The Exercise

A measurement on a principal system $Q$ has its outcomes $m$ in a finite set, and the outcome $m$ acts on $Q$ by a quantum operation $\mathcal{E}_m$. The operations are given by operation elements: operators $E_\sigma$ on the state space of $Q$, one for each index $\sigma$ of a finite set, with $\sum_\sigma E_\sigma^\dagger E_\sigma = 1$, and a map $o$ that assigns to each index its outcome, so that the elements of $\mathcal{E}_m$ are the $E_\sigma$ with $o(\sigma) = m$.

The exercise builds this measurement from a unitary and a projective measurement. An environment $\mathrm{Env}$ with an orthonormal basis $b_\sigma$, one vector for each index, starts in a unit vector $e_0$ and interacts with $Q$ through a unitary $U$ that writes the index of the operation element acting on $Q$ into the environment as $b_\sigma$. The environment is then measured with the projectors $P_m$ onto the basis vectors of outcome $m$. The task is to show that, for $Q$ in a state $\rho$, the outcome $m$ occurs with probability $\operatorname{Tr}\mathcal{E}_m(\rho)$ and leaves $Q$ in the state $\mathcal{E}_m(\rho)/\operatorname{Tr}\mathcal{E}_m(\rho)$.

In the library $U$ is one fixed unitary of the joint system of $Q$ and $\mathrm{Env}$ built from these data, and its action on $Q$ with the environment in $e_0$ is part of the claim, beside the probability and the state after the outcome.

A Quantum System

A quantum system $S$ is a physical system whose state space $\mathcal{H}_S$ is 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 $\mathcal{H}_S \otimes \mathcal{H}_T$.^✦

A state $\rho$ of $S$ is a positive operator of unit trace on $\mathcal{H}_S$, a density operator.^✦ A pure state of $S$ is a unit vector $\psi$ in $\mathcal{H}_S$, and it determines the state $\lvert\psi\rangle\langle\psi\rvert$ of $S$, the projector onto $\psi$. The product state $\rho \otimes \tau$ of a state $\rho$ of $S$ and a state $\tau$ of $T$ is the state of $S \otimes T$ given by the tensor product of the operators $\rho$ and $\tau$.^✦

A unitary $U$ of $S$ is an operator on $\mathcal{H}_S$ with $U^\dagger U = U U^\dagger = 1$, and it evolves a state $\rho$ of $S$ into the state $U \rho U^\dagger$. A linear isometry of $\mathcal{H}_S$ onto itself is a unitary of $S$.^✦

Measurements

A measurement on $S$ with a finite set of outcomes is a family of measurement operators $M_m$ on $\mathcal{H}_S$, one for each outcome $m$, with $\sum_m M_m^\dagger M_m = 1$.^✦ Performed on a state $\rho$ of $S$, it gives the outcome $m$ with the probability $p(m)$ that the Born rule assigns through the positive operator valued measure (POVM) with elements $M_m^\dagger M_m$, and when $p(m) \neq 0$ it leaves $S$ in the state $\rho_m$ after the outcome: $$p(m) = \operatorname{Tr}(M_m \rho M_m^\dagger) = \operatorname{Tr}(M_m^\dagger M_m \rho), \qquad \rho_m = \frac{M_m \rho M_m^\dagger}{p(m)} .$$(1)

A measurement with operators $M_m$ on $T$ is also a measurement on $S \otimes T$, with operators $1_S \otimes M_m$, the measurement of the second factor alone.^✦ The reduced state $\operatorname{Tr}_T \rho$ of a state $\rho$ of $S \otimes T$ is the state of $S$ obtained by tracing out $T$.^✦

The Model

The environment block measurement is the measurement on $\mathrm{Env}$ whose operator for the outcome $m$ is the projector onto the span of the basis vectors $b_\sigma$ with $o(\sigma) = m$, $$P_m = \sum_{\sigma \,:\, o(\sigma) = m} \lvert b_\sigma \rangle\langle b_\sigma \rvert .$$(2)^✦ On $Q \otimes \mathrm{Env}$ it is the measurement with operators $1_Q \otimes P_m$.^✦

The operation $\mathcal{E}_m$ induced by the outcome $m$ takes a state $\rho$ of $Q$ to the operator $$\mathcal{E}_m(\rho) = \sum_{\sigma \,:\, o(\sigma) = m} E_\sigma \rho E_\sigma^\dagger$$(3)^✦ on $\mathcal{H}_Q$.

The joint unitary $U$ of the model is a unitary of $Q \otimes \mathrm{Env}$ fixed by the data, a unitary dilation of the operation elements $E_\sigma$ built from $e_0$ and the basis $b_\sigma$: $$U \colon \mathcal{H}_Q \otimes \mathcal{H}_{\mathrm{Env}} \to \mathcal{H}_Q \otimes \mathcal{H}_{\mathrm{Env}}, \qquad U^\dagger U = U U^\dagger = 1 .$$(4)^✦

The Claim

What Is Checked

The claim is stated over the joint unitary $U$ of (4), the block measurement (2) taken on the environment factor, and the operations (3).

The projectors $P_m$ are self adjoint with $P_m^2 = P_m$ and add up to the identity, $\sum_m P_m = 1$, because the projectors $\lvert b_\sigma\rangle\langle b_\sigma\rvert$ add up to the identity. The operators $1_Q \otimes P_m$ do the same in turn, since the sum passes into the second factor. The projector onto $e_0$ and the product of two states are density operators, and conjugating a density operator by a unitary gives one again, so $\rho'$ is a state. The state after an outcome of nonzero probability is a density operator, and so is its reduced state.