The Exercise

A closed quantum system evolves by a unitary $U$. A pure state of it can be followed in two ways: as a unit vector $|\psi\rangle$, which goes to $U|\psi\rangle$, or as its density operator $\rho = |\psi\rangle\langle\psi|$, which goes to $U\rho U^\dagger$. The exercise asks to show that both routes end at the same state: the density operator of $U|\psi\rangle$ is $U\rho U^\dagger$.^✦

A Quantum System

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ A state $\rho$ of $S$ is a positive operator of unit trace on the state space of $S$, $$\rho \geq 0, \qquad \operatorname{tr}\rho = 1 .$$(1)^✦ Pure and mixed states alike are states in this sense.

A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$, $\langle\psi|\psi\rangle = 1$.^✦ The density operator $\rho_\psi$ of a pure state $|\psi\rangle$ of $S$ is the state of $S$ given by the rank one projector onto $|\psi\rangle$, $$\rho_\psi = |\psi\rangle\langle\psi| .$$(2)^✦

Unitary Evolution

A unitary $U$ of $S$, the evolution of $S$ as a closed system, is an operator on the state space of $S$ with $$U^\dagger U = U U^\dagger = 1 .$$(3)^✦ It carries a pure state $|\psi\rangle$ of $S$ to the pure state $U|\psi\rangle$, $$|\psi\rangle \mapsto U|\psi\rangle .$$(4) Under $U$ a state $\rho$ of $S$ evolves to the state $U\rho U^\dagger$, $$\rho \mapsto U\rho U^\dagger .$$(5)

The Claim

What Is Checked

The claim is stated over a unitary $U$ of a quantum system $S$ and a pure state $|\psi\rangle$ of $S$.

The library proves that a unitary preserves norms, so that $U|\psi\rangle$ is again a unit vector. It proves as well that the projector onto a unit vector is positive with unit trace, and that conjugating a density operator by a unitary gives a density operator.