The Exercise

Alice holds a system $A$ in an unknown state $\rho$ and wants Bob to end up with it. They share a state $\sigma$ of two further systems, $B$ on Alice’s side and $C$ on Bob’s, and $\sigma$ may be any state, maximally entangled or not. Alice measures $A$ and $B$ together and tells Bob her result $i$. Each result occurs with some probability and leaves Bob’s system in a state that depends on $\rho$; Bob then works on $C$ alone, according to $i$, to get $\rho$ back. The probabilities of Alice’s results add up to one. And if Bob’s operation for the result $i$ returns every input, the state that $i$ leaves on his side determines the input.^✦

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 systems $S$ and $T$ is the system whose state space is the tensor product of their state spaces.^✦ A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$, also called a density operator. 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 $\sigma$.^✦

An identification $e$ of a system $S$ with a system $T$ is a linear isometric bijection from the state space of $S$ onto that of $T$; its inverse $e^{-1}$ identifies $T$ with $S$. The relabelling of a state $\rho$ of $S$ along $e$ is the state $e \rho e^{-1}$ of $T$.^✦ The regrouping $\alpha$ of three systems $A$, $B$ and $C$ is the identification of $A \otimes (B \otimes C)$ with $(A \otimes B) \otimes C$ that sends $|u\rangle \otimes (|v\rangle \otimes |w\rangle)$ to $(|u\rangle \otimes |v\rangle) \otimes |w\rangle$.^✦

The reduced states of a state $\rho$ of a composite system $S \otimes T$ are its partial traces: $\operatorname{Tr}_T \rho$, a state of $S$, and $\operatorname{Tr}_S \rho$, a state of $T$. A unitary $U$ of a system $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = 1$, and the evolution of a state $\rho$ of $S$ under $U$ is the state $U \rho U^\dagger$. Given finitely many states $\rho_k$ of $S$ and weights $w_k \geq 0$ adding up to one, their mixture is the state of $S$ with density operator $\sum_k w_k \rho_k$.^✦

Measurements

A measurement $m$ on a system $S$ with a finite set $K$ of outcomes is a family of measurement operators $M_k$ on the state space of $S$, one for each outcome $k$, satisfying the completeness relation $\sum_k M_k^\dagger M_k = 1$.^✦ The positive operator valued measure (POVM) of $m$ is the family of positive operators $E_k = M_k^\dagger M_k$, which add up to the identity.^✦ Performed on a state $\rho$ of $S$, the measurement gives the outcome $k$ with probability $p(k) = \operatorname{Tr}(M_k \rho M_k^\dagger) = \operatorname{Tr}(E_k \rho)$, and when $p(k) \neq 0$ the state after that outcome is $M_k \rho M_k^\dagger / p(k)$.

For a measurement $m$ on $S$ and a further system $T$, the measurement of $m$ on the first factor of $S \otimes T$ is the measurement on $S \otimes T$ with the same outcomes as $m$ and the operators $M_k \otimes 1_T$.^✦ A POVM on $S$ is a family of positive operators $F_i$ on the state space of $S$ that add up to the identity, and the measurement of a POVM $\{F_i\}$ is the measurement on $S$ whose operators are the square roots $\sqrt{F_i}$.^✦ The coarse graining of $m$ along a map $g \colon K \to I$ into a finite set $I$ is the measurement of the POVM $F_i = \sum_{k :\, g(k) = i} E_k$, in which all outcomes with the same image under $g$ are merged into one.^✦

Alice’s Measurement

The three systems start in the product state $\rho \otimes \sigma$ of $A \otimes (B \otimes C)$. Regrouped so that Alice’s two systems stand together, this is the state of $(A \otimes B) \otimes C$ $$\chi = \alpha \, (\rho \otimes \sigma) \, \alpha^{-1} .$$(1)^✦

Alice performs a measurement $m$ on $A \otimes B$ with outcomes $k$ in a finite set $K$; on the three systems it is the measurement on the first factor, with operators $M_k \otimes 1_C$. The outcome $k$ has the probability $$p(k) = \operatorname{Tr}\bigl[ (M_k \otimes 1_C) \, \chi \, (M_k \otimes 1_C)^\dagger \bigr]$$(2) and, when $p(k) \neq 0$, leaves Bob’s system in the reduced state of the state after the outcome, $$\tau_k = \operatorname{Tr}_{A \otimes B} \biggl[ \frac{(M_k \otimes 1_C) \, \chi \, (M_k \otimes 1_C)^\dagger}{p(k)} \biggr] .$$(3)

Bob does not learn $k$. Alice reports the result $i = g(k)$, for a map $g \colon K \to I$ into a finite set $I$ of results, so the measurement Bob hears about is the coarse graining along $g$ of Alice’s measurement on the first factor. The result $i$ has the probability $$p(i) = \operatorname{Tr}(F_i \, \chi) , \qquad F_i = \sum_{k :\, g(k) = i} (M_k \otimes 1_C)^\dagger (M_k \otimes 1_C) ,$$(4) When $p(i) \neq 0$, the state the result $i$ induces on Bob’s system is the mixture of the states $\tau_k$ of those outcomes, each weighted by its probability given $i$: $$\hat{\mathcal{E}}_i(\rho) = \sum_{k :\, g(k) = i} \frac{p(k)}{p(i)} \, \tau_k .$$(5)^✦

Bob’s Recovery

Bob reads his system as Alice’s through an identification $e$ of $A$ with $C$. A recovery $R$ of Bob’s is given by an ancilla, a system $D$ in a state $\omega$, and a unitary $U$ of $C \otimes D$: it adjoins $D$ in the state $\omega$, applies $U$, traces out $D$ and relabels what remains along $e^{-1}$. A state $\tau$ of $C$ goes to the state of $A$ $$R(\tau) = e^{-1} \operatorname{Tr}_D \bigl[ U (\tau \otimes \omega) U^\dagger \bigr] \, e .$$(6)^✦

A recovery $R$ of Bob’s, built from an ancilla $D$, a state $\omega$ of $D$ and a unitary $U$ of $C \otimes D$, succeeds for the result $i$ when it returns every state $\rho$ of $A$ giving the result $i$ a nonzero probability: $$R\bigl( \hat{\mathcal{E}}_i(\rho) \bigr) = \rho .$$(7)^✦

The Claim

What Is Checked

The first theorem is stated over the induced state (5), built from the regrouped state (1), the probabilities (2) and (4) and the states (3), and over Bob’s recovery (6) with its condition of success (7). The second is stated over the probability (4) alone.

About these constructions the library proves that they stay among states and measurements. The induced state is a state of $C$: its weights $p(k)/p(i)$ are nonnegative and add up to one. The measurement on the first factor satisfies the completeness relation because $(M_k \otimes 1)^\dagger (M_k \otimes 1) = M_k^\dagger M_k \otimes 1$, and the measurement of a POVM satisfies it because $\sqrt{F_i}^{\,\dagger} \sqrt{F_i} = F_i$. The state after an outcome of nonzero probability, the partial traces, the product state, the relabelled state and the evolved state $U \rho U^\dagger$ are all density operators.