The Exercise

Two systems $A$ and $B$ together form a joint system $AB$, which starts in a state $\rho$. It is brought into contact with an environment, a second joint system $CD$ prepared in a pure state $\lvert\psi\rangle$, and the four systems interact through a unitary $U$. Afterwards $A$ and $D$ are discarded, which leaves a state $\mathcal{E}(\rho)$ of the joint system $BC$ formed by the two remaining systems. The channel $\mathcal{E}$ thus takes the states of one system to the states of another, and the exercise is to show that it still has an operator sum representation: $\mathcal{E}(\rho) = \sum_k E_k \, \rho \, E_k^\dagger$ with operation elements $E_k$ that map the state space of $AB$ into that of $BC$ and satisfy the completeness relation $\sum_k E_k^\dagger E_k = I$.

Systems and States

A quantum system $S$ is described by its state space, a finite dimensional complex Hilbert space.^✦ The composite system $S \otimes T$ of two quantum systems $S$ and $T$ is the system whose state space is the tensor product of the state spaces of $S$ and $T$.^✦ A pair of letters stands for the composite, $AB = A \otimes B$, and the four systems together form $AB \otimes CD$.

A state $\rho$ of a system $S$ is a positive operator of unit trace on the state space of $S$.^✦ A pure state $\lvert\psi\rangle$ of $S$ is a unit vector in the state space of $S$; as a state it is the projector $\lvert\psi\rangle\langle\psi\rvert$ onto that vector. For a state $\rho$ of $S$ and a state $\sigma$ of $T$, the product state $\rho \otimes \sigma$ is the state of $S \otimes T$ given by the tensor product of the operators $\rho$ and $\sigma$.^✦ The partial trace over $T$ of a state $\rho$ of $S \otimes T$ is the reduced state $\operatorname{Tr}_T \rho$ of $S$, obtained by tracing out the right factor.^✦

A unitary $U$ of $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = I$.^✦ Under it a state $\rho$ of $S$ evolves into $U \rho U^\dagger$.^✦

An identification $e$ of a system $S$ with a system $S'$ is a linear isometry of the state space of $S$ onto that of $S'$; it relabels a state $\rho$ of $S$ as the state $e \rho e^{-1}$ of $S'$. The composite of an identification $e$ of $S$ with $S'$ and an identification $f$ of $S'$ with $S''$ is the identification $f \circ e$ of $S$ with $S''$, first $e$ and then $f$.^✦ The product of an identification $e$ of $S$ with $S'$ and an identification $g$ of $T$ with $T'$ is the identification $e \otimes g$ of $S \otimes T$ with $S' \otimes T'$ by the tensor product of the two isometries.^✦ The identity identification of $S$ is the identification of $S$ with itself by the identity map of its state space.^✦ The swap of $S$ and $T$ is the identification of $S \otimes T$ with $T \otimes S$ by $u \otimes v \mapsto v \otimes u$, for $u$ in the state space of $S$ and $v$ in that of $T$.^✦ The interchange of four systems $S_1$, $S_2$, $S_3$, $S_4$ is the identification of $(S_1 \otimes S_2) \otimes (S_3 \otimes S_4)$ with $(S_1 \otimes S_3) \otimes (S_2 \otimes S_4)$ that exchanges the two inner factors.^✦

The Channel

Discarding $A$ and $D$ means tracing out $AD$, so the four systems are first regrouped as $BC \otimes AD$. The regrouping $\tau$ is the identification of $AB \otimes CD$ with $BC \otimes AD$ composed of the product of the swap of $A$ and $B$ with the identity of $CD$, followed by the interchange of $(B \otimes A) \otimes (C \otimes D)$.^✦ On product vectors, with $a$, $b$, $c$ and $d$ vectors in the state spaces of $A$, $B$, $C$ and $D$, $$(a \otimes b) \otimes (c \otimes d) \;\mapsto\; (b \otimes a) \otimes (c \otimes d) \;\mapsto\; (b \otimes c) \otimes (a \otimes d).$$(1)

For a pure state $\lvert\psi\rangle$ of $CD$ and a unitary $U$ of $AB \otimes CD$, the channel $\mathcal{E}$ of the exercise takes a state $\rho$ of $AB$ to the state of $BC$ $$\mathcal{E}(\rho) = \operatorname{Tr}_{AD} \Bigl[ \tau \, U \bigl( \rho \otimes \lvert\psi\rangle\langle\psi\rvert \bigr) U^\dagger \, \tau^{-1} \Bigr].$$(2)^✦

The Claim

What Is Checked

The claim is stated over the channel (2), with the four systems, the pure state of $CD$ and the unitary all arbitrary.

The library proves that each step of (2) yields a density operator: the projector onto a unit vector is one, and so are the tensor product of two density operators, the conjugate of one by a unitary or by an isometry, and the partial trace of one.