Pure States and Irreducible Representations

A state of a unital C*-algebra is pure exactly when its GNS representation is irreducible. This theorem connects a convex property of expectation rules with an invariant-subspace property of operators. It explains why a representing vector alone does not certify purity: a mixed state also has a GNS vector, but its cyclic representation is reducible.

Required background. GNS Construction supplies the cyclic triple. The Algebraic State Space defines pure states as extreme points. The projection and scalar-commutant criterion is explained in Irreducible CCR Representations; the same star-closed invariant-subspace argument applies here.

Purity excludes nontrivial convex decompositions

A state $\omega$ is pure if

$$\omega=t\omega_1+(1-t)\omega_2,\qquad0<t<1,$$

with states $\omega_1,\omega_2$, implies $\omega_1=\omega_2=\omega$. Purity is relative to the algebra on which the functionals are defined.

Let $(\mathcal H,\pi,\Omega)$ be its GNS triple. Irreducibility means that no proper nonzero closed subspace is invariant under all $\pi(a)$. Since the algebra is star closed, invariant subspaces are reducing and their projections lie in $\pi(\mathcal A)'$.

The equivalent criterion is

$$\pi(\mathcal A)'=\mathbb CI.$$

The spectral projections of a nonscalar self-adjoint commutant element would supply a nontrivial reducing projection. This is the operator-theoretic input; the new ingredient below is how commutant elements encode convex decompositions.

A reducing projection splits the cyclic state

Suppose $P\in\pi(\mathcal A)'$ is a projection distinct from $0,I$. Cyclicity implies $P\Omega\ne0$ and $(I-P)\Omega\ne0$: otherwise commutation would make one projection vanish on the dense set $\pi(\mathcal A)\Omega$.

Set $t=\|P\Omega\|^2$, so $0<t<1$, and define

$$\omega_1(a)= \frac{\langle P\Omega,\pi(a)P\Omega\rangle}{t},$$

with $\omega_2$ defined similarly using $I-P$. The orthogonal reducing subspaces eliminate cross terms, giving $\omega=t\omega_1+(1-t)\omega_2$.

This decomposition is nontrivial. If $\omega_1=\omega$, then

$$\langle\Omega,(P-tI)\pi(a)\Omega\rangle=0 \qquad(a\in\mathcal A).$$

Density implies $(P-tI)\Omega=0$. A projection cannot have a nonzero eigenvector with eigenvalue strictly between zero and one. Thus a pure state cannot have a reducible GNS representation.

A dominated functional becomes a commutant contraction

For the converse, let $\varphi$ be a positive functional satisfying $0\leq\varphi\leq\omega$, meaning that $\omega-\varphi$ is positive. On the GNS quotient define

$$q([a],[b])=\varphi(a^*b).$$

Cauchy–Schwarz for $\varphi$, together with domination, gives

$$|q([a],[b])| \leq\|[a]\|_\omega\|[b]\|_\omega.$$

Consequently the form is well defined on null classes and extends to the Hilbert completion. The Riesz representation theorem supplies a unique operator $0\leq T\leq I$ with $q(\xi,\eta)=\langle\xi,T\eta\rangle$.

The identity $\varphi(a^*cb)=\varphi((c^*a)^*b)$ implies, on the dense quotient,

$$T\pi(c)=\pi(c)T.$$

Hence $T$ belongs to the commutant, and taking $a=1$ shows

$$\varphi(b)=\langle\Omega,T\pi(b)\Omega\rangle.$$

This is the dominated-functional form of the noncommutative Radon–Nikodym theorem needed here. The proof used bounded forms and GNS density; it did not assume that $\omega$ has a density operator in a previously chosen representation.

Irreducibility forces convex components to coincide

Assume now that $\pi$ is irreducible and $\omega=t\omega_1+(1-t)\omega_2$. The positive functional $\varphi=t\omega_1$ is dominated by $\omega$. Its associated contraction must be scalar, $T=cI$. Normalization gives $c=\varphi(1)=t$, so $\omega_1=\omega$; the decomposition then also gives $\omega_2=\omega$.

This completes both directions of the theorem. It also distinguishes irreducibility from faithfulness. An irreducible representation may have a nonzero kernel, and a faithful GNS representation can be reducible.

For a density matrix of rank $r$ on $M_n$, the GNS model has dimension $nr$ and acts by left multiplication. Its commutant contains right multiplication on the $r$-dimensional support. The theorem therefore agrees with the familiar criterion that the density matrix is pure precisely when $r=1$.

Exercise: a cyclic vector need not give a pure state

Represent $\mathbb C\oplus\mathbb C$ diagonally on $\mathbb C^2$ and take $\Omega=(\sqrt p,\sqrt{1-p})$ with $0<p<1$. Show that it is cyclic but its state is not pure.

Solution

Independent choices of the two diagonal entries generate every vector from $\Omega$, so it is cyclic. Its state is $\omega(a,b)=pa+(1-p)b$, the nontrivial mixture of the two evaluation states. The coordinate projections lie in the commutant and implement exactly that decomposition. The vector is pure as a vector state on the larger algebra $M_2$, but its restriction to the diagonal algebra is mixed.

References

• Bratteli, Ola, and Derek W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987, doi:10.1007/978-3-662-02520-8 — dominated functionals, commutants, and the pure-state theorem.
• Naaijkens, Pieter, Quantum Spin Systems on Infinite Lattices, Lecture Notes in Physics 933, Springer, 2017, arXiv:1311.2717 — purity, irreducibility, and state-generated representations.