Math 401, Fall 2025: Thesis notes, S3, Coherent states and POVMs

This section should extends on the reading for

Holomorphic methods in analysis and mathematical physics

Bargmann space (original)

Also known as Segal-Bargmann space or Bargmann-Fock space.

It is the space of holomorphic functions that is square-integrable over the complex plane.

Section belows use Remarks on a Hilbert Space of Analytic Functions as the reference.

A family of Hilbert spaces, $\mathfrak{F}_n(n=1,2,3,\cdots)$ , is defined as follows:

The element of $\mathfrak{F}_n$ are entire analytic functions in complex Euclidean space $\mathbb{C}^n$ . $f:\mathbb{C}^n\to \mathbb{C}\in \mathfrak{F}_n$

Let $f,g\in \mathfrak{F}_n$ . The inner product is defined by

\langle f,g\rangle=\int_{\mathbb{C}^n} \overline{f(z)}g(z) d\mu_n(z)

Let $z_k=x_k+iy_k$ be the complex coordinates of $z\in \mathbb{C}^n$ .

The measure $\mu_n$ is the defined by

d\mu_n(z)=\pi^{-n}\exp(-\sum_{i=1}^n |z_i|^2)\prod_{k=1}^n dx_k dy_k

Example

For $n=2$ ,

\mathfrak{F}_2=\text{ space of entire analytic functions on } \mathbb{C}^2\to \mathbb{C}

\langle f,g\rangle=\int_{\mathbb{C}^2} \overline{f(z)}g(z) d\mu(z),z=(z_1,z_2)

d\mu_2(z)=\frac{1}{\pi^2}\exp(-|z|^2)dx_1 dy_1 dx_2 dy_2

so that $f$ belongs to $\mathfrak{F}_n$ if and only if $\langle f,f\rangle<\infty$ .

This is absolutely terrible early texts, we will try to formulate it in a more modern way.

The section belows are from the lecture notes Holomorphic method in analysis and mathematical physics

Complex function spaces

Holomorphic spaces

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 $U$ .

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.

Let $\alpha$ be a continuous, strictly positive function on $U$ .

\mathcal{H}L^2(U,\alpha)=\left\{F\in \mathcal{H}(U): \int_U |F(z)|^2 \alpha(z) d\mu(z)<\infty\right\},

where $\mu$ is the Lebesgue measure on $\mathbb{C}^d=\mathbb{R}^{2d}$ .

Theorem of holomorphic spaces

For all $z\in U$ , there exists a constant $c_z$ such that $|F(z)|^2\le c_z \|F\|^2_{L^2(U,\alpha)}$ for all $F\in \mathcal{H}L^2(U,\alpha)$ .
$\mathcal{H}L^2(U,\alpha)$ is a closed subspace of $L^2(U,\alpha)$ , and therefore a Hilbert space.

Proof

First we check part 1.

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

P_s(z)=\{v\in \mathbb{C}^d: |v_k-z_k|<s, k=1,2,\cdots,d\}

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.

If $d=1$ . Then by Taylor series at $v=z$ , since $F$ is analytic in $U$ we have

F(v)=F(z)+\sum_{k=1}^{\infty}a_n(v-z)^n

Since the series converges uniformly to $F$ on the compact set $\overline{P_s(z)}$ , we can interchange the integral and the sum.

Using polar coordinates with origin at $z$ , $(v-z)^n=r^n e^{in\theta}$ where $r=|v-z|, \theta=\arg(v-z)$ .

For $n\geq 1$ , the integral over $P_s(z)$ (open disk) is zero (by Cauchy’s theorem).

So,

\begin{aligned} F(z)&=(\pi s^2)^{-1}\int_{P_s(z)}F(z)+\sum_{k=1}^{\infty}a_n(v-z)^n d\mu(v)\\ &=(\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)\\ &=(\pi s^2)^{-1}F(z) \end{aligned}

For $d>1$ , we can use the same argument to show that

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 $P_s(z)$ .

\begin{aligned} 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)\\ &=(\pi s^2)^{-d}\langle \mathbb{I}_{P_s(z)}\frac{1}{\alpha},F\rangle_{L^2(U,\alpha)} \end{aligned}

By definition of inner product.

So $\|F(z)\|^2\leq (\pi s^2)^{-2d}\|\mathbb{I}_{P_s(z)}\frac{1}{\alpha}\|^2_{L^2(U,\alpha)} \|F\|^2_{L^2(U,\alpha)}$ .

All the terms are bounded and finite.

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

|F(z)|^2\leq d_z \|F\|^2_{L^2(U,\alpha)}

Suppose we have a sequence $F_n\in \mathcal{H}L^2(U,\alpha)$ such that $F_n\to F$ , $F\in L^2(U,\alpha)$ .

Then $F_n$ is a cauchy sequence in $L^2(U,\alpha)$ . So,

\sup_{v\in V}|F_n(v)-F_m(v)|\leq \sqrt{d_z}\|F_n-F_m\|_{L^2(U,\alpha)}\to 0\text{ as }n,m\to \infty

So the sequence $F_m$ converges locally uniformly to some limit function which must be $F$ ( $\mathbb{C}^d$ is Hausdorff, unique limit point).

Locally uniform limit of holomorphic functions is holomorphic. (Use Morera’s Theorem to show that the limit is still holomorphic in each variable.) So the limit function $F$ is actually in $\mathcal{H}L^2(U,\alpha)$ , which shows that $\mathcal{H}L^2(U,\alpha)$ is closed.

which shows that $\mathcal{H}L^2(U,\alpha)$ is closed.

Tip

[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.

Reproducing kernel

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:

$K(z,w)$ is holomorphic in $z$ and anti-holomorphic in $w$ .
$K(w,z)=\overline{K(z,w)}$
For each fixed $z\in U$ , $K(z,w)$ is a square integrable $d\alpha(w)$ . For all $F\in \mathcal{H}L^2(U,\alpha)$ ,
$F(z)=\int_U K(z,w)F(w) \alpha(w) dw$
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
$PF(z)=\int_U K(z,w)F(w) \alpha(w) dw$
For all $z,u\in U$ ,
$\int_U K(z,w)K(w,u) \alpha(w) dw=K(z,u)$
For all $z\in U$ ,
$|F(z)|^2\leq K(z,z) \|F\|^2_{L^2(U,\alpha)}$

Proof

For part 1, By Riesz Theorem, the linear functional evaluation at $z\in U$ on $\mathcal{H}L^2(U,\alpha)$ can be represented uniquely as inner product with some $\phi_z\in \mathcal{H}L^2(U,\alpha)$ .

F(z)=\langle F,\phi_z\rangle_{L^2(U,\alpha)}=\int_U F(w)\overline{\phi_z(w)} \alpha(w) dw

And assume part 2 is true, then we have

$K(z,w)=\overline{\phi_z(w)}$

So part 1 is true.

For part 2, we can use the same argument

\phi_z(w)=\langle \phi_z,\phi_w\rangle_{L^2(U,\alpha)}=\overline{\langle \phi_w,\phi_z\rangle_{L^2(U,\alpha)}}=\overline{\phi_w(z)}

… continue if needed.

Construction of reproducing kernel

Let $\{e_j\}$ be any orthonormal basis of $\mathcal{H}L^2(U,\alpha)$ . Then for all $z,w\in U$ ,

\sum_{j=1}^{\infty} |e_j(z)\overline{e_j(w)}|<\infty

and

K(z,w)=\sum_{j=1}^{\infty} e_j(z)\overline{e_j(w)}

Bargmann space

The Bargmann spaces are the holomorphic spaces

\mathcal{H}L^2(\mathbb{C}^d,\mu_t)

where

\mu_t(z)=(\pi t)^{-d}\exp(-|z|^2/t)

For this research, we can tentatively set $t=1$ and $d=2$ for simplicity so that you can continue to read the next section.

Reproducing kernel for Bargmann space

For all $d\geq 1$ , the reproducing kernel of the space $\mathcal{H}L^2(\mathbb{C}^d,\mu_t)$ is given by

K(z,w)=\exp(z\cdot \overline{w}/t)

where $z\cdot \overline{w}=\sum_{k=1}^d z_k\overline{w_k}$ .

This gives the pointwise bounds

|F(z)|^2\leq \exp(\|z\|^2/t) \|F\|^2_{L^2(\mathbb{C}^d,\mu_t)}

For all $F\in \mathcal{H}L^2(\mathbb{C}^d,\mu_t)$ , and $z\in \mathbb{C}^d$ .

Proofs are intentionally skipped, you can refer to the lecture notes for details.

Lie bracket of vector fields

Let $X,Y$ be two vector fields on a smooth manifold $M$ . The Lie bracket of $X$ and $Y$ is an operator $[X,Y]:C^\infty(M)\to C^\infty(M)$ defined by

[X,Y](f)=X(Y(f))-Y(X(f))

This operator is a vector field.

Continue here for quantization of Coherent states and POVMs