Physlib.QuantumMechanics.OperatorAlgebra.Observables.Jordan

For two observables $a$ and $b$ in a complex $C^*$-algebra $A$, their Jordan product $a \odot b$ is defined as the symmetrized product $\frac{1}{2}(ab + ba)$, which corresponds to the real part of their algebraic product in $A$.

The notation $a \odot b$ represents the Jordan product of two observables $a$ and $b$, defined by the symmetrized product $\frac{1}{2}(ab + ba)$.

Let $A$ be a complex $C^*$-algebra. For any two observables $a, b \in A$ (defined as self-adjoint elements where $a^* = a$ and $b^* = b$), their Jordan product $a \odot b$ is given by the symmetrized product $$a \odot b = \frac{1}{2}(ab + ba)$$ where the multiplication and addition are those of the underlying algebra $A$.

For any two observables $a$ and $b$ in a complex $C^*$-algebra $A$, the Jordan product $a \odot b$ is commutative, satisfying $a \odot b = b \odot a$. Here, the Jordan product is the symmetrized product defined by $a \odot b = \frac{1}{2}(ab + ba)$.

Let $A$ be a complex $C^*$-algebra. For any observable $a \in A$ (a self-adjoint element), the Jordan product of $a$ with itself is equal to its square in the algebra, i.e., $a \odot a = a^2$.

Let $A$ be a complex $C^*$-algebra. For any observables $a, b, c \in A$ (self-adjoint elements), the Jordan product $x \odot y = \frac{1}{2}(xy + yx)$ satisfies the distributive law $(a + b) \odot c = a \odot c + b \odot c$.

Let $A$ be a complex $C^*$-algebra. For any observables (self-adjoint elements) $a, b, c$ in $A$, the Jordan product $\odot$ distributes over addition such that $a \odot (b + c) = a \odot b + a \odot c$.

Let $A$ be a complex $C^*$-algebra and let $a, b$ be observables in $A$ (self-adjoint elements such that $a = a^*$ and $b = b^*$). For any real scalar $t \in \mathbb{R}$, the Jordan product $a \odot b = \frac{1}{2}(ab + ba)$ satisfies the scalar homogeneity property: $$a \odot (t \cdot b) = t \cdot (a \odot b)$$

Let $A$ be a complex $C^*$-algebra. For any observables (self-adjoint elements) $a, b \in A$, the Jordan product defined by $a \odot b = \frac{1}{2}(ab + ba)$ satisfies the Jordan identity: $$(a \odot b) \odot (a \odot a) = a \odot (b \odot (a \odot a))$$

Let $A$ be an operator algebra. The type `JordanObservable A` is defined as the space of observables (self-adjoint elements $a \in A$ such that $a = a^*$) equipped with the symmetrized Jordan product $a \odot b = \frac{1}{2}(ab + ba)$. It is a type synonym for `Observable A` used to provide the structure of a commutative Jordan algebra.

For an operator algebra $A$, the space of Jordan observables $\text{JordanObservable}(A)$ is an additive commutative group. This structure is inherited from the additive group structure of the space of observables $\text{Observable}(A)$.

For an operator algebra $A$, the space of Jordan observables $\text{JordanObservable}(A)$ is a module over the real numbers $\mathbb{R}$. This vector space structure is inherited from the space of self-adjoint observables $\text{Observable}(A)$.

For an operator algebra $A$, the space of Jordan observables $\text{JordanObservable } A$ is equipped with the structure of a non-unital non-associative commutative ring. The multiplication in this ring is the Jordan product, defined for any two observables $a, b$ as $a \odot b = \frac{1}{2}(ab + ba)$. This product is commutative ($a \odot b = b \odot a$) and satisfies the distributive laws $a \odot (b + c) = a \odot b + a \odot c$ and $(a + b) \odot c = a \odot c + b \odot c$.

Let $A$ be an operator algebra. The space of Jordan observables $\text{JordanObservable } A$, consisting of the self-adjoint elements of $A$ equipped with the symmetrized product $a \odot b = \frac{1}{2}(ab + ba)$, is a commutative Jordan algebra. This means the product is commutative and satisfies the Jordan identity $(a^2 \odot b) \odot a = a^2 \odot (b \odot a)$ for all $a, b \in \text{JordanObservable } A$.