Properties of ρ

A matrix qualifies as a density operator only if it satisfies three conditions. These conditions are exactly what guarantees it produces sensible probabilities for every possible measurement.

The three defining properties

A density operator $\rho$ on a Hilbert space must be:

1. Unit trace: $\operatorname{Tr}(\rho) = 1$. The trace collects the total probability, so it must equal one.
2. Hermitian: $\rho = \rho^{\dagger}$. Equivalently $\rho_{ij} = \rho_{ji}^*$. Hermiticity guarantees real eigenvalues and real expectation values for observables.
3. Positive semidefinite: $\langle\phi|\rho|\phi\rangle \ge 0$ for every $|\phi\rangle$. Equivalently, all eigenvalues of $\rho$ are $\ge 0$. This makes every measurement probability non-negative.

Conversely, any matrix with these three properties is the density operator of some physical state. This is the complete characterization.

Why each property is forced

Write $\rho = \sum_i p_i |\psi_i\rangle\langle\psi_i|$. The probability of obtaining outcome $|\phi\rangle$ is $\langle\phi|\rho|\phi\rangle = \sum_i p_i |\langle\phi|\psi_i\rangle|^2$. Because each $p_i \ge 0$ and each $|\langle\phi|\psi_i\rangle|^2 \ge 0$, the sum is non-negative — positivity. Each $|\psi_i\rangle\langle\psi_i|$ is Hermitian and the $p_i$ are real, so $\rho$ is Hermitian. And $\operatorname{Tr}\big(|\psi_i\rangle\langle\psi_i|\big) = \langle\psi_i|\psi_i\rangle = 1$, so $\operatorname{Tr}(\rho) = \sum_i p_i = 1$.

The eigenvalue picture

Because $\rho$ is Hermitian, the spectral theorem diagonalizes it:

The eigenvalues $\{\lambda_k\}$ are themselves a probability distribution. So every density operator is a mixture of orthonormal pure states (its eigenvectors) with weights given by its eigenvalues — its spectral decomposition. This is the canonical, ambiguity-free ensemble hiding inside $\rho$.

Try it

Build $\rho = \tfrac12|0\rangle\langle0| + \tfrac12|+\rangle\langle+|$ and check all three properties. Return true only if the trace is $1$, the matrix is Hermitian, and both eigenvalues are non-negative.

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