The Exercise

Two unitaries $g$ and $L$ act on the same quantum system and commute, $gL = Lg$. A state $|\psi\rangle$ of the system is left unchanged by $g$, so $g|\psi\rangle = |\psi\rangle$. Show that $g$ leaves the state $L|\psi\rangle$ unchanged as well.

Unitaries and Their Eigenstates

A quantum system $S$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$, $\langle\psi|\psi\rangle = 1$, defined physically up to a global phase.^✦

A unitary $U$ of $S$ is an operator on the state space of $S$ that satisfies $$U^\dagger U = U U^\dagger = I ;$$(1)^✦ it describes an evolution of $S$ as a closed system. The unitary $U$ carries a pure state $|\psi\rangle$ of $S$ to the state $U|\psi\rangle$.^✦ For two unitaries $U$ and $V$ of $S$, the composite $UV$ is the evolution in which $V$ acts first and $U$ second.^✦

A pure state $|\psi\rangle$ of $S$ is an eigenstate of a unitary $U$ of $S$ with eigenvalue $\mu$, a complex number, when $$U|\psi\rangle = \mu\,|\psi\rangle .$$(2)^✦ An eigenstate of a unitary $U$ of $S$ with eigenvalue $1$ is a pure state of $S$ that $U$ leaves unchanged.

The Claim

What Is Checked

The claim (3) is stated over a quantum system $S$, two commuting unitaries $g$ and $L$ of $S$, their composite $gL$, and a pure state $|\psi\rangle$ of $S$ that is an eigenstate of $g$ with eigenvalue $1$.

The library shows that the composite of two unitaries of $S$ is again a unitary of $S$, as a product of unitaries, and that a unitary carries a pure state to a pure state, since it preserves norms. Both $gL|\psi\rangle$ and $L|\psi\rangle$ are therefore pure states of $S$.