Pure quantum state - Knowlpedia

Let $H$ be a (finite-dimensional) complex Hilbert space (see complex-hilbert-space-finite). A pure quantum state can be specified in either of two equivalent ways:

State vector (ray): a unit vector $\psi \in H$ with $\|\psi\|=1$ , where $\psi$ and $e^{i\theta}\psi$ represent the same physical state (global phase is unobservable).

Density operator (projector): the rank-one projector

\rho_\psi = |\psi\rangle\langle\psi|,

which is a density-operator (positive semidefinite with trace $1$ ).

Equivalent characterizations

For a density operator $\rho$ on $H$ , the following are equivalent:

$\rho$ is pure.
$\rho$ has rank $1$ .
$\rho^2=\rho$ (idempotent projector).
$\operatorname{Tr}(\rho^2)=1$ (maximal “purity” among states).
The eigenvalues of $\rho$ are $\{1,0,\dots,0\}$ .

Expectation values

If $A$ is an observable (a self-adjoint operator; see self-adjoint-operator-observable), then in the pure state $\psi$ ,

\langle A\rangle_\psi = \langle \psi, A\psi\rangle \qquad\text{and equivalently}\qquad \langle A\rangle_\psi = \operatorname{Tr}(\rho_\psi A).

Remarks

Pure states are the extreme points of the convex set of all density operators; mixtures of distinct pure states produce mixed-state-quantum.