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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}?

glS
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user27286
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1 Answers1

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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 $$ \begin{aligned} \mathcal{E}(\rho) &= \mathrm{Tr}_{\mathrm{env}}[(P_0 \otimes I + P_1 \otimes X)(\rho \otimes |0\rangle \langle 0 |) (P_0 \otimes I + P_1 \otimes X)] \\ &= \mathrm{Tr}_{\mathrm{env}}[(P_0 \otimes I + P_1 \otimes X)(\rho P_0 \otimes |0\rangle \langle 0| + \rho P_1 \otimes |1\rangle \langle 0|)] \\ &= \mathrm{Tr}_{\mathrm{env}}[(P_0\rho P_0 \otimes |0\rangle \langle 0| + P_0\rho P_1 \otimes |1\rangle \langle 0| + P_1\rho P_0 \otimes |1\rangle\langle 0| + P_1 \rho P_1 \otimes |1\rangle\langle1|] \\ &= P_0 \rho P_0 + P_1 \rho P_1. \end{aligned} $$

Rammus
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  • 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.

     – user27286 Feb 19 '21 at 17:21
  • 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? – user27286 Feb 19 '21 at 17:26
  • $$ \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1&0&0 \ 0&0&0&1 \ 0&0&1&0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} = |0\rangle\langle 0 |\otimes I + |1\rangle \langle 1 |\otimes X $$ 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 $(A\otimes B)(C\otimes D) = AC \otimes BD$. – Rammus Feb 19 '21 at 18:07
  • .: 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) – user27286 Feb 19 '21 at 18:09
  • 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 $X$). – Rammus Feb 19 '21 at 18:12
  • @user27286 I think that you can imagine $|m\rangle\langle m|$ as indicating which qubit you want to apply the gate to. The gate is indicated by the tensor product. It is helpful to use Dirac notation to do calculations, as it prevents ridiculously large matrix multiplications. – BrockenDuck Feb 20 '21 at 10:01