The Exercise

A system $Q$ starts in a state $\rho$ and passes through a noisy channel $\mathcal{E}_1$, then through a second channel $\mathcal{E}_2$. How much of the quantum information in $\rho$ a channel $\mathcal{E}$ lets through is measured by the coherent information $I(\rho,\mathcal{E})$. The theorem below is the quantum data processing inequality.

A Quantum System

A quantum system $A$ is a physical object whose state space is a finite dimensional complex Hilbert space.^✦ The dimension $\dim A$ of a system $A$ is the complex dimension of its state space.^✦ Any finite dimensional complex Hilbert space $H$ is the state space of a system, the system modelled on $H$.^✦ The composite system $A\otimes B$ of two systems $A$ and $B$ is the system whose state space is the tensor product of their state spaces.^✦

A state $\rho$ of $A$ is a positive operator of unit trace on the state space of $A$.^✦ A pure state $|\psi\rangle$ of $A$ is a unit vector in the state space of $A$^✦; it determines the state $|\psi\rangle\langle\psi|$ of $A$, the projector onto that vector.^✦ For a state $\rho$ of $A$ and a state $\sigma$ of $B$, the product state $\rho\otimes\sigma$ is the state of $A\otimes B$ whose density operator is the tensor product of the two.^✦

A unitary $U$ of $A$ is an operator on the state space of $A$ with $U^\dagger U = UU^\dagger = 1$^✦, and under it a state $\rho$ of $A$ evolves to the state $U\rho U^\dagger$.^✦ The identity unitary of $A$ is the identity operator $1$.^✦ For a unitary $U$ of $A$ and a unitary $V$ of $B$, the product unitary $U\otimes V$ is the unitary of $A\otimes B$ whose operator is the tensor product of the two.^✦

An identification $e$ of a system $A$ with a system $B$ is a linear isometric bijection from the state space of $A$ onto that of $B$.^✦ Along it a state $\rho$ of $A$ is relabelled as the state $e\rho e^{-1}$ of $B$^✦, and a pure state $|\psi\rangle$ of $A$ as the pure state $e|\psi\rangle$ of $B$.^✦ The inverse $e^{-1}$ of $e$ is the identification of $B$ with $A$ by the inverse map.^✦ The swap identifies $A\otimes B$ with $B\otimes A$ by $|u\rangle\otimes|v\rangle\mapsto|v\rangle\otimes|u\rangle$^✦, and the regrouping $\alpha_{A,B,C}$ of three systems identifies $A\otimes(B\otimes C)$ with $(A\otimes B)\otimes C$ by $|u\rangle\otimes(|v\rangle\otimes|w\rangle)\mapsto(|u\rangle\otimes|v\rangle)\otimes|w\rangle$.^✦

For a state $\rho$ of $A\otimes B$, the reduced state $\rho_A=\operatorname{Tr}_B\rho$ of $A$ is obtained by tracing out $B$^✦, and the reduced state $\rho_B=\operatorname{Tr}_A\rho$ of $B$ by tracing out $A$.^✦

Channels

The dilated channel $\mathcal{E}_{U,\omega}$ on a system $Q$, built from an ancilla system $C$ in a pure state $|\omega\rangle$ and a unitary $U$ of $Q\otimes C$, sends a state $\rho$ of $Q$ to the state $$\mathcal{E}_{U,\omega}(\rho) = \operatorname{Tr}_C\bigl(U(\rho\otimes|\omega\rangle\langle\omega|)U^\dagger\bigr)$$(1)^✦ of $Q$.

A channel can act on one factor of a composite system, the identity map $\mathrm{id}$ being applied to the other factor. For a state $\rho$ of $A\otimes B$, an ancilla $C$ in a pure state $|\omega\rangle$ and a unitary $U$ of $B\otimes C$, the state $(\mathrm{id}\otimes\mathcal{E}_{U,\omega})(\rho)$ of $A\otimes B$, obtained by letting the dilated channel act on $B$ alone, is $$(\mathrm{id}\otimes\mathcal{E}_{U,\omega})(\rho) = \operatorname{Tr}_C\bigl((1\otimes U)(\rho\otimes|\omega\rangle\langle\omega|) (1\otimes U)^\dagger\bigr)$$(2)^✦ Here $1\otimes U$ is the product unitary of the identity unitary of $A$ with $U$, acting on $A\otimes(B\otimes C)$.

Coherent Information

The von Neumann entropy of a state $\rho$ of a system is $$S(\rho) = -\operatorname{tr}(\rho\log\rho),$$(3)^✦ measured in nats. For a state $\rho$ of $A\otimes B$, the conditional entropy $S(A|B)$ and the coherent information $I(A\rangle B)$ are $$S(A|B) = S(\rho) - S(\rho_B), \qquad I(A\rangle B) = -S(A|B).$$(4) For a state $\sigma$ we write $I(A\rangle B)_\sigma$.

Let $\rho$ be a state of $Q$, and $R$ the system modelled on $\mathbb{C}^d$ with $d=\dim Q$. Write $\rho=\sum_i p_i|i\rangle\langle i|$ in an orthonormal eigenbasis, and let $|i_R\rangle$ be the standard basis of $\mathbb{C}^d$. The canonical purification of $\rho$ is the pure state $\sum_i\sqrt{p_i}\,|i\rangle|i_R\rangle$ of $Q\otimes R$, and its reduced state on $Q$ is $\rho$.^✦ The reference purification $|RQ\rangle$ of $\rho$ is the pure state of $R\otimes Q$ obtained by relabelling the canonical purification along the swap.^✦

For a dilated channel $\mathcal{E}=\mathcal{E}_{U,\omega}$ on $Q$, the entropy exchange $S(\rho,\mathcal{E})$ is the entropy of the joint state of $R$ and $Q$ once $\mathcal{E}$ has acted on $Q$^✦, and the coherent information $I(\rho,\mathcal{E})$ is the entropy of the output less the entropy exchange: $$S(\rho,\mathcal{E}) = S\bigl((\mathrm{id}\otimes\mathcal{E})(|RQ\rangle\langle RQ|)\bigr), \qquad I(\rho,\mathcal{E}) = S\bigl(\mathcal{E}(\rho)\bigr) - S(\rho,\mathcal{E}).$$(5)^✦

For two dilated channels $\mathcal{E}_1$ and $\mathcal{E}_2$ on $Q$, the coherent information of $\rho$ under $\mathcal{E}_2\circ\mathcal{E}_1$ is that of the state of $R\otimes Q$ the two leave when they act on $Q$ in turn: $$I(\rho,\mathcal{E}_2\circ\mathcal{E}_1) = I(R\rangle Q)_\sigma, \qquad \sigma = (\mathrm{id}\otimes\mathcal{E}_2) \bigl((\mathrm{id}\otimes\mathcal{E}_1)(|RQ\rangle\langle RQ|)\bigr).$$(6)^✦

The Claim

What Is Checked

Both inequalities are stated over a state $\rho$ of a system $Q$ and channels on $Q$ given by their dilations (1). The coherent informations (5) and (6) are built from the reference purification $|RQ\rangle$, the channels acting on its second factor as in (2), and the von Neumann entropy (3).

The library proves that each construction on the way yields a state: the projector onto a unit vector, the tensor product of two density operators, a density operator conjugated by a unitary or by an identification, and a partial trace of a density operator are density operators, and a relabelled pure state is again a unit vector. It proves too that the tensor product of two unitaries is unitary, so $1\otimes U$ is a unitary of $A\otimes(B\otimes C)$.