PhyslibAlpha.QuantumMechanics.OperatorAlgebra.Dynamics.Automorphism

Let $A$ be a unital complex $C^*$-algebra. Given a $\star$-automorphism $\beta: A \xrightarrow{\sim} A$ and an observable $a \in A$ (a self-adjoint element such that $a^* = a$), this definition describes the action of the automorphism on the observable, mapping $a$ to $\beta(a)$. Since $\star$-automorphisms preserve the involution, the resulting element $\beta(a)$ is also an observable.

Let $A$ be a unital complex $C^*$-algebra. For any $\star$-automorphism $\beta : A \xrightarrow{\sim} A$ and any observable $a \in A$ (defined as a self-adjoint element such that $a = a^*$), the action of $\beta$ on $a$ as an observable, when viewed as an element of the underlying algebra $A$, is equal to the result of applying $\beta$ to $a$ directly. That is, $(\beta.observable(a) : A) = \beta(a)$.

Let $A$ be a unital complex $C^*$-algebra. The identity $\star$-automorphism of $A$ acts as the identity map on the space of observables (the set of self-adjoint elements $a \in A$ such that $a^* = a$). That is, applying the identity $\star$-automorphism to any observable $a$ returns the observable $a$ itself.

Let $A$ be a unital complex $C^*$-algebra representing the observables of a physical system. For any two $\star$-automorphisms $\beta, \gamma : A \xrightarrow{\sim} A$ and any observable $a \in A$ (a self-adjoint element such that $a^* = a$), the action of the composed $\star$-automorphism $\gamma \circ \beta$ on the observable $a$ is equal to the successive application of the automorphisms: $(\gamma \circ \beta)(a) = \gamma(\beta(a))$.

Let $A$ be a unital complex $C^*$-algebra. For any $\star$-automorphism $\beta: A \xrightarrow{\sim} A$ and any observable $a \in A$ (a self-adjoint element such that $a^* = a$), applying the inverse $\star$-automorphism $\beta^{-1}$ to the transformed observable $\beta(a)$ recovers the original observable, such that $\beta^{-1}(\beta(a)) = a$.

Let $A$ be a unital complex $C^*$-algebra. For any $\star$-automorphism $\beta$ of $A$ and any observable $a \in A$ (a self-adjoint element such that $a^* = a$), applying the automorphism $\beta$ to the result of its inverse $\beta^{-1}$ acting on $a$ recovers the original observable. That is, $\beta(\beta^{-1}(a)) = a$.

Let $A$ be a unital complex $C^*$-algebra representing the observables of a physical system. For any $\star$-automorphism $\beta : A \xrightarrow{\sim} A$ and any observables $a, b \in A$ (self-adjoint elements), the automorphism preserves the Lie bracket of observables: \[\beta(\unicode{x2045}a, b\unicode{x2046}) = \unicode{x2045}\beta(a), \beta(b)\unicode{x2046}\] where $\unicode{x2045}a, b\unicode{x2046} = \text{Im}(ab)$ denotes the Lie bracket defined on the space of self-adjoint elements.

Let $A$ be an operator algebra. If $\alpha$ and $\beta$ are two one-parameter automorphism groups of $A$ such that for all times $t$ and all elements $a \in A$, the action $\alpha_t(a)$ is equal to $\beta_t(a)$, then the automorphism groups $\alpha$ and $\beta$ are equal.

Given a $\star$-algebra automorphism $\beta : A \simeq_{\star, \text{alg}} A$ and a one-parameter automorphism group $\alpha_t$ of an operator algebra $A$, the conjugated automorphism group is defined such that at any time $t \in \mathbb{R}$, the resulting automorphism is $\beta \circ \alpha_t \circ \beta^{-1}$. For an element $a \in A$, the action is given by $(\text{conj}_\beta \alpha)_t(a) = \beta(\alpha_t(\beta^{-1}(a)))$.

Let $A$ be an operator algebra. For a $\star$-algebra automorphism $\beta : A \simeq_{\star, \text{alg}} A$ and a one-parameter automorphism group $\alpha$ of $A$, the action of the conjugated group $(\text{conj}_\beta \alpha)$ at time $t \in \mathbb{R}$ on an element $a \in A$ is given by the identity $$(\text{conj}_\beta \alpha)_t(a) = \beta(\alpha_t(\beta^{-1}(a)))$$ where $\beta^{-1}$ denotes the inverse automorphism of $\beta$.

Let $A$ be a $\star$-algebra over $\mathbb{C}$ and let $\alpha$ be a one-parameter $\star$-automorphism group of $A$. Let $\text{id}_A$ denote the identity $\star$-automorphism of $A$. The conjugation of the group $\alpha$ by $\text{id}_A$ leaves the group unchanged, i.e., $\text{conj}_{\text{id}_A} \alpha = \alpha$.

Let $A$ be an operator algebra and $\alpha$ be a one-parameter group of $\star$-automorphisms of $A$. Let $\beta, \gamma : A \simeq_{\star, \mathbb{C}} A$ be $\star$-algebra automorphisms. Then the result of conjugating $\alpha$ by $\beta$ and subsequently conjugating the result by $\gamma$ is equal to conjugating $\alpha$ once by the composite automorphism $\gamma \circ \beta$. That is, $$\text{conj}_\gamma (\text{conj}_\beta \alpha) = \text{conj}_{\gamma \circ \beta} \alpha$$ where the conjugation $(\text{conj}_\beta \alpha)_t$ is defined as $\beta \circ \alpha_t \circ \beta^{-1}$.

Let $A$ be an operator algebra and $\alpha$ be a one-parameter automorphism group of $A$. For any $\star$-algebra automorphism $\beta$ of $A$ with inverse $\beta^{-1}$, let $\text{conj}_\beta \alpha$ denote the one-parameter group conjugated by $\beta$, defined by $(\text{conj}_\beta \alpha)_t = \beta \circ \alpha_t \circ \beta^{-1}$ for all $t$. Then, conjugating this resulting flow by the inverse automorphism $\beta^{-1}$ recovers the original flow $\alpha$. That is, $\text{conj}_{\beta^{-1}}(\text{conj}_\beta \alpha) = \alpha$.

Let $A$ be an operator algebra, $\alpha$ be a one-parameter automorphism group of $A$, and $\beta: A \simeq_{\star, \text{alg}} A$ be a $\star$-algebra automorphism of $A$ with inverse $\beta^{-1}$. If the automorphism group $\alpha$ is conjugated by $\beta^{-1}$ and the resulting group is then conjugated by $\beta$, the result is the original automorphism group $\alpha$. That is, $$\text{conj}_{\beta}(\text{conj}_{\beta^{-1}}(\alpha)) = \alpha$$ where $(\text{conj}_{\beta} \alpha)_t = \beta \circ \alpha_t \circ \beta^{-1}$.

Given a unitary one-parameter group $U(t)$ on a complex Hilbert space $\mathcal{H}$, this definition constructs the corresponding automorphism group $\alpha$ of the algebra of bounded linear operators $\mathcal{B}(\mathcal{H})$. For any $t \in \mathbb{R}$ and any operator $a \in \mathcal{B}(\mathcal{H})$, the action of the automorphism group is given by the unitary conjugation $\alpha_t(a) = U(t) a U(t)^*$, where $U(t)^*$ denotes the adjoint of $U(t)$.

For a complex Hilbert space $\mathcal{H}$, let $\mathcal{B}(\mathcal{H})$ be the algebra of bounded linear operators on $\mathcal{H}$. Given a unitary one-parameter group $U(t)$ on $\mathcal{H}$, the induced automorphism group $\alpha$ of $\mathcal{B}(\mathcal{H})$ acts on an operator $a \in \mathcal{B}(\mathcal{H})$ at time $t \in \mathbb{R}$ according to the rule $\alpha_t(a) = U(t) a U(t)^*$, where $U(t)^*$ denotes the adjoint (star) of the unitary operator $U(t)$.

Let $\mathcal{H}$ be a complex Hilbert space and $U$ be a norm-continuous unitary one-parameter group on $\mathcal{H}$ with generator $A$. Let $\alpha_t$ be the automorphism group of the algebra of bounded operators $\mathcal{B}(\mathcal{H})$ induced by unitary conjugation, such that $\alpha_t(a) = U(t) a U(t)^*$ for any operator $a \in \mathcal{B}(\mathcal{H})$ and time $t \in \mathbb{R}$. Then the derivative of this conjugation flow satisfies the Heisenberg equation: $$\frac{d}{dt} \alpha_t(a) = [\alpha_t(a), i A],$$ where $i$ is the imaginary unit and $[X, Y] = XY - YX$ denotes the commutator of operators.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the algebra of bounded linear operators on $\mathcal{H}$. Let $U(t)$ be a norm-continuous unitary one-parameter group on $\mathcal{H}$ with generator $G$. Suppose $\alpha: \mathbb{R} \to (\mathcal{B}(\mathcal{H}) \to \mathcal{B}(\mathcal{H}))$ is a map satisfying the following conditions: 1. **Initial condition:** $\alpha_0 = \text{id}_{\mathcal{B}(\mathcal{H})}$, where $\text{id}$ is the identity map. 2. **Differential equation:** For every operator $a \in \mathcal{B}(\mathcal{H})$, the map $t \mapsto \alpha_t(a)$ is differentiable and satisfies the Heisenberg-type equation $$\frac{d}{dt} \alpha_t(a) = [\alpha_t(a), iG],$$ where $[x, y] = xy - yx$ denotes the commutator and $i$ is the imaginary unit. Then $\alpha$ is the automorphism group defined by unitary conjugation, i.e., $\alpha_t(a) = U(t) a U(t)^*$ for all $t \in \mathbb{R}$ and $a \in \mathcal{B}(\mathcal{H})$.