Normal and Singular States

A nonnormal state need not be wholly singular: it may contain both a normal part and a singular remainder. The distinction is precise in the order structure of positive functionals. On bounded operators, the normal part is determined by the state’s values on compact operators, while the singular part vanishes on all of them.

Required background. Normal States proves the density-operator characterization on $\mathcal B(\mathcal H)$. Von Neumann Algebras defines preduals and ultraweak continuity. This page uses an infinite-dimensional separable $\mathcal H$ for the concrete decomposition.

Normality and singularity are different conditions

A positive functional on a von Neumann algebra $\mathcal M$ is normal if it preserves suprema of bounded increasing nets of positive elements. Equivalently, it belongs to the positive cone of the predual $\mathcal M_*$.

A positive functional $\varphi$ is singular if the only normal positive functional $\nu$ satisfying $0\leq\nu\leq\varphi$ is zero. This is stronger than merely failing normality.

The normal-singular decomposition theorem states that each positive functional has a unique sum

$$\varphi=\varphi_{\rm n}+\varphi_{\rm s},$$

with a normal positive part and a singular positive part. This general von Neumann-algebra theorem is developed by Bratteli and Robinson. We now derive its concrete form on $\mathcal B(\mathcal H)$ using compact operators.

Compact observables determine the normal part

Let $\omega$ be a state on $\mathcal B(\mathcal H)$. Its restriction to $\mathcal K(\mathcal H)$ is a positive bounded functional. The compact-operator duality theorem gives a unique positive trace-class $R$ such that

$$\omega(K)=\operatorname{Tr}(RK) \qquad(K\in\mathcal K(\mathcal H)).$$

Here $c=\operatorname{Tr}R$ can be less than one because the compact algebra has no identity. With finite-rank projections $P_n\uparrow I$,

$$c=\lim_n\omega(P_n)\leq1.$$

Define $\omega_{\rm n}(A)=\operatorname{Tr}(RA)$. To show this normal functional is dominated by $\omega$, take $A\geq0$. The compact operator $A^{1/2}P_nA^{1/2}$ lies between $0$ and $A$, so

$$\operatorname{Tr}(R A^{1/2}P_nA^{1/2}) =\omega(A^{1/2}P_nA^{1/2}) \leq\omega(A).$$

Normality of the trace functional permits the increasing limit $n\to\infty$, giving $\omega_{\rm n}(A)\leq\omega(A)$. Thus $\omega_{\rm s}=\omega-\omega_{\rm n}$ is positive and vanishes on all compact operators.

The result is independent of the chosen basis or cutoff sequence. The restricted functional on the entire compact algebra uniquely fixes $R$.

Vanishing on compacts characterizes the singular remainder

If a normal positive $\nu$ satisfies $\nu\leq\omega_{\rm s}$, then $\nu(P_n)=0$ for every finite-rank cutoff. Normality gives $\nu(I)=\lim_n\nu(P_n)=0$, so $\nu=0$. Hence the remainder is singular.

Conversely, if a positive functional has nonzero restriction to compact operators, the preceding construction extracts a nonzero normal positive functional dominated by it. Therefore, on $\mathcal B(\mathcal H)$,

$$\varphi\text{ is singular} \quad\Longleftrightarrow\quad \varphi(K)=0\quad(K\in\mathcal K(\mathcal H)).$$

This compact-operator criterion is specific to the full bounded-operator algebra. A general von Neumann algebra may contain no nonzero compact operators in its chosen representation, so the criterion must not be exported there unchanged.

When $0<c<1$, one can normalize both parts to write

$$\omega=c\,\widehat\omega_{\rm n} +(1-c)\,\widehat\omega_{\rm s}.$$

This state is nonnormal but not singular. Calling every nonnormal state “singular” loses a meaningful decomposition.

Singular states exist without a hidden density matrix

The quotient $\mathcal B(\mathcal H)/\mathcal K(\mathcal H)$ is the unital Calkin algebra. Choose any state on this nonzero C*-algebra and compose it with the quotient map. The resulting state on $\mathcal B(\mathcal H)$ vanishes on compacts and is singular.

It cannot have a density operator: all finite-rank probabilities vanish while the identity has probability one. The explicit ultrafilter realization on Normal States is a concrete instance of the same phenomenon.

This does not mean ordinary finite-probability experiments become inconsistent. The functional remains positive and finitely additive. What fails is the monotone continuity needed to recover countably complete spectral probabilities in that representation.

Every algebraic state is a vector state in its own GNS representation and therefore extends normally to the represented von Neumann closure. Its singularity relative to another representation is not contradicted by that fact. One must specify the observable algebra, its representation, and the continuity requirement together.

Exercise: recover the mixing weight from finite cutoffs

Let $\omega=\frac34\operatorname{Tr}(\rho\,\cdot)+\frac14\eta$, with $\rho$ a density operator and $\eta$ singular on $\mathcal B(\mathcal H)$. Determine $\lim_n\omega(P_n)$ for increasing finite-rank $P_n\uparrow I$. Is $\omega$ singular?

Solution

The singular state vanishes on each cutoff and normality gives $\operatorname{Tr}(\rho P_n)\to1$. Hence the limit is $3/4$. The normal part has weight $3/4$, so $\omega$ is not singular. It is not normal either, because $\omega(I)=1$ differs from the cutoff limit. This measurement distinguishes the two properties without attempting to invent a density operator for the remainder.

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 — normal and singular functionals and decomposition theory.
• Jones, Vaughan F. R., Von Neumann Algebras, lecture notes, 2010, author’s notes — normal functionals and concrete von Neumann representations.