The Exercise

A system $Q$ is prepared in a state $\rho$ and passes through noise, a quantum operation $\mathcal{E}$ that leaves it in the state $\rho' = \mathcal{E}(\rho)$. A second operation $\mathcal{R}$, the recovery, then acts on $Q$ to undo the noise. What undoing means is fixed by a reference system $R$ that purifies $\rho$: $R$ and $Q$ start in a pure state $|RQ\rangle$ whose reduced state on $Q$ is $\rho$, and neither operation touches $R$. The recovery corrects the noise perfectly when noise followed by recovery returns the joint state $|RQ\rangle\langle RQ|$ unchanged.^✦

The exercise is to show that in that case the entropy of the noisy state $\rho'$ exceeds the entropy of $\rho$ by exactly the entropy exchange of the recovery, taken on its input $\rho'$.^✦

A Quantum System

A quantum system $A$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ Two systems $A$ and $B$ form the composite system $A \otimes B$, 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 $\psi$ of $A$ is a unit vector in the state space of $A$^✦, and it determines the state $|\psi\rangle\langle\psi|$ of $A$, the projector onto $\psi$.^✦ 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.^✦ In the other direction, the partial trace of a state $\tau$ of $A \otimes B$ over $B$ is the reduced state $\operatorname{Tr}_B(\tau)$ of $A$.^✦

The von Neumann entropy of a state $\rho$ of $A$, in nats, is $$S(\rho) = -\operatorname{tr}(\rho \log \rho).$$(1)^✦

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

An identification of systems $e$ between $A$ and $B$ is a linear isometric bijection from the state space of $A$ onto that of $B$^✦; its inverse $e^{-1}$ identifies $B$ with $A$.^✦ Along $e$ a state $\rho$ of $A$ is relabelled as the state $e \rho e^{-1}$ of $B$^✦, and a pure state $\psi$ of $A$ as the pure state $e\psi$ of $B$.^✦ The swap $\mathrm{swap}_{A,B}$, sending $u \otimes v$ to $v \otimes u$, identifies $A \otimes B$ with $B \otimes A$.^✦ The regrouping $\alpha_{A,B,C}$, sending $u \otimes (v \otimes w)$ to $(u \otimes v) \otimes w$, identifies $A \otimes (B \otimes C)$ with $(A \otimes B) \otimes C$.^✦

Noise and Recovery

The dilated channel $\mathcal{E}_{U,\omega}$ on a system $A$, built from an ancilla $C$, a unitary $U$ of $A \otimes C$ and a pure state $\omega$ of $C$, sends a state $\rho$ of $A$ to the state $$\mathcal{E}_{U,\omega}(\rho) = \operatorname{Tr}_C\bigl(U (\rho \otimes |\omega\rangle\langle\omega|) U^\dagger\bigr)$$(2)^✦ of $A$. By Stinespring’s theorem every trace preserving quantum operation on $A$ arises this way.

For a state $\tau$ of a composite system $A \otimes B$, a unitary $U$ of $B \otimes C$ and a pure state $\omega$ of $C$, the operation $\mathcal{I}_A \otimes \mathcal{E}_{U,\omega}$, the channel $\mathcal{E}_{U,\omega}$ acting on the second factor alone and the identity $\mathcal{I}_A$ on the first, sends $\tau$ to the state of $A \otimes B$ $$(\mathcal{I}_A \otimes \mathcal{E}_{U,\omega})(\tau) = \operatorname{Tr}_C\bigl((1_A \otimes U)(\tau \otimes |\omega\rangle\langle\omega|) (1_A \otimes U)^\dagger\bigr).$$(3)^✦

The reference for a state $\rho$ of $Q$ is the system $R$ whose state space is the coordinate space $\mathbb{C}^d$^✦, where the dimension $d$ of $Q$ is the complex dimension of its state space.^✦ Write $\rho = \sum_i p_i |i\rangle\langle i|$ in an orthonormal eigenbasis and $|i_R\rangle$ for the standard basis of $\mathbb{C}^d$. The canonical purification of $\rho$ is the pure state of $Q \otimes R$ with vector $$\sum_i \sqrt{p_i}\, |i\rangle |i_R\rangle ,$$(4)^✦ and the reference purification $|RQ\rangle$ of $\rho$ is the pure state of $R \otimes Q$ obtained from it along $\mathrm{swap}_{Q,R}$, the reference first and the system second.^✦

The entropy exchange of a state $\rho$ of $Q$ under a dilated channel $\mathcal{E}_{U,\omega}$ on $Q$ is the entropy of reference and system once the channel has acted on $Q$ alone, $$S(\rho, \mathcal{E}_{U,\omega}) = S\bigl((\mathcal{I}_R \otimes \mathcal{E}_{U,\omega})(|RQ\rangle\langle RQ|)\bigr).$$(5)^✦

Let the noise be $\mathcal{E} = \mathcal{E}_{U_1,\omega_1}$, with an environment $C$, a unitary $U_1$ of $Q \otimes C$ and a pure state $\omega_1$ of $C$, and the recovery $\mathcal{R} = \mathcal{E}_{U_2,\omega_2}$, with an ancilla $D$, a unitary $U_2$ of $Q \otimes D$ and a pure state $\omega_2$ of $D$. The recovery $\mathcal{R}$ perfectly corrects the noise $\mathcal{E}$ on a state $\rho$ of $Q$ when $$(\mathcal{I}_R \otimes \mathcal{R})\bigl((\mathcal{I}_R \otimes \mathcal{E}) (|RQ\rangle\langle RQ|)\bigr) = |RQ\rangle\langle RQ| .$$(6)^✦

The Claim

What Is Checked

The claim is stated over the dilated channels (2) as noise and recovery, the reference purification of $\rho$ in the condition (6), the von Neumann entropy (1) and the entropy exchange (5).

The library proves that each construction gives a state of its system: the projector onto a unit vector, the product of two states, a state evolved by a unitary or relabelled along an identification, and a partial trace are density operators, so $\rho'$ and both sides of (6) are states. The product of two unitaries is again unitary, and the canonical purification (4) is a unit vector.