PhyslibAlpha.ProbabilisticTheory.HilbertSpace.Unbounded.StoneUnitaryGroup

For a given real parameter $t$, this is the function from $\mathbb{R}$ to $\mathbb{C}$ defined by $x \mapsto e^{itx}$, where $i$ is the imaginary unit.

For any real number $t$, the function $x \mapsto e^{itx}$ from $\mathbb{R}$ to $\mathbb{C}$ is measurable.

For any real number $t$, the exponential multiplier function $x \mapsto e^{itx}$ is bounded. That is, there exists a constant $C$ such that for all $x \in \mathbb{R}$, $\|e^{itx}\| \leq C$, where $i$ is the imaginary unit.

For any real numbers $t$ and $x$, the complex modulus of the exponential function $x \mapsto e^{itx}$ is equal to $1$. That is, $|e^{itx}| = 1$.

For any real number $t$, the exponential multiplier $x \mapsto e^{-itx}$ is equal to the complex conjugate of the exponential multiplier $x \mapsto e^{itx}$.

For any real numbers $t$ and $s$, there exists a constant $C$ such that for all $x \in \mathbb{R}$, $|e^{itx} - e^{isx}| \le C$, where $i$ is the imaginary unit and $e^{itx}$ denotes the complex exponential function.

For any real numbers $t$ and $s$, let $f_t: \mathbb{R} \to \mathbb{C}$ be the exponential multiplier function defined by $x \mapsto e^{itx}$. Then the function $f_{t+s}$ is equal to the pointwise product of $f_t$ and $f_s$, i.e., $f_{t+s} = f_t \cdot f_s$.

Given a spectral measure $\mu$ on $\mathbb{R}$ acting on a Hilbert space $H$ and a real parameter $t$, this definition constructs a bounded linear operator through the spectral integral of the exponential function $x \mapsto e^{itx}$. Specifically, it defines the operator $$U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda),$$ which corresponds to the unitary operator $e^{itT}$ in the one-parameter unitary group generated by the self-adjoint operator $T$ associated with the spectral measure $\mu$.

Let $\mu$ be a spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$. For any $t \in \mathbb{R}$, let $U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda)$ be the bounded linear operator defined by the spectral integral of the function $\lambda \mapsto e^{it\lambda}$. Then for any real numbers $t$ and $s$, the operators satisfy the relation $$U(t + s) = U(t) U(s).$$

Let $\mu$ be a spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$. Let $U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda)$ be the operator defined by the spectral integral of the exponential function for $t \in \mathbb{R}$. Then for $t = 0$, the operator $U(0)$ is equal to the identity operator $I$ on $H$.

Let $\mu$ be a spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$. Let $U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda)$ be the bounded linear operator defined by the spectral integral of the exponential function for a real parameter $t$. Then for any $t \in \mathbb{R}$, the composition of $U(-t)$ and $U(t)$ is the identity operator $I$ on $H$: $$U(-t) U(t) = I$$

Let $\mu$ be a spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$. For any $t \in \mathbb{R}$, let $U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda)$ be the operator defined by the spectral integral of the exponential function. Then the composition of $U(t)$ with $U(-t)$ is the identity operator, i.e., $U(t) U(-t) = I$.

Let $\mu$ be a spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$. For any $t \in \mathbb{R}$, let $U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda)$ be the bounded linear operator defined by the spectral integral of the function $\lambda \mapsto e^{it\lambda}$. Then the adjoint of $U(t)$, denoted by $U(t)^*$, is equal to $U(-t)$.

Let $\mu$ be a spectral measure on $\mathbb{R}$ taking values in the bounded operators on a Hilbert space $H$. For any real number $t$, the operator defined by the spectral integral $$U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda)$$ is a unitary operator on $H$.

Let $H$ be a non-empty Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. For any real numbers $t, s \in \mathbb{R}$ and any vector $x \in H$, the difference between the unitary operators associated with $t$ and $s$ applied to $x$ is equal to the spectral integral of the difference of their corresponding exponential functions applied to $x$: $$\left( \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda) \right) x - \left( \int_{\mathbb{R}} e^{is\lambda} \, d\mu(\lambda) \right) x = \left( \int_{\mathbb{R}} (e^{it\lambda} - e^{is\lambda}) \, d\mu(\lambda) \right) x$$ where $e^{it\lambda}$ is the complex exponential function $x \mapsto e^{itx}$.

Let $H$ be a Hilbert space and $\mu$ be a spectral measure on $\mathbb{R}$ taking values in the space of bounded linear operators on $H$. For any $t \in \mathbb{R}$, let $U(t)$ be the operator defined by the spectral integral $$U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda).$$ Then, for any vector $x \in H$, the mapping $t \mapsto U(t)x$ is continuous. This property is known as the strong continuity of the one-parameter unitary group associated with the spectral measure.

This instance allows a strongly continuous one-parameter unitary group $G$ on a Hilbert space $H$ to be treated as a function from the real numbers $\mathbb{R}$ to the space of continuous linear operators on $H$. Given a group $G$ and a time $t \in \mathbb{R}$, $G(t)$ returns the corresponding unitary operator acting on $H$.

For any strongly continuous one-parameter unitary group $G$ on a Hilbert space $H$, the operator corresponding to the parameter $0$ is the identity operator, i.e., $G(0) = I$.

Let $G$ be a strongly continuous one-parameter unitary group acting on a Hilbert space $H$. For any two real numbers $t$ and $s$, the operator $G(t + s)$ is equal to the composition of the operators $G(t)$ and $G(s)$, such that $G(t + s) = G(t) G(s)$.

Let $H$ be a Hilbert space and $G$ be a strongly continuous one-parameter group. For any $t \in \mathbb{R}$, the operator $G(t)$ is a unitary operator on $H$.

Let $H$ be a Hilbert space and $G$ be a strongly continuous one-parameter unitary group on $H$. For any vector $x \in H$, the mapping $t \mapsto G(t)x$ is continuous.

Given a spectral measure $\mu$ on $\mathbb{R}$ acting on a Hilbert space $H$, this definition constructs the strongly continuous one-parameter unitary group $U: \mathbb{R} \to \mathcal{B}(H)$. For each $t \in \mathbb{R}$, the operator is defined by the spectral integral $$U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda).$$ This construction satisfies the unitary group properties $U(0) = I$ and $U(t+s) = U(t)U(s)$, where each $U(t)$ is a unitary operator. The group is strongly continuous in the sense that for any vector $x \in H$, the mapping $t \mapsto U(t)x$ is continuous.