184 lines
5.7 KiB
Markdown
184 lines
5.7 KiB
Markdown
# Math 401, Fall 2025: Thesis notes, S4, Bargmann space
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## Bargmann space (original)
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Also known as Segal-Bargmann space or Bargmann-Fock space.
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It is the space of [holomorphic functions](../../Math416/Math416_L3#definition-28-holomorphic-functions) that is square-integrable over the complex plane.
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> Section belows use [Remarks on a Hilbert Space of Analytic Functions](https://www.jstor.org/stable/71180) as the reference.
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A family of Hilbert spaces, $\mathfrak{F}_n(n=1,2,3,\cdots)$, is defined as follows:
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The element of $\mathfrak{F}_n$ are [entire](../../Math416/Math416_L13#definition-711) [analytic functions](../../Math416/Math416_L9#definition-analytic) in complex Euclidean space $\mathbb{C}^n$. $f:\mathbb{C}^n\to \mathbb{C}\in \mathfrak{F}_n$
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Let $f,g\in \mathfrak{F}_n$. The inner product is defined by
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$$
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\langle f,g\rangle=\int_{\mathbb{C}^n} \overline{f(z)}g(z) d\mu_n(z)
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$$
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Let $z_k=x_k+iy_k$ be the complex coordinates of $z\in \mathbb{C}^n$.
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The measure $\mu_n$ is the defined by
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$$
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d\mu_n(z)=\pi^{-n}\exp(-\sum_{i=1}^n |z_i|^2)\prod_{k=1}^n dx_k dy_k
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$$
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<details>
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<summary>Example</summary>
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For $n=2$,
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$$
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\mathfrak{F}_2=\text{ space of entire analytic functions on } \mathbb{C}^2\to \mathbb{C}
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$$
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$$
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\langle f,g\rangle=\int_{\mathbb{C}^2} \overline{f(z)}g(z) d\mu(z),z=(z_1,z_2)
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$$
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$$
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d\mu_2(z)=\frac{1}{\pi^2}\exp(-|z|^2)dx_1 dy_1 dx_2 dy_2
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$$
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</details>
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so that $f$ belongs to $\mathfrak{F}_n$ if and only if $\langle f,f\rangle<\infty$.
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This is absolutely terrible early texts, we will try to formulate it in a more modern way.
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> The section belows are from the lecture notes [Holomorphic method in analysis and mathematical physics](https://arxiv.org/pdf/quant-ph/9912054)
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## Complex function spaces
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### Holomorphic spaces
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Let $U$ be a non-empty open set in $\mathbb{C}^d$. Let $\mathcal{H}(U)$ be the space of holomorphic (or analytic) functions on $U$.
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Let $f\in \mathcal{H}(U)$, note that by definition of holomorphic on several complex variables, $f$ is continuous and holomorphic in each variable with the other variables fixed.
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Let $\alpha$ be a continuous, strictly positive function on $U$.
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$$
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\mathcal{H}L^2(U,\alpha)=\left\{F\in \mathcal{H}(U): \int_U |F(z)|^2 \alpha(z) d\mu(z)<\infty\right\},
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$$
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where $\mu$ is the Lebesgue measure on $\mathbb{C}^d=\mathbb{R}^{2d}$.
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#### Theorem of holomorphic spaces
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1. For all $z\in U$, there exists a constant $c_z$ such that
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$$
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|F(z)|^2\le c_z \|F\|^2_{L^2(U,\alpha)}
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$$
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for all $F\in \mathcal{H}L^2(U,\alpha)$.
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2. $\mathcal{H}L^2(U,\alpha)$ is a closed subspace of $L^2(U,\alpha)$, and therefore a Hilbert space.
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<details>
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<summary>Proof</summary>
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First we check part 1.
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Let $z=(z_1,z_2,\cdots,z_d)\in U, z_k\in \mathbb{C}$. Let $P_s(z)$ be the "polydisk"of radius $s$ centered at $z$ defined as
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$$
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P_s(z)=\{v\in \mathbb{C}^d: |v_k-z_k|<s, k=1,2,\cdots,d\}
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$$
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If $z\in U$, we cha choose $s$ small enough such that $\overline{P_s(z)}\subset U$ so that we can claim that $F(z)=(\pi s^2)^{-d}\int_{P_s(z)}F(v)d\mu(v)$ is well-defined.
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If $d=1$. Then by Taylor series at $v=z$, since $F$ is analytic in $U$ we have
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$$
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F(v)=F(z)+\sum_{k=1}^{\infty}a_n(v-z)^n
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$$
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Since the series converges uniformly to $F$ on the compact set $\overline{P_s(z)}$, we can interchange the integral and the sum.
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Using polar coordinates with origin at $z$, $(v-z)^n=r^n e^{in\theta}$ where $r=|v-z|, \theta=\arg(v-z)$.
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For $n\geq 1$, the integral over $P_s(z)$ (open disk) is zero (by Cauchy's theorem).
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So,
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$$
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\begin{aligned}
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F(z)&=(\pi s^2)^{-1}\int_{P_s(z)}F(z)+\sum_{k=1}^{\infty}a_n(v-z)^n d\mu(v)\\
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&=(\pi s^2)^{-1}F(z)+(\pi s^2)^{-1}\sum_{k=1}^{\infty}a_n\int_{P_s(z)}r^n e^{in\theta} d\mu(v)\\
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&=(\pi s^2)^{-1}F(z)
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\end{aligned}
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$$
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For $d>1$, we can use the same argument to show that
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Let $\mathbb{I}_{P_s(z)}(v)=\begin{cases}1 & v\in P_s(z) \\0 & v\notin P_s(z)\end{cases}$ be the indicator function of $P_s(z)$.
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$$
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\begin{aligned}
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F(z)&=(\pi s^2)^{-d}\int_{U}\mathbb{I}_{P_s(z)}(v)\frac{1}{\alpha(v)}F(v)\alpha(v) d\mu(v)\\
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&=(\pi s^2)^{-d}\langle \mathbb{I}_{P_s(z)}\frac{1}{\alpha},F\rangle_{L^2(U,\alpha)}
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\end{aligned}
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$$
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By definition of inner product.
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So $\|F(z)\|^2\leq (\pi s^2)^{-2d}\|\mathbb{I}_{P_s(z)}\frac{1}{\alpha}\|^2_{L^2(U,\alpha)} \|F\|^2_{L^2(U,\alpha)}$.
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All the terms are bounded and finite.
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For part 2, we need to show that $\forall z\in U$, we can find a neighborhood $V$ of $z$ and a constant $d_z$ such that
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$$
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|F(z)|^2\leq d_z \|F\|^2_{L^2(U,\alpha)}
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$$
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</details>
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> [!TIP]
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>
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> [1.] states that point-wise evaluation of $F$ on $U$ is continuous. That is, for each $z\in U$, the map $\varphi: \mathcal{H}L^2(U,\alpha)\to \mathbb{C}$ that takes $F\in \mathcal{H}L^2(U,\alpha)$ to $F(z)$ is a continuous linear functional on $\mathcal{H}L^2(U,\alpha)$. This is false for ordinary non-holomorphic functions, e.g. $L^2$ spaces.
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#### Reproducing kernel
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Let $\mathcal{H}L^2(U,\alpha)$ be a holomorphic space. The reproducing kernel of $\mathcal{H}L^2(U,\alpha)$ is a function $K:U\times U\to \mathbb{C}$, $K(z,w),z,w\in U$ with the following properties:
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1. $K(z,w)$ is holomorphic in $z$ and anti-holomorphic in $w$.
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$$
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K(w,z)=\overline{K(z,w)}
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$$
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2. For each fixed $z\in U$, $K(z,w)$ is a square integrable $d\alpha(w)$. For all $F\in \mathcal{H}L^2(U,\alpha)$,
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$$
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F(z)=\int_U K(z,w)F(w) \alpha(w) dw
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$$
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3. If $F\in L^2(U,\alpha)$, let $PF$ denote the orthogonal projection of $F$ onto closed subspace $\mathcal{H}L^2(U,\alpha)$. Then
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$$
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PF(z)=\int_U K(z,w)F(w) \alpha(w) dw
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$$
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4. For all $z,u\in U$,
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$$
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\int_U K(z,w)K(w,u) \alpha(w) dw=K(z,u)
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$$
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5. For all $z\in U$,
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$$
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|F(z)|^2\leq K(z,z) \|F\|^2_{L^2(U,\alpha)}
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$$
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### Bargmann space
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The Bargmann spaces are the holomorphic spaces
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$$
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\mathcal{H}L^2(\mathbb{C}^d,\mu_t)
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$$
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where
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$$
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\mu_t(z)=\text{ CONTINUE HERE }
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$$ |