How does $\mathcal E(\rho)=\mathrm{Tr}_{env}[U(\rho\otimes\rho_{env})U^\dagger]$ turn into $P_0\rho P_0+P_1\rho P_1$?

Question

In the Quantum Operations section in Nielsen and Chuang, (page 358 in the 2002 edition), they have the following equation: $$\mathcal E(\rho) = \mathrm{Tr}_{env} [U(\rho \otimes \rho_{env})U^\dagger]$$

They show an example with $\rho_{env} = |0\rangle \langle0|$ and $U = \mathrm{CNOT}$, and claim that the final solution is: $$P_0\rho P_0 + P_1\rho P_1,$$ where $P_0=|0\rangle \langle0|$ and $P_1=|1\rangle \langle 1|$.

These are my steps so far to get this, but I don't know how to trace out the environment after this:

Let $\rho$ be $|\psi \rangle \langle \psi |$, so that $\rho \otimes \rho_{env} = |\psi, 0\rangle \langle \psi, 0|$.

Applying the unitary $U$, we have

$$ |00 \rangle \langle 00| \psi, 0 \rangle \langle \psi, 0 | 00 \rangle \langle 00 | + |00 \rangle \langle 00| \psi 0 \rangle \langle \psi 0 | 10 \rangle \langle 11 | \\ + |11 \rangle \langle 10| \psi 0 \rangle \langle \psi 0 | 00 \rangle \langle 00 | + |11 \rangle \langle 10| \psi 0 \rangle \langle \psi 0 | 10 \rangle \langle 11 |. $$

I don't know how to trace out the environment in the above state.

Also, I realize that I have considered only a pure state, if anyone can show it for a general state that would be great.

Answer 1

Let's start with a general state $$ \rho\otimes\rho_0=\sum_{x,y\in\{0,1\}}\langle x|\rho|y\rangle|x\rangle\langle y|\otimes |0\rangle\langle 0|. $$ If we apply the controlled-not, we have $$ \rightarrow\rho_{\text{final}}=\sum_{x,y\in\{0,1\}}\langle x|\rho|y\rangle|x\rangle\langle y|\otimes |x\rangle\langle y|. $$

Now we want to take the partial trace over the second subsystem. This means calculating $$ \sum_k(\mathbb{I}\otimes\langle k|)\rho_{\text{final}}(\mathbb{I}\otimes|k\rangle)=\sum_k\sum_{x,y\in\{0,1\}}\langle x|\rho|y\rangle|x\rangle\langle y|\times \langle k|x\rangle\langle y|k\rangle. $$ If we perform the sums over $x$ and $y$, we find that $x=y=k$, so $$ =\sum_k\langle k|\rho|k\rangle|k\rangle\langle k|, $$ which is entirely equivalent to removing all the off-diagonal elements of $\rho$.

Answer 2

Let $\newcommand{\rmD}{\mathrm{D}}\newcommand{\rmU}{\mathrm{U}}\newcommand{\calU}{\mathcal{U}}\newcommand{\CC}{\mathbb{C}}\rho\in\rmD(\CC^n),\sigma\in\rmD(\CC^m)$ be two arbitrary finite-dimensional quantum states, and let $\calU\in \rmU(\CC^n\otimes\CC^m)$ be a unitary in the total space.

We have

$$[U(\rho\otimes\sigma)U^\dagger]_{ij,k\ell} = \sum_{mnpq} U_{ij,mn} U^*_{k\ell,pq} \rho_{mp} \sigma_{nq}.\tag1 $$

Suppose $\sigma=|0\rangle\!\langle0|$. Then $$ [U(\rho\otimes|0\rangle\!\langle0|)U^\dagger]_{ij,k\ell} = \sum_{mp} U_{ij,m0} U^*_{k\ell,p0} \rho_{mp}.\tag2$$

Remember now the definition of partial trace. Let $X:\CC^n\otimes\CC^m\to\CC^{n'}\otimes\CC^{m}$ be some linear operator. Its matrix elements can be written as $X_{ij,k\ell}$, where $i\in [n'], j\in[m], k\in[n], \ell\in[m]$, with $n,n',m\in\mathbb N$ positive integers, and $[n]\equiv\{1,...,n\}$ etc. Its partial trace with respect to the second space is the linear operator $\mathrm{Tr}_2(X):\CC^n\to\CC^{n'}$ with matrix elements $$[\mathrm{Tr}_2(X)]_{ik} = \sum_j X_{ij,kj}.$$

Therefore, the partial trace of (2) wrt the second space reads $$ \Big[\mathrm{Tr}_2[U(\rho\otimes|0\rangle\!\langle0|)U^\dagger]\Big]_{ik} = \sum_{jmp} U_{ij,m0} U^*_{kj,p0} \rho_{mp}. $$ Define the operators $A_j:\CC^n\to\CC^{n'}$ via their matrix elements $(A_j)_{im} = U_{ij,m0}$. With these, we can rewrite the above equation as $$ \mathrm{Tr}_2[U(\rho\otimes|0\rangle\!\langle0|)U^\dagger] = \sum_j A_j \rho A_j^\dagger. $$

It's worth noting that $A_j$ can be understood as sub-blocks of the matrix $\mathcal U$ corresponding to the columns specified by $j$. See also my answer here about this.