Density Operators: Trace Norm and Integrability

A density operator on a Hilbert space is a positive trace-class operator of trace one. In infinite dimensions, the trace-class condition gives convergent probabilities for bounded measurements; it does not ensure finite energy, entropy, or variance. This page establishes those distinctions. The preparation interpretation and basic matrix rules belong to Density Operators in Core Formalism.

Helpful background. Trace-class operators supplies absolute convergence; positive operators supplies operator order.

The density-operator criterion

Let $\mathcal H$ be a separable complex Hilbert space. The state set in the normal Hilbert-space formulation is

$$\mathcal D(\mathcal H) =\{\rho\in\mathcal S_1(\mathcal H):\rho\geq0,\ \operatorname{Tr}\rho=1\}.$$

Compactness and the positive spectral theorem give

$$\rho=\sum_np_n|u_n\rangle\langle u_n|, \qquad p_n>0,\qquad\sum_np_n=1,$$

with convergence in trace norm. The vectors span the support of $\rho$; the kernel can be nontrivial. Every eigenvalue is at most one, so $0\leq\rho\leq I$ and $\|\rho\|\leq1$.

For a bounded operator $A$, define $\omega_\rho(A)=\operatorname{Tr}(\rho A)$. This is linear, normalized, and positive, with $|\omega_\rho(A)|\leq\|A\|$. For an effect $E$, $0\leq\omega_\rho(E)\leq1$. For a countable orthogonal resolution of the identity, the probabilities sum to one by monotone convergence. The representation theorem in normal states explains why these are precisely the normal states on $\mathcal B(\mathcal H)$.

Countable mixtures converge in trace norm

Suppose $\psi_n$ are normalized, not necessarily orthogonal, and $q_n\geq0$ with $\sum_nq_n=1$. Then

$$\rho=\sum_nq_n|\psi_n\rangle\langle\psi_n|$$

exists in trace norm because each summand has trace norm $q_n$. Positivity survives the limit, and trace continuity gives normalization. The error after keeping the first $N$ unnormalized terms is exactly $\sum_{n>N}q_n$, since the remainder is positive.

Thus an ensemble expansion is not a formal infinite matrix identity. Its convergence has a topology, and that topology controls every bounded-observable prediction. The expansion is highly nonunique, even though the resulting operator and all its probabilities are fixed. A spectral expansion is one convenient ensemble, not a unique record of preparation.

Holevo treats this operator framework and its extension to infinite systems. In nonseparable $\mathcal H$, a trace-class state still has separable support: there are at most countably many nonzero eigenvalues. The countable presentation describes the state support, without proving that every physically chosen Hilbert space must be separable.

Unbounded observables require integrability

Let $A$ be self-adjoint with spectral measure $E_A$. The state defines a probability measure

$$\mu_\rho^A(\Delta) =\operatorname{Tr}\bigl(\rho E_A(\Delta)\bigr).$$

A finite expectation exists when

$$\int_{\mathbb R}|\lambda|\,d\mu_\rho^A(\lambda)<\infty.$$

Then $\langle A\rangle_\rho=\int\lambda\,d\mu_\rho^A(\lambda)$. For a positive observable the same integral is meaningful as an extended value in $[0,\infty]$. For a general signed observable, divergent positive and negative parts cannot be canceled to define a physical mean.

The condition is equivalently finite quadratic-form weight,

$$\sum_np_n\bigl\||A|^{1/2}u_n\bigr\|^2<\infty,$$

where every $u_n$ with $p_n>0$ must lie in the form domain. It is weaker than $\sum_np_n\|Au_n\|^2<\infty$, which gives a finite second moment. Writing $\operatorname{Tr}(\rho A)$ for a first moment is customary but must not silently assert that the unbounded product is everywhere defined and trace class. The spectral integral is the safe definition.

Finite normalization and infinite energy

On $\ell^2(\mathbb N)$, take a Hamiltonian $He_n=nE_0e_n$ with $E_0>0$ and

$$\rho=\frac6{\pi^2}\sum_{n=1}^{\infty} \frac{|e_n\rangle\langle e_n|}{n^2}.$$

This is a valid state, but its mean energy is

$$\langle H\rangle_\rho =\frac{6E_0}{\pi^2}\sum_{n=1}^{\infty}\frac1n =\infty.$$

Every eigenvector in its support belongs to $D(H)$; nevertheless the weighted ensemble energy diverges. Thus checking the domain separately for each ensemble vector is insufficient. A preparation with finite energy needs a quantitative tail condition.

With probabilities proportional to $n^{-3}$, the mean energy is finite while the second moment diverges. These examples distinguish three questions: existence of a density operator, finite mean, and finite variance. They also explain why trace-norm convergence of states alone need not control an unbounded observable.

Exercises

1. Let $\rho_N$ be the normalized first-$N$ spectral truncation of a state, and let $t_N=\sum_{n>N}p_n$. Show that $\|\rho-\rho_N\|_1=2t_N$.

Solution

On the retained support the eigenvalues change from $p_n$ to $p_n/(1-t_N)$, giving total absolute change $t_N$. The discarded support contributes another $t_N$. The two supports are orthogonal, so the norms add. This supplies an exact truncation error for density-matrix computations.

2. A normalized vector $\psi$ lies in $D(|A|^{1/2})$ but not in $D(A)$. Can its pure state have a finite mean of the self-adjoint observable $A$?

Solution

Yes. Form-domain membership means $\int|\lambda|\,d\mu_\psi^A<\infty$, exactly the first absolute moment. Membership in $D(A)$ instead requires $\int\lambda^2\,d\mu_\psi^A<\infty$. The vector expression $\langle\psi,A\psi\rangle$ is unavailable, but the spectral-integral expectation exists.

References

• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd revised and expanded ed., De Gruyter, 2019, Chapters 2 and 11 — density operators, observables, and constrained infinite systems. Publisher.
• Simon, Barry, Trace Ideals and Their Applications, 2nd ed., Mathematical Surveys and Monographs 120, American Mathematical Society, 2005 — trace-norm convergence and positive traces. Publisher.