PhyslibAlpha.AlgebraicFramework.HilbertSpace.Unbounded.AnalyticVector.Basic

For a partially defined linear operator $T$ on a complex Hilbert space $H$ and a vector $x \in H$, a sequence $v: \mathbb{N} \to \text{dom}(T)$ is the iterate sequence of $x$ under $T$ if $v_0 = x$ and $v_{n+1} = T v_n$ for all $n \in \mathbb{N}$. This definition implies that $x$ must be in the domain of every power $T^n$ for such a sequence to exist.

Let $T$ be a partially defined linear operator on a complex Hilbert space $H$. For any vector $x \in H$, if $v, w : \mathbb{N} \to \text{dom}(T)$ are both iterate sequences of $x$ under $T$ (meaning $v_0 = w_0 = x$ and $v_{n+1} = T v_n$, $w_{n+1} = T w_n$ for all $n \in \mathbb{N}$), then $v_n = w_n$ for all $n \in \mathbb{N}$.

For a partially defined linear operator $T$ on a complex Hilbert space $H$, a vector $x \in H$ is an **analytic vector** for $T$ if $x$ is in the domain of $T^n$ for all $n \in \mathbb{N}$ and there exists a positive real number $t > 0$ such that the power series $$\sum_{n=0}^{\infty} \frac{\|T^n x\|}{n!} t^n$$ is summable (converges).

Let $T$ be a partially defined linear operator on a complex Hilbert space $H$. If $W$ is an `AnalyticVectorWitness` for $T$, then the vector $x = W.\text{state}$ is an analytic vector for $T$. A vector $x$ is defined as an analytic vector for $T$ if $x$ is in the domain of $T^n$ for all $n \in \mathbb{N}$ and there exists a positive real number $t > 0$ such that the power series $$\sum_{n=0}^{\infty} \frac{\|T^n x\|}{n!} t^n$$ is summable.

For a partially defined linear operator $T$ on a complex Hilbert space $H$ and a vector $x \in H$, this function extracts an explicit witness structure from a proof that $x$ is an analytic vector. The resulting witness contains the vector $x$, the sequence of iterates $v_n = T^n x$ (where $v_n \in \text{dom}(T)$ for all $n$), and a positive real number $t > 0$ such that the power series $$\sum_{n=0}^{\infty} \frac{\|v_n\|}{n!} t^n$$ is summable.

For a partially defined linear operator $T$ on a complex Hilbert space $H$, a vector $x \in H$ is an **entire vector** if there exists a sequence $v: \mathbb{N} \to \text{dom}(T)$ of iterates of $x$ under $T$ (where $v_0 = x$ and $v_{n+1} = T v_n$ for all $n \in \mathbb{N}$) such that for every positive real number $t > 0$, the series $$\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$$ is summable. This condition implies that $x$ is in the domain of all powers $T^n$ and the power series of the exponential orbit $\exp(tT)x$ has an infinite radius of convergence.

For a partially defined linear operator $T$ on a complex Hilbert space $H$, a sequence $v$ of iterates (where $v_n$ represents the $n$-th power $T^n x$ acting on a vector $x$), and a real parameter $s$, the $n$-th term of the formal exponential orbit is defined as: $$\frac{(is)^n}{n!} v_n$$ where $i$ is the imaginary unit and $n!$ denotes the factorial of $n$. This term corresponds to the $n$-th component of the series expansion for $e^{isT}x$, following the convention used in Stone's theorem.

For a partially defined linear operator $T$ on a complex Hilbert space $H$, a sequence of iterates $v: \mathbb{N} \to \text{dom}(T)$ (where $v_n$ represents the $n$-th power $T^n$ acting on an initial vector), and a real parameter $s$, the **local exponential orbit** is defined as the sum of the infinite series: $$\sum_{n=0}^\infty \frac{(is)^n}{n!} v_n$$ where $i$ is the imaginary unit and $n!$ denotes the factorial of $n$. This series represents the formal expansion of the operator exponential $e^{isT}$ applied to the vector $v_0$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Let $x \in H$ be a vector, and let $v, w: \mathbb{N} \to \text{dom}(T)$ be two sequences of iterates of $x$ under $T$ (that is, $v_0 = w_0 = x$ and $v_{n+1} = T v_n$, $w_{n+1} = T w_n$ for all $n \in \mathbb{N}$). Then for any real parameter $s$, the local exponential orbit series defined by $v$ and $w$ are equal: $$\sum_{n=0}^\infty \frac{(is)^n}{n!} v_n = \sum_{n=0}^\infty \frac{(is)^n}{n!} w_n$$

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Given a sequence of vectors $v: \mathbb{N} \to \text{dom}(T)$ representing the iterates of $T$ (where $v_n$ corresponds to $T^n x$), a real parameter $s$, and a natural number $n$, the norm of the $n$-th term of the exponential series, defined as $\frac{(is)^n}{n!} v_n$, is given by: $$\left\| \frac{(is)^n}{n!} v_n \right\| = \frac{\|v_n\| \cdot |s|^n}{n!}$$ where $i$ is the imaginary unit and $n!$ is the factorial of $n$.

For a partially defined linear operator $T$ on a complex Hilbert space $H$, given a sequence of vectors $v: \mathbb{N} \to \text{dom}(T)$ (representing iterates of $T$), a real parameter $s \in \mathbb{R}$, and a natural number $n$, the $n$-th term of the formal derivative of the exponential series with respect to $s$ is defined as: $$\frac{n \cdot i \cdot (i \cdot s)^{n-1}}{n!} v_n$$ where $i$ is the imaginary unit. This corresponds to the derivative of the $n$-th term of the Taylor series $\sum_{n=0}^{\infty} \frac{(i \cdot s \cdot T)^n}{n!}x$ evaluated at $s$. For $n=0$, the term evaluates to $0$ because of the factor $n$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Let $v: \mathbb{N} \to \text{dom}(T)$ be a sequence of vectors (representing iterates of $T$ acting on a vector $x$). For any natural number $n$ and any real number $s$, the function mapping $r \in \mathbb{R}$ to the $n$-th term of the exponential series $$\frac{(ir)^n}{n!} v_n$$ is differentiable at $s$, and its derivative is given by $$\frac{n \cdot i \cdot (is)^{n-1}}{n!} v_n$$ where $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Given a sequence of vectors $v: \mathbb{N} \to \text{dom}(T)$, a real number $s$, and a natural number $n$, the norm of the $n$-th term of the derivative of the analytic exponential series, $\text{analyticExpDerivTerm}(T, v, s, n)$, is given by: $$\|\text{analyticExpDerivTerm}(T, v, s, n)\| = \frac{\|v_n\| \cdot n \cdot |s|^{n-1}}{n!}$$ where $\|v_n\|$ is the norm of the $n$-th vector in the sequence and $n!$ denotes the factorial of $n$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Let $v: \mathbb{N} \to \text{dom}(T)$ be a sequence of vectors (typically iterates of $T$ acting on some $x \in H$). Suppose that for some radius $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ is summable. Then for any real number $s$ such that $s \in (-t/2, t/2)$, the local exponential orbit mapping $r \mapsto \text{analyticExp}(T, v, r)$ is differentiable at $s$, and its derivative is given by the sum of the term-wise derivative series: $$\frac{d}{dr} \left( \sum_{n=0}^\infty \frac{(ir)^n}{n!} v_n \right) \bigg|_{r=s} = \sum_{n=0}^\infty \frac{n \cdot i \cdot (is)^{n-1}}{n!} v_n$$ where $i$ is the imaginary unit and $n!$ denotes the factorial of $n$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Let $x \in H$ and $v: \mathbb{N} \to \text{dom}(T)$ be a sequence of iterates of $x$ under $T$, meaning $v_0 = x$ and $v_{n+1} = T v_n$ for all $n \in \mathbb{N}$. Suppose there exists a radius $t > 0$ such that the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ is summable. Then the local exponential orbit $s \mapsto \text{analyticExp}(T, v, s) = \sum_{n=0}^\infty \frac{(is)^n}{n!} v_n$ is differentiable at $s = 0$ with derivative $i T x$, where $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ a partially defined linear operator on $H$. Let $v: \mathbb{N} \to \text{dom}(T)$ be a sequence of vectors. Suppose that for some radius $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ is summable. Then for any $s \in \mathbb{R}$ such that $|s| < t/2$, the series of derivative terms $$\sum_{n=0}^{\infty} \frac{n \cdot i \cdot (is)^{n-1}}{n!} v_n$$ is summable in $H$, where $i$ denotes the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ be a closable partially defined linear operator on $H$ with closure $\overline{T}$. Let $x \in H$ and $v: \mathbb{N} \to \text{dom}(T)$ be the sequence of iterates such that $v_0 = x$ and $v_{n+1} = T v_n$ for all $n$. Suppose that for some radius $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ is summable. Then for any $s \in \mathbb{R}$ such that $|s| < t/2$, the pair $$\left( \sum_{n=0}^\infty \frac{(is)^n}{n!} v_n, \quad -i \cdot \sum_{n=0}^\infty \frac{ni(is)^{n-1}}{n!} v_n \right)$$ belongs to the graph of the closure $\overline{T}$, where $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ a closable partially defined linear operator on $H$. Let $v: \mathbb{N} \to \text{dom}(T)$ be the sequence of iterates of a vector $x \in H$ under $T$ (satisfying $v_0 = x$ and $v_{n+1} = T v_n$). Suppose that for some $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ is summable. Then for any $s \in (-t/2, t/2)$, the local exponential orbit defined by the series $$\sum_{n=0}^\infty \frac{(is)^n}{n!} v_n$$ belongs to the domain of the closure of $T$ (denoted $\text{dom}(\overline{T})$).

Let $H$ be a complex Hilbert space and $T$ be a closable partially defined linear operator on $H$ with closure $\overline{T}$. Let $v: \mathbb{N} \to \text{dom}(T)$ be the sequence of iterates of a vector $x \in H$ under $T$, defined by $v_0 = x$ and $v_{n+1} = T v_n$. Suppose that for some $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ is summable. Then for any $s \in \mathbb{R}$ such that $|s| < t/2$, the local exponential orbit $\sum_{n=0}^\infty \frac{(is)^n}{n!} v_n$ belongs to the domain of $\overline{T}$, and the application of $\overline{T}$ to this vector satisfies: $$\overline{T} \left( \sum_{n=0}^\infty \frac{(is)^n}{n!} v_n \right) = -i \sum_{n=0}^\infty \frac{ni(is)^{n-1}}{n!} v_n$$ where $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ be a closable partially defined linear operator on $H$. Let $v: \mathbb{N} \to \text{dom}(T)$ be the sequence of iterates of a vector $x \in H$ under $T$, such that $v_0 = x$ and $v_{n+1} = T v_n$ for all $n$. Suppose there exists a real number $t > 0$ such that the power series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ is summable. Then for any $s \in (-t/2, t/2)$, the local exponential orbit defined by $$\phi(s) = \sum_{n=0}^\infty \frac{(is)^n}{n!} v_n$$ is differentiable at $s$, and its derivative is given by $$\frac{d}{ds} \phi(s) = i \overline{T} \phi(s)$$ where $i$ is the imaginary unit and $\overline{T}$ denotes the closure of the operator $T$.

Let $\mathcal{H}$ be a complex Hilbert space and $T$ be a symmetric partially defined linear operator on $\mathcal{H}$. For any vector $x$ in the domain of $T$, the real part of the inner product of $x$ with $iTx$ is zero: $$\text{Re}(\langle x, iTx \rangle) = 0$$ where $i$ is the imaginary unit and $\langle \cdot, \cdot \rangle$ denotes the inner product on $\mathcal{H}$.

Let $T$ be a symmetric partially defined linear operator on a complex Hilbert space $H$, and suppose that the span of the set of analytic vectors for $T$ is dense in $H$. Let $x \in H$ be a vector and $v_n$ be its iterate sequence under $T$ (where $v_0 = x$ and $v_{n+1} = T v_n$). If there exists a real number $t > 0$ such that the power series $\sum_{n=0}^\infty \frac{\|v_n\|}{n!} t^n$ is summable, then for any $s \in (-t/2, t/2)$, the derivative of the squared norm of the local exponential orbit $$r \mapsto \left\| \sum_{n=0}^\infty \frac{(ir)^n}{n!} v_n \right\|^2$$ at $r = s$ is equal to $0$.

Let $H$ be a complex Hilbert space and $T$ be a symmetric partially defined linear operator on $H$. Suppose that the set of analytic vectors of $T$ spans a dense subspace of $H$. Let $x \in H$ be a vector and $v: \mathbb{N} \to \text{dom}(T)$ be its sequence of iterates, where $v_n = T^n x$. If for some $t > 0$ the series $\sum_{n=0}^{\infty} \frac{\|v_n\|}{n!} t^n$ converges, then for any $s \in (-t/2, t/2)$, the squared norm of the local exponential orbit satisfies: $$\left\| \sum_{n=0}^\infty \frac{(is)^n}{n!} v_n \right\|^2 = \|x\|^2$$ where $i$ is the imaginary unit and $\|\cdot\|$ denotes the norm on $H$.

Let $H$ be a complex Hilbert space and $T$ be a symmetric partially defined linear operator on $H$. Suppose that the subspace spanned by the analytic vectors of $T$ is dense in $H$. Let $x \in H$ be a vector and $v_n$ be the sequence of its iterates under $T$, i.e., $v_0 = x$ and $v_{n+1} = T v_n$. If there exists a real number $t > 0$ such that the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ converges, then for any $s \in (-t/2, t/2)$, the norm of the local exponential orbit of $x$ is equal to the norm of $x$: $$\left\| \sum_{n=0}^\infty \frac{(is)^n}{n!} v_n \right\| = \|x\|$$ where $i$ is the imaginary unit and $n!$ is the factorial of $n$.

Let $H$ be a complex Hilbert space and $T$ a partially defined symmetric linear operator on $H$. Suppose that the span of the analytic vectors of $T$ is dense in $H$. Let $x \in H$ be a vector and $v_n$ be its sequence of iterates under $T$ (i.e., $v_0 = x$ and $v_n = T^n x$). Suppose there exists a real number $t > 0$ such that the series $$\sum_{n=0}^{\infty} \frac{\|v_n\|}{n!} t^n$$ is summable. Then for any $s \in (-t/2, t/2)$, the local exponential orbit series $$\sum_{n=0}^\infty \frac{(is)^n}{n!} v_n$$ is equal to zero if and only if $x = 0$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. If a vector $x \in H$ is an entire vector for $T$, then $x$ is also an analytic vector for $T$.

Let $T$ be a partially defined linear operator on a complex Hilbert space $H$ and let $x \in H$ be a vector. Let $v: \mathbb{N} \to \text{dom}(T)$ be the sequence of iterates such that $v_0 = x$ and $v_{n+1} = T v_n$ for all $n \in \mathbb{N}$. If the operator $T$ is closable and the vector $x$ is an entire vector (i.e., the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ converges for all $t > 0$), then for any $s \in \mathbb{R}$, the local exponential orbit defined by the series $$\sum_{n=0}^\infty \frac{(is)^n}{n!} v_n$$ belongs to the domain of the closure of $T$.

Let $H$ be a complex Hilbert space and $T$ be a closable partially defined linear operator on $H$ with closure $\overline{T}$. Suppose $x \in H$ is an **entire vector** for $T$, meaning that $x$ lies in the domain of all powers $T^n$ and the series $\sum_{n=0}^\infty \frac{\|T^n x\| t^n}{n!}$ converges for all $t > 0$. Then for any $s \in \mathbb{R}$, the exponential orbit defined by the series $$e^{isT}x := \sum_{n=0}^\infty \frac{(is)^n}{n!} T^n x$$ belongs to the domain of the closure $\overline{T}$.

Let $H$ be a complex Hilbert space and $T$ be a symmetric, partially defined linear operator on $H$. Suppose that the subspace spanned by the analytic vectors of $T$ is dense in $H$. If $x \in H$ is an entire vector for $T$, let $v: \mathbb{N} \to \text{dom}(T)$ be the sequence of iterates such that $v_0 = x$ and $v_{n+1} = T v_n$. Then for any real number $s$, the norm of the local exponential orbit is equal to the norm of $x$: $$\left\| \sum_{n=0}^\infty \frac{(is)^n}{n!} v_n \right\| = \|x\|$$ where $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ be a closable partially defined linear operator on $H$ with closure $\overline{T}$. Let $x \in H$ be an entire vector for $T$, such that the sequence of iterates $v_n = T^n x$ satisfies the condition that $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ converges for all $t > 0$. For any $s \in \mathbb{R}$, define the local exponential orbit by the series: $$\psi(s) = \sum_{n=0}^\infty \frac{(is)^n}{n!} v_n$$ Then the function $s \mapsto \psi(s)$ is differentiable at $s$, and its derivative is given by: $$\frac{d}{ds} \psi(s) = i \overline{T} \psi(s)$$ where $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ be a closable, partially defined linear operator on $H$ with closure $\bar{T}$. Let $x \in H$ be an entire vector for $T$ with a corresponding sequence of iterates $v: \mathbb{N} \to \text{dom}(T)$ such that $v_0 = x$ and $v_{n+1} = T v_n$. Let $y \in H$ be a vector in the deficiency space $(\text{ran}(\bar{T} - iI))^\perp$, where $i$ is the imaginary unit and $I$ is the identity operator. For any $s \in \mathbb{R}$, define the local exponential orbit $e^{isT}x$ as the sum $\sum_{n=0}^\infty \frac{(is)^n}{n!} v_n$. Then the derivative of the complex inner product $f(s) = \langle y, e^{isT}x \rangle$ with respect to $s$ is given by: $$\frac{d}{ds} \langle y, e^{isT}x \rangle = -\langle y, e^{isT}x \rangle$$

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Suppose that $T$ is symmetric, closable, and the subspace spanned by its analytic vectors is dense in $H$. Let $x \in H$ be an entire vector for $T$, meaning that $x$ is in the domain of $T^n$ for all $n \in \mathbb{N}$ and for any $t > 0$, the series $\sum_{n=0}^{\infty} \frac{\|T^n x\|}{n!} t^n$ is summable. If $y \in H$ is a vector in the orthogonal complement of the range of $\bar{T} - iI$ (where $\bar{T}$ is the closure of $T$ and $I$ is the identity operator), then the inner product of $y$ and $x$ is zero: $$\langle y, x \rangle = 0$$

Let $H$ be a complex Hilbert space and $T$ be a closable partially defined linear operator on $H$. Let $x \in H$ and $v: \mathbb{N} \to \text{dom}(T)$ be the sequence of iterates of $x$ under $T$ (i.e., $v_0 = x$ and $v_{n+1} = T v_n$). Suppose $x$ is an entire vector, meaning that for all $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ is summable. Let $y \in H$ be a vector in the orthogonal complement of the range of the operator $\overline{T} + iI$, where $\overline{T}$ is the closure of $T$ and $I$ is the identity operator. Then, for any $s \in \mathbb{R}$, the function $f(r) = \langle y, \sum_{n=0}^\infty \frac{(ir)^n}{n!} v_n \rangle$ is differentiable at $s$ with derivative $$f'(s) = \left\langle y, \sum_{n=0}^\infty \frac{(is)^n}{n!} v_n \right\rangle$$

Let $H$ be a complex Hilbert space and $T$ be a symmetric, closable partially defined linear operator on $H$. Assume that the span of the set of analytic vectors for $T$ is dense in $H$. Let $x \in H$ be a vector with an iterate sequence $v_n$ (defined by $v_0 = x$ and $v_{n+1} = T v_n$) such that for every $t > 0$, the series $$\sum_{n=0}^{\infty} \frac{\|v_n\|}{n!} t^n$$ is summable. Let $\overline{T}$ be the closure of $T$. If $y \in H$ belongs to the orthogonal complement of the range of the operator $\overline{T} + iI$ (where $i$ is the imaginary unit and $I$ is the identity operator), then the inner product of $y$ and $x$ is zero: $$\langle y, x \rangle = 0$$