$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$ $\newcommand{\sgn}{\operatorname{sgn}}$ $\newcommand{\rank}{\operatorname{rank}}$

Chapter 7. WKB in dimension $\ge 2$

Preliminaries

1. Elements of symplectic geometry
2. Fourier transform of exponent

Introduction.

Recall that construction of Chapter 5 works as long as $S(x,t)$¹ exists; in other words as long as projection $\pi_x:\Lambda_t\ni (x,p)\to x \in \bR^d$ is a diffeomorphism.

In the previous Chapter 6 we considered $1$-dimensional case. Now we need a bit more of sophistication.

Definition 1. Lagrangian manifold is a smooth $d$-dimensional manifold $\Lambda \subset \bR^{2d}$ on which symplectic form vanishes: \begin{equation} \sigma :=\sum_{1\le j\le d} dx_j\wedge dp_j =0 \label{eq-7.1.1} \end{equation}

The following statements could be proven:

Lemma 1.

1. $\Lambda = \{(x,p): \, p= \nabla S(x)\}$ is a Lagrangian manifold such that $\pi_x$ is a local diffeomorphism.
2. If $\Lambda$ is a Lagrangian manifold such that $\pi_x$ is a local diffeomorphism then $\Lambda = \{(x,p): \, p= \nabla S(x)\}$ for some function $S(x)$.

Recall that $\Lambda_t$ is a Lagrangian manifold constructed in the following way:

• $\Lambda_0=\{ (x,S_{0\, x} \}$ is defined as $t=0$.
• $\Lambda_t$ is an evolution of $\Lambda_0$ along Hamiltonian trajectories \begin{align} &\frac{dx}{dt}=H_{p},\label{eq-7.1.2}\\ &\frac{dp}{dt}=-H_{x}.\label{eq-7.1.3} \end{align} Recall that $S(x,t)$ is defined from \begin{align} &\frac{dS}{dt}=p\cdot x-H(x,p),\label{eq-7.1.4}\\ &S|_{t=0}=S_0.\label{eq-7.1.5} \end{align} We skip subscript $_0$ at $H$.

Lemma 2. Hamiltonian flow (\ref{eq-7.1.2})--(\ref{eq-7.1.3}) preserves symplectic form and therefore $\Lambda_t$ is a Lagrangian manifold.

Lemma 3. At each point $(x,p)$ there exists a partition $(I,J)$ of the set $\{1,\ldots, d\}$ such that $\pi_I:\Lambda \ni (x,p)\to (x_I, p_J)$ is a local diffeomorphism.

Without any loss of the generality we can assume that $I=\{1,\ldots, m\}, $J=(m+1,\ldots, d\}$.

Fourier transform of exponent

To do so we need to describe what does it mean exactly:

Definition 2. Partial $h$-Fourier transform is \begin{equation} (F_J u)(x_I,p_J)= (2\pi h)^{-\frac{d-m}{2}} \int e^{-ih^{-1} p_J\cdot x_J } u(x)\,dx_j. \label{eq-7.1.6} \end{equation}

Then from the theory of Fourier transform \begin{equation} u (x) = F_J^{-1}F_Ju = (2\pi h)^{-\frac{m-d}{2}} \int e^{ih^{-1} p_j\cdot x_J } (F_Ju)(x_I, p_J)\,dp_J. \label{eq-7.1.7} \end{equation} and \begin{align} &(F_J hD_{x_j} u )(x_I,p_J)= p_(F_Ju)(x_I,p_J),\label{eq-7.1.8}\\ &(F_J x_J u )(x_I,p_J)= -hD_{p_j} (F_Ju)(x_I,p_J),\label{eq-7.1.9} \end{align} as $j\in J$.

Theorem 1. Let $u(x)=e^{ih^{-1}S(x)}A(x)$ where $\rank (S_{x_Jx_J})=d$. Then \begin{equation} (F u)(p) = e^{-ih^{-1}\tilde{S}(p)} \tilde{A}(p,h) \label{eq-7.1.10} \end{equation} where $\tilde{S}(p)$ is a Legendre transform of $S(x)$: \begin{equation} \tilde{S}(p)= p\cdot x(p) - S(x(p)) \label{eq-7.1.11} \end{equation} where $x(p)$ is defined from $\nabla S (x)=p$, and $\tilde{A}(p,h)\sim \sum_n \tilde{A}_n(p) h^n$ with \begin{equation} \tilde{A}_0(p)= \frac{1}{\sqrt{|\det S_{xx}|}} e^{-\frac{i\pi}{4}\sgn(S_{xx})} A(x(p)) \label{eq-7.1.12} \end{equation} where $\sgn(S_{xx})=n_+-n_-$, $n_\pm$ is a number of positive/negative eigenvalues of $S_{xx}$.

Definition 3. $\sgn S_{xx}$ is a signature of $S_{xx}$.

Proof. Immediately from the Stationary point principle Theorem 2.3.4. Indeed, $\phi(x)= p\cdot x - S(x)$ and we integrate by $x$.

Corollary 1. Let $v(p)=e^{-ih^{-1}\tilde{S}(p)}A(p)$ where $\tilde{S}_{pp}\ne 0$. Then \begin{equation} (F^{-1}u)(x) = e^{ih^{-1}S(x)} A(x,h) \label{eq-7.1.13} \end{equation} where $S(x)$ is a Legendre transform of $\tilde{S}(p)$: \begin{equation} S(x)= p(x)\cdot x - \tilde{S}(p(x)) \label{eq-7.1.14} \end{equation} where $p(x)$ is defined from $\nabla \tilde{S} (p)=x$, and $A(x,h)\sim \sum_n A_n(x) h^n$ with \begin{equation} A_0(x)= \frac{1}{\sqrt{|\det \tilde{S}_{pp}|}} e^{\frac{i\pi}{4}\sgn(\tilde{S}_{pp})} \tilde{A}(p(x)). \label{eq-7.1.15} \end{equation}

Remark 1.

1. Corollary 1 follows from Theorem 1 and revers it.
2. Legendre transformation applied twice restores function.
3. $S_{xx}=J_x p$ and $\tilde{S}_{pp}=J_p x$ on Lagrange manifold where $J$ denotes Jacobi matrix.
4. In particular, $\sgn(\tilde{S}_{pp})=\sgn(S_{xx})$. where $\sgn(S_{xx})$ is a sign of $S_{xx}$.
5. For partial Fourier transform we arrive to the similar formulae.

$\Leftarrow$  $\Uparrow$  $\Rightarrow$