Positive Operators

A positive operator is the operator counterpart of a nonnegative random variable: every vector assigns it a nonnegative quadratic value. For bounded operators this condition supplies self-adjointness, a positive square root, and an order relation. These facts explain why density operators and measurement effects produce probabilities, but positivity alone supplies neither a finite trace nor a bounded inverse.

Helpful background. Density operators supplies the physical role of positivity; bounded operators supplies the operator norm.

Positivity as a quadratic condition

Let $\mathcal H$ be a complex Hilbert space and $A\in\mathcal B(\mathcal H)$ an everywhere-defined bounded linear operator. Write $A\geq0$ when

$$\langle\psi,A\psi\rangle\geq0 \qquad(\psi\in\mathcal H).$$

The inequality asserts that the quadratic value is real as well as nonnegative. Polarization then gives $A=A^*$. By the bounded spectral theorem, positivity is equivalent to self-adjointness together with $\sigma(A)\subset[0,\infty)$. Indeed, a spectral projection onto a negative interval would provide a vector with negative expectation; conversely integrating a nonnegative spectral variable gives a nonnegative quadratic value.

The phrase positive definite needs qualification in infinite dimensions. The pointwise condition $\langle\psi,A\psi\rangle>0$ for $\psi\ne0$ says that $A$ has trivial kernel. The stronger statement $A\geq cI$ for some $c>0$ says that $A$ has a uniform lower bound. Only the latter ensures a bounded inverse on all of $\mathcal H$. Finite-dimensional intuition conflates them because the unit sphere is compact there.

These statements concern bounded operators. For an unbounded self-adjoint operator, positivity is tested on its domain, and square roots have their own domains. A positive symmetric operator need not be self-adjoint; the extension problem belongs to quadratic forms.

Square roots and factorization

The continuous functional calculus gives a unique positive bounded operator $A^{1/2}$ satisfying $(A^{1/2})^2=A$. Consequently,

$$\langle\psi,A\psi\rangle=\|A^{1/2}\psi\|^2.$$

Conversely every operator $B^*B$ is positive, so positivity is equivalent to admitting a bounded factorization of this form. Uniqueness here is uniqueness among positive square roots: even $I$ has many nonpositive square roots.

The factorization also yields the Cauchy–Schwarz inequality for the positive form:

$$|\langle\phi,A\psi\rangle|^2 \leq \langle\phi,A\phi\rangle\langle\psi,A\psi\rangle.$$

In particular, a zero quadratic value forces $A^{1/2}\psi=0$ and hence $A\psi=0$. This implication is useful when proving that a density operator supported on a one-dimensional subspace is pure, and when controlling matrix elements from diagonal bounds.

Simon develops these positivity and factorization tools as the starting point for trace ideals. Their use for states and effects is described by Holevo.

Operator order and probability effects

Define $A\leq B$ by $B-A\geq0$. If $A\geq0$ and $C$ is bounded, then $C^*AC\geq0$. This order-preserving congruence operation is valid even when $A$ and $C$ fail to commute. By contrast, the product $AB$ of two positive operators need not be positive, since generally $(AB)^*=BA\ne AB$.

An effect is a bounded operator $0\leq E\leq I$. Its expectation in any normalized vector lies in $[0,1]$. Positivity alone gives the lower bound; $E\leq I$ supplies the upper one. For a density operator $\rho$, the same bounds follow from

$$\operatorname{Tr}(\rho E) =\operatorname{Tr}(\rho^{1/2}E\rho^{1/2}), \qquad 0\leq\rho^{1/2}E\rho^{1/2}\leq\rho.$$

The trace expressions are justified because $\rho^{1/2}$ is Hilbert–Schmidt. See trace-class operators for the product theorem that makes this argument legitimate.

A positive operator without a spectral gap

On $\ell^2(\mathbb N)$, let $Ae_n=n^{-1}e_n$. Then

$$\langle\psi,A\psi\rangle =\sum_{n=1}^{\infty}\frac{|\psi_n|^2}{n}>0 \quad\text{for }\psi\ne0.$$

Nevertheless $\langle e_n,Ae_n\rangle\to0$, so no constant $c>0$ satisfies $A\geq cI$. Its inverse multiplies the $n$th coordinate by $n$ and is unbounded on its natural dense domain. Moreover $\operatorname{Tr}A=\sum_n1/n=\infty$: neither strict pointwise positivity nor boundedness makes $A$ a density operator.

Replacing $1/n$ by $6/(\pi^2n^2)$ produces a positive trace-one operator. Its kernel remains trivial, but its eigenvalues still approach zero. Thus a faithful infinite-dimensional density operator does not have a bounded inverse. Arguments using $\rho^{-1}$ or $\log\rho$ must keep track of this fact.

Exercises

1. Let $E$ be an effect in the measurement formalism. Prove $E-E^2\geq0$, and identify when equality holds.

Solution

The spectrum of $E$ lies in $[0,1]$, on which $x-x^2\geq0$. Functional calculus gives the inequality. Equality holds exactly when the spectrum is contained in $\{0,1\}$, equivalently $E^2=E$. Since $E=E^*$, this is an orthogonal projection: a sharp yes–no observable.

2. Suppose $A\geq cI$ with $c>0$. Show that $A^{-1}$ exists and satisfies $\|A^{-1}\|\leq1/c$.

Solution

The spectral condition is $\sigma(A)\subset[c,\|A\|]$. The function $x\mapsto1/x$ is bounded there. Its functional calculus defines the inverse and gives the norm bound. The diagonal example above shows why replacing the uniform bound by a trivial kernel would invalidate the argument.

References

• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd revised and expanded ed., De Gruyter, 2019, Chapters 1–2 — positivity in the operator formulation of states and observables. Publisher.
• Simon, Barry, Trace Ideals and Their Applications, 2nd ed., Mathematical Surveys and Monographs 120, American Mathematical Society, 2005 — positive operators and operator-ideal factorizations. Publisher.