Normal States

A normal state preserves increasing limits of bounded positive observables. On $\mathcal B(\mathcal H)$ this condition is equivalent to representation by a density operator. Normality is therefore the precise additional hypothesis behind the statement that every quantum state is a density matrix; positivity and normalization alone do not imply it in infinite dimensions.

Helpful background. Density operators supplies trace-class states; states as positive functionals supplies the larger algebraic state space.

Monotone continuity

A state $\omega$ on $\mathcal B(\mathcal H)$ is a positive normalized linear functional. It is normal if, for every increasing norm-bounded net of positive operators $A_\alpha$ with supremum $A$,

$$\omega(A)=\sup_\alpha\omega(A_\alpha).$$

The supremum exists in $\mathcal B(\mathcal H)$ and is the strong-operator limit. Nets are the general definition; on a separable Hilbert space the sequential formulations used below suffice for the density-operator criterion. Normality is also equivalent to ultraweak continuity.

For orthogonal projections $P_n$, the sum $\sum_nP_n$ exists strongly as the projection onto the closed span of their ranges. Normality gives countable additivity:

$$\omega\!\left(\sum_nP_n\right)=\sum_n\omega(P_n).$$

Finite additivity follows from linearity for every state. Countable additivity requires the continuity condition because the infinite sum need not converge in operator norm.

Density operators give normal states

Let $\rho=\sum_kp_k|u_k\rangle\langle u_k|$ be a density operator. For a bounded increasing sequence $A_n\uparrow A$,

$$\omega_\rho(A_n)=\sum_kp_k\langle u_k,A_nu_k\rangle.$$

Each summand increases to its value at $A$. Monotone convergence exchanges the sum and the limit. For a general increasing net the same conclusion follows by first controlling the trace-weight tail and then handling a finite set of vectors. Hence $\omega_\rho$ is normal.

This argument shows more than formal trace notation: it explains why a countably complete measurement has total probability one. The same normality extends to the bounded Borel functional calculus of a self-adjoint observable, producing the usual countably additive spectral probability measure.

Recovering a density operator from normality

Assume now that $\mathcal H$ is separable and $\omega$ is normal. Let $P_N$ project onto the first $N$ vectors of an orthonormal basis. Then $P_N\uparrow I$ strongly, so $\omega(P_N)\uparrow1$.

Restriction of $\omega$ to the finite matrix algebra $P_N\mathcal B(\mathcal H)P_N$ determines a unique positive matrix $\rho_N$ through the finite-dimensional trace pairing. The matrices are compatible:

$$P_N\rho_MP_N=\rho_N \quad(M\geq N), \qquad \operatorname{Tr}\rho_N=\omega(P_N)\leq1.$$

Their entries define a positive bounded quadratic form on finite-support vectors, with norm at most one. It extends to a positive operator $\rho$ whose finite diagonal sums approach one. Thus $\rho$ is trace class and $\operatorname{Tr}\rho=1$.

The functionals agree on finite-rank operators by construction. To pass to arbitrary bounded $A$, control the cutoff error using the Cauchy–Schwarz inequality for positive functionals. Writing $Q_N=I-P_N$ gives

$$|\omega(A)-\omega(P_NAP_N)| \leq2\|A\|\sqrt{\omega(Q_N)}\longrightarrow0.$$

The corresponding density-operator cutoff error also vanishes. Hence $\omega(A)=\operatorname{Tr}(\rho A)$ for every bounded $A$. Uniqueness follows because the rank-one matrix units determine every matrix element of $\rho$. Bratteli and Robinson give the general von Neumann-algebra formulation.

A state that fails normality

On $\ell^2(\mathbb N)$ choose a free ultrafilter $\mathcal U$ and define

$$\omega_{\mathcal U}(A) =\lim_{n\to\mathcal U}\langle e_n,Ae_n\rangle.$$

Every bounded scalar sequence has an ultrafilter limit. The resulting functional is linear, positive, and normalized. For every finite-rank projection $P_N$, however, $\omega_{\mathcal U}(P_N)=0$, while $\omega_{\mathcal U}(I)=1$. Thus it fails monotone continuity.

It cannot equal $\operatorname{Tr}(\rho\,\cdot)$ for any density operator: all diagonal entries of such a $\rho$ would vanish, forcing trace zero. This example describes probability escaping every fixed finite-dimensional subspace. It is not a normal state with an unusual density matrix.

Normality is relative to a chosen von Neumann algebra and representation. In more general algebras a normal state need not be represented by an intrinsic trace-class element of that algebra. The theorem proved here specifically concerns the full algebra $\mathcal B(\mathcal H)$.

Exercises

1. Let $H$ have an orthonormal energy eigenbasis. Why does a normal state assign probabilities summing to one to the complete energy measurement?

Solution

The spectral projections sum strongly to $I$. Normality gives the countable sum of their probabilities as $\omega(I)=1$. This is the infinite-outcome extension of the trace rule, and does not imply a finite mean energy.

2. Show that the ultrafilter state above vanishes on every compact operator.

Solution

An orthonormal sequence converges weakly to zero. Compactness implies $\|Ke_n\|\to0$, so $|\langle e_n,Ke_n\rangle|\leq\|Ke_n\|\to0$. Its ultrafilter limit is therefore zero.

References

• Bratteli, Ola, and Derek W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1: C- and W*-Algebras, Symmetry Groups, Decomposition of States*, 2nd ed., Springer, 1987 — normal functionals and representation theory. Publisher.
• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd revised and expanded ed., De Gruyter, 2019 — the density-operator state formulation in infinite dimensions. Publisher.