PhyslibAlpha.AlgebraicFramework.HilbertSpace.Unbounded.AnalyticVector.Local

For a partially defined linear operator $T$ on a complex Hilbert space $H$ and a vector $x \in H$, an instance of a local analytic orbit $U$ (of type `LocalAnalyticOrbit T x`) can be treated as a function from the real numbers $\mathbb{R}$ to the Hilbert space $H$. This allows the notation $U(t)$ to represent the state of the orbit at "time" $t$.

Given an unbounded linear operator $T$ on a Hilbert space $H$ and a local analytic orbit $U$ of $T$ starting at a vector $x \in H$ with radius $R$, for any $a \in \mathbb{R}$ satisfying $|a| < R$, the translation defines a new local analytic orbit starting at $U(a)$. This new orbit has a radius $R - |a|$ and is defined by the mapping $s \mapsto U(a + s)$.

Let $H$ be a Hilbert space and $T$ be a complex linear operator on $H$. Let $x \in H$ be a vector and $U$ be a local analytic orbit of $T$ starting at $x$ with radius $R$. For any $a \in \mathbb{R}$ such that $|a| < R$, there exists a local analytic orbit of $T$ starting at the point $U(a)$.

Let $H$ be a complex Hilbert space and $T$ be an unbounded linear operator on $H$. Given a local analytic orbit $U$ of $T$ starting at $x \in H$ with radius $R$, and a positive real number $r$ such that $r \leq R$, the restriction $U|_r$ is a local analytic orbit starting at $x$ with radius $r$, defined by restricting the domain of the original orbit $U$ to the interval $(-r, r)$.

Let $H$ be a Hilbert space and $T$ be a partially defined linear operator on $H$. Let $x \in H$ be a vector and $U$ be a local analytic orbit for $T$ starting at $x$ with a radius of convergence $R$. For any real number $r$ such that $0 < r \leq R$, there exists a local analytic orbit for $T$ starting at $x$ restricted to the smaller symmetric radius $r$.

Let $\mathcal{H}$ be a complex Hilbert space and $T$ be a partially defined linear operator on $\mathcal{H}$ such that its closure $\overline{T}$ is symmetric. Let $x \in \mathcal{H}$ and let $U$ be a local analytic orbit for $T$ starting at $x$ with a radius of convergence denoted by $U.\text{radius}$. Suppose $\delta$ and $R$ are positive real numbers such that $\delta + R \leq U.\text{radius}$. If $V$ is a local analytic orbit for $T$ starting at the state $U(\delta)$ with a radius of convergence at least $R$ (i.e., $V.\text{radius} \geq R$), then for any $s \in \mathbb{R}$ satisfying $|s| < R$, the orbits agree: $$U(\delta + s) = V(s)$$

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ such that its closure $\overline{T}$ is symmetric. Let $x \in H$ and let $U$ be a local analytic orbit of $x$ under $T$ with radius $r_U$. Given $a \in \mathbb{R}$ and $R > 0$ such that $|a| + R \leq r_U$, let $V$ be a local analytic orbit of the state $U(a)$ under $T$ with radius $r_V \geq R$. Then for any $s \in \mathbb{R}$ such that $|s| < R$, it holds that $$U(a + s) = V(s)$$

Let $H$ be a Hilbert space and $T$ be a partially defined linear operator on $H$. If $T$ is closable and there exists a local analytic orbit of a vector $x \in H$ under the closure $\overline{T}$, then there exists a local analytic orbit of $x$ under $T$.

Given a global analytic orbit $U$ of a vector $x$ under a partially defined linear operator $T$ on a complex Hilbert space $H$, this instance allows $U$ to be treated as a function that maps a real number $t \in \mathbb{R}$ to a vector $U(t) \in H$.

For any real number $r$ such that $0 \le r < 1$ and any natural number $k$, there exists a constant $C \in \mathbb{R}$ such that for all $n \in \mathbb{N}$, the inequality $\binom{n+k}{k} r^n \le C$ holds.

Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of non-negative real numbers. Let $t$ and $q$ be real numbers such that $t > 0$ and $0 \leq q < t$. For any natural number $k$, if the series $\sum_{n=0}^{\infty} \frac{a_n t^n}{n!}$ converges, then the shifted series $\sum_{n=0}^{\infty} \frac{a_{n+k} q^n}{n!}$ also converges.

Given a sequence $v: \mathbb{N} \to \text{dom}(T)$ of vectors in the domain of a partial linear operator $T$ and a natural number $k$, this function defines the sequence shifted by $k$ steps, mapping $n$ to $v(n+k)$.

Let $H$ be a complex Hilbert space and $T: \text{dom}(T) \to H$ be a partial linear operator. Let $x \in H$ and $v: \mathbb{N} \to \text{dom}(T)$ be a sequence of vectors. If $v$ is the sequence of iterates of $T$ starting from $x$ (that is, $v_0 = x$ and $v_{n+1} = T v_n$ for all $n \in \mathbb{N}$), then for any natural number $k$, the shifted sequence $(v_{n+k})_{n \in \mathbb{N}}$ is also a sequence of iterates of $T$, starting from the initial vector $v_k$.

Let $H$ be a complex Hilbert space and $T$ a partial linear operator on $H$. Let $(v_n)_{n \in \mathbb{N}}$ be a sequence of vectors in the domain of $T$. If $t > 0$ and the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ converges, then for any natural number $k$ and any real number $0 \le q < t$, the series of shifted iterates $\sum_{n=0}^\infty \frac{\|v_{n+k}\| q^n}{n!}$ also converges.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Let $(v_n)_{n \in \mathbb{N}}$ be the sequence of iterates of $T$ starting from a vector $x \in H$, such that $v_0 = x$ and $v_{n+1} = T v_n$. Suppose that for some radius $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ converges. For any natural number $k$ and any real number $s \in (-t/2, t/2)$, let $\mathcal{E}_k(s) = \sum_{n=0}^\infty \frac{(is)^n}{n!} v_{n+k}$ be the analytic expansion (orbit) starting from the $k$-th iterate. Then the sum of the term-by-term derivatives of the expansion $\mathcal{E}_k(s)$ satisfies: $$-i \sum_{n=0}^\infty \frac{d}{ds} \left( \frac{(is)^n}{n!} v_{n+k} \right) = \mathcal{E}_{k+1}(s)$$ where $i$ is the imaginary unit and $\mathcal{E}_{k+1}(s)$ is the analytic expansion starting from the $(k+1)$-th iterate.

Let $H$ be a complex Hilbert space and $T$ be a symmetric partially defined linear operator on $H$. Let $x \in H$ be a vector and $v_n = T^n x$ be the sequence of its iterates under $T$. Suppose that the set of analytic vectors of $T$ is dense in $H$. If for some radius $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ converges, then for any $a \in \mathbb{R}$ satisfying $|a| < \frac{t}{2}$, there exists a local analytic orbit (a `LocalAnalyticOrbit`) for $T$ starting at the vector $\sum_{n=0}^\infty \frac{a^n}{n!} v_n$.

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 $v_n = T^n x$ be the sequence of its iterates under $T$. Suppose that for some radius $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ converges. Then for any $a \in \mathbb{R}$ satisfying $|a| < t/2$, the vector $$y = \sum_{n=0}^\infty \frac{a^n}{n!} v_n$$ is an analytic vector for the closure $\overline{T}$ of the operator $T$.

Let $H$ be a complex Hilbert space and $T$ be a symmetric partially defined linear operator on $H$ such that the subspace of analytic vectors for $T$ is dense in $H$. Let $x \in H$ be a vector and $v_n = T^n x$ be the sequence of its iterates under $T$. Suppose that for some radius $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ converges. For any translation $a \in \mathbb{R}$ satisfying $|a| < t/2$, let $y = \exp(aT)x = \sum_{n=0}^\infty \frac{a^n v_n}{n!}$. Then for any $0 < q < t$, the vector $y$ is an analytic vector for the closure $\overline{T}$ with radius at least $q$. Specifically, there exists a sequence $w_n = \overline{T}^n y$ such that the series $\sum_{n=0}^\infty \frac{\|w_n\| q^n}{n!}$ converges.

Let $H$ be a complex Hilbert space and $T$ be a symmetric partially defined linear operator on $H$ such that the subspace of analytic vectors for $T$ is dense in $H$. Let $x \in H$ be a vector and $v_n = T^n x$ be the sequence of its iterates under $T$. Suppose that for some radius $t > 0$, the series $\sum_{n=0}^\infty \frac{\|v_n\| t^n}{n!}$ converges. For any translation $a \in \mathbb{R}$ satisfying $|a| < t/2$ and any radius $0 < q < t$, the vector $y = \exp(aT)x = \sum_{n=0}^\infty \frac{a^n v_n}{n!}$ is an analytic vector for the closure $\overline{T}$ of the operator $T$.

Let $\delta, R$, and $s$ be real numbers. If $0 < \delta < R$, then there exists an integer $n$ such that $|s - n\delta| < R$.

Given an unbounded linear operator $T$ on a complex Hilbert space $H$ and an initial vector $x \in H$, this definition constructs a `LocalOrbitCoreCover` for $T$ and $x$ using a family of local orbits indexed by the integers $\mathbb{Z}$. Suppose we are given a sequence of states $s_n \in H$, center points $c_n \in \mathbb{R}$, and core radii $r_n \in \mathbb{R}$ for each $n \in \mathbb{Z}$, along with local analytic orbits $\phi_n$ originating at $s_n$. The construction requires that: 1. Each core radius is positive ($r_n > 0$) and does not exceed the radius of convergence of the corresponding local orbit $\phi_n$. 2. The sequence is rooted at $x$: $c_0 = 0$ and $s_0 = x$. 3. The norm is preserved: $\|s_n\| = \|x\|$ for all $n$. 4. The family of open intervals $(c_n - r_n, c_n + r_n)$ covers the real line $\mathbb{R}$. 5. The charts are compatible: if a time $s \in \mathbb{R}$ lies in the core of both chart $m$ and chart $n$, then $\phi_m(s - c_m) = \phi_n(s - c_n)$. The resulting object reindexes these integer-indexed components into a sequence indexed by natural numbers $\mathbb{N}$ using a standard bijection between $\mathbb{Z}$ and $\mathbb{N}$.

Let $T$ be an unbounded linear operator on a complex Hilbert space $H$ and let $x \in H$. This definition constructs a `LocalOrbitCoreCover` for the vector $x$ under $T$ by gluing together a sequence of local analytic orbits indexed by integers $n \in \mathbb{Z}$. Given a step size $\delta$ and a core radius $R$ such that $0 < \delta < R$, let $s_n \in H$ be a sequence of states and $\phi_n$ be local analytic orbits of $T$ starting at $s_n$ with radius of convergence at least $R$. If the following conditions are met: 1. The sequence starts at $x$: $s_0 = x$. 2. The norm is preserved: $\|s_n\| = \|x\|$ for all $n \in \mathbb{Z}$. 3. Adjacent orbits are compatible: for every $n \in \mathbb{Z}$, $\phi_n(\delta + z) = \phi_{n+1}(z)$ holds for all $|z| < R$. Then these local orbits can be extended to cover the entire real line $\mathbb{R}$ using centers $c_n = n\delta$, forming a global certificate for the analytic evolution of $x$.

Let $H$ be a complex Hilbert space and $T$ be an unbounded linear operator on $H$. Given a `LocalOrbitCoreCover` $C$ for a vector $x \in H$—which consists of a sequence of states $s_n \in H$, centers $c_n \in \mathbb{R}$, local analytic orbits $\phi_n$ with radii $R_n$, and core radii $r_n$ satisfying $0 < r_n \leq R_n$—this function constructs a `LocalOrbitCover`. It achieves this by restricting each chart $\phi_n$ to the symmetric interval $(-r_n, r_n)$, effectively narrowing the radius of each local orbit to match its core radius $r_n$ while preserving the underlying states, centers, and compatibility conditions.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Given a local orbit cover $C$ for a vector $x \in H$, which consists of a collection of local analytic orbits $\phi_n$ centered at points $c_n$ that together cover the real line $\mathbb{R}$, this definition constructs a global analytic orbit $\Phi: \mathbb{R} \to H$. For any $s \in \mathbb{R}$, the value $\Phi(s)$ is obtained by choosing a chart $n$ such that $|s - c_n| < R_n$ (where $R_n$ is the radius of the $n$-th chart) and evaluating $\Phi(s) = \phi_n(s - c_n)$. The compatibility conditions of the cover ensure that this function is well-defined, satisfies the initial condition $\Phi(0) = x$, remains in the domain of $T$, is differentiable, and preserves the norm $\|\Phi(s)\| = \|x\|$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Given a `LocalOrbitCoreCover` $C$ for a vector $x \in H$—which is a collection of local analytic orbits and associated symmetric cores $(-r_n, r_n)$ that together cover the real line $\mathbb{R}$—this definition constructs a `GlobalAnalyticOrbit` for $T$ starting at $x$. The construction proceeds by first restricting each chart in the core cover to its symmetric core (thereby forming a `LocalOrbitCover`) and then gluing these restricted local orbits into a single continuous, norm-preserving global function $\Phi: \mathbb{R} \to H$ that satisfies the initial condition $\Phi(0) = x$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. For any vector $x \in H$, if there exists a `LocalOrbitCoreCover` $C$ for $T$ at $x$—defined as a consistent collection of local analytic orbits on symmetric cores that cover the real line $\mathbb{R}$—then there exists at least one global analytic orbit for $T$ starting at $x$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. For any vector $x \in H$, if there exists a `LocalOrbitCover` $C$ for $T$ at $x$—defined as a consistent collection of local analytic orbits that cover the real line $\mathbb{R}$—then the set of global analytic orbits `GlobalAnalyticOrbit T x` is non-empty.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. If $T$ is closable and there exists a global analytic orbit for the closure $\bar{T}$ starting at a vector $x \in H$, then there exists a global analytic orbit for $T$ starting at $x$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Suppose $x \in H$ is a vector that admits a global analytic orbit $U$ for $T$. If $y \in H$ is a vector in the orthogonal complement of the range of the operator $\bar{T} - iI$, where $\bar{T}$ is the closure of $T$ and $i$ is the imaginary unit, then the inner product $\langle y, x \rangle = 0$.

Let $H$ be a complex Hilbert space and $T$ a partially defined linear operator on $H$. Suppose $x \in H$ is a vector that admits a global analytic orbit for $T$. If $y \in H$ is a vector in the orthogonal complement of the range of the operator $\bar{T} + iI$ (where $\bar{T}$ is the closure of $T$, $i$ is the imaginary unit, and $I$ is the identity operator), then the inner product of $y$ and $x$ is zero, i.e., $\langle y, x \rangle = 0$.