WKB in dimension 1. 1


### Preliminaries

#### Introduction

Recall that construction of previous Chapter 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. Recall that $\Lambda_t$ is a Lagrangian manifold (the definition we will introduce later) 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-6.1.1}\\ &\frac{dp}{dt}=-H_{x}.\label{eq-6.1.2} \end{align} Recall that $S(x,t)$ is defined from \begin{align} &\frac{dS}{dt}=p\cdot x-H(x,p),\label{eq-6.1.3}\\ &S|_{t=0}=S_0.\label{eq-6.1.4} \end{align} We skip subscript $_0$ at $H$.

We start from $1$-dimensional case. Then locally either $\pi_x$ or $\pi_p:\Lambda_t\ni (x,p)\to p$ are diffeomorphisms. So idea: to use $p$-representation as long as $\pi_p$ is a local diffeomorphism.

#### Fourier transform of exponent

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

Definition 1. $h$-Fourier transform is $$(F u)(p)= (2\pi h)^{-\frac{d}{2}} \int e^{-ih^{-1} p\cdot x } u(x)\,dx. \label{eq-6.1.5}$$

Then from the theory of Fourier transform $$u (x) = F^{-1}Fu = (2\pi h)^{-\frac{d}{2}} \int e^{ih^{-1} p\cdot x } (Fu)(p)\,dp. \label{eq-6.1.6}$$ and \begin{align} &(F hD_{x} u )(p)= p(Fu)(p),\label{eq-6.1.7}\\ &(F x u )(p)= -hD_p (Fu)(p),\label{eq-6.1.8} \end{align}

Theorem 1. Let $d=1$ and $u(x)=e^{ih^{-1}S(x)}A(x)$ where $S_{xx}\ne 0$. Then $$(Fu)(p) = e^{-ih^{-1}\tilde{S}(p)} \tilde{A}(p,h) \label{eq-6.1.9}$$ where $\tilde{S}(p)$ is a Legendre transform of $S(x)$: $$\tilde{S}(p)= p\cdot x(p) - S(x(p)) \label{eq-6.1.10}$$ where $x(p)$ is defined from $S_x (x)=p$, and $\tilde{A}(p,h)\sim \sum_n \tilde{A}_n(p) h^n$ with $$\tilde{A}_0(p)= \frac{1}{\sqrt{|S_{xx}|}} e^{-\frac{i\pi}{4}\sgn(S_{xx})} A(x(p)) \label{eq-6.1.11}$$ where $\sgn(S_{xx})$ is a sign 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 $d=1$ and $v(p)=e^{-ih^{-1}\tilde{S}(p)}A(p)$ where $\tilde{S}_{pp}\ne 0$. Then $$(F^{-1}u)(x) = e^{ih^{-1}S(x)} A(x,h) \label{eq-6.1.12}$$ where $S(x)$ is a Legendre transform of $\tilde{S}(p)$: $$S(x)= p(x)\cdot x - \tilde{S}(p(x)) \label{eq-6.1.13}$$ where $p(x)$ is defined from $\tilde{S}_p (p)=x$, and $A(x,h)\sim \sum_n A_n(x) h^n$ with $$A_0(x)= \frac{1}{\sqrt{|\tilde{S}_{pp}|}} e^{\frac{i\pi}{4}\sgn(\tilde{S}_{pp})} \tilde{A}(p(x)). \label{eq-6.1.14}$$

Remark 1.

1. Corollary 1 follows from Theorem 1 and revers it.
2. Legendre transformation applied twice restores function.
3. $S_{xx}=\frac{dp}{dx}$ and $\tilde{S}_{pp}=\frac{dx}{dp}$ on Lagrange manifold.
4. In particular, $\sgn(\tilde{S}_{pp})=\sgn(S_{xx})$. where $\sgn(S_{xx})$ is a sign of $S_{xx}$.

1. Here we do not include $t$ in $x$.