Exercise PLAN-003

Probability mixtures are density states and pure states are precisely their extreme points. Relative boundary in the trace-one Hermitian affine space means singular density states, whereas the full Hermitian/complex ambient boundary is every density state; the original unrestricted boundary=pure request is false.

1. Nonnegative real weights preserve every nonnegative quadratic form, establishing positivity without the unit-sum hypothesis.
2. Trace linearity turns trace one of each component into trace equal to the sum of weights, so unit sum establishes the normalization and completes convexity.
3. Spectral decomposition and the pure-support argument characterize extreme points as rank-one projectors; positive-definite and kernel-direction perturbations separately classify relative/ambient boundaries, refuting the old clause with explicit d=0,1,2 and d>=3 cases.

The scalars of the Hilbert space are complex, but convex weights and the vector-space structure used for Herm_d are real. Positivity in the original quadratic-form definition already implies Hermiticity in finite complex dimension: writing a matrix as B+iC with B,C Hermitian, real quadratic forms force <v,Cv>=0 for every v, hence C=0 by polarization. Thus using Herm_d is not an extra state promise. For d>=1, Herm_d has real dimension d^2 (d real diagonal entries and two real parameters per off-diagonal pair); its trace-one affine hyperplane A_d has dimension d^2-1. All norms on these finite real vector spaces induce the same topology, and subspaces/affine translates carry the induced topology.

For any v, <v,rho v>=sum_i p_i<v,rho_i v>>=0 because p_i>=0 and each summand's quadratic form is nonnegative. This proof uses no unit-sum condition. Linearity gives tr(rho)=sum_i p_i tr(rho_i)=sum_i p_i=1; this identity uses unit sum, not nonnegativity. Together they establish convexity. These are the roles of the hypotheses in universal mixture arguments, not a necessity claim for individual coefficients: if all rho_i are equal, some negative coefficients summing to one can still give a state. Conversely, in d>=2 weights (2,-1) on orthogonal pure states give trace one but not positivity. With trace-one summands, unit trace is equivalent to the weights' sum being one.

The finite-dimensional spectral theorem gives rho=sum_j lambda_j P_j, where P_j=|u_j><u_j| for an orthonormal eigenbasis, lambda_j>=0 and sum_j lambda_j=1. This is a convex pure-state decomposition. Rank one therefore means rho=P for a unit vector. If rank(rho)>=2, choose positive eigenvalues lambda_u,lambda_v and 0<t<min(lambda_u,lambda_v). Both rho_+=rho+t(P_u-P_v) and rho_-=rho-t(P_u-P_v) are positive trace-one, distinct, and rho=(rho_++rho_-)/2. So rho is not extreme. Conversely let P=|psi><psi|=s sigma+(1-s)tau for states sigma,tau and 0<s<1. For any z perpendicular to psi, zero=<z,Pz>=s<z,sigma z>+(1-s)<z,tau z>. Both terms are nonnegative, so each vanishes. Spectral decomposition of a positive matrix shows <z,sigma z>=0 implies sigma z=0, and likewise tau z=0. Hermiticity puts their ranges in span(psi); trace one then forces sigma=tau=P. Thus extreme points are exactly the pure rank-one projectors, including when d=1. This is not a definition or characterization of all boundary points.

The PSD cone is closed, since every quadratic form is continuous and nonnegativity is a closed condition. Trace one is also closed. If rho is positive definite, let lambda_min>0. Every trace-zero Hermitian E with ||E||_op<lambda_min satisfies <v,(rho+E)v>>= (lambda_min-||E||_op)||v||^2>=0, and preserves trace. Thus every positive-definite state is interior RELATIVE TO A_d. In particular I_d/d has an A_d-neighborhood inside D_d, proving aff(D_d)=A_d: any trace-zero Hermitian direction has sufficiently small positive and negative multiples in that neighborhood. This also covers the zero-dimensional affine space A_1. For d>=2, if rho is singular, choose a unit kernel vector z and a unit w perpendicular to z. The trace-zero Hermitian perturbation E_t=t(|w><w|-|z><z|) tends to zero as t decreases to zero, but <z,(rho+E_t)z>=-t<0. Hence arbitrarily nearby points of A_d fail positivity. Closedness now implies that rho is on the relative boundary. Consequently, for every d>=1, int_(A_d)(D_d)={rho in D_d:rho is positive definite}, and boundary_(A_d)(D_d)={rho in D_d:rank(rho)<d}; in d=1 the latter set is empty. One can also approach every state from the relative interior by (1-epsilon)rho+epsilon I_d/d for 0<epsilon<1.

For d=1, D_1=A_1={ [1] }; its sole state is pure and extreme, its relative interior is the entire singleton, and its relative boundary is empty. For d=2, singular trace-one positive matrices have spectrum (1,0), so the relative boundary happens to equal the pure states. For d>=3, rho_*=diag(1/2,1/2,0,...,0) is rank two, mixed and nonextreme, but relatively boundary. It is (P_1+P_2)/2, and rho_*+t(P_1-P_3) has trace one with a negative third eigenvalue for every t>0, tending to rho_*. Thus the original boundary-equals-pure clause is false even in the affine-relative topology outside its special qubit coincidence; d=1 is a separate counterexample.

In the full Herm_d, D_d is closed with empty interior for every d>=1: rho+t I_d is arbitrarily close to rho but has trace 1+td, not one, for nonzero real t. Therefore boundary_(Herm_d)(D_d)=D_d. The same perturbation and closedness apply in the whole M_d(C), viewed with its usual real norm topology, so boundary_(M_d(C))(D_d)=D_d too. For d>=2 these ambient boundaries contain mixed positive-definite states such as I_d/d. In d=1 their only boundary point is [1], unlike the empty relative boundary in A_1. None of these facts identifies topological boundary with convex extremality in general.

Finally d=0 has only the zero operator, whose trace is zero. Thus D_0 and A_0 are empty, there are no pure states or extreme points, and D_0 has empty boundary in either full matrix ambient space (each is the singleton zero space), as well as empty interior/boundary in the empty A_0. There is no probability mixture of existing states to test and no normalized vector or I_d/d. No affine-dimension formula for a nonempty trace-one hyperplane is asserted at d=0. The correction preserves the convexity/extremality target but explicitly changes the old false requested conclusion into its assessment and a topology-specific classification; it is not an equivalent answer to the old question.

1. Derives positivity by quadratic forms with nonnegative real weights, trace one by trace linearity and unit sum, and concludes convexity; distinguishes these hypothesis roles from false necessity for every individual positive combination. (1 weight)
2. Uses the spectral decomposition and proves both directions of extreme iff rank-one projector: higher-rank perturbation/decomposition and positivity-supported uniqueness for a pure-state mixture. (1 weight)
3. Specifies real Herm_d and A_d with their finite-dimensional induced norm topology; proves aff(D_d)=A_d for d>=1, positive-definite relative interior and singular relative boundary via closedness and trace-preserving perturbations. Gives d=1 empty relative boundary, d=2 coincidence and the explicit mixed rank-two d>=3 witness. (1 weight)
4. Explicitly refutes the old unrestricted boundary=pure clause without changing definitions or restricting dimensions; proves full Hermitian and complex-matrix ambient boundaries equal D_d for d>=1, and treats d=0 directly without I_d/d or fictitious states. Keeps topological boundary distinct from extremality. (1 weight)

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