# The Heisenberg group and Hermite functions

Jordan Bell
August 6, 2015

## 1 The Heisenberg group

For $(z,t),(w,s)\in\mathbb{C}^{n}\times\mathbb{R}$, define the operation

 $(z,t)(w,s)=\left(z+w,t+s+\frac{1}{2}\mathrm{Im}\,(z\cdot\overline{w})\right),$

which satisfies

 $(z,t)(0,0)=(z,t),$

and because $\mathrm{Im}\,(z\cdot\overline{z})=0$,

 $(z,t)^{-1}=(-z,-t).$

We denote $\mathbb{C}^{n}\times\mathbb{R}$ with this operation by $H^{n}$. This is a Lie group of dimension $2n+1$, called the Heisenberg group.

Writing $z=x+iy$ define

 $X_{j}=\frac{\partial}{\partial x_{j}}-\frac{1}{2}y_{j}\frac{\partial}{\partial t% },\qquad 1\leq j\leq n,$

and

 $Y_{j}=\frac{\partial}{\partial y_{j}}+\frac{1}{2}x_{j}\frac{\partial}{\partial t% },\qquad 1\leq j\leq n,$

and

 $T=\frac{\partial}{\partial t}.$

We calculate the Lie brackets of these vector fields. For $X_{j}$ and $X_{k}$,

 $\displaystyle X_{j}X_{k}$ $\displaystyle=\left(\frac{\partial}{\partial x_{j}}-\frac{1}{2}y_{j}\frac{% \partial}{\partial t}\right)\left(\frac{\partial}{\partial x_{k}}-\frac{1}{2}y% _{k}\frac{\partial}{\partial t}\right)$ $\displaystyle=\frac{\partial^{2}}{\partial x_{j}\partial x_{k}}-\frac{1}{2}y_{% k}\frac{\partial}{\partial x_{j}}\frac{\partial}{\partial t}-\frac{1}{2}y_{j}% \frac{\partial}{\partial t}\frac{\partial}{\partial x_{k}}+\frac{1}{4}y_{j}y_{% k}\frac{\partial^{2}}{\partial t^{2}},$

yielding

 $[X_{j},X_{k}]=X_{j}X_{k}-X_{k}X_{j}=0.$

For $Y_{j}$ and $Y_{k}$,

 $\displaystyle Y_{j}Y_{k}$ $\displaystyle=\left(\frac{\partial}{\partial y_{j}}+\frac{1}{2}x_{j}\frac{% \partial}{\partial t}\right)\left(\frac{\partial}{\partial y_{k}}+\frac{1}{2}x% _{k}\frac{\partial}{\partial t}\right)$ $\displaystyle=\frac{\partial^{2}}{\partial y_{j}\partial y_{k}}+\frac{1}{2}x_{% k}\frac{\partial}{\partial y_{j}}\frac{\partial}{\partial t}+\frac{1}{2}x_{j}% \frac{\partial}{\partial t}\frac{\partial}{\partial y_{k}}+\frac{1}{4}x_{j}x_{% k}\frac{\partial^{2}}{\partial t^{2}},$

yielding

 $[Y_{j},Y_{k}]=Y_{j}Y_{k}-Y_{k}Y_{j}=0.$

For $X_{j}$ and $Y_{j}$,

 $\displaystyle X_{j}Y_{j}$ $\displaystyle=\left(\frac{\partial}{\partial x_{j}}-\frac{1}{2}y_{j}\frac{% \partial}{\partial t}\right)\left(\frac{\partial}{\partial y_{j}}+\frac{1}{2}x% _{j}\frac{\partial}{\partial t}\right)$ $\displaystyle=\frac{\partial^{2}}{\partial x_{j}\partial y_{j}}+\frac{1}{2}% \frac{\partial}{\partial t}+\frac{1}{2}x_{j}\frac{\partial}{\partial x_{j}}% \frac{\partial}{\partial t}-\frac{1}{2}y_{j}\frac{\partial}{\partial t}\frac{% \partial}{\partial y_{j}}-\frac{1}{4}y_{j}x_{j}\frac{\partial^{2}}{\partial t^% {2}},$

and

 $\displaystyle Y_{j}X_{j}$ $\displaystyle=\left(\frac{\partial}{\partial y_{j}}+\frac{1}{2}x_{j}\frac{% \partial}{\partial t}\right)\left(\frac{\partial}{\partial x_{j}}-\frac{1}{2}y% _{j}\frac{\partial}{\partial t}\right)$ $\displaystyle=\frac{\partial^{2}}{\partial y_{j}\partial x_{j}}-\frac{1}{2}% \frac{\partial}{\partial t}-\frac{1}{2}y_{j}\frac{\partial}{\partial y_{j}}% \frac{\partial}{\partial t}+\frac{1}{2}x_{j}\frac{\partial}{\partial t}\frac{% \partial}{\partial x_{j}}-\frac{1}{4}x_{j}y_{j}\frac{\partial^{2}}{\partial t^% {2}},$

yielding

 $[X_{j},Y_{j}]=X_{j}Y_{j}-Y_{j}X_{j}=\frac{\partial}{\partial t}=T.$

For $X_{j}$ and $Y_{k}$ with $j\neq k$,

 $\displaystyle X_{j}Y_{k}$ $\displaystyle=\left(\frac{\partial}{\partial x_{j}}-\frac{1}{2}y_{j}\frac{% \partial}{\partial t}\right)\left(\frac{\partial}{\partial y_{k}}+\frac{1}{2}x% _{k}\frac{\partial}{\partial t}\right)$ $\displaystyle=\frac{\partial^{2}}{\partial x_{j}\partial y_{k}}+\frac{1}{2}x_{% k}\frac{\partial}{\partial x_{j}}\frac{\partial}{\partial t}-\frac{1}{2}y_{j}% \frac{\partial}{\partial t}\frac{\partial}{\partial y_{k}}-\frac{1}{4}y_{j}x_{% k}\frac{\partial^{2}}{\partial t^{2}}$

and

 $\displaystyle Y_{k}X_{j}$ $\displaystyle=\left(\frac{\partial}{\partial y_{k}}+\frac{1}{2}x_{k}\frac{% \partial}{\partial t}\right)\left(\frac{\partial}{\partial x_{j}}-\frac{1}{2}y% _{j}\frac{\partial}{\partial t}\right)$ $\displaystyle=\frac{\partial^{2}}{\partial y_{k}\partial x_{j}}-\frac{1}{2}y_{% j}\frac{\partial}{\partial y_{k}}\frac{\partial}{\partial t}+\frac{1}{2}x_{k}% \frac{\partial}{\partial t}\frac{\partial}{\partial x_{j}}-\frac{1}{4}x_{k}y_{% j}\frac{\partial^{2}}{\partial t^{2}},$

yielding

 $[X_{j},Y_{k}]=0.$

For $X_{j}$ and $T$,

 $X_{j}T=\left(\frac{\partial}{\partial x_{j}}-\frac{1}{2}y_{j}\frac{\partial}{% \partial t}\right)\frac{\partial}{\partial t}=\frac{\partial}{\partial x_{j}}% \frac{\partial}{\partial t}-\frac{1}{2}y_{j}\frac{\partial^{2}}{\partial t^{2}% }=TX_{j},$

yielding

 $[X_{j},T]=0.$

For $Y_{j}$ and $T$,

 $Y_{j}T=\left(\frac{\partial}{\partial y_{j}}+\frac{1}{2}x_{j}\frac{\partial}{% \partial t}\right)\frac{\partial}{\partial t}=\frac{\partial}{\partial y_{j}}% \frac{\partial}{\partial t}+\frac{1}{2}x_{j}\frac{\partial^{2}}{\partial t^{2}% }=TY_{j},$

yielding

 $[Y_{j},T]=0.$

We summarize the above calculations in the following theorem.

###### Theorem 1.

The Lie brackets of the vector fields $X_{j},Y_{j}$, $1\leq j\leq n$, and $T$ are:

• $[X_{j},X_{k}]=0$

• $[Y_{j},Y_{k}]=0$

• $[X_{j},Y_{j}]=T$

• $[X_{j},Y_{k}]=0$ for $j\neq k$

• $[X_{j},T]=0$

• $[Y_{j},T]=0$

The Lie algebra of the $H^{n}$ is called the Heisenberg Lie algebra and is denoted $\mathfrak{h}^{n}$. The above vector fields are left-invariant and are a basis for $\mathfrak{h}^{n}$.11 1 Sundaram Thangavelu, An Introduction to the Uncertainty Principle: Hardy’s Theorem on Lie Groups, p. 47, §2.1.

## 2 Representation theory

For a Hilbert space $H$, we denote by $\mathscr{B}(H)$ the set of bounded linear operators $H\to H$, which is a Banach algebra with the operator norm. We denote by $\mathscr{B}_{0}(H)$ the set of compact operators $H\to H$, which is a closed ideal of the Banach algebra $\mathscr{B}(H)$. We denote by $\mathscr{B}_{\mathrm{HS}}(H)$ the collection of Hilbert-Schmidt operators $H\to H$: if $\{e_{i}:i\in I\}$ is an orthonormal basis of $H$, a linear map $A:H\to H$ is called a Hilbert-Schmidt operator if

 $\left\|A\right\|_{\mathrm{HS}}^{2}=\sum_{i\in I}\left\|Ae_{i}\right\|^{2}<\infty.$

This satisfies $\left\|A\right\|\leq\left\|A\right\|_{\mathrm{HS}}$. A Hilbert-Schmidt operator is a compact operator. A linear map $U:H\to H$ is called a unitary operator if it is a bijection and satisfies

 $\left\langle Ux,Uy\right\rangle={x}{y},\qquad x,y\in H.$

We denote the set of unitary operators $H\to H$ by $\mathscr{U}(H)$.

For $\lambda\in\mathbb{R},\lambda\neq 0$, for $(x+iy,t)\in H^{n}$, and for $f\in L^{2}(\mathbb{R}^{n})$, define

 $\pi_{\lambda}(x+iy,t)f(\xi)=e^{i\lambda t}e^{i\lambda\left(x\cdot\xi+\frac{1}{% 2}x\cdot y\right)}f(\xi+y),\qquad\xi\in\mathbb{R}^{n}.$

It is apparent that $\pi_{\lambda}(z,t)$ is a linear map $L^{2}(\mathbb{R}^{n})\to L^{2}(\mathbb{R}^{n})$.

For $(x+iy,t),(u+iv,s)\in H^{n}$ we calculate

 $\displaystyle\pi_{\lambda}(x+iy,t)\pi_{\lambda}(u+iv,s)f(\xi)$ $\displaystyle=\pi_{\lambda}(x+iy,t)e^{i\lambda s}e^{i\lambda\left(u\cdot\xi+% \frac{1}{2}u\cdot v\right)}f(\xi+v)$ $\displaystyle=e^{i\lambda t}e^{i\lambda\left(x\cdot\xi+\frac{1}{2}x\cdot y% \right)}e^{i\lambda s}e^{i\lambda\left(u\cdot(\xi+y)+\frac{1}{2}u\cdot v\right% )}f(\xi+y+v)$ $\displaystyle=e^{i\lambda(t+s)}e^{i\lambda\left((x+u)\cdot\xi+\frac{1}{2}x% \cdot y+u\cdot y+\frac{1}{2}u\cdot v\right)}f(\xi+y+v).$

On the other hand, with $z=x+iy$ and $w=u+iv$,

 $\displaystyle(z,t)(w,s)$ $\displaystyle=\left(z+w,t+s+\frac{1}{2}\mathrm{Im}\,(z\cdot\overline{w})\right)$ $\displaystyle=\left(x+iy+u+iv,t+s+\frac{1}{2}\mathrm{Im}\,((x+iy)\cdot(u-iv))\right)$ $\displaystyle=\left(x+u+i(y+v),t+s+\frac{1}{2}\mathrm{Im}\,(x\cdot u-ix\cdot v% +iy\cdot u+y\cdot v)\right)$ $\displaystyle=\left(x+u+i(y+v),t+s-\frac{1}{2}x\cdot v+\frac{1}{2}y\cdot u% \right),$

for which

 $\displaystyle\pi_{\lambda}((z,t)(w,s))f(\xi)$ $\displaystyle=e^{i\lambda\left(t+s-\frac{1}{2}x\cdot v+\frac{1}{2}y\cdot u% \right)}e^{i\lambda\left((x+u)\cdot\xi+\frac{1}{2}(x+u)\cdot(y+v)\right)}f(\xi% +y+v)$ $\displaystyle=e^{i\lambda(t+s)}e^{i\lambda\left((x+u)\cdot\xi+\frac{1}{2}x% \cdot y+y\cdot u+\frac{1}{2}u\cdot v\right)}f(\xi+y+v),$

and therefore

 $\pi_{\lambda}(x+iy,t)\pi_{\lambda}(u+iv,s)=\pi_{\lambda}((z,t)(w,s)).$

We calculate

 $\pi_{\lambda}(0,0)f(\xi)=f(\xi)$

and

 $\pi_{\lambda}(x+iy,t)\pi_{\lambda}((x+iy,t)^{-1})f=\pi_{\lambda}(0,0)f=f.$

For $f,g\in L^{2}(\mathbb{R}^{n})$,

 $\begin{split}&\displaystyle\left\langle\pi_{\lambda}(x+iy,t)f,\pi_{\lambda}(x+% iy,t)g\right\rangle\\ &\displaystyle\int_{\mathbb{R}^{n}}\pi_{\lambda}(x+iy,t)f(\xi)\overline{\pi_{% \lambda}(x+iy,t)g(\xi)}d\xi\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}}e^{i\lambda t}e^{i\lambda% \left(x\cdot\xi+\frac{1}{2}x\cdot y\right)}f(\xi+y)e^{-i\lambda t}e^{-i\lambda% \left(x\cdot\xi+\frac{1}{2}x\cdot y\right)}\overline{g(\xi+y)}d\xi\\ \displaystyle=&\displaystyle\int_{\mathbb{R}^{n}}f(\xi+y)\overline{g(\xi+y)}d% \xi\\ \displaystyle=&\displaystyle\left\langle f,g\right\rangle.\end{split}$

Therefore $\pi_{\lambda}(z,t)$ is a unitary operator $L^{2}(\mathbb{R}^{n})\to L^{2}(\mathbb{R}^{n})$, and

 $\pi_{\lambda}:H^{n}\to\mathscr{U}(L^{2}(\mathbb{R}^{n}))$

is a group homomorphism, namely, $\pi_{\lambda}$ is a unitary representation of $H^{n}$ on $L^{2}(\mathbb{R}^{n})$.2 Furthermore, using that $y\mapsto f(\cdot+y)$ is continuous $\mathbb{R}^{n}\to L^{2}(\mathbb{R}^{n})$,

 $\displaystyle\left\|\pi_{\lambda}(x+iy,t)f-f\right\|^{2}$ $\displaystyle=\int_{\mathbb{R}^{n}}|e^{i\lambda t}e^{i\lambda\left(x\cdot\xi+% \frac{1}{2}x\cdot y\right)}f(\xi+y)-f(\xi)|^{2}d\xi\to 0$

as $(z,t)\to 0$, showing that $\pi_{\lambda}:H^{n}\to\mathscr{U}(L^{2}(\mathbb{R}^{n}))$ is strongly continuous. (That is, it is continuous when $\mathscr{U}(L^{2}(\mathbb{R}^{n}))$ is assigned the strong operator topology.)

###### Theorem 2.

For $\lambda\in\mathbb{R}$, $\lambda\neq 0$, the map $\pi_{\lambda}$ defined by

 $\pi_{\lambda}(x+iy,t)f(\xi)=e^{i\lambda t}e^{i\lambda\left(x\cdot\xi+\frac{1}{% 2}x\cdot y\right)}f(\xi+y),$

for $(x+iy,t)\in H^{n}$, $f\in L^{2}(\mathbb{R}^{n})$, and $\xi\in\mathbb{R}^{n}$, is a strongly continuous unitary representation of $H^{n}$ on $L^{2}(\mathbb{R}^{n})$.

We call $\pi_{1}$ the Schrödinger representation. Its kernel is

 $\Gamma=\{(0,2\pi k):k\in\mathbb{Z}\}.$

For $f\in L^{1}(H^{n}/\Gamma)$ we define

 $\pi_{1}(f)=\int_{H^{n}/\Gamma}f(z,t)\pi_{1}(z,t)dzdt.$

For $f,g\in L^{1}(H^{n}/\Gamma)$,

 $(f*g)(z,t)=\int_{H^{n}/\Gamma}f((z,t)\cdot(w,s)^{-1})g(w,s)dwds,\qquad(z,t)\in H% ^{n}/\Gamma.$

It is a fact that Lebesgue measure on $\mathbb{C}^{n}\times\mathbb{R}$ is a bi-invariant Haar measure on $H^{n}$, and using this we calculate

 $\begin{split}&\displaystyle\pi_{1}(f*g)\\ \displaystyle=&\displaystyle\int_{H^{n}/\Gamma}\left(\int_{H^{n}/\Gamma}f((z,t% )\cdot(w,s)^{-1})g(w,s)dwds\right)\pi_{1}(z,t)dzdt\\ \displaystyle=&\displaystyle\int_{H^{n}/\Gamma}g(w,s)\left(\int_{H^{n}/\Gamma}% f((z,t)\cdot(w,s)^{-1})\pi_{1}((z,t)\cdot(w,s)^{-1})dzdt\right)\pi_{1}(w,s)% dwds\\ \displaystyle=&\displaystyle\int_{H^{n}/\Gamma}g(w,s)\pi_{1}(f)dwds\\ \displaystyle=&\displaystyle\pi_{1}(f)\pi_{1}(g).\end{split}$
###### Lemma 3.

For $f,g\in L^{1}(H^{n}/\Gamma)$,

 $\pi_{1}(f*g)=\pi_{1}(f)\pi_{1}(g).$

We define

 $W(z)=\pi_{1}(z,0),$

with which

 $\pi_{1}(z,t)=e^{it}W(z).$

Define

 $f_{1}(z)=(2\pi)^{-1/2}\int_{0}^{2\pi}f(z,t)e^{it}dt.$

Then

 $\displaystyle\pi_{1}(f)$ $\displaystyle=\int_{H^{n}/\Gamma}f(z,t)e^{it}W(z)dzdt$ $\displaystyle=\int_{\mathbb{C}^{n}}W(z)\left(\int_{0}^{2\pi}f(z,t)e^{it}dt% \right)dz$ $\displaystyle=(2\pi)^{1/2}\int_{\mathbb{C}^{n}}f_{1}(z)W(z)dz.$

For $f\in L^{1}(\mathbb{C}^{n})$, define

 $f^{\#}(z,t)=(2\pi)^{-1}e^{-it}f(z).$

$f^{\#}\in L^{1}(H^{n}/\Gamma)$, and

 $f^{\#}_{1}(z)=(2\pi)^{-1/2}\int_{0}^{2\pi}f^{\#}(z,t)e^{it}dt=(2\pi)^{-1/2}f(z),$

thus

 $\pi_{1}(f^{\#})=(2\pi)^{1/2}\int_{\mathbb{C}^{n}}f^{\#}(z)W(z)dz=\int_{\mathbb% {C}^{n}}f(z)W(z)dz.$

We define $W:L^{1}(\mathbb{C}^{n})\to\mathscr{U}(L^{2}(\mathbb{R}^{n}))$ by

 $W(f)=\pi_{1}(f^{\#}),$

called the Weyl transform.

For $f,g\in L^{1}(\mathbb{C}^{n})$ and for $(z,t)\in H^{n}/\Gamma$,

 $\begin{split}&\displaystyle(f^{\#}*g^{\#})(z,t)\\ \displaystyle=&\displaystyle\int_{H^{n}/\Gamma}f^{\#}((z,t)\cdot(w,s)^{-1})g^{% \#}(w,s)dwds\\ \displaystyle=&\displaystyle\int_{H^{n}/\Gamma}f^{\#}((z,t)\cdot(-w,-s))g^{\#}% (w,s)dwds\\ \displaystyle=&\displaystyle\int_{H^{n}/\Gamma}f^{\#}\left(z-w,t-s-\frac{1}{2}% \mathrm{Im}\,(z\cdot\overline{w})\right)g^{\#}(w,s)dwds\\ \displaystyle=&\displaystyle\int_{H^{n}/\Gamma}(2\pi)^{-2}e^{-i\left(t-s-\frac% {1}{2}\mathrm{Im}\,(z\cdot\overline{w})\right)}f(z-w)e^{-is}g(w)dwds\\ \displaystyle=&\displaystyle(2\pi)^{-1}e^{-it}\int_{\mathbb{C}^{n}}f(z-w)g(w)e% ^{\frac{i}{2}\mathrm{Im}\,(z\cdot\overline{w})}dw\\ \displaystyle=&\displaystyle(f\times g)^{\#}(z,t),\end{split}$

for

 $(f\times g)(z)=\int_{\mathbb{C}^{n}}f(z-w)g(w)e^{\frac{i}{2}\mathrm{Im}\,(z% \cdot\overline{w})}dw,$

called the twisted convolution. Using what we have established so far gives the following.

###### Lemma 4.

For $f,g\in L^{1}(\mathbb{C}^{n})$,

 $W(f\times g)=\pi_{1}((f\times g)^{\#})=\pi_{1}(f^{\#}*g^{\#})=\pi_{1}(f^{\#})% \pi_{1}(g^{\#})=W(f)W(g)$

For $\phi\in L^{1}(\mathbb{C}^{n})$, we define

 $K_{\phi}(\xi,\eta)=\int_{\mathbb{R}^{n}}\phi(x+i(\eta-\xi))e^{\frac{i}{2}(\xi+% \eta)\cdot x}dx,\qquad(\xi,\eta)\in\mathbb{R}^{n}\times\mathbb{R}^{n},$

which satisfies, for $f\in L^{2}(\mathbb{R}^{n})$ and $\xi\in\mathbb{R}^{n}$,

 $\displaystyle W(\phi)f(\xi)$ $\displaystyle=\int_{\mathbb{C}^{n}}\phi(z)W(z)f(\xi)dz$ $\displaystyle=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\phi(x+iy)e^{i\left(x% \cdot\xi+\frac{1}{2}x\cdot y\right)}f(\xi+y)dydx$ $\displaystyle=\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\phi(x+i(y-\xi))e^{% \frac{i}{2}(x\cdot\xi+x\cdot y)}f(y)dydx$ $\displaystyle=\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}\phi(x+i(y-\xi))% e^{\frac{i}{2}(\xi+y)\cdot x)}dx\right)f(y)dy$ $\displaystyle=\int_{\mathbb{R}^{n}}K_{\phi}(\xi,y)f(y)dy.$

Thus $K_{\phi}$ is an integral kernel for the operator $W(\phi)$.

We show in the following theorem that the Weyl transform sends elements of $L^{1}(\mathbb{C}^{n})$ to compact operators on $L^{2}(\mathbb{R}^{n})$, and that it sends square integrable functions to Hilbert-Schmidt operators.33 3 Sundaram Thangavelu, Lectures on Hermite and Laguerre Expansions, p. 13, Theorem 1.2.1.

###### Theorem 5.

$W:L^{1}(\mathbb{C}^{n})\to\mathscr{B}_{0}(L^{2}(\mathbb{R}^{n}))$, and for $\phi\in L^{1}(\mathbb{C}^{n})\cap L^{2}(\mathbb{C}^{n})$ we have $W(\phi)\in\mathscr{B}_{\mathrm{HS}}(L^{2}(\mathbb{R}^{n}))$ and

 $\left\|\phi\right\|_{L^{2}(\mathbb{R}^{n})}=(2\pi)^{-n/2}\left\|W(\phi)\right% \|_{\mathrm{HS}}.$
###### Proof.

First take $\phi\in L^{1}(\mathbb{C}^{n})\cap L^{2}(\mathbb{C}^{n})$. It follows from this that $K_{\phi}\in L^{2}(\mathbb{R}^{n}\times\mathbb{R}^{n})$, and because $K_{\phi}$ is the integral kernel of $W(\phi)$ this implies44 4 Michael Reed and Barry Simon, Methods of Modern Mathematical Physics, volume I: Functional Analysis, revised and enlarged edition, p. 210, Theorem VI.23. that $W(\phi)\in\mathscr{B}_{\mathrm{HS}}(L^{2}(\mathbb{R}^{n}))$ and

 $\left\|W(\phi)\right\|_{\mathrm{HS}}^{2}=\int_{\mathbb{R}^{n}\times\mathbb{R}^% {n}}|K(\xi,\eta)|^{2}d\xi d\eta.$

## 3 Hermite functions

For $\phi\in\mathscr{S}(\mathbb{R}^{n})$, define

 $\widehat{\phi}(\xi)=(\mathscr{F}\phi)(\xi)=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}% \phi(x)e^{-ix\cdot\xi}dx,\qquad\xi\in\mathbb{R}^{n}.$

$\mathscr{S}(\mathbb{R}^{n})$ is a dense linear subspace of $L^{2}(\mathbb{R}^{n})$, and the Fourier transform extends to a unique Hilbert space isomorphism $L^{2}(\mathbb{R}^{n})\to L^{2}(\mathbb{R}^{n})$. For $f,g\in L^{2}(\mathbb{R})$,

 $\left\langle f,g\right\rangle=\int_{\mathbb{R}^{n}}f(x)\overline{g(x)}dx.$

For $\phi\in\mathscr{S}(\mathbb{R})$, let

 $(D\phi)(x)=\phi^{\prime}(x),\qquad(M\phi)(x)=x\phi(x),\qquad x\in\mathbb{R},$

and let

 $A=-D+M,\qquad B=D+M.$

Let

 $H=\sum_{j=1}^{n}(-D_{j}^{2}+M_{j}^{2})=\frac{1}{2}\sum_{j=1}^{n}(A_{j}B_{j}+B_% {j}A_{j}),$

which satisfies

 $(H\phi)(x)=-(\Delta\phi)(x)+|x|^{2}\phi(x),$

called the Hermite operator.

For $k\geq 0$, define

 $H_{k}(x)=(-1)^{k}e^{x^{2}}D^{k}e^{-x^{2}}$

and

 $h_{k}(x)=(2^{k}k!\sqrt{\pi})^{-1/2}e^{-x^{2}/2}H_{k}(x).$

The Hermite functions are an orthonormal basis for $L^{2}(\mathbb{R})$. Let $\mathbb{N}$ be the nonnegative integers, and for $\alpha\in\mathbb{N}^{n}$ let

 $\Phi_{\alpha}=h_{\alpha_{1}}\otimes\cdots\otimes h_{\alpha_{n}},$

which are an orthonormal basis for $L^{2}(\mathbb{R}^{n})$. It is a fact that

 $A_{j}\Phi_{\alpha}=(2\alpha_{j}+2)^{1/2}\Phi_{\alpha+e_{j}},\qquad B_{j}\Phi_{% \alpha}=(2\alpha_{j})^{1/2}\Phi_{\alpha-e_{j}}$

and

 $H\Phi_{\alpha}=(2|\alpha|+n)\Phi_{\alpha}.$

It is a fact that

 $\widehat{h}_{k}=(-i)^{k}h_{k},$

whence

 $\widehat{\Phi}_{\alpha}=(-i)^{|\alpha|}\Phi_{\alpha}.$

Because $\{\Phi_{\alpha}:\alpha\in\mathbb{N}^{n}\}$ is an orthonormal basis for $L^{2}(\mathbb{R}^{n})$, for $f\in L^{2}(\mathbb{R}^{n})$,

 $f=\sum_{\alpha}\left\langle f,\Phi_{\alpha}\right\rangle\Phi_{\alpha}.$

and then

 $\widehat{f}=\sum_{\alpha}\left\langle f,\Phi_{\alpha}\right\rangle(-i)^{|% \alpha|}\Phi_{\alpha}.$

Let $E_{k}$ be the linear span of $\{\Phi_{\alpha}:|\alpha|=k\}$, which has dimension $\binom{k+n-1}{k}$. For $f\in E_{k}$, $Hf=(2k+n)f$. Let $P_{k}:L^{2}(\mathbb{R}^{n})\to E_{k}$ be the projection:

 $P_{k}f=\sum_{|\alpha|=k}\left\langle f,\Phi_{\alpha}\right\rangle\Phi_{\alpha}% ,\qquad f\in L^{2}(\mathbb{R}^{n}).$

Let

 $\Phi_{k}(x,y)=\sum_{|\alpha|=k}\Phi_{\alpha}(x)\Phi_{\alpha}(y),\qquad x,y\in% \mathbb{R}^{n}.$

For $x\in\mathbb{R}^{n}$ we calculate

 $\displaystyle\int_{\mathbb{R}^{n}}\Phi_{k}(x,y)f(y)dy$ $\displaystyle=\sum_{|\alpha|=k}\Phi_{\alpha}(y)\int_{\mathbb{R}^{n}}f(y)\Phi_{% \alpha}(y)dy$ $\displaystyle=\sum_{|\alpha|=k}\Phi_{\alpha}(y)\left\langle f,\Phi_{\alpha}\right\rangle$ $\displaystyle=(P_{k}f)(y),$

thus $\Phi_{k}$ is a kernel for the projection operator $P_{k}$.

Using the $1$-dimensional Mehler’s formula we obtain the $n$-dimensional Mehler’s formula:

 $\sum_{\alpha}r^{|\alpha|}\Phi_{\alpha}(x)\Phi_{\alpha}(y)=\pi^{-\frac{n}{2}}(1% -r^{2})^{-\frac{n}{2}}\exp\left(-\frac{1}{2}\frac{1+r^{2}}{1-r^{2}}(|x|^{2}+|y% |^{2})+\frac{2r}{1-r^{2}}x\cdot y\right).$

## 4 Special Hermite functions

We first define the Fourier-Wigner transform. For $f,g\in L^{2}(\mathbb{R}^{n})$ and $z=x+iy\in\mathbb{C}^{n}$,

 $V(f,g)(z)=(2\pi)^{-n/2}\int_{\mathbb{R}^{n}}e^{ix\cdot\xi}f\left(\xi+\frac{1}{% 2}y\right)\overline{g\left(\xi-\frac{1}{2}y\right)}d\xi.$

The following theorem relates the inner product on $L^{2}(\mathbb{R}^{n})$ and the inner product on $L^{2}(\mathbb{C}^{n})$.55 5 Sundaram Thangavelu, Lectures on Hermite and Laguerre Expansions, p. 14, Proposition 1.3.1.

###### Theorem 6.

For $f,g,\phi,\psi\in L^{2}(\mathbb{R}^{n})$,

 $\int_{\mathbb{C}^{n}}V(f,g)(z)\overline{V(\phi,\psi)(z)}dz=\left\langle f,\phi% \right\rangle\left\langle\psi,g\right\rangle.$

We now define the special Hermite functions on $\mathbb{C}^{n}$. For $\alpha,\beta\in\mathbb{N}^{n}$, let

 $\Phi_{\alpha\beta}(z)=V(\Phi_{\alpha},\Phi_{\beta})(z).$

We calculate

 $\displaystyle\left\langle W(z)\Phi_{\alpha},\Phi_{\beta}\right\rangle$ $\displaystyle=\int_{\mathbb{R}^{n}}W(z)\Phi_{\alpha}(\xi)\Phi_{\beta}(\xi)d\xi$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{i\left(x\cdot\xi+\frac{1}{2}x\cdot y% \right)}\Phi_{\alpha}(\xi+y)\Phi_{\beta}(\xi)d\xi$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{ix\cdot\xi}\Phi_{\alpha}\left(\xi+\frac{% 1}{2}y\right)\Phi_{\beta}\left(\xi-\frac{1}{2}y\right)d\xi$ $\displaystyle=(2\pi)^{n/2}V(\Phi_{\alpha},\Phi_{\beta}).$
###### Lemma 7.

For $\alpha,\beta\in\mathbb{N}^{n}$ and $z\in\mathbb{C}^{n}$,

 $\Phi_{\alpha\beta}(z)=(2\pi)^{-n/2}\left\langle W(z)\Phi_{\alpha},\Phi_{\beta}% \right\rangle.$

Using that the Hermite functions $\Phi_{\alpha}$ are an orthonormal basis for $L^{2}(\mathbb{R}^{n})$, it is proved that the special Hermite functions $\Phi_{\alpha\beta}$ are an orthonormal basis for $L^{2}(\mathbb{C}^{n})$.66 6 Sundaram Thangavelu, Lectures on Hermite and Laguerre Expansions, p. 16, Theorem 1.3.2.