PhyslibAlpha.AlgebraicFramework.HilbertSpace.Trace

For an element $c$ in a star-algebra $A$ over a ring $R$, this definition represents the positive linear map from $A$ to $A$ given by $x \mapsto c x c^*$, where $c^*$ denotes the involution (star operator) of $c$.

Let $E$ be a complex Hilbert space (or a finite-dimensional complex vector space). This definition represents the trace functional $\text{Tr}: \mathcal{L}(E) \to \mathbb{C}$ on the space of continuous linear operators $\mathcal{L}(E)$ as a positive linear map. It maps each operator $T \in \mathcal{L}(E)$ to its trace $\text{Tr}(T)$, such that for any positive operator $T \geq 0$, the result is a non-negative real number $\text{Tr}(T) \geq 0$.

Let $E$ be a complex Hilbert space. The trace functional $\text{Tr} : \mathcal{L}(E) \to \mathbb{C}$, considered as a linear map, is a tracial functional. This means it satisfies the cyclic property $\text{Tr}(AB) = \text{Tr}(BA)$ for all continuous linear operators $A, B \in \mathcal{L}(E)$.

Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the space of continuous linear operators on $E$. For an operator $\rho \in \mathcal{L}(E)$, this definition provides the positive linear functional $x \mapsto \text{Tr}(\sqrt{\rho} x \sqrt{\rho}^*)$, where $\text{Tr}$ denotes the trace and $\sqrt{\rho}$ is the square root of $\rho$ as defined via continuous functional calculus. In the context of the definition, $\sqrt{\rho}^*$ represents the adjoint of the square root. If $\rho$ is not a positive operator, the functional evaluates to zero.

Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the space of continuous linear operators on $E$. For any positive operator $\rho \in \mathcal{L}(E)$ (such that $0 \le \rho$) and any operator $x \in \mathcal{L}(E)$, the positive linear functional $\text{traceMulOp}_{\rho}$ evaluated at $x$ is equal to the trace of the product $x\rho$: $$\text{traceMulOp}_{\rho}(x) = \text{Tr}(x\rho)$$ where $\text{Tr}$ denotes the trace of a linear map.

Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the space of continuous linear operators on $E$. If $\rho \in \mathcal{L}(E)$ is not a non-negative operator (i.e., $\neg (0 \le \rho)$), then for any operator $x \in \mathcal{L}(E)$, the positive linear functional $\text{traceMulOpₚ}(\rho)$ applied to $x$ is equal to $0$.