States as Positive Linear Functionals

In the algebraic formulation, a state is not initially a vector or an operator. It is a rule that assigns an expectation value to every element of the observable algebra. The rule must respect linear combinations, return nonnegative values on positive elements, and assign value one to the identity.

For a unital C*-algebra $\mathcal A$, a state is a complex-linear functional

$$\omega:\mathcal A\to\mathbb C$$

such that

$$\omega(A^*A)\geq0 \quad \text{for every }A\in\mathcal A,$$

and

$$\omega(I)=1.$$

The first condition is positivity and the second is normalization. Together they turn an abstract algebra into a predictive theory: self-adjoint elements acquire real expectation values, projections and effects acquire probabilities, and variances are nonnegative.

Density matrices provide every such state for finite matrix algebras. The functional viewpoint is nevertheless more general. It separates physical expectation-value assignments from any particular Hilbert-space representation and remains meaningful when inequivalent representations occur.

Expectation Values Define States

Suppose an experimenter can assign an ensemble average to each observable $A$. Denote that assignment by

$$\omega(A)=\langle A\rangle_\omega.$$

If two preparations agree on $\omega(A)$ for every observable in $\mathcal A$, then the algebraic formulation regards them as the same state on $\mathcal A$. This is an operational equivalence statement: a state is determined by all observable statistics available in the chosen algebra.

The phrase “chosen algebra” matters. Restricting a state from a larger algebra to a subalgebra can erase distinctions. For example, two global states may induce the same state on all observables local to one subsystem. A vector state in a large representation can also restrict to a mixed state on a smaller algebra.

Although only self-adjoint elements are interpreted as real-valued observables, it is convenient and mathematically necessary to define $\omega$ on the full complex algebra. Every element has the decomposition

$$A = \frac{A+A^*}{2} + i\frac{A-A^*}{2i},$$

where both terms on the right are self-adjoint. Expectation values on self-adjoint elements therefore determine the complex-linear functional on all of $\mathcal A$.

Linearity

A state is complex linear:

$$\omega(\alpha A+\beta B) = \alpha\omega(A)+\beta\omega(B), \qquad \alpha,\beta\in\mathbb C.$$

For self-adjoint $A$ and $B$ and real $a,b$, this reproduces the familiar rule

$$\langle aA+bB\rangle_\omega = a\langle A\rangle_\omega + b\langle B\rangle_\omega.$$

Linearity does not require $A$ and $B$ to commute. The observable $A+B$ is well defined even when the two terms cannot be measured sharply in the same experimental run.

Linearity also encodes classical randomization of preparations. If preparation $k$ is selected with probability $p_k$, then the resulting state is

$$\omega = \sum_k p_k\omega_k, \qquad p_k\geq0, \qquad \sum_kp_k=1,$$

meaning

$$\omega(A) = \sum_kp_k\omega_k(A).$$

This convex combination describes a statistical mixture of states. It should not be confused with a coherent superposition of vectors; the distinction is developed in Classical Mixtures vs Quantum Superpositions.

Positivity

Every element of the form $A^*A$ is positive. Requiring

$$\omega(A^*A)\geq0$$

ensures that the state assigns nonnegative values to the positive cone of $\mathcal A$. In a C*-algebra, every positive element $X$ has a positive square root, so $X=B^*B$ for some $B$. The condition above is therefore equivalent to

$$X\geq0 \quad\Longrightarrow\quad \omega(X)\geq0.$$

Positivity has several consequences that need not be imposed separately.

Adjoint Compatibility

A positive linear functional is Hermitian:

$$\omega(A^*) = \overline{\omega(A)}.$$

Hence a self-adjoint element has a real expectation value:

$$A^*=A \quad\Longrightarrow\quad \omega(A)\in\mathbb R.$$

Cauchy–Schwarz Inequality

The sesquilinear form

$$(A,B)_\omega = \omega(A^*B)$$

is positive semidefinite and obeys

$$\left|\omega(A^*B)\right|^2 \leq \omega(A^*A)\,\omega(B^*B).$$

This inequality is central to the GNS construction and to uncertainty relations. It also shows that elements with zero $\omega$-norm are orthogonal to every element.

Nonnegative Variance

For a self-adjoint $A$, define the centered element

$$\Delta_\omega A = A-\omega(A)I.$$

Positivity gives

$$\begin{aligned} 0 &\leq \omega\!\left((\Delta_\omega A)^2\right)\\ &= \omega(A^2)-\omega(A)^2. \end{aligned}$$

Thus the usual nonnegative variance is already contained in the positivity axiom.

Positivity is stronger than requiring nonnegative expectations for a few preferred observables. It must hold throughout the algebra’s positive cone. A normalized linear functional that violates this condition can assign negative probabilities or negative squared norms and is not a physical state.

Normalization and Probabilities

The identity represents the constant observable one. The normalization

$$\omega(I)=1$$

therefore says that total probability is one.

If $P=P^*=P^2$ is a projection, then both $P$ and $I-P$ are positive. Consequently,

$$0\leq\omega(P)\leq1.$$

The number $\omega(P)$ is the probability of the corresponding sharp event. If mutually orthogonal projections resolve the identity,

$$\sum_jP_j=I,$$

then linearity and normalization give

$$\sum_j\omega(P_j)=1.$$

The same logic applies to an effect $E$ satisfying $0\leq E\leq I$. A POVM $\{E_j\}$ with $\sum_jE_j=I$ has probabilities

$$p_j=\omega(E_j).$$

The measurement formalism is treated in POVMs: First Encounter. The algebraic lesson is narrower: probability rules follow by evaluating a normalized positive functional on effects.

For a positive functional on a unital C*-algebra,

$$\lVert\omega\rVert=\omega(I).$$

A state therefore has norm one and is automatically continuous. Normalization is not only probabilistic; it also fixes the functional’s analytic size.

Density Matrices as Examples

Let $\mathcal A=M_n(\mathbb C)$. If $\rho$ is positive and has unit trace, then

$$\omega_\rho(A) = \operatorname{Tr}(\rho A)$$

defines a state. Linearity follows from the trace, normalization follows from $\operatorname{Tr}\rho=1$, and positivity follows from

$$\begin{aligned} \omega_\rho(A^*A) &= \operatorname{Tr}(\rho A^*A)\\ &= \operatorname{Tr} \!\left[ (A\rho^{1/2})^* (A\rho^{1/2}) \right]\\ &\geq0. \end{aligned}$$

Conversely, every state on $M_n(\mathbb C)$ has this form for a unique density matrix $\rho$. Finite-dimensional duality first gives a unique matrix $\rho$ satisfying

$$\omega(A)=\operatorname{Tr}(\rho A).$$

Hermiticity and positivity of $\omega$ imply $\rho=\rho^\dagger$ and $\rho\geq0$, while $\omega(I)=1$ implies $\operatorname{Tr}\rho=1$.

Thus finite-dimensional algebraic states and density matrices are equivalent descriptions:

$$\left\{ \begin{array}{c} \text{states on }M_n(\mathbb C) \end{array} \right\} \quad\longleftrightarrow\quad \left\{ \begin{array}{c} \rho\geq0,\\ \operatorname{Tr}\rho=1 \end{array} \right\}.$$

The detailed density-operator construction belongs to Density Operators, and explicit calculations belong to the Trace Rule for Expectation Values.

Qubit Example

For a qubit,

$$\rho_{\mathbf r} = \frac{1}{2} \left( I+\mathbf r\cdot\boldsymbol{\sigma} \right), \qquad \lVert\mathbf r\rVert\leq1.$$

If

$$A=a_0I+\mathbf a\cdot\boldsymbol{\sigma},$$

then

$$\omega_{\mathbf r}(A) = a_0+\mathbf r\cdot\mathbf a.$$

The condition $\lVert\mathbf r\rVert\leq1$ is exactly the positivity condition for $\rho_{\mathbf r}$. Boundary vectors with $\lVert\mathbf r\rVert=1$ give pure states, while interior vectors give mixed states.

The Infinite-Dimensional Caveat

On an infinite-dimensional Hilbert space, every positive trace-class operator $\rho$ with unit trace still defines

$$\omega_\rho(A) = \operatorname{Tr}(\rho A), \qquad A\in\mathcal B(\mathcal H).$$

These are the normal states of $\mathcal B(\mathcal H)$. Normality can be characterized by suitable continuity under increasing limits of positive operators, or equivalently by countable additivity on orthogonal projections.

Not every state on $\mathcal B(\mathcal H)$ is normal when $\mathcal H$ is infinite-dimensional. Nonnormal states exist and cannot be represented by trace-class density operators in that representation. Therefore the finite-dimensional slogan “every state is a density matrix” must not be promoted without qualification.

This distinction is one reason the functional definition is valuable. It defines a state before asking whether a density operator represents it in a chosen Hilbert space. In many-body theory and QFT, physically relevant states may lead through GNS reconstruction to representations inequivalent to the one used for another phase or sector.

Pure and Mixed States

The state space

$$S(\mathcal A) = \{\omega:\omega\text{ is a state on }\mathcal A\}$$

is convex. If $\omega_1,\omega_2\in S(\mathcal A)$ and $0\leq\lambda\leq1$, then

$$\omega = \lambda\omega_1 + (1-\lambda)\omega_2$$

is also a state.

A state is pure if it is an extreme point of this convex set: whenever

$$\omega = \lambda\omega_1 + (1-\lambda)\omega_2, \qquad 0\lt\lambda\lt1,$$

one must have $\omega_1=\omega_2=\omega$. A state that is not pure is mixed.

For $M_n(\mathbb C)$, the pure states are exactly the rank-one density matrices

$$\rho = \lvert\psi\rangle\langle\psi\rvert.$$

For a commutative algebra of functions, pure states behave like point evaluations. These examples show that purity is fundamentally an extremality property of the state on the specified algebra, not merely the presence of a ket in some larger Hilbert space.

The operational and density-matrix criteria are developed in Pure vs Mixed States. Here the canonical point is that convex geometry is intrinsic to the functional state space.

Classical Probability as a Commutative Example

Consider the commutative algebra

$$\mathcal A=\mathbb C^n$$

with componentwise operations and involution

$$(f_1,\ldots,f_n)^* = (\overline f_1,\ldots,\overline f_n).$$

Let $e_j$ be the function that equals one at outcome $j$ and zero elsewhere. A state determines

$$p_j=\omega(e_j), \qquad p_j\geq0, \qquad \sum_jp_j=1.$$

For any $f=(f_1,\ldots,f_n)$,

$$\omega(f) = \sum_jp_jf_j.$$

Thus a state on a finite commutative algebra is exactly an ordinary probability distribution. Noncommutative state functionals generalize this familiar expectation map while retaining positivity and normalization.

For a suitable algebra of continuous functions on a compact space, the Riesz–Markov theorem similarly represents states by probability measures. Pure states reduce to evaluation at individual points. The quantum difference lies in the noncommutative product, not in abandoning the logic of positive normalized expectations.

GNS Construction Preview

The Gelfand–Naimark–Segal construction shows how a state produces a Hilbert-space representation. Begin with the positive semidefinite form

$$\langle A\vert B\rangle_\omega = \omega(A^*B).$$

Some nonzero algebra elements may have zero length. Define the null space

$$\mathcal N_\omega = \{A\in\mathcal A: \omega(A^*A)=0\}.$$

The Cauchy–Schwarz inequality ensures that the inner product descends to equivalence classes in the quotient

$$\mathcal A/\mathcal N_\omega.$$

Completing this quotient gives a Hilbert space $\mathcal H_\omega$. Left multiplication defines a representation:

$$\pi_\omega(C)[A] = [CA].$$

The class of the identity,

$$\Omega_\omega=[I],$$

is cyclic because vectors of the form

$$\pi_\omega(A)\Omega_\omega=[A]$$

are dense in $\mathcal H_\omega$. Most importantly,

$$\omega(A) = \langle\Omega_\omega\vert \pi_\omega(A) \Omega_\omega\rangle.$$

The abstract expectation functional has become a vector expectation value, but on a Hilbert space built from that functional. Up to unitary equivalence, the cyclic representation is determined by the state. For C*-algebras, pure states correspond to irreducible GNS representations.

This construction does not make the original Hilbert-space formulation irrelevant. It explains where a Hilbert space, representation, and cyclic reference vector can come from when the algebra and state are taken as primary.

Common Mistakes

• Treating a state as an element of the observable algebra. A state is a functional on the algebra.
• Requiring only real linearity. The extension to the full complex algebra is complex linear.
• Demanding that $\omega(A)$ be real for every $A$. Reality is guaranteed for self-adjoint elements; non-self-adjoint elements can have complex expectations.
• Checking positivity only on a preferred list of observables instead of the full positive cone.
• Interpreting $\omega(A)$ as a probability for arbitrary $A$. Probabilities arise from projections or effects.
• Confusing a convex mixture of states with a coherent superposition of vectors.
• Assuming every infinite-dimensional algebraic state is represented by a density operator in a preselected Hilbert space.
• Assuming every vector functional is pure on every subalgebra.
• Calling a state faithful merely because it is normalized. Faithfulness is the stronger condition $\omega(A^*A)=0\Rightarrow A=0$.
• Treating the GNS Hilbert space as independent of the state used to construct it.

References

• O. Bratteli and D. W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed., Springer, 1987.
• R. V. Kadison and J. R. Ringrose, Fundamentals of the Theory of Operator Algebras, Volume I, AMS, 1997.
• G. J. Murphy, C-Algebras and Operator Theory*, Academic Press, 1990.
• R. Haag, Local Quantum Physics: Fields, Particles, Algebras, 2nd ed., Springer, 1996.
• F. Strocchi, An Introduction to the Mathematical Structure of Quantum Mechanics, 2nd ed., World Scientific, 2008.
• I. E. Segal, “Irreducible representations of operator algebras,” Bulletin of the American Mathematical Society 53, 73–88, 1947.

Exercises

1. Show that a state assigns a number between zero and one to every effect.

Solution

An effect satisfies

$$0\leq E\leq I.$$

Positivity gives

$$\omega(E)\geq0.$$

Because $I-E$ is also positive,

$$\omega(I-E)\geq0.$$

Using linearity and normalization,

$$1-\omega(E)\geq0.$$

Therefore

$$0\leq\omega(E)\leq1.$$

2. Derive the Cauchy–Schwarz inequality for a positive linear functional.

Solution

For any $\lambda\in\mathbb C$, positivity gives

$$0 \leq \omega\!\left[ (A+\lambda B)^* (A+\lambda B) \right].$$

Expanding,

$$\begin{aligned} 0\leq{}& \omega(A^*A) + \lambda\omega(A^*B)\\ &+ \overline\lambda\omega(B^*A) + |\lambda|^2\omega(B^*B). \end{aligned}$$

If $\omega(B^*B)\gt0$, choose

$$\lambda = - \frac{\omega(B^*A)} {\omega(B^*B)}.$$

Substitution yields

$$\left|\omega(A^*B)\right|^2 \leq \omega(A^*A)\omega(B^*B).$$

If $\omega(B^*B)=0$, positivity of the quadratic expression for arbitrary $\lambda$ forces $\omega(A^*B)=0$, so the same inequality holds.

3. Show that states on $\mathbb C^n$ are probability distributions.

Solution

Let $e_j$ be the standard coordinate functions. Each $e_j$ is positive, so

$$p_j=\omega(e_j)\geq0.$$

Because

$$\sum_je_j=I,$$

normalization gives

$$\sum_jp_j = \omega(I) =1.$$

Every $f\in\mathbb C^n$ has the expansion

$$f=\sum_jf_je_j.$$

Linearity then gives

$$\omega(f) = \sum_jp_jf_j.$$

Conversely, any probability vector defines a positive normalized functional by this formula.

4. Determine the positivity and purity conditions for the qubit functional

$$\omega_{\mathbf r} \left( a_0I+\mathbf a\cdot\boldsymbol{\sigma} \right) = a_0+\mathbf r\cdot\mathbf a.$$

Solution

The representing density matrix is

$$\rho_{\mathbf r} = \frac{1}{2} \left( I+\mathbf r\cdot\boldsymbol{\sigma} \right).$$

Its eigenvalues are

$$\lambda_\pm = \frac{1}{2} \left( 1\pm\lVert\mathbf r\rVert \right).$$

Both are nonnegative exactly when

$$\lVert\mathbf r\rVert\leq1.$$

The state is pure exactly when the density matrix has rank one, which occurs when one eigenvalue vanishes:

$$\lVert\mathbf r\rVert=1.$$

If $\lVert\mathbf r\rVert\lt1$, both eigenvalues are positive and the state is mixed.

5. Work out the GNS construction for a vector state

$$\omega_\psi(A) = \langle\psi\vert A\vert\psi\rangle$$

on $\mathcal B(\mathcal H)$, where $\lVert\psi\rVert=1$.

Solution

The GNS seminorm is

$$\lVert[A]\rVert_\omega^2 = \omega_\psi(A^*A) = \lVert A\psi\rVert^2.$$

Therefore

$$\mathcal N_\omega = \{A:A\psi=0\}.$$

Define

$$U[A]=A\psi.$$

If two representatives differ by an element of $\mathcal N_\omega$, they act identically on $\psi$, so $U$ is well defined. The norm identity above shows that it is an isometry.

Every vector $\phi\in\mathcal H$ can be reached from $\psi$ by a bounded rank-one operator, for example

$$A_\phi = \lvert\phi\rangle\langle\psi\rvert.$$

Thus $U$ has dense, in fact full, range and extends to a unitary map from $\mathcal H_\omega$ to $\mathcal H$. Under this map,

$$U\pi_\omega(C)[A] = CA\psi = CU[A],$$

so the GNS representation is unitarily equivalent to the defining representation. The cyclic vector $[I]$ maps to $\psi$.