The Exercise

A quantum system is in the state $\rho$ and a measurement is made on it, but nobody reads the outcome. The state $\rho'$ assigned to the system afterwards is the mixture of the states that follow the possible outcomes, each weighted by its probability. The exercise asks for this state when the measurement is projective, with projectors $P_i$. It then takes a system made of two parts, with the measurement acting on the second part alone, and asks to show that the reduced state of the first part is the same before the measurement and after it.

The library states both results for any two quantum systems $S$ and $T$, the second for every measurement on $T$, projective or not.

Systems and States

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ The composite system $S \otimes T$ of two quantum systems $S$ and $T$ is the quantum 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 $S$.^✦

The reduced state $\mathrm{tr}_2(\rho)$ of a state $\rho$ of $S \otimes T$ is the state of $S$ obtained from $\rho$ by the partial trace over $T$, taken in an orthonormal basis $\{|k\rangle\}$ of the state space of $T$, where $I_S$ is the identity operator on the state space of $S$: $$\mathrm{tr}_2(\rho) = \sum_k \left(I_S \otimes \langle k|\right) \rho \left(I_S \otimes |k\rangle\right).$$(1)^✦

Measurements

A measurement on $S$ with a finite set of outcomes is a family of operators $M_i$ on the state space of $S$, one for each outcome $i$, satisfying the completeness relation, where $I$ is the identity operator on that space, $$\sum_i M_i^\dagger M_i = I.$$(2) A projective measurement on $S$ is a measurement on $S$ each of whose operators is an orthogonal projector $P_i$, a Hermitian operator equal to its own square.^✦

The extension of a measurement $\{M_i\}$ on $T$ to the composite system $S \otimes T$ is the measurement on $S \otimes T$ with the same outcomes whose operator for the outcome $i$ is $I_S \otimes M_i$.^✦

The state $\rho'$ after an unread measurement $\{M_i\}$ of a state $\rho$ of $S$ is the state of $S$ given by $$\rho' = \sum_i M_i \rho M_i^\dagger .$$(3)^✦

The Claim

What Is Checked

The claim is stated over the constructions defined above: the state (3) after an unread measurement, the projective measurements, the extension of a measurement on $T$ to $S \otimes T$ and the reduced state (1), all built on measurements satisfying (2) and on states of quantum systems and their composites.

The library proves that $\rho'$ is a state of $S$: the operator (3) is positive, and its trace is one by (2). It proves that the operators $I_S \otimes M_i$ satisfy (2) on $S \otimes T$, since $(I_S \otimes M_i)^\dagger (I_S \otimes M_i) = I_S \otimes M_i^\dagger M_i$ and, with $I_T$ the identity operator on the state space of $T$, $I_S \otimes I_T$ is the identity operator $I$ on the state space of $S \otimes T$, so the extension is a measurement. The partial trace of a density operator is a density operator, so $\mathrm{tr}_2(\rho)$ is a state of $S$.