The Heisenberg group and Hermite functions

1 The Heisenberg group

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

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

which satisfies

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

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

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

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

Writing $z=x+iy$ define

$$X_j=\frac{\partial }{\partial x_j}-\frac{1}{2}{y}_j\frac{\partial }{\partial t},1\le j\le n,$$

and

$$Y_j=\frac{\partial }{\partial y_j}+\frac{1}{2}{x}_j\frac{\partial }{\partial t},1\le j\le n,$$

and

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

We calculate the Lie brackets of these vector fields. For $X_j$ and $X_k$,

$X_j{X}_k$ $=\left({\displaystyle \frac{\partial }{\partial x_j}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial x_k}}-{\displaystyle \frac{1}{2}}{y}_k{\displaystyle \frac{\partial }{\partial t}}\right)$

$={\displaystyle \frac{{\partial }^2}{\partial x_j\partial x_k}}-{\displaystyle \frac{1}{2}}{y}_k{\displaystyle \frac{\partial }{\partial x_j}}{\displaystyle \frac{\partial }{\partial t}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial t}}{\displaystyle \frac{\partial }{\partial x_k}}+{\displaystyle \frac{1}{4}}{y}_j{y}_k{\displaystyle \frac{{\partial }^2}{\partial t^2}},$

yielding

$$[X_j,X_k]=X_j{X}_k-X_k{X}_j=0.$$

For $Y_j$ and $Y_k$,

$Y_j{Y}_k$ $=\left({\displaystyle \frac{\partial }{\partial y_j}}+{\displaystyle \frac{1}{2}}{x}_j{\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial y_k}}+{\displaystyle \frac{1}{2}}{x}_k{\displaystyle \frac{\partial }{\partial t}}\right)$

$={\displaystyle \frac{{\partial }^2}{\partial y_j\partial y_k}}+{\displaystyle \frac{1}{2}}{x}_k{\displaystyle \frac{\partial }{\partial y_j}}{\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{1}{2}}{x}_j{\displaystyle \frac{\partial }{\partial t}}{\displaystyle \frac{\partial }{\partial y_k}}+{\displaystyle \frac{1}{4}}{x}_j{x}_k{\displaystyle \frac{{\partial }^2}{\partial t^2}},$

yielding

$$[Y_j,Y_k]=Y_j{Y}_k-Y_k{Y}_j=0.$$

For $X_j$ and $Y_j$,

$X_j{Y}_j$ $=\left({\displaystyle \frac{\partial }{\partial x_j}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial y_j}}+{\displaystyle \frac{1}{2}}{x}_j{\displaystyle \frac{\partial }{\partial t}}\right)$

$={\displaystyle \frac{{\partial }^2}{\partial x_j\partial y_j}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{1}{2}}{x}_j{\displaystyle \frac{\partial }{\partial x_j}}{\displaystyle \frac{\partial }{\partial t}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial t}}{\displaystyle \frac{\partial }{\partial y_j}}-{\displaystyle \frac{1}{4}}{y}_j{x}_j{\displaystyle \frac{{\partial }^2}{\partial t^2}},$

and

$Y_j{X}_j$ $=\left({\displaystyle \frac{\partial }{\partial y_j}}+{\displaystyle \frac{1}{2}}{x}_j{\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial x_j}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial t}}\right)$

$={\displaystyle \frac{{\partial }^2}{\partial y_j\partial x_j}}-{\displaystyle \frac{1}{2}}{\displaystyle \frac{\partial }{\partial t}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial y_j}}{\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{1}{2}}{x}_j{\displaystyle \frac{\partial }{\partial t}}{\displaystyle \frac{\partial }{\partial x_j}}-{\displaystyle \frac{1}{4}}{x}_j{y}_j{\displaystyle \frac{{\partial }^2}{\partial t^2}},$

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 $j\ne k$,

$X_j{Y}_k$ $=\left({\displaystyle \frac{\partial }{\partial x_j}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial y_k}}+{\displaystyle \frac{1}{2}}{x}_k{\displaystyle \frac{\partial }{\partial t}}\right)$

$={\displaystyle \frac{{\partial }^2}{\partial x_j\partial y_k}}+{\displaystyle \frac{1}{2}}{x}_k{\displaystyle \frac{\partial }{\partial x_j}}{\displaystyle \frac{\partial }{\partial t}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial t}}{\displaystyle \frac{\partial }{\partial y_k}}-{\displaystyle \frac{1}{4}}{y}_j{x}_k{\displaystyle \frac{{\partial }^2}{\partial t^2}}$

and

$Y_k{X}_j$ $=\left({\displaystyle \frac{\partial }{\partial y_k}}+{\displaystyle \frac{1}{2}}{x}_k{\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial x_j}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial t}}\right)$

$={\displaystyle \frac{{\partial }^2}{\partial y_k\partial x_j}}-{\displaystyle \frac{1}{2}}{y}_j{\displaystyle \frac{\partial }{\partial y_k}}{\displaystyle \frac{\partial }{\partial t}}+{\displaystyle \frac{1}{2}}{x}_k{\displaystyle \frac{\partial }{\partial t}}{\displaystyle \frac{\partial }{\partial x_j}}-{\displaystyle \frac{1}{4}}{x}_k{y}_j{\displaystyle \frac{{\partial }^2}{\partial t^2}},$

yielding

$$[X_j,Y_k]=0.$$

For $X_j$ and $T$,

$$X_{j}T=\left(\frac{\partial }{\partial x_j}-\frac{1}{2}{y}_j\frac{\partial }{\partial t}\right)\frac{\partial }{\partial t}=\frac{\partial }{\partial x_j}\frac{\partial }{\partial t}-\frac{1}{2}{y}_j\frac{{\partial }^2}{\partial t^2}=T{X}_j,$$

yielding

$$[X_j,T]=0.$$

For $Y_j$ and $T$,

$$Y_{j}T=\left(\frac{\partial }{\partial y_j}+\frac{1}{2}{x}_j\frac{\partial }{\partial t}\right)\frac{\partial }{\partial t}=\frac{\partial }{\partial y_j}\frac{\partial }{\partial t}+\frac{1}{2}{x}_j\frac{{\partial }^2}{\partial t^2}=T{Y}_j,$$

yielding

$$[Y_j,T]=0.$$

We summarize the above calculations in the following theorem.

Theorem 1.

The Lie brackets of the vector fields $X_j,Y_j$, $1\le j\le n$, and $T$ are:

• $[X_j,X_k]=0$

• $[Y_j,Y_k]=0$

• $[X_j,Y_j]=T$

• $[X_j,Y_k]=0$ for $j\ne k$

• $[X_j,T]=0$

• $[Y_j,T]=0$

The Lie algebra of the $H^n$ is called the Heisenberg Lie algebra and is denoted ${𝔥}^n$. The above vector fields are left-invariant and are a basis for ${𝔥}^n$.

2 Representation theory

For a Hilbert space $H$, we denote by $ℬ(H)$ the set of bounded linear operators $H\to H$, which is a Banach algebra with the operator norm. We denote by ${ℬ}_0(H)$ the set of compact operators $H\to H$, which is a closed ideal of the Banach algebra $ℬ(H)$. We denote by ${ℬ}_{\mathrm{HS}}(H)$ 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

This satisfies $\parallel A\parallel \le {\parallel A\parallel }_{\mathrm{HS}}$. 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

$$⟨Ux,Uy⟩=xy,x,y\in H.$$

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

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

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

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

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

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

$=e^{i\lambda t}{e}^{i\lambda \left(x\cdot \xi +\frac{1}{2}x\cdot y\right)}{e}^{i\lambda s}{e}^{i\lambda \left(u\cdot (\xi +y)+\frac{1}{2}u\cdot v\right)}f(\xi +y+v)$

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

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

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

$=(x+iy+u+iv,t+s+{\displaystyle \frac{1}{2}}\mathrm{Im}((x+iy)\cdot (u-iv)))$

$=(x+u+i(y+v),t+s+{\displaystyle \frac{1}{2}}\mathrm{Im}(x\cdot u-ix\cdot v+iy\cdot u+y\cdot v))$

$=(x+u+i(y+v),t+s-{\displaystyle \frac{1}{2}}x\cdot v+{\displaystyle \frac{1}{2}}y\cdot u),$

for which

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

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

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 $f,g\in L^2({ℝ}^n)$,

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

$${\pi }_{\lambda }:H^n\to 𝒰(L^2({ℝ}^n))$$

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

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

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

Theorem 2.

For $\lambda \in ℝ$, $\lambda \ne 0$, 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 $(x+iy,t)\in H^n$, $f\in L^2({ℝ}^n)$, and $\xi \in {ℝ}^n$, is a strongly continuous unitary representation of $H^n$ on $L^2({ℝ}^n)$.

We call ${\pi }_1$ the Schrödinger representation. Its kernel is

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

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

$${\pi }_1(f)={\int }_{H^n/\mathrm{\Gamma }}f(z,t){\pi }_1(z,t)𝑑z𝑑t.$$

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

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

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

Lemma 3.

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

$${\pi }_1(f*g)={\pi }_1(f){\pi }_1(g).$$

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

${\pi }_1(f)$ $={\displaystyle {\int }_{H^n/\mathrm{\Gamma }}}f(z,t)e^{it}W(z)𝑑z𝑑t$

$={\displaystyle {\int }_{{ℂ}^n}}W(z)\left({\displaystyle {\int }_0^{2\pi }}f(z,t)e^{it}𝑑t\right)𝑑z$

$={(2\pi )}^{1/2}{\displaystyle {\int }_{{ℂ}^n}}{f}_1(z)W(z)𝑑z.$

For $f\in L^1({ℂ}^n)$, define

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

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

$$f_1^{\mathrm{\#}}(z)={(2\pi )}^{-1/2}{\int }_0^{2\pi }{f}^{\mathrm{\#}}(z,t)e^{it}𝑑t={(2\pi )}^{-1/2}f(z),$$

thus

$${\pi }_1(f^{\mathrm{\#}})={(2\pi )}^{1/2}{\int }_{{ℂ}^n}{f}^{\mathrm{\#}}(z)W(z)𝑑z={\int }_{{ℂ}^n}f(z)W(z)𝑑z.$$

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

$$W(f)={\pi }_1(f^{\mathrm{\#}}),$$

called the Weyl transform.

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

for

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

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

Lemma 4.

For $f,g\in L^1({ℂ}^n)$,

$$W(f\times g)={\pi }_1({(f\times g)}^{\mathrm{\#}})={\pi }_1(f^{\mathrm{\#}}*g^{\mathrm{\#}})={\pi }_1(f^{\mathrm{\#}}){\pi }_1(g^{\mathrm{\#}})=W(f)W(g)$$

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

$$K_{\varphi }(\xi ,\eta )={\int }_{{ℝ}^n}\varphi (x+i(\eta -\xi ))e^{\frac{i}{2}(\xi +\eta )\cdot x}𝑑x,(\xi ,\eta )\in {ℝ}^n\times {ℝ}^n,$$

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

$W(\varphi )f(\xi )$ $={\displaystyle {\int }_{{ℂ}^n}}\varphi (z)W(z)f(\xi )𝑑z$

$={\displaystyle {\int }_{{ℝ}^n}}{\displaystyle {\int }_{{ℝ}^n}}\varphi (x+iy)e^{i\left(x\cdot \xi +\frac{1}{2}x\cdot y\right)}f(\xi +y)𝑑y𝑑x$

$={\displaystyle {\int }_{{ℝ}^n}}{\displaystyle {\int }_{{ℝ}^n}}\varphi (x+i(y-\xi ))e^{\frac{i}{2}(x\cdot \xi +x\cdot y)}f(y)𝑑y𝑑x$

$={\displaystyle {\int }_{{ℝ}^n}}\left({\displaystyle {\int }_{{ℝ}^n}}\varphi (x+i(y-\xi ))e^{\frac{i}{2}(\xi +y)\cdot x)}𝑑x\right)f(y)𝑑y$

$={\displaystyle {\int }_{{ℝ}^n}}{K}_{\varphi }(\xi ,y)f(y)𝑑y.$

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

We show in the following theorem that the Weyl transform sends elements of $L^1({ℂ}^n)$ to compact operators on $L^2({ℝ}^n)$, and that it sends square integrable functions to Hilbert-Schmidt operators.

Theorem 5.

$W:L^1({ℂ}^n)\to {ℬ}_0(L^2({ℝ}^n))$, and for $\varphi \in L^1({ℂ}^n)\cap L^2({ℂ}^n)$ we have $W(\varphi )\in {ℬ}_{\mathrm{HS}}(L^2({ℝ}^n))$ and

$${\parallel \varphi \parallel }_{L^2({ℝ}^n)}={(2\pi )}^{-n/2}{\parallel W(\varphi )\parallel }_{\mathrm{HS}}.$$

Proof.

First take $\varphi \in L^1({ℂ}^n)\cap L^2({ℂ}^n)$. It follows from this that $K_{\varphi }\in L^2({ℝ}^n\times {ℝ}^n)$, and because $K_{\varphi }$ is the integral kernel of $W(\varphi )$ this implies that $W(\varphi )\in {ℬ}_{\mathrm{HS}}(L^2({ℝ}^n))$ and

$${\parallel W(\varphi )\parallel }_{\mathrm{HS}}^2={\int }_{{ℝ}^n\times {ℝ}^n}{|K(\xi ,\eta )|}^2𝑑\xi 𝑑\eta .$$

∎

3 Hermite functions

For $\varphi \in 𝒮({ℝ}^n)$, define

$$\widehat{\varphi }(\xi )=(ℱ\varphi )(\xi )={(2\pi )}^{-n/2}{\int }_{{ℝ}^n}\varphi (x)e^{-ix\cdot \xi }𝑑x,\xi \in {ℝ}^n.$$

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

$$⟨f,g⟩={\int }_{{ℝ}^n}f(x)\overline{g(x)}𝑑x.$$

For $\varphi \in 𝒮(ℝ)$, let

$$(D\varphi )(x)={\varphi }^{\prime }(x),(M\varphi )(x)=x\varphi (x),x\in ℝ,$$

and let

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

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

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

called the Hermite operator.

For $k\ge 0$, 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 $L^2(ℝ)$. Let $\mathbb{N}$ be the nonnegative integers, and for $\alpha \in {\mathbb{N}}^n$ let

$${\mathrm{\Phi }}_{\alpha }=h_{{\alpha }_1}\otimes \mathrm{\cdots }\otimes h_{{\alpha }_n},$$

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

$$A_j{\mathrm{\Phi }}_{\alpha }={(2{\alpha }_j+2)}^{1/2}{\mathrm{\Phi }}_{\alpha +e_j},B_j{\mathrm{\Phi }}_{\alpha }={(2{\alpha }_j)}^{1/2}{\mathrm{\Phi }}_{\alpha -e_j}$$

and

$$H{\mathrm{\Phi }}_{\alpha }=(2|\alpha |+n){\mathrm{\Phi }}_{\alpha }.$$

It is a fact that

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

whence

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

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

$$f=\sum _{\alpha }⟨f,{\mathrm{\Phi }}_{\alpha }⟩{\mathrm{\Phi }}_{\alpha }.$$

and then

$$\widehat{f}=\sum _{\alpha }⟨f,{\mathrm{\Phi }}_{\alpha }⟩{(-i)}^{|\alpha |}{\mathrm{\Phi }}_{\alpha }.$$

Let $E_k$ be the linear span of $\{{\mathrm{\Phi }}_{\alpha }:|\alpha |=k\}$, which has dimension $\left(\genfrac{}{}{0pt}{}{k+n-1}{k}\right)$. For $f\in E_k$, $Hf=(2k+n)f$. Let $P_k:L^2({ℝ}^n)\to E_k$ be the projection:

$$P_{k}f=\sum _{|\alpha |=k}⟨f,{\mathrm{\Phi }}_{\alpha }⟩{\mathrm{\Phi }}_{\alpha },f\in L^2({ℝ}^n).$$

Let

$${\mathrm{\Phi }}_k(x,y)=\sum _{|\alpha |=k}{\mathrm{\Phi }}_{\alpha }(x){\mathrm{\Phi }}_{\alpha }(y),x,y\in {ℝ}^n.$$

For $x\in {ℝ}^n$ we calculate

${\displaystyle {\int }_{{ℝ}^n}}{\mathrm{\Phi }}_k(x,y)f(y)𝑑y$ $={\displaystyle \sum _{|\alpha |=k}}{\mathrm{\Phi }}_{\alpha }(y){\displaystyle {\int }_{{ℝ}^n}}f(y){\mathrm{\Phi }}_{\alpha }(y)𝑑y$

$={\displaystyle \sum _{|\alpha |=k}}{\mathrm{\Phi }}_{\alpha }(y)⟨f,{\mathrm{\Phi }}_{\alpha }⟩$

$=(P_{k}f)(y),$

thus ${\mathrm{\Phi }}_k$ is a kernel for the projection operator $P_k$.

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

$$\sum _{\alpha }{r}^{|\alpha |}{\mathrm{\Phi }}_{\alpha }(x){\mathrm{\Phi }}_{\alpha }(y)={\pi }^{-\frac{n}{2}}{(1-r^2)}^{-\frac{n}{2}}\exp \left(-\frac{1}{2}\frac{1+r^2}{1-r^2}({|x|}^2+{|y|}^2)+\frac{2r}{1-r^2}x\cdot y\right).$$

4 Special Hermite functions

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

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

The following theorem relates the inner product on $L^2({ℝ}^n)$ and the inner product on $L^2({ℂ}^n)$.

Theorem 6.

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

$${\int }_{{ℂ}^n}V(f,g)(z)\overline{V(\varphi ,\psi )(z)}𝑑z=⟨f,\varphi ⟩⟨\psi ,g⟩.$$

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

$${\mathrm{\Phi }}_{\alpha \beta }(z)=V({\mathrm{\Phi }}_{\alpha },{\mathrm{\Phi }}_{\beta })(z).$$

We calculate

$⟨W(z){\mathrm{\Phi }}_{\alpha },{\mathrm{\Phi }}_{\beta }⟩$ $={\displaystyle {\int }_{{ℝ}^n}}W(z){\mathrm{\Phi }}_{\alpha }(\xi ){\mathrm{\Phi }}_{\beta }(\xi )𝑑\xi$

$={\displaystyle {\int }_{{ℝ}^n}}{e}^{i\left(x\cdot \xi +\frac{1}{2}x\cdot y\right)}{\mathrm{\Phi }}_{\alpha }(\xi +y){\mathrm{\Phi }}_{\beta }(\xi )𝑑\xi$

$={\displaystyle {\int }_{{ℝ}^n}}{e}^{ix\cdot \xi }{\mathrm{\Phi }}_{\alpha }\left(\xi +{\displaystyle \frac{1}{2}}y\right){\mathrm{\Phi }}_{\beta }\left(\xi -{\displaystyle \frac{1}{2}}y\right)𝑑\xi$

$={(2\pi )}^{n/2}V({\mathrm{\Phi }}_{\alpha },{\mathrm{\Phi }}_{\beta }).$

Lemma 7.

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

$${\mathrm{\Phi }}_{\alpha \beta }(z)={(2\pi )}^{-n/2}⟨W(z){\mathrm{\Phi }}_{\alpha },{\mathrm{\Phi }}_{\beta }⟩.$$

Using that the Hermite functions ${\mathrm{\Phi }}_{\alpha }$ are an orthonormal basis for $L^2({ℝ}^n)$, it is proved that the special Hermite functions ${\mathrm{\Phi }}_{\alpha \beta }$ are an orthonormal basis for $L^2({ℂ}^n)$.