The Segal-Bargmann transform and the Segal-Bargmann space

Jordan Bell

July 31, 2015

1 The Fourier transform

Let $dm_{n}(x)=(2\pi)^{-n/2}dx$ . For Borel measurable functions $f,g:\mathbb{R}^{n}\to\mathbb{C}$ , when $y\mapsto f(x-y)g(y)$ is integrable we define

(f*g)(x)=\int_{\mathbb{R}^{n}}f(x-y)g(y)dm_{n}(y).

For $f\in L^{1}$ ,

\hat{f}(\xi)=(\mathscr{F}f)(\xi)=\int_{\mathbb{R}^{n}}f(x)e^{-i\left\langle\xi% ,x\right\rangle}dm_{n}(x),\qquad\xi\in\mathbb{R}^{n}.

For $f,g\in L^{1}$ , for almost all $x\in\mathbb{R}^{n}$ , $y\mapsto f(x-y)g(y)$ is integrable,¹¹ 1 Walter Rudin, Real and Complex Analysis, third ed., p. 170, Theorem 8.14. and using Fubini’s theorem one checks that

\widehat{f*g}=\widehat{f}\widehat{g}.

Let $\mathscr{S}$ be the Schwartz functions $\mathbb{R}^{n}\to\mathbb{C}$ . For a multi-index $\alpha$ and $\phi\in\mathscr{S}$ define $X^{\alpha}\phi:\mathbb{R}^{n}\to\mathbb{C}$ by

(X^{\alpha}\phi)(x)=x^{\alpha}\phi(x).

Define $\Delta\phi:\mathbb{R}^{n}\to\mathbb{C}$ by

(\Delta\phi)(x)=\sum_{j=1}^{n}(\partial_{j}^{2}\phi)(x).

One proves that

\mathscr{F}D^{\alpha}=i^{|\alpha|}X^{\alpha}\mathscr{F},\qquad D^{\alpha}% \mathscr{F}=(-i)^{|\alpha|}\mathscr{F}X^{\alpha}

and

\mathscr{F}(\Delta\phi)(\xi)=-|\xi|^{2}(\mathscr{F}\phi)(\xi).

Parseval’s formula states that for $f,g\in L^{2}$ ,

\left\langle f,g\right\rangle_{L^{2}}=\int_{\mathbb{R}^{n}}f\overline{g}dm_{n}% =\int_{\mathbb{R}^{n}}(\mathscr{F}f)\overline{(\mathscr{F}g)}dm_{n}=\left% \langle\mathscr{F}f,\mathscr{F}g\right\rangle_{L^{2}},

thus

\left\|f\right\|_{L^{2}}^{2}=\int_{\mathbb{R}^{n}}|f|^{2}dm_{n}=\int_{\mathbb{% R}^{n}}|\mathscr{F}f|^{2}dm_{n}=\left\|\mathscr{F}f\right\|_{L^{2}}^{2}.

For $z\in\mathbb{C}^{n}$ , using Cauchy’s integral theorem we obtain

\int_{\mathbb{R}^{n}}F(x+iy)e^{-i\left\langle\xi,x\right\rangle}dx=e^{-\left% \langle\xi,y\right\rangle}\int_{\mathbb{R}^{n}}F(x)e^{-i\left\langle\xi,x% \right\rangle}dx.

(1)

2 The heat kernel

For $t\geq 0$ and $f\in L^{2}$ , define $H_{t}f:\mathbb{R}^{n}\to\mathbb{C}$ by

(H_{t}f)(x)=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}\widehat{f}(\xi)e^{-t|\xi|^{2}}e% ^{i\left\langle\xi,x\right\rangle}dm_{n}(\xi).

For $t\in\mathbb{R}_{>0}$ let

h_{t}(x)=(4\pi t)^{-n/2}e^{-\frac{|x|^{2}}{4t}},\qquad x\in\mathbb{R}^{n},

and we calculate

\partial_{t}h_{t}=(4\pi t)^{-n/2}e^{-\frac{|x|^{2}}{4t}}\left(-\frac{n}{2t}+% \frac{|x|^{2}}{4t^{2}}\right)=\Delta h_{t},

which yields

\partial_{t}(f*h_{t})=f*(\partial_{t}h_{t})=f*(\Delta h_{t})=\Delta(f*h_{t}).

The Fourier transform of $h_{t}$ is²² 2 http://individual.utoronto.ca/jordanbell/notes/stationaryphase.pdf, Theorem 2.

	$\displaystyle\widehat{h}_{t}(\xi)$	$\displaystyle=\int_{\mathbb{R}^{n}}(4\pi t)^{-n/2}e^{-\frac{\|x\|^{2}}{4t}}e^{-i% \left\langle\xi,x\right\rangle}dm_{n}(x)$
		$\displaystyle=(4\pi t)^{-n/2}\cdot(2\pi)^{-n/2}(4\pi t)^{n/2}\exp(-t\|\xi\|^{2})$
		$\displaystyle=(2\pi)^{-n/2}\exp(-t\|\xi\|^{2}).$

Using $\widehat{h_{t}*f}=\widehat{p}_{t}\cdot\widehat{f}$ and the Fourier inversion theorem,

	$\displaystyle(h_{t}*f)(x)$	$\displaystyle=\int_{\mathbb{R}^{n}}\widehat{h_{t}*f}(\xi)e^{i\left\langle\xi,x% \right\rangle}dm_{n}(\xi)$
		$\displaystyle=\int_{\mathbb{R}^{n}}\widehat{p}_{t}(\xi)\widehat{f}(\xi)e^{i% \left\langle\xi,x\right\rangle}dm_{n}(\xi)$
		$\displaystyle=\int_{\mathbb{R}^{n}}(2\pi)^{-n/2}\exp(-t\|\xi\|^{2})\widehat{f}(% \xi)e^{i\left\langle\xi,x\right\rangle}dm_{n}(\xi)$
		$\displaystyle=(H_{t}f)(x).$

For $t>0$ and for $z\in\mathbb{C}^{n}$ ,

(H_{t}f)(z)=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}\widehat{f}(\xi)e^{-t|\xi|^{2}}e% ^{i\left\langle\xi,z\right\rangle}dm_{n}(\xi)

and

h_{t}(z)=(4\pi t)^{-n/2}\exp\left(-\frac{z_{1}^{2}+\cdots+z_{n}^{2}}{4t}\right).

It is apparent that $h_{t}:\mathbb{C}^{n}\to\mathbb{C}$ is holomorphic. By the dominated convergence theorem,

\frac{dH_{t}f}{dz_{j}}(z)=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}\widehat{f}(\xi)e^% {-t|\xi|^{2}}i\xi_{j}e^{i\left\langle\xi,z\right\rangle}dm_{n}(\xi),

and $H_{t}f:\mathbb{C}^{n}\to\mathbb{C}$ is holomorphic.

3 The Segal-Bargmann transform and the Segal-Bargmann space

Let $\lambda_{n}$ be Lebesgue measure on $\mathbb{R}^{n}$ , for $t>0$ let

\omega_{t}(y)=t^{-n/2}e^{-\frac{|y|^{2}}{2t}},

and let $\mu_{t}$ be the Borel measure on $\mathbb{C}^{n}=\mathbb{R}^{n}\times\mathbb{R}^{n}$ whose density with respect to $\lambda_{n}\times\lambda_{n}$ is $x+iy\mapsto\omega_{t}(y)$ . We define $\mathcal{H}_{t}(\mathbb{C}^{n})$ to be the set of those holomorphic functions $F:\mathbb{C}^{n}\to\mathbb{C}$ satisfying

\left\|F\right\|_{\mathcal{H}_{t}}^{2}=\int_{\mathbb{C}^{n}}|F|^{2}d\mu_{t}<\infty,

and for $G,H\in\mathcal{H}_{t}$ we define

\left\langle F,G\right\rangle_{\mathcal{H}_{t}}=\int_{\mathbb{C}^{n}}F% \overline{G}d\mu_{t}=\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}F(x+iy)% \overline{G(x+iy)}\omega_{t}(y)dy\right)dx.

We call $\mathcal{H}_{t}$ the Segal-Bargmann space. It can be proved that it is a Hilbert space.

For $y\in\mathbb{R}^{n}$ write $g(x)=(H_{t}f)(x+iy)$ , and applying Parseval’s formula and (1) yields

	$\displaystyle\int_{\mathbb{R}^{n}}\|(H_{t}f)(x+iy)\|^{2}dm_{n}(x)$	$\displaystyle=\int_{\mathbb{R}^{n}}\|g(x)\|^{2}dm_{n}(x)$
		$\displaystyle=\int_{\mathbb{R}^{n}}\|\widehat{g}(\xi)\|^{2}dm_{n}(\xi)$
		$\displaystyle=\int_{\mathbb{R}^{n}}\|e^{-\left\langle\xi,y\right\rangle}% \widehat{H_{t}f}(\xi)\|^{2}dm_{n}(\xi).$

Using this with $\widehat{H_{t}f}=\widehat{h}_{t}\widehat{f}$ and then using Fubini’s theorem and an identity for Gaussian integrals³³ 3 http://individual.utoronto.ca/jordanbell/notes/stationaryphase.pdf, Theorem 3. we get

	$\displaystyle\left\\|H_{t}f\right\\|_{\mathcal{H}_{t}}^{2}$	$\displaystyle=\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}\|(H_{t}f)(x+iy)\|% ^{2}\omega_{t}dy\right)dx$
		$\displaystyle=(2\pi)^{n}\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}\|(H_{t% }f)(x+iy)\|^{2}dm_{n}(x)\right)\omega_{t}(y)dm_{n}(y)$
		$\displaystyle=(2\pi)^{n}\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}e^{-2% \left\langle\xi,y\right\rangle}\|\widehat{H_{t}f}(\xi)\|^{2}dm_{n}(\xi)\right)% \omega_{t}(y)dm_{n}(y)$
		$\displaystyle=(2\pi)^{n}\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}e^{-2% \left\langle\xi,y\right\rangle}(2\pi)^{-n}\exp(-2t\|\xi\|^{2})\|\widehat{f}(\xi)\|% ^{2}dm_{n}(\xi)\right)\omega_{t}(y)dm_{n}(y)$
		$\displaystyle=\int_{\mathbb{R}^{n}}\|\widehat{f}(\xi)\|^{2}\exp(-2t\|\xi\|^{2})% \left(t^{-n/2}\int_{\mathbb{R}^{n}}e^{-2\left\langle\xi,y\right\rangle}e^{-% \frac{\|y\|^{2}}{2t}}dm_{n}(y)\right)dm_{n}(x)$
		$\displaystyle=\int_{\mathbb{R}^{n}}\|\widehat{f}(\xi)\|^{2}\exp(-2t\|\xi\|^{2})% \cdot\exp(2t\|\xi\|^{2})dm_{n}(x)$
		$\displaystyle=\left\\|\mathscr{F}f\right\\|_{L^{2}}^{2}$
		$\displaystyle=\left\\|f\right\\|_{L^{2}}^{2}.$

Therefore $H_{t}:L^{2}(\mathbb{R}^{n})\to\mathcal{H}_{t}(\mathbb{C}^{n})$ is a linear isometry. We call $H_{t}$ the Segal-Bargmann transform. It can be proved that $H_{t}$ is a Hilbert space isomorphism.⁴⁴ 4 cf. https://www.math.lsu.edu/~olafsson/pdf_files/ht.pdf

For $F\in\mathcal{H}_{t}$ and $z\in\mathbb{C}^{n}$ , write

\mathrm{ev}_{z}(F)=F(z)

and

(T_{w}F)(z)=F(z-w).

For $f\in L^{2}(\mathbb{R}^{n})$ and $t>0$ let $F=H_{t}f\in\mathcal{H}_{t}(\mathbb{C}^{n})$ , and for $w\in\mathbb{C}^{n}$ , using

\overline{h_{t}(w-x)}=\overline{h_{t}(x-w)}=h_{t}(x-\overline{w})=(T_{% \overline{w}}h_{t})(x),

we get

	$\displaystyle\mathrm{ev}_{w}(F)$	$\displaystyle=(f*h_{t})(w)$
		$\displaystyle=\int_{\mathbb{R}^{n}}f(x)\overline{(T_{\overline{w}}h_{t})(x)}dm% _{n}(x)$
		$\displaystyle=\left\langle f,T_{\overline{w}}h_{t}\right\rangle_{L^{2}}$
		$\displaystyle=\left\langle H_{t}f,H_{t}T_{\overline{w}}h_{t}\right\rangle_{L^{% 2}}$
		$\displaystyle=\left\langle F,H_{t}T_{\overline{w}}h_{t}\right\rangle_{L^{2}}.$

Then $(w,z)\mapsto(H_{t}T_{\overline{w}}h_{t})(z)$ is a reproducing kernel for the Hilbert space $\mathcal{H}_{t}$ .

	$\displaystyle\widehat{h}_{t}(\xi)$	$\displaystyle=\int_{\mathbb{R}^{n}}(4\pi t)^{-n/2}e^{-\frac{\|x\|^{2}}{4t}}e^{-i% \left\langle\xi,x\right\rangle}dm_{n}(x)$
		$\displaystyle=(4\pi t)^{-n/2}\cdot(2\pi)^{-n/2}(4\pi t)^{n/2}\exp(-t\|\xi\|^{2})$
		$\displaystyle=(2\pi)^{-n/2}\exp(-t\|\xi\|^{2}).$

	$\displaystyle\int_{\mathbb{R}^{n}}\|(H_{t}f)(x+iy)\|^{2}dm_{n}(x)$	$\displaystyle=\int_{\mathbb{R}^{n}}\|g(x)\|^{2}dm_{n}(x)$
		$\displaystyle=\int_{\mathbb{R}^{n}}\|\widehat{g}(\xi)\|^{2}dm_{n}(\xi)$
		$\displaystyle=\int_{\mathbb{R}^{n}}\|e^{-\left\langle\xi,y\right\rangle}% \widehat{H_{t}f}(\xi)\|^{2}dm_{n}(\xi).$

	$\displaystyle\left\\|H_{t}f\right\\|_{\mathcal{H}_{t}}^{2}$	$\displaystyle=\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}\|(H_{t}f)(x+iy)\|% ^{2}\omega_{t}dy\right)dx$
		$\displaystyle=(2\pi)^{n}\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}\|(H_{t% }f)(x+iy)\|^{2}dm_{n}(x)\right)\omega_{t}(y)dm_{n}(y)$
		$\displaystyle=(2\pi)^{n}\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}e^{-2% \left\langle\xi,y\right\rangle}\|\widehat{H_{t}f}(\xi)\|^{2}dm_{n}(\xi)\right)% \omega_{t}(y)dm_{n}(y)$
		$\displaystyle=(2\pi)^{n}\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}e^{-2% \left\langle\xi,y\right\rangle}(2\pi)^{-n}\exp(-2t\|\xi\|^{2})\|\widehat{f}(\xi)\|% ^{2}dm_{n}(\xi)\right)\omega_{t}(y)dm_{n}(y)$
		$\displaystyle=\int_{\mathbb{R}^{n}}\|\widehat{f}(\xi)\|^{2}\exp(-2t\|\xi\|^{2})% \left(t^{-n/2}\int_{\mathbb{R}^{n}}e^{-2\left\langle\xi,y\right\rangle}e^{-% \frac{\|y\|^{2}}{2t}}dm_{n}(y)\right)dm_{n}(x)$
		$\displaystyle=\int_{\mathbb{R}^{n}}\|\widehat{f}(\xi)\|^{2}\exp(-2t\|\xi\|^{2})% \cdot\exp(2t\|\xi\|^{2})dm_{n}(x)$
		$\displaystyle=\left\\|\mathscr{F}f\right\\|_{L^{2}}^{2}$
		$\displaystyle=\left\\|f\right\\|_{L^{2}}^{2}.$