PhyslibAlpha.ProbabilisticTheory.CStarAlgebra.JordanDecomposition

Let $a$ be an observable (a self-adjoint element) in a $C^*$-algebra $A$. Then $a$ is non-negative ($0 \le a$) if and only if there exists an observable $b \in A$ such that $a = b^2$.

Given an observable $a$ (a self-adjoint element) in a $C^*$-algebra $A$, this function returns its positive part $a^+$. The result $a^+$ is a positive observable, meaning it is both self-adjoint ($(a^+)^* = a^+$) and non-negative ($a^+ \ge 0$). This corresponds to the part of the Jordan decomposition $a = a^+ - a^-$ where $a^+$ is obtained via continuous functional calculus.

For an observable $a$ (a self-adjoint element) in a $C^*$-algebra $A$, the negative part $a^-$ is defined as a positive observable—that is, an element of $A$ which is both self-adjoint and non-negative ($a^- \ge 0$). This forms part of the unique Jordan decomposition $a = a^+ - a^-$, where $a^+$ and $a^-$ are orthogonal ($a^+ a^- = 0$).

For any observable $a$ (a self-adjoint element) in a $C^*$-algebra $A$, let $a^+$ and $a^-$ denote its positive part and negative part, respectively. Then $a$ can be expressed as the difference of these parts: $a = a^+ - a^-$.

Let $A$ be a $C^*$-algebra and $a$ be an observable (a self-adjoint element) in $A$. Let $a^+$ and $a^-$ denote the positive part and the negative part of $a$, respectively, obtained from its unique Jordan decomposition. Then the product of the positive part and the negative part is zero: $$a^+ a^- = 0.$$

Let $A$ be a $C^*$-algebra and $a \in A$ be an observable (a self-adjoint element). If $a^+$ and $a^-$ denote the positive and negative parts of $a$ respectively, then their product in that order is zero, i.e., $a^- a^+ = 0$.

Let $A$ be a $C^*$-algebra. For any observable $a \in A$ and any two positive observables $b, c \in A$, if $a = b - c$ and $bc = 0$, then $b = a^+$ and $c = a^-$, where $a^+$ and $a^-$ denote the positive and negative parts of $a$ respectively.