Density operator

Let $H$ be a nonzero finite-dimensional complex inner-product space, with the root library's convention that the inner product is linear in its first argument. A density operator on $H$ is a non-negative endomorphism $\rho :H\to H$ with $\mathrm{tr}(\rho )=1$; non-negative and trace have the meanings of Non-negative and positive operators and The basis-independent trace of an endomorphism of a finite-dimensional vector space. The zero-dimensional space is excluded: its only endomorphism has trace $0$.

Here $\ker \rho =\{x\in H:\rho x=0\}$, using the kernel of a linear map from Kernel and image of a linear map. The support of $\rho$ is the orthogonal complement of its kernel in the root convention:

$\mathrm{supp}(\rho ):=\{x\in H:⟨x,y⟩=0\text{ for every }y\in \ker \rho \}\mathrm{.}$

For a subspace $S\subseteq H$, $\rho$ is faithful on $S$ when $\mathrm{supp}(\rho )=S$ and the restriction $\rho {\mathrm{\mid }}_S:S\to S$ is positive definite in the sense of Non-negative and positive operators. This restriction is an endomorphism: if $x\in S$ and $y\in \ker \rho$, self-adjointness gives $⟨\rho x,y⟩=⟨x,\rho y⟩=0$, so $\rho x\in (\ker \rho {)}^{\perp }=S$.

The definition separates the finite-dimensional mathematical object from any preparation story. An ensemble representation of a density operator is additional data and need not be unique: Bell-state Schmidt decomposition and reduced density ↗ gives two distinct equal-weight decompositions of $I_2\mathrm{/}2$, using the computational and $\{\mathrm{\mid }+⟩,\mathrm{\mid }-⟩\}$ bases. No measurement or physical postulate is part of this definition.