Proposition. Finite states are probability weights [fdip-0026]
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  • October 23, 2026

Let \(\Omega \) be finite, \(K=\mathbb R\) or \(\mathbb C\), and give \(A_\Omega =\operatorname {Map}(\Omega ,K)\) the pointwise involution \(f^*(x)=\overline {f(x)}\). The weight functional \(\omega _p(f)=\sum _xp(x)f(x)\) is a real or complex algebraic state exactly when every \(p(x)\) is real and nonnegative and \(\sum _xp(x)=1\).

Proof. 

  • October 23, 2026

Pointwise conjugation is additive, conjugates scalars, fixes the unit and is involutive. It reverses products because \(K\) is commutative. For necessity, \(\delta _x^*\delta _x=\delta _x\), so square positivity gives \(p(x)=\omega _p(\delta _x)\in \mathbb R_{\geq 0}\). Normalization gives the sum condition. Conversely, real weights imply \(\omega _p(f^*)=\sum _xp(x)\overline {f(x)}=\overline {\omega _p(f)}\). Furthermore \[\omega _p(f^*f)=\sum _xp(x)|f(x)|^2\geq 0,\] and \(\omega _p(1)=\sum _xp(x)=1\). For a real pointwise nonnegative function, each summand \(p(x)f(x)\) is also nonnegative, so its expectation is nonnegative.

There is no state on the empty configuration set over \(\mathbb R\) or \(\mathbb C\), because the empty sum is zero. On two points, the weights \(p(0)=p(1)=1/2\) give a state that is not multiplicative: \(\omega _p(\delta _0\delta _1)=0\), whereas \(\omega _p(\delta _0)\omega _p(\delta _1)=1/4\). A state is not required to be an algebra homomorphism.

The positive-functional convention is [waldmann2004states, Definition 3.9, p. 11]; the indicator calculation above identifies it with finite probability and expectation on this particular algebra. It does not identify positivity cones on arbitrary star algebras; the distinctions on [waldmann2004states, p. 12] remain separate.