} $$

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  • $\begingroup$ Omg...how do you know all these? 1. I didn’t know CNOT can be represented like this? (This is entirely because of the density matrix notation right? Do you have any idea where can I find more information regarding this?” 2. In the second line derivation you applied an expansion, where from do you know this formula? I am scared that I still don’t know all these. $\endgroup$
    – user27286
    Feb 19 at 17:21
  • $\begingroup$ Ok I guess for second question you applied 2.46 of NC. But how do you know about this idea of representing CNOT like this? $\endgroup$
    – user27286
    Feb 19 at 17:26
  • $\begingroup$ $$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} = |0\rangle\langle 0 |\otimes I + |1\rangle \langle 1 |\otimes X $$ For 2. I don't have a copy of NC so I don't know what 2.46 is but the second line is just expanding brackets and using $(A\otimes B)(C\otimes D) = AC \otimes BD$. $\endgroup$
    – Rammus
    Feb 19 at 18:07
  • $\begingroup$ .: But how do you know this thing? Can you represent Hadamard in same form? I just want to learn this from you. (You did express CNOT thinking this? I want to know the general idea) $\endgroup$
    – user27286
    Feb 19 at 18:09
  • $\begingroup$ This representation is arguably how you would define $\mathrm{CNOT}$, if first qubit is $0$ then do nothing to second qubit (apply identity) and if first qubit is $1$ then flip the second qubit (apply $X$). $\endgroup$
    – Rammus
    Feb 19 at 18:12

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