The Ornstein-Uhlenbeck process ${X={(X_t)}_{t\in[0,\infty)}}$ on ${\mathbb{R}^n}$ is the solution of the stochastic differential equation

$dX_t=\sqrt{2}dB_t-X_tdt$

where ${{(B_t)}_{t\in[0,\infty)}}$ is a standard Brownian motion. Since the diffusion coefficient is constant and the drift is affine, it follows that ${X}$ is a Gaussian process. The computation of the mean and of the variance of ${X_t}$ conditional on ${\{X_0=x\}}$ yields

$\mathrm{Law}(X_t\mid X_0=x)=\mathcal{N}(xe^{-t},\sqrt{1-e^{-2t}}I_n).$

This shows that for any ${x}$ and conditional on ${\{X_0=x\}}$, ${X}$ converges in distribution:

$X_t\underset{t\rightarrow\infty}{\overset{d}{\longrightarrow}}\gamma_n$

where ${\gamma_n=\mathcal{N}(0,I_n)}$ has density ${(2\pi)^{-\frac{n}{2}}e^{-\frac{1}{2}|x|^2}}$. This shows also that ${\gamma_n}$ is invariant:

$X_0\sim\gamma_n\quad\Rightarrow\quad\forall t\geq0,\quad X_t\sim\gamma_n.$

Actually a stronger property holds true: the law ${\gamma_n}$ is reversible in the sense that

$X_0\sim\gamma_n\quad\Rightarrow\quad\forall t\geq0,\quad (X_0,X_t)\overset{d}{=}(X_t,X_0).$

The explicit law of the process allows computations, for instance for any ${s,t\geq0}$,

$\mathrm{Cov}(X_s,X_t)=e^{-|t-s|}(1-e^{-2\min(s,t)}).$

For any bounded and measurable ${f:\mathbb{R}^n\rightarrow\mathbb{R}}$, any ${x\in\mathbb{R}}$ and ${t\in[0,+\infty)}$, we set

$P_t(f)(x)=\mathbb{E}(f(X_t)\mid X_0=x).$

We have ${P_t(\mathbf{1}_A)(x)=\mathbb{P}(X_t\in A\mid X_0=x)}$. The family ${{(P_t)}_{t\in[0,\infty)}}$ is a semigroup of linear operators acting on continuous and bounded functions, in the sense that

$P_0=id, \quad \forall s,t\geq0, \quad P_t\circ P_s = P_{t+s}.$

These operators are Markov operators, in the sense that for any ${t\in[0,\infty)}$,

$\forall c\in\mathbb{R}, P_t(c)=c, \quad\text{and}\quad \forall f, f\geq0 \Rightarrow P_t(f)\geq0.$

The explicit law of the process provides the Mehler formula for the semigroup

$P_t(f)(x) =\int\!f(xe^{-t}+\sqrt{1-e^{-2t}}y)\gamma_n(dy)\\ =\mathbb{E}(f(xe^{-t}+\sqrt{1-e^{-2t}}Z)).$

This gives the following commutation with the gradient when ${f}$ is smooth:

$(\nabla P_t f)(x)=e^{-t}P_t(\nabla f)(x),$

where in the right hand side, ${P_t}$ acts on each coordinates of the vector ${\nabla f}$.

The infinitesimal generator is the unbounded operator in ${L^2(\gamma_n)}$ given by

$Af =\partial_{t=0^+}P_tf =\lim_{t\rightarrow0^+}\frac{P_tf-f}{t} =\Delta f-\langle x,\nabla f\rangle.$

The Chapman-Kolmogorov evolution equation writes

$\partial_tP_t=AP_t=P_tA.$

If we fix ${f}$ and write ${u_t(x)=P_t(f)(x)}$ for any ${x}$ and ${t}$ then

$u_0=f\quad\text{and}\quad \partial_t u_t=Au_t=\Delta u_t-\langle x,\nabla u_t\rangle.$

The operator ${A}$ (and ${P_t}$ for any ${t\geq0}$) is symmetric in ${L^2(\gamma_n)}$, in other words an integration by parts holds, meaning that for any ${f}$ and ${g}$,

$-\int\!fAg\,d\gamma_n =\int\!\nabla f\cdot\nabla g\,d\gamma_n.$

If ${X_0}$ has density ${f_0}$ with respect to ${\gamma_n}$ then ${X_t}$ has also a density with respect to ${\gamma_n}$ given by ${f_t=P_tf_0}$. If ${g}$ is the Lebesgue density of ${\gamma_n}$, then ${g_t=f_tg}$ is the Lebesgue density of ${X_t}$. The evolution of ${g_t}$ with respect to ${t}$ is described by the Fokker-Planck equation, dual of the Chapman-Kolmogorov equation,

$\partial_tg_t=\Delta g_t+\mathrm{div}(xg_t).$

If ${\mu}$ and ${\nu}$ are probability measures on ${\mathbb{R}^n}$ with ${\nu\ll\mu}$ then the Kullback-Leibler divergence or relative entropy of ${\nu}$ with respect to ${\mu}$ is defined by

$H(\nu\mid\mu)=\int\!f\log f\,d\mu=\int\!\log f\,d\nu \quad\text{where}\quad f=\frac{d\nu}{d\mu}.$

We take the convention ${H(\nu\mid\mu)=+\infty}$ if ${f\log f\not\in L^1(\mu)}$ or if ${\nu\not\ll\mu}$. Note that Jensen’s inequality shows that ${H(\nu\mid\mu)\geq0}$ with equality iff ${\mu=\nu}$.

In the case where ${\mu}$ is a Boltzmann-Gibbs measure with Lebesgue density ${g(x)=e^{-V(x)}}$, the quantity ${H(\nu\mid\mu)}$ becomes a Helmholtz free energy, in the sense that

$H(\nu\mid\mu)=\int\!V\,d\nu-S(\nu)$

where the first term in the right hand side is the mean energy of ${\nu}$ while the second term in the right hand side is the Boltzmann-Shannon entropy

$S(\nu)=\displaystyle\int\!fg\log(fg)\,dx.$

Suppose that the law ${\mu_0}$ of ${X_0}$ has density ${f_0}$ with respect to ${\gamma_n}$. Then the law ${\mu_t}$ of ${X_t}$ has density ${f_t=P_tf_0}$ with respect to ${\gamma_n=\mu_\infty}$. The free energy decays along the time. Namely, using the evolution equation and the integration by parts,

$\begin{array}{rcl} \frac{d}{dt}H(\mu_t\mid\gamma_n) &=&\displaystyle\int\!\partial_t(f_t\log f_t)\,d\gamma_n\\ &=&\displaystyle\int\!(1+\log f_t)Af_t\,d\gamma_n\\ &=&\displaystyle-\int\!\frac{|\nabla f_t|^2}{f_t}\,d\gamma_n\\ &=&-J(\mu_t\mid\gamma_n) \leq0. \end{array}$

This is know as the de Bruijn identity:

$\frac{d}{dt}H(\mu_t\mid\gamma_n)=-J(\mu_t\mid\gamma_n)\leq0.$

The quantity

$J(\nu\mid\mu) =\int\!\frac{|\nabla f|^2}{f}\,d\mu=\int\!|\nabla\log f|^2\,d\nu \quad\text{where}\quad f=\frac{d\nu}{d\mu}$

is the Fisher information. How it behaves along the O.-U. dynamics? Well, using commutation, two times Jensen’s inequality, and the invariance of ${\gamma_n}$, we get

$\begin{array}{rcl} J(\mu_t\mid\gamma_n) &=&\int\!\frac{|\nabla f_t|^2}{f_t}\,d\gamma_n\\ &=&e^{-2t}\int\!\frac{|(\nabla f)_t|^2}{f_t}\,d\gamma_n\\ &\leq& e^{-2t}\int\!\frac{(|\nabla f|)_t^2}{f_t}\,d\gamma_n\\ &\leq& e^{-2t}\int\!\left(\frac{|\nabla f_0|^2}{f_0}\right)_t\,d\gamma_n\\ &=& e^{-2t}J(\mu_0\mid\gamma_n), \end{array}$

in other words the Fisher information decays exponentially:

$\forall \mu_0\ll\gamma_n, \forall t\geq0,\quad J(\mu_t\mid\gamma_n)\leq e^{-2t}J(\mu_0\mid\gamma_n).$

In particular we get \begin{align*} H(\mu_0\mid\gamma_n) =-\int_0^\infty\!\frac{d}{dt}H(\mu_t\mid\gamma_n)\,dt =\int_0^\infty\!J(\mu_t\mid\gamma_n)\,dt \leq \frac{1}{2}J(\mu_0\mid\gamma_n). \end{align*} This inequality is known as a logarithmic Sobolev inequality:

$\forall \nu\ll\gamma_n,\quad H(\nu\mid\gamma_n)\leq\frac{1}{2}J(\nu\mid\gamma_n).$

This inequality is optimal in the sense that equality is achieved when ${d\nu(x)/d\gamma_n(x)=e^{ax}}$ for some ${a\in\mathbb{R}}$. Using this inequality for ${\nu=\mu_t}$ yields

$\frac{d}{dt}H(\mu_t\mid\gamma_n) =-J(\mu_t\mid\gamma_n) \leq -\frac{1}{2}H(\mu_t\mid\gamma_n).$

which gives, by Gronwall’s lemma, an exponential decay of the free energy, namely

$\forall\mu_0\ll\gamma_n, \forall t\geq0, H(\mu_t\mid\gamma_n)\leq e^{-2t}H(\mu_0\mid\gamma_n).$

Since both sides are equal for ${t=0}$, taking the derivative at time ${t=0}$ allows to recover from this exponential decay the logarithmic Sobolev inequality!

Hypercontractivity. For any ${t\in[0,\infty)}$ and any ${p\in[1,\infty]}$, Mehler’s formula shows immediately that ${P_t}$ can be extended into a linear operator on ${L^p(\gamma_n)}$. In fact ${P_t}$ is always a contraction:

$\forall p\geq1, \forall t\in[0,\infty), \forall f\in L^p(\gamma_n),\quad \Vert P_tf\Vert_p\leq\Vert f\Vert_p.$

Namely, using Jensen’s inequality and the invariance of ${\gamma_n}$,

$\begin{array}{rcl} \Vert P_tf\Vert_p^p &=&\int\!|\mathbb{E}(f(X_t)\mid X_0=x)|^p\,d\gamma_n(x)\\ &\leq& \int\!\mathbb{E}(|f(X_t)|^p\mid X_0=x)\,d\gamma_n(x)\\ &=&\int\!P_t(|f|^p)\,d\gamma_n(x)\\ &=&\int\!|f|^p\,d\gamma_n(x)\\ &=&\Vert f\Vert_p^p. \end{array}$

Since equality is achieved for constant functions, it follows that ${\Vert P_t\Vert_{p\rightarrow p}=1}$. The semigroup ${{(P_t)}_{t\in[0,\infty)}}$ is in fact hypercontractive:

$\forall p\geq1, \forall t\geq0, \forall f\in L^p(\gamma_n), \quad \Vert P_t f \Vert_{p(t)} \leq \Vert f \Vert_p,$

where ${p(t) = 1 + (p-1)e^{2t}}$, in other words ${\Vert P_t\Vert_{p\rightarrow p(t)}=1}$, and moreover this value ${p(t)}$ is critical in the sense that if ${q > p(t)}$ then ${\Vert P_t\Vert_{p\rightarrow q}=+\infty}$.

Let us give a proof. One can assume that ${f\geq0}$ since ${|P_t f|\leq P_t|f|}$ by Jensen’s inequality. Note that ${p(0)=0}$ and ${p(t)>p}$ if ${t>0}$. Set ${\alpha(t)=\log\Vert P_t f\Vert_{p(t)}}$. To lighten the notation, let us set ${f_t=P_tf}$. We have, for any ${t\geq0}$,

$\begin{array}{rcl} \alpha'(t) &=&\left(\frac{1}{p(t)}\log\int\!f_t^{p(t)}\,d\gamma_n\right)’\\ &=&-\frac{p'(t)}{p(t)^2}\log\int\!f_t^{p(t)}\,d\gamma_n +\frac{1}{p(t)}\frac{\left(\displaystyle\int\!f_t^{p(t)}\,d\gamma_n\right)’}{\displaystyle\int\!f_t^{p(t)}\,d\gamma_n}\\ &=&-\frac{p'(t)}{p(t)^2}\log\int\!(f_t)^{p(t)}\,d\gamma_n +\frac{1}{p(t)}\frac{\displaystyle\int\!\left(p'(t)\log f_t+p(t)\frac{Af_t}{f_t}\right)f_t^{p(t)}\,d\gamma_n}{\displaystyle\int\!f_t^{p(t)}\,d\gamma_n}\\ &=&-\frac{p'(t)}{p(t)^2}\log\int\!f_t^{p(t)}\,d\gamma_n +\frac{p'(t)}{p(t)^2}\frac{\displaystyle\int\!f_t^{p(t)}\log f_t^{p(t)}\,d\gamma_n}{\displaystyle\int\!f_t^{p(t)}\,d\gamma_n} +\frac{\displaystyle\int\!(Af_t)f_t^{p(t)-1}\,d\gamma_n}{\displaystyle\int\!f_t^{p(t)}\,d\gamma_n}\\ &=&\frac{p'(t)}{p(t)^2}\left(H(h_t^{p(t)}\gamma_n\mid\gamma_n)+\frac{p(t)^2}{p'(t)}\int\!(Ah_t)h_t^{p(t)-1}\,d\gamma_n\right) \end{array}$

where ${h_t=f_t/\Vert f_t\Vert_{p(t)}}$. Now the logarithmic Sobolev inequality and the integration by parts give, for any ${h\geq0}$ such that ${h^p}$ is a probability density with respect to ${\gamma_n}$,

$\begin{array}{rcl} \mathrm{H}(h^p\gamma_n\mid\gamma_n) &\leq&\frac{1}{2}\int\!\frac{|\nabla h^p|^2}{h^p}\,d\gamma_n\\ &=&\frac{p^2}{2}\int\!|\nabla h|^2h^{p-2}\,d\gamma_n\\ &=&\frac{p^2}{2(p-1)}\int\!\left<\nabla h,\nabla h^{p-1}\right>\,d\gamma_n\\ &=&-\frac{p^2}{2(p-1)}\int\!(Ah)h^{p-1}\,d\gamma_n. \end{array}$

Using this inequality for ${h=h_t}$ and ${p=p(t)}$, and using ${2(p(t)-1)=p'(t)}$, we obtain that ${\alpha'(t)\leq0}$ for any ${t\geq0}$, and as a consequence

$\log\Vert P_tf\Vert_{p(t)} = \alpha(t) \leq\alpha(0)=\log\Vert f\Vert_p.$

Finally, if now ${q>p(t)}$ then taking ${f_\lambda(x)=e^{\langle\lambda,x\rangle}}$ for some ${\lambda\in\mathbb{R}^n}$ gives

$\Vert f_\lambda\Vert_p=e^{p|\lambda|^2/2} \quad\text{and}\quad P_t f_\lambda=e^{|\lambda|^2(1-e^{-2t})/2}f_{\lambda e^{-t}}$

and therefore

$\frac{\Vert P_t f_\lambda\Vert_{q}}{\Vert f_\lambda\Vert_p} =e^{|\lambda|^2(e^{-2t}(q-1)+1-p)/2},$

a quantity which tends to ${+\infty}$ as ${|\lambda|\rightarrow\infty}$ since ${q>p(t)=1+(p-1)e^{2t}}$.

The proof shows that conversely, from the hypercontractive statement, one can extract the logarithmic Sobolev inequality by taking the derivative at ${t=0}$.

Polynomials. The set of polynomials ${\mathbb{R}[X]}$ is dense in ${L^2(\gamma_1)}$. To see it, let us take ${f\in L^2(\gamma_1)}$, then the Laplace transform ${\varphi_\mu}$ of the signed measure ${\mu(dx)=f(x)\gamma_1(dx)}$ is finite on ${\mathbb{R}}$ since for any ${\theta\in\mathbb{R}}$, by the Cauchy-Schwarz inequality,

$(\varphi_\mu(\theta))^2=\left(\int\! \exp(\theta x)\,\mu(dx)\right)^2 \leq \int\!f^2\,d\gamma_1\int\!\exp(2\theta x)\,\gamma_1(dx)<+\infty,$

and in particular, ${\varphi_\mu}$ is analytic on a neighborhood of ${0}$. Now since for any ${k\in\mathbb{N}}$,

$\varphi_\mu^{(k)}(0) =\int\!x^kf(x)\,\gamma_1(dx) =\langle P_k,f\rangle_{L^2(\gamma_1)}\quad\text{where}\quad P_{k}(x) =x^{k},$

and if ${f\perp\mathbb{R}[X_1,\ldots,X_n]}$ in ${L^2(\mathbb{R})}$, then the derivatives of any order of ${\varphi_\mu}$ vanish at ${0}$, and since ${\varphi_\mu}$ is analytic, we get ${\varphi_\mu\equiv0}$ and then ${\mu=0}$ and then ${f=0}$ in ${L^2(\gamma_n)}$.

Hermite polynomials. Hermite’s polynomials ${{(H_k)}_{k\in\mathbb{N}}}$ are the orthogonal polynomials obtained using the Gram-Schmidt algorithm in ${L^2(\gamma_1)}$ from the canonical basis of ${\mathbb{R}[X]}$. They are normalized in such a way that the coefficient of the term of highest degree in ${H_k}$ is ${1}$ for any ${k\geq0}$. We find

$H_0(x)=1,\quad H_1(x)=x,\quad H_2(x)=x^2-1,\quad\ldots$

It can be checked that Hermite’s polynomials ${{(H_k)}_{k\geq0}}$ satisfy

• Generating series: for any ${k\geq0}$ and ${x\in\mathbb{R}}$,

$H_k(x)=\partial^k_1G(0,x) \quad\text{where}\quad G(s,x)=e^{sx-\frac{1}{2}s^2}=\sum_{k=0}^\infty\frac{s^k}{k!}H_k(x);$

• Three terms recursion formula: for any ${k\geq0}$ and ${x\in\mathbb{R}}$,

$H_{k+1}(x)= xH_{k}(x) – kH_{k-1}(x);$

• Recursive differential equation: for any ${k\geq0}$ and ${x\in\mathbb{R}}$,

$H_k'(x)=kH_{k-1}(x);$

• Differential equation: for any ${k\geq0}$ and ${x\in\mathbb{R}}$,

$H_k”(x)-xH_k'(x)+kH_k(x)=0.$

Using the generating series and Plancherel’s formula, we get

$\sum_{k=0}^\infty \frac{s^{2k}}{k!^2}\Vert H_k\Vert_2^2 =\int\!G(s,x)^2\,\gamma_1(dx)=\exp(-s^2)\int\!e^{2sx}\,\gamma_1(dx) =e^{s^2} =\sum_{k=0}^\infty\frac{s^{2k}}{k!},$

which gives ${\Vert H_k\Vert_2^2=k!}$ by identifying the series coefficients. It follows that ${{(H_k/\sqrt{k!})}_{k\in\mathbb{N}}}$ is a dense orthonormal sequence in the Hilbert space ${L^2(\gamma_1)}$.

For any ${f\in L^2(\gamma_1)}$, we have

$f=\sum_{k\geq0}a_kH_k\quad\text{where}\quad k!a_k=\int\!fH_k\,d\gamma_1.$

In particular

$\Vert f\Vert_2^2=\int\!f^2\,d\gamma_1=\sum_{k\geq0}k!a_k^2.$

Note that ${a_0=\displaystyle\int\!f\,d\gamma_1=\gamma_1(f)}$ is the mean of ${f}$ under ${\gamma_1}$.

Hermite’s polynomials and Ornstein-Uhlenbeck process. Hermite’s polynomials are eigenvectors of the operators ${P_t}$ and ${A}$, namely for any ${k\in\mathbb{N}}$ and ${t\geq0}$,

$P_tH_k=e^{-kt}H_k \quad\text{and}\quad AH_k=-kH_k.$

The property for ${A}$ is immediate from the differential equation satisfied by Hermite’s polynomials. To establish the property for ${P_t}$, we note that for any ${Z\sim\gamma_1}$,

$P_t(G(s,\cdot))(x) =e^{se^{-t}x-\frac{1}{2}s^2}\mathbb{E}\bigr(e^{s\sqrt{1-e^{-2t}}Z}\bigr),$

and since the Laplace transform of ${Z}$ is given by ${\mathbb{E}(e^{\theta Y})=e^{\frac{1}{2}\theta^2}}$ we get

$P_t(G(s,\cdot))(x)=G(se^{-t},x),$

therefore, by the generating series property of Hermite’s polynomials,

$\begin{array}{rcl} P_t(H_k)(x) &=&P_t(\partial_1^kG(0,\cdot))(x)\\ &=&\partial_{s}^k P_t(G(s,\cdot))(x)_{\vert s=0}\\ &=&\partial_{s}^k G(se^{-t},x)_{\vert s=0}\\ &=&e^{-kt}\partial_{1}^k G(se^{-t},x)_{\vert s=0}\\ &=&e^{-kt}H_k(x). \end{array}$

This shows that Hermite’s polynomials are eigenvectors of ${P_t}$.

Exponential decay. If ${f=\sum_{k\geq0}a_kH_k\in L^2(\gamma_1)}$ then for any ${t\geq0}$,

$P_tf=\sum_{k\geq 0}e^{-kt}a_kH_k,$

and thus

$\Vert P_t f-\gamma_1(f)\Vert_2^2 =\sum_{k\geq 1}a_k^2 e^{-2kt}k! \leq e^{-2t}\sum_{k\geq 1}a_k^2 k! =e^{-2t}\Vert f-\gamma_1(f)\Vert_2^2.$

We have obtained the exponential decay in ${L^2(\gamma_1)}$: for any ${t\geq0}$ and ${f\in L^2(\gamma_1)}$,

$\Vert P_t f-\gamma_1(f)\Vert_2 \leq e^{-t}\Vert f-\gamma_1(f)\Vert_2.$

Using the invariance of ${\gamma_1}$, we get ${\displaystyle\int\!P_tf\,d\gamma_1=\int\!f\,d\gamma_1=\gamma_1(f)}$ and therefore

$\mathrm{Var}_{\gamma_1}(P_tf)\leq e^{-2t}\mathrm{Var}_{\gamma_1}(f),$

which is is equivalent to the Poincaré inequality with constant ${1}$ (optimal for ${H_1}$):

$\mathrm{Var}_{\gamma_1}(f)\leq-\int\!fAf\,d\gamma_1=\int\!f’^2\,d\gamma_1.$

This inequality is the linearization at ${h=1+\varepsilon f}$ of the logarithmic Sobolev inequality

$\int\!h^2\log(h^2)\,d\gamma_1 -\int\!h^2\,d\gamma_1\log\int\!h^2\,d\gamma_1 \leq -2\int\!hAh\,d\gamma_1=2\int\!h’^2\,d\gamma_1.$

The gap between the first eigenvalue ${0}$ and the second eigenvalue ${-1}$ of ${A}$ is of length ${1}$. This spectral gap produces the exponential convergence. More generally, the semigroup preserves the spectral decomposition. If ${f\perp\mathrm{Vect}\{H_1,\ldots,H_{k-1}\}}$ in ${L^2(\gamma_1)}$ then ${P_t(f)\perp\mathrm{Vect}\{H_1,\ldots,H_{k-1}\}}$ for any ${t\geq0}$ and for any ${t\geq0}$,

$\Vert P_t f-\gamma_1(f)\Vert_2 \leq e^{-k t}\Vert f-\gamma_1(f)\Vert_2.$

Dimension ${n}$. The operator ${A}$ is a sum of operators acting on one variable:

$Af =\Delta f-\langle x,\nabla f\rangle =A_1f+\cdots+A_nf \quad\text{where}\quad A_kf=\partial_k^2f-x_k\partial_kf.$

The eigenvectors of ${A}$ are products of univariate Hermite’s polynomials. Namely, for any ${k\in\mathbb{N}^n}$, if we denote, for any ${x\in\mathbb{R}^n}$,

$H_k(x)=H_{k_1}(x_1)\cdots H_{k_n}(x_n),$

then

$AH_k=(k_1+\cdots+k_n)H_k.$

Quantum harmonic oscillator. Let ${g_n}$ be the density of ${\gamma_n}$. Consider the isometry

$\Phi:f\in L^2(dx)\rightarrow \Phi(f)=g_n^{-1/2}f\in L^2(\gamma_n).$

One can define the operators ${K}$ on ${L^2(dx)}$ from the operator ${A}$ on ${L^2(\gamma_n)}$, namely

$Kf=(\Phi^{-1}\circ A\circ\Phi)(f) =g_n^{1/2}A(fg_n^{-1/2}) =\Delta f+\Bigr(\frac{n}{2}-\frac{1}{4}|x|^2\Bigr)f.$

This is the quantum harmonic oscillator, a special kind of Schrödinger operator. We have ${\partial_t Q_t=KQ_t}$ where ${{(Q_t)}_{t\in[0,\infty)}}$ is the semigroup of operators defined by

$Q_t(f) =(\Phi^{-1}\circ P_t\circ \Phi)(f) =g_n^{1/2}P_t(g_n^{-1/2}f)$

The eigenvectors of ${K}$ are Hermite’s wave functions: for any ${k\in\mathbb{N}^n}$,

$\psi_k(x)=g_n^{1/2}(x)H_k(x)=(2\pi)^{-\frac{n}{2}}e^{-\frac{1}{4}|x|^2}H_k(x).$

For instance, for ${k=(0,1,\ldots,n-1)}$, we get the wave function

$\psi(x_1,\ldots,x_n)=g_n^{1/2}(x)e^{-\frac{1}{4}|x|^2}H_0(x_1)\cdots H_{n-1}(x_n).$

A bosonic wave function is obtained by symmetrization over ${x_1,\ldots,x_n}$. A fermionic wave function is obtained by anti-symmetrization (implies nullity on the diagonal):

$\begin{array}{rcl} \psi_{\mathrm{fermions}}(x_1,\ldots,x_n) &=&g_n^{1/2}(x)\sum_{\sigma\in\Sigma_n}(-1)^{\mathrm{signature}(\sigma)}H_{\sigma(1)-1}(x_1)\cdots H_{\sigma(n)-1}(x_n)\\ &=&g_n^{1/2}(x)\det \begin{pmatrix} H_0(x_1)&\ldots&H_0(x_n)\\ \vdots &\vdots&\vdots\\ H_{n-1}(x_1)&\ldots&H_{n-1}(x_n) \end{pmatrix}\\ &=&g_n^{1/2}(x)\det \begin{pmatrix} x_1^0&\ldots&x_n^0\\ \vdots &\vdots&\vdots\\ x_1^{n-1}&\ldots&x_n^{n-1} \end{pmatrix}\\ &=&g_n^{1/2}\prod_{1\leq i<j\leq n}(x_i-x_j). \end{array}$

The Slater determinant is here proportional to a Vandermonde determinant. Now

$|\psi_{\mathrm{fermions}}(x_1,\ldots,x_n)|^2 =(2\pi)^{-\frac{n}{2}}e^{-\frac{1}{2}(x_1^2+\cdots+x_n^2)}\prod_{1\leq i<j\leq n}(x_i-x_j)^2.$

We recognize up to normalization the formula of the density of the Gaussian Unitary Ensemble (GUE) namely the density of the eigenvalues of a Gaussian ${n\times n}$ Hermitian random matrix with Lebesgue density in ${\mathbb{R}^{n+n^2-n}=\mathbb{R}^{n^2}}$ proportional to

$H\mapsto e^{-\frac{1}{2}\mathrm{Tr}(H^2)}.$

Notes. By pure provocation, we used the Cauchy-Schwarz inequality only once. We have learned the link with the GUE during a talk by Satya Majumdar.