Pure States as Rank-One Projectors

A pure normal state on the full observable algebra $\mathcal B(\mathcal H)$ is represented by a rank-one orthogonal projection. This description removes the phase of a vector representative and places pure and mixed states in the same operator space. The equivalence between a ray and its projection is established in Physical States as Rays. Here the aim is different: recognize purity from operator identities, prove extremality within density operators, and quantify how nearby pure preparations differ.

Required background. Physical States as Rays identifies rays with rank-one projections. Density Operators supplies positivity, unit trace, and mixtures.

Projection identities and the trace-one condition

For a unit vector $\psi$, set $P_\psi=|\psi\rangle\langle\psi|$. Its action is $P_\psi\phi=\psi\langle\psi,\phi\rangle$. It obeys

$$P_\psi^*=P_\psi, \qquad P_\psi^2=P_\psi, \qquad \operatorname{Tr}P_\psi=1.$$

Conversely, an orthogonal projection with trace one has one-dimensional range. Indeed, a projection has eigenvalues zero and one, and its trace is its rank, possibly infinite. Thus the trace-one qualification cannot be dropped: an arbitrary projection represents a sharp event, but need not represent a normalized state.

For a finite-rank projection $Q$ of rank $r>1$, the density operator $Q/r$ is the maximally mixed state on its range. It is not a projection, since $(Q/r)^2=Q/r^2$. Likewise, an infinite-rank projection cannot be normalized by division by its trace.

For a bounded observable $A$,

$$\operatorname{Tr}(P_\psi A)=\langle\psi,A\psi\rangle.$$

This follows by expanding the trace in an orthonormal basis containing $\psi$. It identifies a vector expectation with a phase-independent operator expression. Unbounded observables require the moment and domain qualifications in Expectation Functionals.

Purity tests for a density operator

Let $\rho\geq0$ be trace class with $\operatorname{Tr}\rho=1$. Its nonzero eigenvalues $\lambda_j$ form a finite or countable sequence, with $\lambda_j>0$ and $\sum_j\lambda_j=1$. Therefore

$$\operatorname{Tr}(\rho^2)=\sum_j\lambda_j^2\leq1.$$

Equality holds exactly when a single eigenvalue equals one. One way to see this is to sum the nonnegative quantities $\lambda_j(1-\lambda_j)$. A zero sum requires each nonzero eigenvalue to equal one, and normalization permits only one such eigenvalue.

Consequently the following conditions are equivalent within the density-operator state space: $\rho$ has rank one; $\rho^2=\rho$; and $\operatorname{Tr}\rho^2=1$. The spectral decomposition used here is the compact positive-operator theorem, not an assumption that every observable has a discrete spectrum.

In infinite dimensions there is no universal positive lower bound on purity. Equal mixtures of $r$ orthogonal states have purity $1/r$, tending to zero as $r$ increases. Zero itself is not attained by a trace-one positive operator.

Why a rank-one state is an extreme point

Suppose

$$P_\psi=t\rho_1+(1-t)\rho_2, \qquad 0<t<1,$$

with both $\rho_i$ density operators. For any $v\perp\psi$,

$$0=t\langle v,\rho_1v\rangle+(1-t)\langle v,\rho_2v\rangle.$$

Positivity forces each term to vanish. Since $\langle v,\rho_iv\rangle=\|\rho_i^{1/2}v\|^2$, each $\rho_i$ annihilates $\psi^\perp$. Self-adjointness then confines its range to the span of $\psi$. Unit trace gives $\rho_1=\rho_2=P_\psi$.

Conversely, if a density operator has at least two positive eigenvalues, shift a sufficiently small amount of weight from one eigenvalue to the other, with opposite shifts in a second density operator. Their equal mixture is the original $\rho$, and the two components are distinct. A mixed density operator is therefore not extreme.

This is a statement about normal states on the full bounded-operator algebra. On a restricted observable algebra, the same vector may induce a mixed functional; the sector example in Direct Sums and Superselection explains why purity must name the algebra.

Comparing two pure preparations

Let $\psi,\phi$ be unit vectors and $c=|\langle\psi,\phi\rangle|$. Rephase $\phi$ so its overlap is nonnegative. If $c<1$, choose a unit vector $\eta\perp\psi$ such that

$$\phi=c\psi+s\eta, \qquad s=\sqrt{1-c^2}.$$

The difference $P_\psi-P_\phi$ vanishes outside their two-dimensional span. In the basis $(\psi,\eta)$ its matrix is

$$\begin{pmatrix} s^2&-cs\\ -cs&-s^2 \end{pmatrix}.$$

The trace is zero and the determinant is $-s^2$, so its eigenvalues are $s$ and $-s$. Hence

$$\|P_\psi-P_\phi\|=\sqrt{1-c^2}, \qquad \frac12\|P_\psi-P_\phi\|_1=\sqrt{1-c^2}.$$

The case $c=1$ follows directly. For pure-state projections, operator-norm and trace-norm convergence thus express the same overlap condition. This does not make those norms interchangeable for arbitrary infinite-rank operators.

In particular, the pure-state embedding of projective Hilbert space into trace-class operators preserves its norm-quotient topology. It also bounds every effect probability difference by the displayed trace distance. An effect projecting onto the positive eigenspace of $P_\psi-P_\phi$ attains that bound.

Exercise: a mixture that resembles no projector

Let $\rho=pP_\psi+(1-p)P_\phi$ for $0<p<1$. Express $\operatorname{Tr}\rho^2$ in terms of $p$ and $c=|\langle\psi,\phi\rangle|$. When is the mixture pure?

Solution

Using $\operatorname{Tr}(P_\psi P_\phi)=c^2$ gives

$$\operatorname{Tr}\rho^2 =p^2+(1-p)^2+2p(1-p)c^2 =1-2p(1-p)(1-c^2).$$

With strictly positive weights, this equals one only if $c=1$, meaning both projectors describe the same ray. Orthogonal inputs instead give purity $p^2+(1-p)^2$, with minimum $1/2$ at equal weights.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — pure and mixed states in Hilbert-space quantum mechanics.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Chapters 1 and 6 — projections, compact positive operators, and trace ideals.