Extreme Points

The extreme points of the density-operator state space are exactly the rank-one projections. The proof has two parts: positivity prevents a rank-one state from hiding components on an orthogonal subspace, while any two positive eigenvalues allow a small probability transfer that splits a mixed state into distinct neighbors. Neither argument requires finite dimension.

Helpful background. Convex state space introduces the geometric setting; positive operators supplies the implication from a vanishing positive quadratic form to a kernel vector.

Extremality of a density operator

A point $\rho$ of a convex set is extreme if

$$\rho=t\sigma+(1-t)\tau,\qquad0<t<1,$$

with $\sigma,\tau$ in that set implies $\sigma=\tau=\rho$. For the normal state set $\mathcal D(\mathcal H)$,

$$\rho\text{ is extreme} \quad\Longleftrightarrow\quad \rho=|\psi\rangle\langle\psi| \text{ for a unit vector }\psi.$$

The statement concerns extremality in the full state set, with no additional energy or expectation-value constraints. Intersecting a convex set with a constraint surface can introduce new extreme points that were not extreme in the original set. Purity should not be redefined by extremality in an arbitrary constrained optimization problem.

A rank-one projection cannot split

Suppose $P_\psi=t\sigma+(1-t)\tau$ with density operators $\sigma,\tau$. For any $\phi\perp\psi$,

$$0=\langle\phi,P_\psi\phi\rangle =t\langle\phi,\sigma\phi\rangle +(1-t)\langle\phi,\tau\phi\rangle.$$

Both terms are nonnegative, so both vanish. The positive square-root identity gives $\sigma^{1/2}\phi=0$ and $\tau^{1/2}\phi=0$, hence $\sigma\phi=\tau\phi=0$.

Self-adjointness now forces the ranges of $\sigma,\tau$ to lie in $\operatorname{span}\{\psi\}$. Thus each is a nonnegative multiple of $P_\psi$. Trace-one normalization fixes both multiples to one. This proves extremality without assuming a finite matrix representation.

The decisive step is positivity: a general trace-one self-adjoint operator can have off-diagonal or signed terms that cancel. Such operators are not admissible states and cannot enter a convex state decomposition.

A mixed state admits a probability-transfer split

If $\rho$ is not rank one, its compact positive spectral decomposition has at least two positive eigenvalues $p_1,p_2$ with orthonormal eigenvectors $u_1,u_2$. Choose $0<\epsilon<\min(p_1,p_2)$ and set

$$\rho_\pm=\rho\pm\epsilon \left(|u_1\rangle\langle u_1|-|u_2\rangle\langle u_2|\right).$$

The two affected eigenvalues remain nonnegative, all other eigenvalues are unchanged, and the traces remain one. Therefore $\rho_\pm$ are distinct states and

$$\rho=\tfrac12\rho_++\tfrac12\rho_-.$$

This proves nonextremality even when $\rho$ has infinitely many nonzero eigenvalues. It also constructs an explicit direction along which probability can be redistributed without leaving the state space.

More generally every density operator is a countable mixture of extreme points via its spectral expansion, with convergence in trace norm. This direct theorem does not require an unjustified appeal to finite-dimensional compact convex geometry: the whole infinite-dimensional density-operator set is not trace-norm compact.

Support subspaces define faces

For an orthogonal projection $P$, define

$$F_P=\{\rho\in\mathcal D(\mathcal H):\rho=P\rho P\}.$$

This is a face: if a convex combination lies in $F_P$, each positive component must also lie in $F_P$. Apply the rank-one proof’s vanishing-expectation argument to every vector in $P^\perp$. The set $F_P$ is naturally the state space on $P\mathcal H$.

For rank-one $P$, the face has one point. For higher-dimensional $P$, it contains nontrivial mixtures. A state can be full rank within its support and still lie in a proper face of the larger Hilbert-space state space.

In finite dimensions the face associated with the support of $\rho$ describes the directions available near $\rho$. In infinite dimensions, a faithful state can have support $I$ and still have no trace-norm interior neighborhood, because its eigenvalues approach zero. Support and interior are different properties.

A constrained qubit warning

Represent a qubit state by its Bloch vector $r$, with $|r|\leq1$. Restrict to diagonal states with a fixed population $p\in(0,1)$. The allowed set contains the single matrix $\operatorname{diag}(p,1-p)$. It is extreme in that restricted set because there are no other allowed points, yet it is mixed in the full state space.

This distinction matters in variational optimization, superselection, and restricted measurements. Bratteli and Robinson define purity for states on the chosen observable algebra; Holevo develops the density-operator geometry. A statement about extreme points must specify its convex set.

Exercises

1. Prove that every vector occurring with positive weight in a countable ensemble decomposition of $P_\psi$ belongs to the ray of $\psi$.

Solution

For any $\phi\perp\psi$, positivity gives $0=\sum_kq_k|\langle\phi,\psi_k\rangle|^2$. Every positive-weight term vanishes. Each $\psi_k$ is therefore orthogonal to the whole orthogonal complement of $\psi$, so it lies in its one-dimensional span. This sharpens the preparation statement in pure versus mixed states.

2. For $\rho=\operatorname{diag}(1/2,1/3,1/6)$, construct two distinct states whose average is $\rho$.

Solution

Transfer $\epsilon=1/12$ between the first two entries: $\rho_+=\operatorname{diag}(7/12,1/4,1/6)$ and $\rho_-=\operatorname{diag}(5/12,5/12,1/6)$. Both are positive trace-one matrices, and their average is $\rho$.

References

• Bratteli, Ola, and Derek W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987 — pure states, convexity, and decomposition theory. Publisher.
• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd revised and expanded ed., De Gruyter, 2019 — convex geometry of density operators. Publisher.