\p_t \p_\zeta^2 \tH\left(0,x,\dfrac{\pi}{2}\right) = \lambda(x), \qquad \p_t \p_\zeta^3 \tH\left(0,x,\dfrac{\pi}{2}\right) = 0.
\end{equation}

Therefore, using that $\p_\zeta^j H(0,x,\zeta) = 0$ for $j \geq 1$, we deduce that

\begin{equation}
\left| \p_\zeta^2 \tH\left(t,x,\dfrac{\pi}{2}\right) - \lambda(x)t \right| \leq Ct^2, \qquad \left| \p_\zeta^3 \tH\left(t,x,\dfrac{\pi}{2}\right) \right| \leq Ct^2.
\end{equation}

Plugging these in (<ref>), we deduce that

\begin{equation}
\left| \p_\zeta \tH (t,x,\zeta) - \lambda(x) t \left( \zeta-\dfrac{\pi}{2}\right) \right| \leq C \left|\zeta-\dfrac{\pi}{2}\right|^3 + Ct^2  \left|\zeta-\dfrac{\pi}{2}\right| + Ct^2 \left|\zeta-\dfrac{\pi}{2}\right|^2 \leq C \left|\zeta-\dfrac{\pi}{2}\right|^3 + Ct^2 \left|\zeta-\dfrac{\pi}{2}\right|.
\end{equation}

It follows that for $(t,x,\zeta) \in X_0 \times (0,\pi)$,

\begin{equation}
\left| \p_\zeta \tH (t,x,\zeta) + \lambda(x) t \cos \zeta \right| \leq C\cos \zeta\left( \cos^2 \zeta + t^2\right).
\end{equation}

This inequality shows that (<ref>) is valid as long as $\inf_\R \lambda \cdot t \geq C\left( \cos^2 \zeta + t^2\right)$, that is $t$ sufficiently small and $t \geq C\cos^2 \zeta$. This region of validity combined with that identified in Step 3 cover a whole strip of times near $0$, see Figure <ref>. This proves the lemma.

[capbesideposition =right,center]figure[]
Step 3 shows that (<ref>) holds within  the blue region. It therefore misses positive times for $\zeta$ near $\pm \pi/2$. Step 4 shows that
(<ref>) holds in the red-enclosed region, hence covers points $(t,\zeta)$ missed by Step 3.

at (0,0) ;

Lemma <ref> provides the bound (<ref>) and Lemma <ref> provides the second estimate in (<ref>). To complete the proof of Theorem <ref>, it remains to show the first estimate in (<ref>):

There exists a constant $c > 0$ such that

|_ξφ(t,x,ξ)| ≥c ( t ξ^-2 + ( x , I_t ) ),     (t,x,ξ) ∈Ω,     x ∉I_t.

We assume that $x \geq x_t^+$; the case $x \leq x_t^-$ is treated similarly. We first note that

\begin{equation}\label{eq-8n}
\lim_{\xi \rightarrow + \infty} \p_\xi \varphi(t,x_t^+,\xi) = \lim_{\xi \rightarrow +\infty} H(t,x_t^+,\xi) = \lim_{\zeta \rightarrow \pi/2} \tH(t,x_t^\pm,\zeta) = \tH\big(t,x_t^+,\pi/2\big) = 0.
\end{equation}

Thanks to (<ref>):

_ξφ(t,x,ξ) = _ξφ(t,x,ξ) - _ξφ(t,x_t^+,ξ) + _ξφ(t,x_t^+,ξ)

= ∫_x_t^+^x _xξ^2  φ(t,s,ξ) ds - ∫_ξ^+∞ _ξ^2 φ(t,x_t^+,s)ds.

We note that $\p_{x\xi}^2  \varphi(0,s,\xi) = 1$ and $\p_{x\xi}^2  \varphi \in C^\infty_b(\Omega)$, hence $\p_{x\xi}^2  \varphi(t,s,\xi) = 1 + O(t)$ for $t$ sufficiently small, uniformly in $(x,\xi) \in \R^2$. For the term $\p_\xi^2  \varphi(t,x,s)$, we use Lemma <ref>. We deduce that for $t$ sufficiently small,

_ξφ(t,x,ξ) ≥(x-x_t^+)( 1 + O(t) ) + ct ∫_ξ^∞s^-3 ds ≥c ( x-x_t + tξ^2 ).

This proves (<ref>) when $x \geq x_t^+$. When $x \leq x_t^-$ instead, we use the identity

_ξφ(t,x,ξ) = ∫_x_t^-^x _xξ^2  φ(t,s,ξ) ds + ∫_-∞^ξ_ξ^2 φ(t,x_t^+,s)ds,

and a similar reasoning. This completes the proof.

The next result shows that $I_t$ is the smallest possible set such that (<ref>) holds. Standard techniques imply that superpositions of WKB waves are essentially supported within the cone $\{(t,x) : x\in I_t\}$; see <ref>.

For $t \in (0,T)$ and $x \in I_t = (x_t^-,x_t^+)$, the equation $\p_\xi \varphi(t,x,\xi) = 0$ has a unique solution denoted $\Xi(t,x)$. It depends smoothly on $(t,x) \in (0,T) \times \R$; moreover, there exists $c > 0$ such that

\begin{equation}\label{eq-8l}
\big|\Xi(t,x)\big| \geq c \left| \dfrac{t}{x - x_t^\pm} \right|^{1/2}, \qquad t \in (0,T), \quad x \in I_t.
\end{equation}

Since $\p_\xi \varphi(t,x,\cdot)$ is decreasing (see Lemma <ref>), the equation $\p_\xi \varphi(t,x,\xi) = 0$ has a unique solution. Moreover, using that $\p_\xi \varphi = H$, the equation $\p_\xi \varphi(t,x,\xi) = 0$ is equivalent to $x = F(t,0, \xi)$. But for $x \in I_t$:

\begin{equation}
\lim_{\xi \rightarrow -\infty} F(t,0,\xi) = x_t^- < x < x_t^+ = \lim_{\xi \rightarrow +\infty} F(t,0,\xi).
\end{equation}

The intermediate value theorem produces then a solution $\Xi(t,x)$ to $x = F(t,0,\xi)$, which is unique. Smoothness with respect to $(t,x)$ follows from the implicit function theorem, noting that $\p_\xi^2 \varphi\big(t,x,\Xi(t,x)\big) \neq 0$.

It remains to prove (<ref>). This is a statement about the behavior of $\Xi(t,x)$ as $x$ approaches $x_t^\pm$. we use (<ref>) specialized at $\xi = \Xi(t,x)$:

0 = _ξφ(t,x,Ξ(t,x)) = ∫_x_t^+^x ^2_xξ φ(t,s,Ξ(t,x)) ds -∫_Ξ(t,x)^+∞ _ξ^2 φ(t,x_t^+,s) ds

≥∫_x_t^+^x ^2_xξ φ(t,s,Ξ(t,x)) ds + ct Ξ(t,x)^-2.

where we used Lemma <ref> for the lower bound.
In particular, using that $\p_{x\xi}^2 \varphi$ is uniformly bounded:

\begin{equation}
C (x-x_t^+) \geq \int_x^{x_t^+} \p^2_{x\xi}\varphi\big(t,s,\Xi(t,x)\big) ds  \geq ct \Xi(t,x)^{-2}; \qquad \text{equivalently} \qquad \Xi(t,x) \geq c \sqrt{\dfrac{t}{x_t^+-x}}.
\end{equation}

A similar estimate holds as $x \rightarrow x_t^-$.

Later, we will need bound on the function $\vartheta$ defined by:

\begin{equation}
\vartheta(t,x) = \varphi\big(t,x,\Xi(t,x)\big), \qquad t \in [0,T], \qquad x \in I_t.
\end{equation}

There exists $C > 0$ such that for $j \in [1,3]$:

\begin{equation}
\big| \p_x^j \vartheta(t,x) \big| \leq C \dist(x,\R \setminus I_t)^{-3j}, \qquad t \in [0,T], \qquad x \in I_t.
\end{equation}

1. Since $\p_\xi \varphi(t,x,\Xi) = 0$, we have $\p_x \vartheta(t,x) = \p_x \varphi(t,x,\Xi)$. Taking then successive derivatives with respect to $x$ of this expression, noting that $|\p_x^j \varphi(t,x,\xi)| \leq C \lr{\xi}$ and $\p_\xi^k \p_x^\ell \varphi$ is uniformly bounded for $k \geq 1$ produces the inequalities:

\begin{equation}\label{eq-10z}
\big| \p_x \vartheta \big| \leq C \lr{\Xi}, \qquad
\big| \p_x^2 \vartheta \big| \leq C \big(\lr{\Xi} + \lr{\p_x \Xi}\big), \qquad
\big| \p_x^3 \vartheta \big| \leq C \big(\lr{\Xi} + \lr{\p_x \Xi}^2 + \lr{\p_x^2\Xi}\big).\end{equation}

It remains now to bound $\Xi$ and its derivatives.

2. Taking derivatives of the expression $\p_\xi \varphi(t,x,\Xi) = 0$ with respect to $\xi$, we obtain – with all identities evaluated at $\xi = \Xi(t,x)$:

\begin{equation}
\p_x \Xi \cdot \p_\xi^2 \varphi + \p_x \varphi = 0, \qquad
\p_x^2 \Xi  \cdot \p_\xi^2 \varphi + (\p_x \Xi)^2 \cdot \p_\xi^3 \varphi + \p_x^2 \varphi + \p_x \Xi \cdot \p^2_{x\xi} \varphi = 0.
\end{equation}

Using the lower bound on $\p_\xi^2 \varphi$ (Lemma <ref>) and the upper bounds on $\p^\alpha \varphi$ (Lemma <ref>), we deduce that

\begin{equation}\label{eq-10y}
\big| \p_x \Xi \big| \leq C\lr{\Xi}^3, \qquad \big| \p_x^2 \Xi \big| \leq \lr{\Xi}^8.
\end{equation}

Plugging the bounds (<ref>) in (<ref>), we deduce that

\begin{equation}\label{eq-11b}
\big| \p_x^j \vartheta \big| \leq C \lr{\Xi}^{3j}, \qquad j \in [1,3].
\end{equation}

3. We now prove an upper bound on $\Xi$. We recall that $\p_\xi \varphi = H$, hence $H(t,x,\Xi) = 0$. It follows that

\begin{equation}
H_+(t,x) = \lim_{\xi \rightarrow +\infty} H(t,x,\xi) - H(t,x,\Xi) = \int_{\Xi(t,x)}^{+\infty} \p_\xi H(t,x,s) ds.
\end{equation}

Moreover, $\p_\xi H = \lr{\xi}^{-2} \p_\xi \tH$, and $\p_\xi \tH$ is uniformly bounded: $\tH$ is the reciprocal of $\tF$, whose derivative with respect to $x$ is bounded below, see Lemmas <ref>-<ref>. We deduce that when $\Xi > 0$,

\begin{equation}
\big| H_+(t,x) \big| \leq C\int_\Xi^{+\infty} s^{-2} ds = \dfrac{C}{\Xi},
\end{equation}

therefore $\Xi(t,x) \leq C \big| H_+(t,x) \big|^{-1}$. We likewise have $-\Xi(t,x) \geq C \big| H_-(t,x) \big|^{-1}$ when $\Xi(t,x) < 0$. Therefore an upper bound on $\Xi$ follows from a lower bound on $H_+$ and $H_-$. Thanks to Lemma <ref> and (<ref>), $\p_x \tH$ hence $\p_x H$ are bounded below on $\Omega$, hence

\begin{equation}
H_+(t,x) = H_+(t,x) - H_+(t,x_t^+) = \int^x_{x_t^+} \p_x H_+(t,y)dy \geq c (x_t^+-x).
\end{equation}

A similar bound holds for $\p_x H_-$. We conclude that $|\Xi(t,x)| \leq C \dist(x,\R \setminus I_t)^{-1}$, which plugged in (<ref>) completes the proof.

§.§ Transport equation
We focus now on solving the transport equation that arises from the strategy outlined in <ref>:

\begin{equation}\label{eq-4y}
\lr{L_1 \alpha u, u} = 0, \qquad L_1 =  D_t  + \left(\lambda(x) D_x + \dfrac{\lambda'(x)}{2i}\right) \sigma_1 + 2n \mu(x) \p_x\varphi  \big( \sigma_1 + \sigma_3 \p_x\varphi \big) + s(x).
\end{equation}

In (<ref>), $u$ is a real-valued normalized eigenvector $u$ of the matrix $\sigma_1 + \sigma_3 \p_x\varphi$, for the eigenvalue $\lr{\p_x\varphi}$; specifically:

\begin{equation}\label{eq-7k}
u = \matrice{\cos(\te/2) \\ \sin(\te/2)}, \qquad \te = \dfrac{\pi}{2} - \arctan \p_x \varphi.
\end{equation}

The solution to (<ref>) with initial data $\alpha(0,x,\xi) = 1$ is

\begin{equation}\label{eq-7g}
\alpha = \big| \p_x H \big|^{1/2} \cdot \exp\left( -2n i
\lr{\xi}^2\cdot \lambda^2 \circ H \cdot \int_{H}^x \dfrac{\mu}{\lambda^3} - \int_0^t s \circ F(\tau,H,\xi) d\tau \right),
\end{equation}

with all functions in (<ref>) evaluated at $(t,x,\xi)$. In particular, $\alpha$ is an admissible function of order $0$ (see Definition <ref>).

1. Since $|u|^2 = 1$, we have $\lr{\p_t u,u} = 0$. Moreover, we observe that $\p_x \lr{\sigma_3u,u} = 2 \lr{\sigma_3 \p_x u,u}$. Also, since $u$ is normalized eigenvector of $\sigma_1 + \sigma_3 \p_x\varphi$ for the eigenvalue $\lr{\p_x\varphi}$,

( σ_1 + σ_3 _xφ) u,u = _xφ,

σ_3 u, u = σ_3 (σ_1 + σ_3 _xφ)u,u_xφ = σ_3 (σ_1 + σ_3 _xφ)u,u_xφ = _xφ_x φ,

where in the second identity we used that $\sigma_3^2 = \Id$ and $\sigma_3 \sigma_1$ is antisymmetric.

2. Thanks to the relations (<ref>), we deduce that, with all functions evaluated at $(t,x,\xi)$:

iL_1 αu, u = ( t + νx
+ is + 12 νx +  2ni μ_xφ_x φ) α,
νλ_xφ_x φ.

3. We now solve the equation (<ref>) along the bicharacteristic of $H_p$ (i.e. the projection along the $x$-axis of the Hamiltonian trajectories of $p$). We remark that

\begin{equation}
\nu(t,x_t,\xi) = \dfrac{\lambda(x_t)\p_x\varphi(t,x_t,\xi)}{\lr{\p_x \varphi(t,x_t,\xi)}} = \dfrac{\lambda(x_t) \xi_t}{\lr{\xi_t}} = \dot{x_t}.
\end{equation}

Thanks to (<ref>), we deduce that $\alpha(t,x_t,\xi)$ satisfies the linear, homogeneous first order ODE

\begin{align}
\dfrac{d}{dt} \big(\alpha(t,x_t,\xi)\big) & =  \dd{\alpha(t,x_t,\xi)}{t} + \dot{x_t} \dd{\alpha(t,x_t,\xi)}{x}  =  \dd{\alpha(t,x_t,\xi)}{t} + \nu(t,x_t,\xi) \dd{\alpha(t,x_t,\xi)}{x}
\\
\label{eq-7d} &  = -\left(is(x_t) + \dfrac{1}{2} \dd{\nu}{x}(t,x_t,\xi)
+ 2ni \mu(x_t) \xi_t \lr{\xi_t}\right) \alpha(t,x_t,\xi).
\end{align}

To solve this equation, we first compute an antiderivative of $\p_x \nu(t,x_t,\xi)$. Thanks to $\nu(t,x_t,\xi_t) = \dot{x_t} = \p_t F(t,x,\xi)$ and $H(t,x_t,\xi) = x$, we have, with all expressions evaluated at $(t,x,\xi)$:

\begin{equation}
\dd{\nu}{x}  = \left(\dd{H}{x} \cdot \dd{^2 F}{x \p t}\right)\big(t,H,\xi\big) = \left(\dfrac{1}{\p_x F} \cdot \dd{\p_x F}{t}\right)\big(t,H,\xi\big) = \dd{ \ln|\p_x F|}{t}(t,H,\xi),
\end{equation}

where we used that $\p_x H \cdot \p_x F(t,H,\xi) = 1$. It follows that

\begin{equation}\label{eq-6v}
\dd{\nu}{x}(t,x_t,\xi) = \dd{\ln|\p_x F|}{t}(t,x,\xi).
\end{equation}

We now focus on the term  $\mu(x_t) \xi_t \lr{\xi_t}$ from (<ref>). Thanks to $p(x_t,\xi_t) = p(x,\xi) = \lambda(x) \lr{\xi}$:

\begin{equation}\label{eq-6w}
\mu(x_t) \xi_t \lr{\xi_t} = \dfrac{\dot{x_t} \mu(x_t) \lr{\xi_t}^2}{\lambda(x_t)} = \dfrac{\dot{x_t} \mu(x_t)}{\lambda(x_t)^3}  p(x,\xi)^2 = p(x,\xi)^2 \dd{}{t} \left( \int_x^{x_t} \dfrac{\mu}{\lambda^3} \right).
\end{equation}

4. We now integrate the ODE (<ref>), using the formulas (<ref>) and (<ref>):

\begin{equation}
\alpha(t,x_t,\xi) = \exp\left( -\dfrac{1}{2}\ln \big| \p_x F(t,x,\xi) \big| -2ni p(x,\xi)^2 \cdot \int_x^{x_t} \dfrac{\mu}{\lambda^3} - i\int_0^t s(x_\tau) d\tau\right).
\end{equation}

Using that $x = H(t,x_t,\xi)$, and using $\p_x H \cdot \p_x F (t,H,\xi) = 1$, we conclude that

\begin{equation}\label{eq-7h}
\alpha = \big| \p_x H \big|^{1/2} \cdot \exp\left( -2n i
\lr{\xi}^2\cdot \lambda^2 \circ H \cdot \int_{H}^x \dfrac{\mu}{\lambda^3} - i\int_0^t s \circ F(\tau,H,\xi) d\tau \right),
\end{equation}

with all functions evaluated at $(t,x,\xi)$. This completes the proof of (<ref>).

5. It remains to show that $\alpha$ is an admissible function of order $0$. The term involving $s$ does not cause problem because derivatives of $F$ and $H$ are uniformly bounded.  By construction of $T$, the function $\p_x H$ is in $C^\infty_b(X)$ and is bounded below by a positive constant – see Lemma <ref>. Therefore $\big| \p_x H(t,x,\xi) \big|^{1/2} \in C^\infty_b(X)$, hence it is admissible.

We now look at the factor

\begin{equation}\label{eq-7j}
\exp\left( -2n i
\lr{\xi}^2\cdot \gamma(t,x,\xi) \right), \qquad \gamma(t,x,\xi) \de \lambda^2 \circ H(t,x,\xi) \cdot \int_{H(t,x,\xi)}^x \dfrac{\mu}{\lambda^3}
\end{equation}

arising in (<ref>). We note that $H(t,x,\xi) - x$ is bounded together with all its derivatives on $X$; and so are $\lambda^2 \circ H$ and $\mu \lambda^{-3}$. Therefore $\gamma \in C^\infty_b(X)$. Taking $k$ derivatives of (<ref>) produces then growth at most $\lr{n,\xi^2}^{k}$. Therefore (<ref>), hence $\alpha$, are admissible of order $0$.

We then set $b_0 = \alpha u$, where $\alpha$ is given by (<ref>), so that $L_1 b_0$ is orthogonal to the kernel of $L_0$. The equation $L_0 b_1 + L_1 b_0 = 0$ has then a unique solution $b_1 \in (\ker L_0)^\perp$. We then set

\begin{equation}\label{eq-5a}
b = b_0+\epsilon b_1
\end{equation}

The function $b$ is admissible of order $1$.

Since $b = b_0 + \epsilon b_1 = \alpha u+\epsilon b_1$, and $\alpha$ is admissible of order $0$ (see Lemma <ref>), it suffices to show that $u$ and $b_1$ are both admissible of order (at most) $1$. Since $L_0$ is a $2 \times 2$ matrix with eigenvalues $0$ and $2\p_t \varphi = - 2\lambda \lr{\p_x\varphi}$, we have

\begin{equation}\label{eq-7l}
b_1 = \dfrac{L_1(\alpha u)}{2 \lambda \lr{\p_x\varphi}}.
\end{equation}

The estimate (<ref>) implies that $\p_x \varphi$ is admissible of order $1/2$ (much better, in fact); hence from (<ref>), $u$ is admissible of order $0$. The coefficients in the non-differential parts of $L_1$ are admissible of order $3/2$, while the differential coefficients are just $C^\infty_b$, see (<ref>). Hence $L_1(\alpha u)$ is admissible of order $3/2$. Moreover, since $\p_x\varphi(0,x,\xi) = \xi$, we have from (<ref>) that

\begin{equation}
\p_x\varphi(t,x,\xi) = \xi + \int_0^t \p_{tx}^2 \varphi(\tau,x,\xi) d\tau \geq \left( 1 - Ct\right) \xi.
\end{equation}

From the expression (<ref>) of $b_1$, we deduce that $b_1$ is admissible of order $1$. Therefore $b$ is admissible of order $1$.

With $b$ and $\varphi$ defined as above, we have

\begin{equation}\label{eq-7n}
\big(\epsilon D_t + \Dii_{n,\epsilon}\big) b e^{i\varphi / \epsilon} = \epsilon^2 r e^{i\varphi / \epsilon},
\end{equation}

for an admissible function $r$ of order $4$.

We recall that

\begin{equation}
e^{-i\varphi / \epsilon} \big(\epsilon D_t + \Dii_{n,\epsilon}\big) e^{i\varphi / \epsilon} = L_0 + \epsilon L_1 + \epsilon^2 L_2
\end{equation}

where $L_0, L_1, L_2$ are given in (<ref>). Using that $b = b_0+\epsilon b_1$, $L_0 b_0 = 0$ and $L_1 b_0 + L_1 b_0 = 0$, we deduce that

\begin{equation}
\big(\epsilon D_t + \Dii_{n,\epsilon}\big) b e^{i\varphi / \epsilon} =  \epsilon^2 \big(L_2 b + L_1 b_1 \big) e^{i\varphi / \epsilon},
\end{equation}

and therefore the function $r$ emerging in (<ref>) is given by $r = L_2 b + L_1 b_1$. We use the formulas (<ref>) and the fact that $b, b_1$ are admissible of order $0$ to check that $r$ is admissible and compute its order. The worst term is $2n \p_x \varphi \mu(x) D_x b$, which is admissible of order $3/2+1+1=7/2 \leq 4$. This completes the proof.

§.§ Conter-propagating WKB wave
We briefly explain how to construct a WKB wave starting from the other branch of (<ref>). One can check that the WKB solution constructed in <ref> is also defined for short negative time, and satisfies the same estimates; in fact, one has

\begin{equation}
\vp(t,x,\xi) = -\vp(-t,x,-\xi).
\end{equation}

Consider then, for $t < 0$:

\begin{equation}\label{eq-0s}
\TT (be^{i\varphi/\epsilon})(t,x,\xi) = i\sigma_2 \ove{b(t,x,\xi)} e^{-i\varphi(t,x,\xi)}.
\end{equation}

Because $\Dii_{n,\epsilon}$ anticommutes with $\TT$, see <ref>, (<ref>) is another WKB solution for $(\epsilon D_t + \Dii_{n,\epsilon})$. It has orientation orthogonal to $be^{i\varphi}$. We refer to it as the counter-propagating WKB wave to $be^{i\varphi}$.

§ PARAMETRIX

In this section we complete the proof of Theorem <ref>. Our proof strategy goes as follows:

1. We produce a parametrix $\fE$ for the operator $\epsilon D_t + \Dii_{n,\epsilon}$. It takes the form of an integral superposition of the WKB waves constructed in <ref>.

2. We prove an asymptotic expansion, $L^\infty$, and $L^2$ estimates for $\frak{E}_{n,\epsilon}$ as $\epsilon \rightarrow 0$.

3. We sum the parametrices, asymptotic expansions and estimates for each component in the block-diagonal decomposition of $\Dii$ (see <ref>).

The result is a parametrix for $h D_t + \Dii$, with associated $L^2 / L^\infty$ estimates that imply Theorem <ref>.

§.§ Preliminary estimates
We start by studying integral quantities of the form

\begin{equation}\label{eq-8f}
Q(t,x) = \int_\R e^{i\varphi(t,x,\xi) / \epsilon} \cdot q(t,x,\xi) \dfrac{d\xi}{\sqrt{\epsilon}},
\end{equation}

where $q$ is an admissible function of order $-\infty$, see Definition <ref>. Quantities of the form (<ref>) are typically superposition of WKB waves constructed in <ref>. We prove here that (<ref>) expands in terms of the critical point $\Xi(t,x)$ of $\p_\xi \varphi(t,x,\cdot)$.

Before stating our next result, we comment that while technically speaking, the quantity $q\big(t,x,\Xi(t,x)\big)$ is only defined for $x \in I_t$, it extends smoothly for $x \in \R$ by:

\begin{equation}\label{eq-8o}
q \big(t,x,\Xi(t,x)\big) = 0, \qquad x \notin I_t.
\end{equation}

This is a consequence of the rapid decay of $\xi \mapsto q(t,x,\xi)$ ($q$ is an admissible function of order $-\infty$) and of the bound (<ref>) for $\Xi(t,x)$ as $x$ approaches $x_t^\pm$.

If $q$ is admissible of order $-\infty$, then:

(i) Uniformly for $t$ in compact subsets of $(0,T)$, $\epsilon \in (0,1)$ and $n \in \N$:

\begin{equation}\label{eq-8v}
Q(t,x) = \left( e^{i \varphi / \epsilon}  \sqrt{\dfrac{2\pi i}{ \p_\xi^2 \varphi }} \cdot q \right)\big( t,x, \Xi(t,x) \big) + O_{L^2(\R)}\big(\epsi^{1/2} n^{-\infty}\big).
\end{equation}

(ii) Uniformly for $t \in (0,T)$, $\epsilon \in (0,1)$ and $n \geq 1$:

\begin{equation}\label{eq-8i}
Q(t,x) = O_{L^\infty(\R)} \big( t^{-1/2}n^{-\infty}  \big),
\qquad Q(t,x)   = O_{L^2(\R)} \big( t^{-1/2}n^{-\infty}  \big).
\end{equation}

1. In this proof we will occasionally drop the variables $t, x$ to keep equations under reasonable length. Assume that $(t,x)$ are such that $\p_\xi \varphi (t,x,\cdot)$ does not vanish on the support of $q(t,x,\cdot)$. We integrate (<ref>) by parts:

\begin{equation}
Q(t,x) = i \epsilon^{1/2} \int_\R e^{i\varphi / \epsilon} \dd{}{\xi}\dfrac{q}{\p_\xi \varphi}  d\xi.
\end{equation}

Using that $\p_\xi^2 \varphi$ is uniformly bounded above, we deduce there exists $C > 0$ such that for every $(t,x)$ with $\p_\xi \varphi (t,x,\cdot)$ not vanishing on the support of $q(t,x,\cdot)$:

\begin{equation}\label{eq-8g}
\big| Q(t,x) \big| \leq C \epsilon^{1/2} \sum_{j=0}^1 \int_\R |\p_\xi \varphi|^{j-2} |\p_\xi^j q|  d\xi.
\end{equation}

2. If $x \notin I_t$ then according to Lemma <ref>:

\begin{equation}\label{eq-8h}
\big|\p_\xi \varphi(\xi)\big| \geq c  \big( t\lr{\xi}^{-2} +  \dist(x,I_t) \big).
\end{equation}

We then apply (<ref>):

\begin{equation}\label{eq-8w}
\big| Q(t,x) \big| \leq C \epsilon^{1/2} \cdot \sum_{j=0}^1 \big( t +  \dist(x,I_t) \big)^{j-2} \int_\R \lr{\xi}^{4-2j}  |\p_\xi^j q| d\xi, \qquad x \notin I_t.
\end{equation}

By assumption $q$ and its derivatives decay faster than any polynomial in $\xi$, uniformly for $(t,x) \in \Omega_0$. Therefore, we deduce that

\begin{equation}\label{eq-8j}
\int_{x \notin I_t} \big| Q(t,x) \big|^2 dx \leq  C \epsilon \sum_{j=0}^1 \int_{x \notin I_t} \big( t +  \dist(x,I_t) \big)^{2j-4} dx \leq C t^{-3} \epsilon.
\end{equation}

3. We now work with $x \in I_t$. Set $C_0 = ct (2+2|\p_\xi^3 \varphi|_\infty)^{-1}$, where $c$ is the constant involved in (<ref>), and assume that

\begin{equation}
\supp \ q(t,x,\cdot) \subset \big\{ \xi : \ |\xi - \Xi(t,x)| \geq C_0	 \lr{\Xi(t,x)}^{-3} \big\}.
\end{equation}

Then on the support of $q(t,x,\cdot)$, we have (dropping the dependence in $t,x$):

| _ξφ(ξ) | = | _ξφ(ξ) - _ξφ(Ξ) | = | ∫_Ξ^ξ_ξ^2 φ(θ) dθ| ≥ct | ∫_Ξ^ξθ^-3 dθ|.

Assume first that $(t,x)$ are such that $C_0 \lr{\Xi}^{-3} \geq 2$. Then the interval between $\xi$ and $\Xi$ contains either $[\xi-2,\xi]$ or $[\xi,\xi+2]$. In particular, using that $\lr{\xi-2}, \lr{\xi}$ and $\lr{\xi+2}$ are comparable, we deduce that

| _ξφ(ξ) | ≥ct ξ^-3,     (t,x,ξ) ∈(q).

Applying (<ref>) produces

\begin{equation}
\big| Q(t,x) \big| \leq C \epsilon^{1/2} \sum_{j=0}^1 \int_\R   \lr{\xi}^{6-3j} |\p_\xi^j q|  d\xi.
\end{equation}

Assume now that $(t,x)$ are such that $C_0 \lr{\Xi}^{-3} \leq 2$. Then if $\xi$ is such that $|\xi - \Xi(t,x)| \geq C_0 \lr{\Xi(t,x)}^{-3}$, then $\lr{\xi}$ and $\lr{\Xi}$ are comparable. It follows from (<ref>) that

\begin{equation}
\big| \p_\xi \varphi(\xi) \big| \geq ct |\xi-\Xi| \lr{\xi}^{-3} \geq ct \lr{\Xi}^{-3} \lr{\xi}^{-3} \geq ct \lr{\xi}^{-6}.
\end{equation}

Again, applying (<ref>) produces

\begin{equation}
\big| Q(t,x) \big| \leq C \epsilon^{1/2} \sum_{j=0}^1 \int_\R \lr{\xi}^{12-6j}  |\p_\xi^j q| d\xi.
\end{equation}

4. We now assume that $x \in I_t$, and that  $(t,x)$ are such that

\begin{equation}
\supp \ q(t,x,\cdot) \subset \{ \xi : \ |\xi - \Xi(t,x)| \leq 2C_0 \lr{\Xi(t,x)}^{-3}\}.
\end{equation}

Following Hörmander's approach to stationary phase estimates – see <cit.> – we write (again dropping the dependence in $t,x$ to shorten expressions):

φ(ξ) = φ(Ξ) + _ξ^2φ(Ξ)2(ξ-Ξ)^2 +  r(ξ)(ξ-Ξ)^3,     r(ξ) ∫_0^1 (1-)^22 _ξ^3 φ(ξ+(1-) Ξ) dθ.

φ_(ξ) φ(Ξ) + _ξ^2φ(Ξ)2(ξ-Ξ)^2 + r(ξ)(ξ-Ξ)^3,
I() ∫_e^iφ_/ ϵ q dξϵ^1/2.

We note that $\varphi_1 = \varphi$, in particular

\begin{equation}
Q(t,x) = I(1) = I(0) + \int_0^1 I'(\te) d\te, \qquad I'(\te) = \dfrac{1}{h^{3/2}} \int_\R (\xi-\Xi)^3 e^{i\varphi_\te / \epsilon} r q d\xi.
\end{equation}

We show that $I'(\te) = O(\epsilon^{1/2})$. For that, we note that

\begin{equation}
\p_\xi \varphi_\te(\xi) = (\xi-\Xi) \tr_\te(\xi), \qquad \tr_\te(\xi) \de \p_\xi^2\varphi(\Xi) + \te(\xi-\Xi) \int_0^1 \p_\xi^3 \varphi\big(\te'\xi+(1-\te') \Xi\big) d\theta'.
\end{equation}

In particular,

\begin{equation}\label{eq-8s}
\big| \tr_\te(\xi) \big| \geq \big| \p_\xi^2\varphi(\Xi) \big| - \dfrac{1}{2}|\p_\xi^3 \varphi|_\infty |\xi-\Xi| \geq \dfrac{1}{2} ct \lr{\Xi}^{-3},
\end{equation}

where we used $|\xi-\Xi| \leq 2C_0 \lr{\Xi}^{-3}$ and the above value of $C_0$. We now integrate by parts twice the expression for $I'(s)$ using the expression

\begin{equation}
(\xi-\Xi) e^{i\varphi_\te / \epsilon} = \dfrac{\p_\xi \varphi_\te}{\tr_\te} e^{i\varphi_\te / \epsilon} = \dfrac{\epsilon}{i \tr_\te} \dd{e^{i\varphi_\te / \epsilon}}{\xi} .
\end{equation}

This gives:

I'() = 1h^1/2 ∫_1i _ e^iφ_/ ϵξ  ·(ξ-Ξ)^2 r q dξ= ih^1/2 ∫_e^iφ_/ ϵ ·ξ (ξ-Ξ)^2 r q_ dξ

= ih^1/2 ∫_(ξ-Ξ) e^iφ_/ ϵ ·( 2 + (ξ-Ξ) ξ) r q_ dξ= -ϵ^1/2 ∫_e^iφ_/ ϵ ·ξ 1_( 2 + (ξ-Ξ) ξ) r q_ dξ.

Thanks to the bound (<ref>), and that $\xi$ and $\Xi$ are comparable, we conclude that

\begin{equation}
\big| I'(\te) \big| \leq C\epsilon^{1/2} \cdot \sum_{j=0}^2 \int_\R \lr{\xi}^{12-3j} \big|\p_\xi^j q \big| d\xi.
\end{equation}

It remains to evaluate $I(0)$. We follow the standard calculation for oscillatory integrals with quadratic phase. From Plancherel's formula:

I(0) = ∫_e^iφ_0 / ϵ q dξϵ^1/2 = e^iφ(Ξ) / ϵ √(2πi_ξ^2 φ(Ξ)) ·∫_e^ - i ϵy^2/2 _ξ^2 φ(Ξ) e^i Ξyq(y) dy

= e^iφ(Ξ) / ϵ √(2πi_ξ^2 φ(Ξ)) ·( q(Ξ) + ∫_( e^ - i ϵy^2/2 _ξ^2 φ(Ξ) - 1) e^i Ξyq(y) dy ).

We control the integral emerging in the right-hand side by:

ϵ^1/2 ∫_ϵ|y|/| 2 _ξ^2 φ(Ξ)|^1/2 | q(y) | dy ≤∫_|y|/| 2 _ξ^2 φ(Ξ)|^1/2 (1+y^2) | (q-_ξ^2 q)^ (y) | dy

≤C ϵ^1/2 ( ∫_ξ^3 | (1-_ξ^2) q |^2 )^1/2,

where we use the Cauchy–Schwartz inequality, Plancherel's formula and that $\lr{\xi}, \lr{\Xi}$ are comparable.

5. To conclude, we first observe that when $x \notin I_t$, (<ref>) provides $L^2$-decay. On the other hand, when $x \in I_t$, $\big| Q(t,x) \big|$ is uniformly bounded. We then split the integral by mean of a cutoff function,
$\chi\big( |\xi-\Xi| \lr{\Xi}^{3} \big)$, where $\chi$ is even, smooth and has compact support, and $\chi(0)=1$. Taking at most two derivatives of $\chi$ produces a factor $\lr{\Xi}^6$, which is comparable to $\lr{\xi}^6$ on the support of $\chi'$. Since $q$ is of order $-\infty$, $L^1$-norms of $\lr{\xi}^k \p_\xi^j q$ decay faster than any power of $n$. This completes the proof of (<ref>).

To limit the length of expressions, we occasionally drop the variables $(t,x)$ in the expressions below. We first assume that $q(t,x,\cdot)$ has support in for $\xi \in [j,j+2]$ for some $j \in \Z$. According to Lemma <ref>, $\varphi$ satisfies the non-stationnary estimate

\begin{equation}
\left| \dd{^2}{\xi^2} \dfrac{\varphi(\xi)}{ct \lr{j}^{-3}} \right| \geq 1, \qquad \xi \in [j,j+2]
\end{equation}

where $c$ is independent of $j$ and $t$. We can therefore apply <cit.>. With the notations of this book, the large parameter is $\epsilon^{-1} ct \lr{j}^{-3}$ and the constant involved is $c_2 = 5 \cdot 2^{2-1}-2 = 8$, see the formula on <cit.>. This implies that

| 1ϵ^1/2∫_j^j+2 e^iφ(ξ)/h q(ξ) dξ| ≤8ϵ^1/2 ( c t j^-3ϵ )^-1/2 ∫_j^j+2 | _ξq (ξ)  | dξ≤C√(t) ∫_j^j+2 ξ^3/2 | _ξq  | dξ,

where we used that $\lr{j}^{3/2}$ and $\lr{\xi}^{3/2}$ are comparable for $\xi \in [j,j+2]$. The constant $C$ on the RHS is independent of $j, t, \epsilon$.

If now $q$ has full support, we decompose it into pieces supported in $\xi \in [j,j+2]$, $j \in \Z$, and we sum the resulting estimates over $j$. It produces the bound

| 1ϵ^1/2∫_e^iφ(ξ)/h q(ξ) dξ| ≤C√(t) ∫_ξ^3/2 | _ξq  | dξ.

Since $q$ is admissible of order $-\infty$, the integral on the RHS decays faster than any power of $n$. This completes the proof.

The bound (<ref>) implies that

\begin{equation}
\big| Q(t,x) \big| \leq C \epsilon^{1/2} \cdot \sum_{j=0}^1 \dist(x,I_t)^{j-2} \int_\R \lr{\xi}^{4-2j}  |\p_\xi^j q| d\xi.
\end{equation}

We square then integrate this expression for $\dist(x,I_t) \geq 1$. It gives

\begin{equation}\label{eq-8y}
\int_{\dist(x,I_t) \geq 1} \big| Q(t,x) \big|^2 dx \leq C \epsilon  \left(  \sum_{j=0}^1 \int_\R \lr{\xi}^{4-2j}  |\p_\xi^j q| d\xi \right)^2.
\end{equation}

When $\dist(x,I_t) \leq 1$, we use the $L^\infty$ bound and get – using that $|I_t| = O(t)$:

\begin{equation}\label{eq-8x}
\int_{\dist(x,I_t) \leq 1} \big| Q(t,x) \big|^2 dx \leq Ct^{-1} (t+ 1) \left(\int_\R \lr{\xi}^{3/2} \big| \p_\xi q  \big| d\xi\right)^2 \leq \dfrac{C}{t} \left(\int_\R \lr{\xi}^{3/2} \big| \p_\xi q  \big| d\xi\right)^2.
\end{equation}

Summing (<ref>) and (<ref>) yields

\begin{equation}
\big|Q(t,\cdot)\big|^2_{L^2} \leq \dfrac{C}{t} \left(  \sum_{j=0}^1 \int_\R \lr{\xi}^{4-2j}  |\p_\xi^j q| d\xi \right)^2.
\end{equation}

Since $q$ is admissible of order $-\infty$, the integral on the RHS decays faster than any power of $n$. This completes the proof of (<ref>).

§.§ Parametrix
We now construct a parametrix for $\epsilon D_t + \Dii_{n,\epsilon}$ as an integral operator of the form (<ref>). Let $a_n$ depending and $x$ and $n$ (but not on $\epsilon$) such that

\begin{equation}\label{eq-9d}
(t,x,\xi) \mapsto \ha_n(\xi) \de \int_\R e^{-ix\xi} a_n(x) dx
\end{equation}

is admissible of order $-\infty$. Let $b_n^+ = b$ given by (<ref>), $b_n^- =\ TT b =  i\sigma_2 \cdot \ove{b}$ (note that $b$ depends implicitly on $n$ and $\epsilon$). Define then

^+ + ^-,     where

^±a_n(t,x) 12πϵ^1/2 ∫_e^±iφ( t,x, ±ξ) / ϵ ·b^±_n,ϵ(t,x,ξ) b^±_n,ϵ(0,0,ξ)^⊤_n(ξ) dξ.

Note that $\fE^-$ is asssociated to the counter-propagating WKB wave constructed in <ref>. Moreover the integral (<ref>) is of the form (<ref>), in particular if $a$ is an admissible function of order $-\infty$ then uniformly in $t \in (0,T)$, $n \geq 1$ and $\epsilon \in (0,1)$:

\begin{equation}\label{eq-13f}
\fE a(t,\cdot) = O_{L^2(\R)}(\epsilon n^{-\infty} t^{-1/2}).
\end{equation}

Our next theorem shows that $\fE$ is a parametrix for $\epsilon D_t + \Di_{n,\epsilon}$: $\fE a_n$ approximately solves the equation

\begin{equation}\label{eq-7f}
\left( \epsilon D_t + \Dii_{n,\epsilon} \right) f = 0, \qquad f(0,x_1) = \dfrac{1}{\epsilon^{1/2}} \cdot a_n\left(  \dfrac{x}{\epsilon}\right).
\end{equation}

The solution $f$ to (<ref>) satisfies,  uniformly in $t \in (0,T)$, $n \geq 1$ and $\epsilon \in (0,1)$:

f(t,x) = a(t,x) + O_L^2()( ϵn^-∞ ).

Moreover, uniformly for $t$ in compact subsets of $(0,T)$:

f(t,x) =  ∑_± ( e^±i φ/ ϵ  √(±i 2π_ξ^2 φ) ·b^±_n,ϵ  b^±_n,ϵ(0,0,·)^⊤_n)( t,x, ±Ξ(t,x) ) + O_L^2()(ϵ^1/2n^-∞).

The formula (<ref>) implies that up to a small remainder, $f$ is also essentially supported within the cone $\{(t,x) : x \in I_t\}$ and uniformly bounded as $\epsilon \rightarrow 0$.  This is a particularly strong form of dispersion: the $L^\infty$-norm passes from $\epsilon^{-1/2}$ at time $0$ to $O(1)$ as soon as $t > 0$. The initial wavepacket structure is immediately lost.

1. We first verify that $f(0,x) = \fE a(0,x)$. Using $\varphi(0,x,\xi) = x\xi$ and that $b_\pm(0,x,\xi)$ does not depend on $x$, see (<ref>), we have

\begin{equation}\label{eq-6y}
\fE a(0,x) = \dfrac{1}{2\pi\epsilon^{1/2}} \int_\R e^{ \frac{ix\xi}{\epsilon} } \left( \sum_\pm b_\pm(0,0,\xi) \cdot b_\pm(0,0,\xi)^\top \right) \ha(\xi) d\xi\end{equation}

Moreover, $b_\pm(0,0,\xi)$ are orthonormal eigenvectors of $\sigma_1+\xi \sigma_3$. It follows that the matrix in (<ref>) is the $2 \times 2$ identity, therefore
\begin{equation}
\fE a(0,x) = \dfrac{1}{2\pi\epsilon^{1/2}} \int_\R e^{ \frac{ix\xi}{\epsilon} } \ha(\xi) d\xi = \dfrac{1}{\epsilon^{1/2}} \cdot a\left(  \dfrac{x}{\epsilon}\right) = f(0,x).
\end{equation}

2. Thanks to Theorem <ref>, we have $\big( \epsilon D_t + \Dii_{n,\epsilon} \big) \fE  = \epsilon^2 R$, where

\begin{equation}
Ra(t,x) = \dfrac{1}{2\pi\epsilon^{1/2}} \sum_\pm \int_\R e^{\pm i\varphi(t,x,\pm \xi) / \epsilon} r_\pm(t,x,\xi) b_\pm(0,0,\xi) \ha(\xi) d\xi,
\end{equation}

for two admissible functions $r_\pm$ of order $4$. In particular, recalling that $\ha$ is admissible of order $-\infty$, $R a$ is an integral of the form (<ref>). From Theorem <ref>(ii), we deduce that uniformly for $t \in (0,T)$:

\begin{equation}\label{eq-13n}
\left\| \big( \epsilon D_t + \Dii_{n,\epsilon} \big) \fE  a(t,\cdot) \right\|_{L^2(\R)} \leq C t^{-1/2} n^{-\infty} \epsilon^2.
\end{equation}

Moreover, the function $t \mapsto t^{-1/2}$ is integrable near $0$. Applying Duhamel's principle as in <cit.>, we conclude that

\begin{equation}
\big\| f(t,\cdot) - \fE a(t,\cdot) \big\|_{L^2} \leq \dfrac{C n^{-\infty}}{\epsilon} \int_0^t \epsilon^2 \tau^{-1/2} d\tau \leq Cn^{-\infty} t^{1/2} \epsilon.
\end{equation}

This proves (<ref>). To deduce (<ref>), we observe that (<ref>) precisely has the form of the integral studied in <ref>; therefore it suffices to apply Part (i) in Theorem <ref>. This completes the proof.

§.§ Summability.
We now produce a leading order expansion for the solution to:

\begin{equation}\label{eq-9b}
(h D_t + \Dii) F = 0, \qquad F(0,x) = \dfrac{1}{\sqrt{h}} \cdot A\left( \dfrac{x}{\sqrt{h}} \right), \qquad A \in \SSS(\R^2).
\end{equation}

Given the block-decomposition of $\Dii$ exhibited in <ref>, and Theorem <ref> about traveling waves, a natural guess for the solution $F$ to (<ref>) is:

E A(t,x) = ∑_n=0^∞(E_n a_n)(t,x_1) G_n,h(x_2),     where:

(E_0 a_0)(t,x_1) e^iS_t√(ρ_th) ·(e^-iπ/4|4πν_t|^1/2 ·e^i x_1^2/4 ν_t ⋆a_0) ( x_1-x_t^+ ρ_t√(h) );     and

E_n E_n,h_n,     h_n = √(h/2n),     a_n(x_1) 1(2n)^1/4 ∫_A( x_1√(2n), x_2 ) G_n(x_2) dx_2.

The solution $F$ to (<ref>) satisfies, uniformly for $t \in (0,T)$:

\begin{equation}\label{eq-9e}
F(t,x) = \frak{E}A(t,x) + O_{L^2(\R^2)}\big(h^{1/2}\big).
\end{equation}

Moreover, uniformly for $t$ in compact sets of $(0,T)$:

\begin{align}\label{eq-9ee}
&  = \dfrac{e^{iS_t}}{\sqrt{\rho_th}} \cdot \left(\dfrac{e^{-i\pi/4}}{|4\pi\nu_t|^{1/2}} \cdot e^{i \frac{x_1^2}{4 \nu_t}} \star a_0\right) \left( \dfrac{x_1-x_t^+}{ \rho_t\sqrt{h}} \right) \cdot e^{-\frac{x_2^2}{2h}} \matrice{1 \\ 0}
\\
& \quad + \sum_{\substack{n=1, \pm}}^\infty \left( e^{\pm i \varphi / h_n}  \sqrt{\dfrac{\pm i}{ 2\pi\p_\xi^2 \varphi }} \cdot b^\pm_{n,h_n}  b^\pm_{n,h_n}(0,0,\cdot)^\top \ha_n\right)\big( t,x, \pm \Xi(t,x) \big) \cdot G_{n,h}(x_2) + O_{L^2}\big(h^{1/4}\big).
\end{align}

[capbesideposition =right,center]figure
Space-time diagram for propagation starting at $(0,0)$. The blue region represents all accessible positions. The red curve is the trajectory of a traveling mode. Away from this curve the propagation is dispersive.

at (0,0) ;

Thanks to Theorem <ref>, it suffices to prove the theorem when $a_0 = 0$. We decompose the solution $F$ to (<ref>) as $F = \sum_{n=1}^\infty f_n \otimes G_{n,h}$, where each $f_n$ evolves according to $\left(h_n D_t + \Dii_{n,h_n} \right) f_n = 0$. In particular, we have

\begin{align}
f_n(0,x_1) & = \int_\R F(0,x_1,x_2) G_{n,h}(x_2) dx_2 = \int_\R \dfrac{1}{h^{3/4}}  A\left( \dfrac{x_1}{\sqrt{h}}, \dfrac{x_2}{\sqrt{h}} \right) G_n\left(\dfrac{x_2}{\sqrt{h}}\right) dx_2
\\
& = \int_\R \dfrac{1}{h^{1/4}} A\left( \dfrac{x_1}{\sqrt{h}}, x_2 \right) G_n(x_2) dx_2 = \int_\R \dfrac{1}{(2n)^{1/4} \sqrt{h_n}} A\left( \dfrac{x_1}{\sqrt{2n} h_n}, x_2 \right) G_n(x_2) dx_2
\\
& = \dfrac{1}{\sqrt{h_n}} \cdot a_n \left( \dfrac{x_1}{h_n} \right),
\label{eq-8c}
\end{align}

where we defined

\begin{equation}\label{eq-9c}
a_n(x_1) \de \dfrac{1}{(2n)^{1/4}} \int_\R A\left( \dfrac{x_1}{\sqrt{2n}}, x_2 \right) G_n(x_2) dx_2.
\end{equation}

We claim that the functions $a_n$ have Schwartz seminorms decaying faster than any power of $n$. Indeed, thanks to (<ref>):

\begin{align}\label{eq-8b}
\big\| x_1^j  & \p_1^k a_n \big\|_{L^2}  = \left\| x_1^j  \p_1^k \int_\R A\left( \dfrac{x_1}{\sqrt{2n}},x_2 \right) G_n(x_2)dx_2 \right\|_{L^2}
\\
& \leq (3n)^{1+k} \left\|  \int_\R x_1^j  \p_1^k A(x) G_n(x_2)dx_2 \right\|_{L^2} = (2n)^{1+k} \left\|  \int_\R x_1^j  \p_1^k A(x) \dfrac{(-\p_2^2+x_2^2)^\ell G_n(x_2)}{(2n)^\ell} dx_2 \right\|_{L^2}
\\
\leq (3n)^{1+k-\ell} \left\| \int_\R  (-\p_2^2+x_2^2)^\ell x_1^j  \p_1^k A(x) G_n(x_2) dx_2 \right\|_{L^2}  \leq (3n)^{1+k-\ell} \left\| \big(-\p_2^2+x_2^2\big)^\ell x_1^j  \p_1^k A \right\|_{L^2}.
\end{align}

In the second line we made the substitution $x_1 \mapsto \sqrt{2n} x_1$, then we used that the components of $G_n$ are eigenvectors of $-\p_2^2 + x_2^2$ with eigenvalues $2n+1, 2n-1$. In the third line we used that $-\p_2^2 + x_2^2$ is selfadjoint, then we used the Cauchy–Schwarz inequality, noting that $|G_n|_{L^2}=1$. By the Sobolev embedding theorem, (<ref>) implies that $a_n \in \SSS(\R)$, with its Schwartz seminorms decaying faster than any power of $n$.

In particular, $\ha_n$ is admissible of order $-\infty$, see (<ref>).
Theorem <ref> produces then a leading-order description of $f_n$ in terms of $a_n$:

\begin{equation}\label{eq-10l}
f_n(t,x_1) - \frak{E}_{n,h_n} a_n(t,x_1) = O_{L^2(\R)} \big( h_n n^{-\infty}\big) =  O_{L^2(\R)} \big( h^{1/2} n^{-\infty}\big).
\end{equation}

To deduce (<ref>), it suffices to take tensor products of (<ref>) with $G_{n,h}$ and to sum over $n$. Using that the functions $G_{n,h}$ form an orthonormal basis of $L^2(\R)$, we have:

\begin{equation}
F(t,x) = \sum_{n=1}^\infty f_n(t,x_1) G_{n,h}(x_2) = \sum_{n=1}^\infty \frak{E}_{n,h_n} a_n(t,x_1) G_{n,h}(x_2) + O_{L^2(\R)} \big( h^{1/2} \big) = \frak{E}A(t,x) + O_{L^2(\R)} \big( h^{1/2} \big).
\end{equation}

To deduce (<ref>), we simply apply (<ref>). This completes the proof.

Theorem <ref> is now a quick consequence of (<ref>). We recall Cramer's inequality for Hermite functions: $|g_n|_{L^\infty} \leq \pi^{-1/4}$, see e.g. [49]; therefore $|G_{n,h}|_{L^\infty} \leq 2(\pi h)^{-1/4}$. We deduce that

\begin{align}
\big| \frak{E} A(t,x) - \frak{E}_0 a_0(t,x_1) G_{0,h}(x_2) \big| & = \left| \sum_{n=1}^\infty \frak{E}_{n,h_n} a_n(t,x_1) G_{n,h}(x_2) \right| \\
& \leq 2(\pi h)^{-1/4} \sum_{n=1}^\infty \big| \frak{E}_{n,h_n} a_n(t,x_1) \big| \leq \dfrac{C}{h^{1/4} \sqrt{t}},
\end{align}

where we eventually used the uniform bound in (<ref>) for $\frak{E}_{n,h_n}$. This completes the proof.

§ THE CASE $\LAMBDA$ CONSTANT

When $\lambda$ is constant, then we can find explicitly the flow (<ref>):

x_t = x_0 + λξξt,     ξ_t = ξ_0.

⇒  F(t,x,ξ) = x+ λξξt,     H(t,x,ξ) = x -  λξξt.

Hence the solution to the eikonal equation is explicit, given by

\begin{equation}\label{eq-3b}
\varphi(t,x,\xi) = x\xi - \lambda \sqrt{1+\xi^2}.
\end{equation}

It follows that $\varphi$ is globally defined and satisfies the estimate (<ref>) for all $t > 0$. Likewise, $\alpha, b_0, b_1, b$ are defined for all $t > 0$ (and admissible of order $-\infty$). It follows that the parametrices $\fE$ are defined for all $t > 0$ and satisfy

\begin{equation}
\fE a(t,x) =
\sum_{\pm} \left( e^{\pm i \varphi / \epsilon}  \sqrt{\pm \dfrac{i}{ 2\pi\p_\xi^2 \varphi }} \cdot b^\pm_{n,\epsilon}  b^\pm_{n,\epsilon}(0,0,\cdot)^\top \ha_n\right)\big( t,x, \pm \Xi(t,x) \big) + O_{L^2(\R)}\big(\epsilon^{1/2}n^{-\infty}\big).
\end{equation}

Therefore, we deduce that Theorem <ref> holds for all fixed $t > 0$.

This is valid for fixed time; we briefly discuss long-time alterations when $\mu$ is non-zero. One get sufficiently instruction by looking at the case $\lambda = 1, \mu = 1$, $s = 0$. A calculation shows that the function $\alpha$ from (<ref>) depends on $\xi$:

\begin{equation}\label{eq-3g}
\alpha(t,x,\xi) = e^{ -2nit \lr{\xi} \xi }.
\end{equation}

When $t, \xi$ are bounded, we do not expect this difference to induce any qualitative change of behavior of the packet. As $t \rightarrow \infty$ however, (<ref>) oscillates rapidly for fixed $\xi$; we therefore expect modified dispersive estimates in long time. We tracked down such effects when studying dynamical edge states in magnetic field, see [7]:  the wavepacket approximation there is valid for $T \sim h^{-1/8}$, smaller than $T \sim h^{-1/2}$ for the magnetic free case [6].

CHAPTER: FOURIER RECONSTRUCTION

We now return to the equation $(hD_t + \Di) \Psi_t = 0$, where $\Di$ is no longer the model operator of Chapter <ref> but instead is the Weyl quantization of a symbol $\di$ satisfying $\bm{(\AAA_1})-\bm{(\AAA_2)}$ (see <ref>). We prove Theorem <ref>: semiclassically near $\Gamma_\di$, waves split in traveling and dispersive parts.

We follow a standard strategy: we reduce the equation $(hD_t + \Di) \Psi_t = 0$ to the canonical model $(hD_t + \Dii) \Psi_t = 0$ studied in <ref> using the Fourier integral operator $\FF$ constructed in Theorem <ref>. We then confirm that the traveling  / dispersive waves emerging in <ref> keep their qualitative features under backward application of $\FF$.

§ PRELIMINARY ESTIMATES
Our program starts with some preliminary estimates that study how Fourier integral operators of the form (<ref>) act on the quantities emerging in (<ref>).

§.§ Traveling wave
We study integrals of the form

\begin{equation}\label{eq-12n}
L(x) \de \int_{\R^4} e^{i\frac{\phi(x,\xi) - y\xi}{h}} c(x,\xi) \cdot \dfrac{1}{\sqrt{h}} q\left( \dfrac{y}{\sqrt{h}}\right)  \dfrac{dy d\xi}{(2\pi h)^2}, \qquad \text{where:}
\end{equation}

* $q$ is a Schwartz function on $\R^2$;

* $c$ is a symbol bounded together with all its derivatives;

* $\phi$ generates the generating function of the symplectomorphism $\kappa$ produced in Lemma <ref>.

In particular, the equation $\p_\xi \phi(x,0) = 0$ has a unique solution, denoted $x_\star$ below. With $\xi_\star = \p_x \phi(x_\star, 0)$, the pair $(x_\star,\xi_\star)$ is uniquely characterized by the relation

\begin{equation}
\kappa(0,0) = (x_\star,\xi_\star).
\end{equation}

Our next lemma is (a version of) a standard result about action of Fourier integral operators on wavepackets. Essentially, (<ref>) is a wavepacket concentrated at $(x_\star,\xi_\star)$ (see Definition <ref>).

Under the above assumptions and notations, we have $L(x) = W(x)+O_{L^2}(\sqrt{h})$, where $W$ is a wavepacket concentrated at $(x_\star,\xi_\star)$.

1. We first realize the integral in $y$ and obtain

\begin{equation}\label{eq-3h}
L(x) = \int_{\R^2} e^{i\frac{\phi(x,\xi)}{h}} c(x,\xi) \cdot \dfrac{1}{\sqrt{h}} \hq\left( \dfrac{\xi}{\sqrt{h}}\right)  \dfrac{d\xi}{(2\pi)^2h} = \dfrac{1}{\sqrt{h}} \int_{\R^2} e^{i\frac{\phi(x,\sqrt{h}\xi)}{h}} c(x,\sqrt{h}\xi) \cdot \hq(\xi) \dfrac{d\xi}{(2\pi)^2}.
\end{equation}

We then write a Taylor expansion of $c$ near $\xi=0$:

c(x,ξ) = c(x,0) + ∑_j=1^2 ξ_j ·c_j(x,ξ),     c_j ∈C^∞_b(^4).

Plugging (<ref>) in (<ref>), we end up with $L = L_1 + \sqrt{h} \cdot L_2$, where:

\begin{align}
L_1(x) & \de \dfrac{c(x,0)}{\sqrt{h}} \cdot \int_{\R^2} e^{i\frac{\phi(x,\sqrt{h}\xi)}{h}}  \cdot \hq(\xi) \dfrac{d\xi}{(2\pi)^2}.
\\
L_2(x) & \de \dfrac{1}{\sqrt{h}} \sum_{j=1}^2 \int_{\R^2} e^{i\frac{\phi(x,\sqrt{h}\xi)}{h}} c_j(x,\sqrt{h}\xi) \cdot \xi_j \hq(\xi) \dfrac{d\xi}{(2\pi)^2}.
\end{align}

We observe that $L_2$ is the sum of two terms of the form (<ref>). Each of these term corresponds to a semiclassical Fourier integral operator applied to a function with fixed $L^2$-norm. We deduce that $L_2 = O_{L^2}(1)$, uniformly as $h \rightarrow 0$; hence $L = L_1 + O_{L^2}(\sqrt{h})$.

2. We now write a Taylor expansion of $\phi$ near $\xi=0$:

ϕ(x,ξ) = ϕ(x,0) + _ξϕ(x,0) ξ+ 12 ξ, _ξ^2 ϕ(x,0) ξ +  ∑_|α|=3ξ_αψ_α(x,ξ).

The functions $\psi_\alpha$ are smooth, compactly supported because $\phi$ is quadratic outside a compact set (see (i) from Lemma <ref>). As in <cit.>, we define for $\te \in [0,1]$:

ϕ_(x,ξ) = ϕ(x,0) + _ξϕ(x,0) ξ+ 12 ξ, _ξ^2 ϕ(x,0) ξ + ·∑_|α|=3ξ_αψ_α(x,ξ),

L_(x) = c(x,0)√(h) ·∫_^2 e^iϕ_(x,√(h)ξ)/h  ·(ξ) dξ(2π)^2.

We then have $\phi_1 = \phi$ and therefore:

\begin{align}
L_1(x) & = L_0(x) + \int_0^1 \dfrac{dL_\te(x)}{d\te} d\te
\\
& = L_0(x) + \sqrt{h} \cdot \dfrac{c(x,0)}{\sqrt{h}} \cdot \int_0^1 \sum_\alpha \left(\int_{\R^2} e^{i\frac{\phi_\te(x,\sqrt{h}\xi)}{h}} \psi_\alpha(x,\sqrt{h}\xi) \cdot \xi^\alpha\hq(\xi) \dfrac{d\xi}{(2\pi)^2}\right) d\te.
\end{align}

Again, the second term corresponds to a (integral over $\te$ of sums of) semiclassical Fourier integral operator applied to a normalized function. In particular, owing to the term $\sqrt{h}$ in front of it, we deduce that $L_1 = L_0 + O_{L^2}(\sqrt{h})$.

3. It remains to estimate the expression

\begin{equation}\label{eq-12s}
L_0(x) = \dfrac{c(x,0) e^{\frac{i}{h}\phi(x,0)}}{\sqrt{h}} \cdot \int_{\R^2} e^{i\xi \frac{\p_\xi \phi(x,0)}{\sqrt{h}}} \cdot e^{ \frac{i}{2} \lr{\xi, \p_\xi^2 \phi(x,0) \xi }}  \cdot \hq(\xi) \dfrac{d\xi}{(2\pi)^2}.
\end{equation}

We write the integral in (<ref>) as

∫_^2 e^iξ_ξϕ(x,0)/√(h) ·ρ(x,ξ) dξ(2π)^2 = ρ (x, _ξϕ(x,0)/√(h)),

ρ(x,ξ) e^ i/2 ξ, _ξ^2 ϕ(x,0) ξ  (ξ),
ρ(x,y) ∫_^2 e^iξy ·ρ(x,ξ) dξ(2π)^2.

The function $\rho(x,\cdot)$ is Schwartz, with seminorms uniformly bounded for $x \in \R^2$: $\p_\xi^2 \phi$ is constant where $\phi$ is quadratic, hence outside a compact set. Hence so is its inverse Fourier transform. Recall that $x_\star$ is the solution to $\p_\xi \phi(x,0) = 0$ and Taylor-expand $\p_\xi \phi(x,0)$ near $x_\star$:

\begin{equation}
\p_\xi \phi(x,0) = \p_{x\xi}^2 \phi(x_\star,0)(x-x_\star) + \sum_{|\alpha| = 3} \Psi_\alpha(x) (x-x_\star)^\alpha, \qquad \Psi_\alpha \in C^\infty_0(\R^2).
\end{equation}

Using that $\p_{x\xi}^2 \phi(x_\star,0)$ is not singular, we can perform successive Taylor expansions of (<ref>) near $x_\star$. We first obtain

\begin{align}
\widecheck{\rho} \left( x,\frac{\p_\xi \phi(x,0)}{\sqrt{h}}\right) & = \widecheck{\rho} \left( x,\p_{x\xi}^2 \phi(x_\star,0) \frac{x-x_\star}{\sqrt{h}}\right) + O_{L^2}(\sqrt{h})
\\
\label{eq-12u} & = \widecheck{\rho} \left( x_\star,\p_{x\xi}^2 \phi(x_\star,0) \frac{x-x_\star}{\sqrt{h}}\right) + O_{L^2}(\sqrt{h}),
\end{align}

We then go back to (<ref>). We recall that $\xi_\star = \p_x \phi(x_\star,0)$. By replacing $c(x,0)$ by $c(x_\star,0)$ and $\phi(x,0)$ by its Taylor expansion at order $2$, we deduce that

\begin{equation}\label{eq-12w}
L_0(x) = \dfrac{c(x_\star,0) e^{\frac{i}{h}\phi(x_\star,0) + \frac{i}{h} \xi_\star(x-x_\star)}}{\sqrt{h}}  \cdot e^{i \lr{z,\p_x^2 \phi(x_\star,0)z}} \widecheck{\rho} \left(x_\star, \p_{x\xi}^2 \phi(x_\star,0) z\right) \Bigg|_{z = \frac{x-x_\star}{\sqrt{h}}} + O_{L^2}(\sqrt{h}),
\end{equation}

Hence, $L_0$ and $L$, are wavepackets concentrated at $(x_\star,\xi_\star)$ up to a remainder $O_{L^2}(\sqrt{h})$.

§.§ Dispersive wave
Now we fix $n \geq 1$, $h > 0$ and we prove upper bounds on

\begin{equation}\label{eq-9k}
B(x) \de \int_{\R^4} e^{i\frac{\phi(x,\xi) - y\xi}{h}} c(x,\xi) e^{i \vartheta_n(y_1) / \sqrt{h}}q(y_1) g_n\left( \dfrac{y_2}{\sqrt{h}}\right)  \dfrac{dy d\xi}{h^2}, \qquad \text{where:}
\end{equation}

* $q$ is admissible of order $-\infty$ (it particular it can depend implicitly on $n,h$) and has support in a bounded interval $I$;

* $g_n$ is a Hermite function, see <ref>;

* $\vartheta_n$ satisfies the bound

\begin{equation}\label{eq-11c}
\big| \p_{y_1}^j \vartheta_n(y_1) \big| \leq Cn^{1/2} \dist(y,\R \setminus I)^{-3j};
\end{equation}

* $\phi$ generates the generating function of the symplectomorphism $\kappa$ produced in Lemma <ref>.

There exists $C$ independent of $n, h$ such that under the above assumptions,

\begin{equation}
\big|B(x)\big| \leq C n^{-\infty}, \qquad x \in \R^2.
\end{equation}

1. We first prepare the integral $B(x)$. Up to multiplying by a phase, we may assume that $\phi(x,0) = 0$. By translating $y_1$, we may assume that $\p_{y_1} \phi(x,0) = 0$. We now make the substitution $\xi \mapsto \sqrt{h} \xi$ and $y_2 \mapsto \sqrt{h} y_2$ so that

\begin{equation}
B(x) = \int_{\R^4} e^{i\frac{\phi(x,\sqrt{h}\xi)}{h} + i\frac{\vartheta_n(y_1)-y_1\xi_1}{\sqrt{h}} - iy_2\xi_2} \cdot c(x,\sqrt{h} \xi) q(y_1) g_n(y_2)  \dfrac{dy d\xi}{\sqrt{h}}.
\end{equation}

We now realize the integral over $y_2$ and obtain:

\begin{equation}\label{eq-9j}
B(x) = \int_{\R^3} e^{i\frac{\phi(x,\sqrt{h}\xi)}{h} + i\frac{\vartheta_n(y_1)-y_1\xi_1}  {\sqrt{h}} } c(x,\sqrt{h} \xi) q(y_1) \hg_n(\xi_2)  \dfrac{dy_1 d\xi}{\sqrt{h}}.
\end{equation}

We comment that since the Fourier transform of the harmonic oscillator is itself, $\hg_n$ equals $g_n$ (up to a phase).

2. Fix $\ell \in \Z$; in Steps $2-4$ we assume that $q$ has support in the set $\big\{ |\vartheta_n'-\ell| \leq 1 \big\}$. Let $\chi \in C^\infty_0(\R)$, equal to $1$ on $[-3,3]$ and to $0$ outside $[-4,4]$. We write $B(x) = B_1(x) + B_2(x)$ with:

B_1(x) ∫_^3 e^iϕ(x,√(h)ξ)/h + iϑ_n(y_1)-y_1ξ_1/√(h)  c(x,√(h) ξ) q(y_1) χ(ξ_1-ℓ) _n(ξ_2)  dy_1 dξ√(h),

B_2(x) ∫_^3 e^iϕ(x,√(h)ξ)/h + iϑ_n(y_1)-y_1ξ_1/√(h)  c(x,√(h) ξ) q(y_1) (1-χ(ξ_1-ℓ)) _n(ξ_2)  dy_1 dξ√(h).

We focus on the integral in $y_1$ emerging in $B_2(x)$. For $(y_1,\xi_1)$ in the domain of integration,

\begin{equation}
\big| \xi_1-\vartheta_n'(y_1) \big| \geq |\xi_1-\ell| - |\vartheta_n'(y_1)-\ell| \geq \dfrac{|\xi_1-\ell|}{2} + \dfrac{3}{2} - 1 \geq \dfrac{\lr{\xi_1-\ell}}{2}.
\end{equation}

We integrate by parts twice according to the identity

\begin{equation}
\dfrac{-i\sqrt{h}}{\vartheta_n'(y_1)-\xi_1} \dd{}{y_1} e^{i\frac{  \vartheta_n(y_1)-\xi_1 y_1}{\sqrt{h}}}  = e^{i\frac{  \vartheta_n(y_1)-\xi_1 y_1}{\sqrt{h}}} .
\end{equation}

This yields

| ∫_e^iϑ(y_1)-y_1ξ_1/√(h)  q(y_1) dy_1√(h) | = √(h) | ∫_e^iϑ(y_1)-y_1ξ_1/√(h)  (y_11ϑ'(y_1)-ξ_1  )^2 q(y_1) dy_1 |

≤C √(nh) ∑_j=0^2 ξ_1-ℓ^j-4 ∫_(y_1,I)^-9 | _y_1^j q(y_1) | dy_1 ≤C√(nh)ξ_1-ℓ^2 ∑_j=0^2  (y_1,I)^-9 _y_1^j q _L^1.

In the second line we used the bound (<ref>) on derivatives of $\vartheta$. We note that since $q$ is smooth with support in $I$, so is $\dist(y_1,I)^{-9} \p_{y_1}^j q$. We then bound the quantity $B_2(x)$ defined in (<ref>) using (<ref>) and the integrability of $\lr{\xi_1}^{-2}$:

|B_2(x)| ≤C ∫_^2 (1-χ(ξ_1)) | _n(ξ_2) | ·| ∫_e^iϑ(y_1)-y_1ξ_1/√(h)  q(y_1) dy_1√(h) | dξ

≤C√(nh)  _n _L^1 ·∫_dξ_1ξ_1-ℓ^2 ·∑_j=0^2  (y_1,I)^-9 _y_1^j q _L^1 ≤C √(n h)  g_n _L^1 ·∑_j=0^2  (y_1,I)^-9 _y_1^j q _L^1.

4. We now bound $B_1(x)$:

\begin{equation}\label{eq-9m}
\big|B_1(x)\big| \leq \| q \|_{L^\infty} \int_{\R^2}  \big| \hg_n(\xi_2)\big| \cdot\left| \int_\R e^{i\frac{\phi(x,\sqrt{h}\xi)}{h} - i\frac{y_1\xi_1}  {\sqrt{h}} } c(x,\sqrt{h} \xi) \chi(\xi_1-\ell)  d\xi_1 \right| \dfrac{dy_1 d\xi_2}{\sqrt{h}}.
\end{equation}

We focus on the integral over $\xi_1$. We Taylor-expand $\phi$ in $\xi_1$ near $0$: there exists a smooth function $\psi$, bounded together with all its derivatives, such that

\begin{equation}\label{eq-13c}
\phi(x,\xi) = \phi(x,0,\xi_2) + \p_{y_1} \phi(x,0,\xi_2) \xi_1 + \xi_1^2 \psi(x,\xi).
\end{equation}

We emphasize that $\psi$ is constant outside a compact set because $\phi$ is quadratic outside a compact set. From plugging (<ref>) in (<ref>), we deduce

| ∫_e^iϕ(x,√(h)ξ)/h - iy_1ξ_1/√(h)   c(x,√(h) ξ) χ(ξ_1-ℓ)  dξ_1 | =
| ∫_e^i_y_1 ϕ(0,ξ_2) - y_1/√(h)ξ_1   ·f_h,x,ξ_2(ξ_1) dξ_1 |

|_h,x,ξ_2( y_1-_y_1 ϕ(0,ξ_2)  √(h) )|,
f_h,x,ξ_2(ξ_1) c(x,√(h) ξ) e^iξ_1^2 ψ(√(h)ξ)  χ(ξ_1-ℓ).

Plugging (<ref>) in (<ref>), we obtain

\begin{equation}
\big| B_1(x) \big| \leq \| q \|_{L^\infty} \int_{\R^2}  \big| \hg_n(\xi_2)\big| \cdot \left|\hf_{h,\xi_2}\left( \dfrac{y_1-\p_{y_1} \phi(0,\xi_2)}  {\sqrt{h}} \right)\right| \dfrac{dy_1 d\xi_2}{\sqrt{h}} \leq \| q \|_{L^\infty} \| \hg_n \|_{L^1} \cdot \sup_{\xi_2 \in \R} \big\| \hf_{h,\xi_2} \big\|_{L^1}.
\end{equation}

We control $\hf_{h,x,\xi_2}$ using Holder's $L^1-L^\infty$ bound and that the Fourier transform is bounded from $L^1$ to $L^\infty$:

\begin{equation}
\big\| \hf_{h,x,\xi_2} \big\|_{L^1} = \big\| \lr{y_1}^{-2} \lr{y_1}^2 \hf_{h,x,\xi_2} \big\|_{L^1} \leq \big\| (1+\p_{\xi_1}^2) f_{h,x,\xi_2} \big\|_{L^1} \leq C\lr{\ell}^4,
\end{equation}

where the constant $C$ is independent of $h, x, \xi_2$.
We recall that we assumed that the support of $q$ is contained in the set $\{ |\vartheta'-\ell| \leq 1\}$. Therefore, $\ell$ and $\vartheta'$ are comparable on the support of $q$; moreover $\lr{\vartheta'}$ is controlled by $\dist(y_1,I)^{-3}$, see (<ref>). We deduce that

\begin{equation}
\big| B_1(x) \big| \leq C \| q \|_{L^\infty} \| \hg_n \|_{L^1} \cdot \lr{\ell}^4 \leq C \big\| \dist(y_1,I)^{-12} q \big\|_{L^\infty} \| g_n \|_{L^1}.
\end{equation}

We conclude that whenever $\supp(q) \subset \{ |\vartheta'-\ell| \leq 1\}$,

\begin{equation}\label{eq-11d}
\big| B(x) \big| \leq C \sqrt{n} \| g_n \|_{L^1} \left( \big\| \dist(y_1,I)^{-12} q \big\|_{L^\infty} + \sum_{j=0}^2 \big\| \dist(y_1,I)^{-9} \p_{y_1}^j q  \big\|_{L^1} \right).
\end{equation}

5. We now prove a bound on $\| g_n \|_{L^1}$. From Cauchy–Schwarz and integrability of $\lr{x_2}^{-2}$ over $\R$, we have:

\begin{equation}
\| g_n \|_{L^1} \leq \| \lr{x_2} g_n \|_{L^2} = \lr{(1+x_2^2) g_n,g_n} \leq \lr{1 - \p^2_2 + x_2^2) g_n,g_n} = 2n+2,
\end{equation}

where we used that $g_n$ is an eigenvector of $-\p_2^2+x_2^2$ for the eigenvalue $2n+1$. Hence, (<ref>) becomes

\begin{equation}
\big| B(x) \big| \leq C n \left( \big\| \dist(y_1,I)^{-12} q \big\|_{L^\infty} + \sum_{j=0}^2 \big\| \dist(y_1,I)^{-9} \p_{y_1}^j q  \big\|_{L^1} \right).
\end{equation}

6. To conclude, we must take off the restriction on the support of $q$. Let $\tchi \in C_0^\infty(\R)$ with support in $[-1,1]$ such that $\sum_{\ell \in \Z} \tchi(y_1-\ell) = 1$. We decompose $q$ as

\begin{equation}
q(y_1) = \sum_{\ell \in \Z} \tchi\big(\vartheta'(y_1)-\ell\big) q(y_1).
\end{equation}

The terms in the sum satisfy the support condition of Steps 2-4. The additional price to pay is related to the growth of (at most two) derivatives of $\tchi\big(\vartheta'(y_1)-\ell\big)$, see (<ref>). These are controlled by $\vartheta''^2 + \vartheta'''$, hence by $n$ times $12$ additional moments of $\dist(y_1,I)^{-1}$. Summing these estimates produces

\begin{equation}
\big| B(x) \big| \leq C n^2 \left( \big\| \dist(y_1,I)^{-12} q \big\|_{L^\infty} + \sum_{j=0}^2 \big\| \dist(y_1,I)^{-21} \p_{y_1}^j q  \big\|_{L^1} \right).
\end{equation}

We can then bound these moments by sufficiently many derivatives of $q$, and use that $q$ is admissible of order $-\infty$ to conclude.

§.§ Pseudodifferential action
We recall that $g_{n,h}$ are semiclassical Hermite functions (see <ref>) and $\frak{E}_n$ are the block parametrices defined in (<ref>).

Let $a$ be an admissible function of order $-\infty$; let $r = \OO(3) + \OO(1) O(h) + O(h^2)$. Then uniformly in $t \in (0,T)$, $h \in (0,1)$ and $n \in \N$:

\begin{equation}\label{eq-13a}
r^w \big( \frak{E}_n a \otimes g_{n,h} \big) = O_{L^2} \big( h^{3/2} n^{-\infty} t^{-1/2} \big).
\end{equation}

1. First assume that $r = O(h^2)$. Then there is nothing to prove, because $r^w$ has $L^2$-norm controlled by $h^2$ and the $L^2$-norm of $\frak{E}_n a$ is controlled by $n^{-\infty} t^{-1/2}$, see (<ref>).

2. We now assume that $r = O(h) \OO(1)$. Then we can write

r   = h (x_2 r_1 + ξ_1 r_2 + ξ_2 r_3) = h (x_2 # r_1 + ξ_1 #  r_2 + ξ_2 # r_3) + O(h^2),

r^w   = h ( r_1^w ·h D_1  +  r_2^w ·x_2 +  r_3^w ·h D_2) + O(h^2)^w.

Above, $\#$ denotes the operation induced at the level of symbols by the composition of two pseudodifferential operators.
Again, for the purpose of proving estimates of the form (<ref>), we can ignore the term $O(h^2)^w$ in (<ref>). Moreover, from $2x_2 = \frak{a}+\frak{a^*}$ and $2 \p_2 = \frak{a}^*-\frak{a^*}$, we have the relations

2x_2 g_n,h = h^1/2 (√(2n+2) g_n+1,h + √(2n) g_n-1,h),

2h _2 g_n,h = h^1/2 (√(2n) g_n-1,h +  √(2n+2) g_n+1,h),

where we recall that $g_{-1} = 0$ by convention.
Therefore, using that $r_2^w$ is bounded on $L^2$, independently of $h$, and the bound (<ref>) for the $L^2$-norm of $\frak{E}_n a$ when $n \geq 1$:

\begin{equation}\label{eq-13l}
\big\| r_2^w \cdot x_2 \big( \frak{E}_n a \otimes  g_{n,h} \big) \big\|_{L^2} \leq C (nh)^{1/2} \sum_\pm \big\| \frak{E}_n a \otimes g_{n \pm 1} \big\|_{L^2} = O(h^{1/2} t^{-1/2} n^{-\infty}).
\end{equation}

This is true also when $n=0$, as can be directly checked from the formula (<ref>) for $\frak{E}_0$. Likewise, the same approach yields

\begin{equation}
\big\| r_3^w \cdot (h D_2) \big( \frak{E}_n a \otimes  g_{n,h} \big) \big\|_{L^2} \leq C (nh)^{1/2} \sum_\pm \big\| \frak{E}_n a \otimes g_{n \pm 1} \big\|_{L^2} = O(h^{1/2} t^{-1/2} n^{-\infty}).
\end{equation}

We now focus on the term $r_1^w \cdot h D_1 \frak{E}_n a \otimes g_{n,h}$. If $n \geq 1$, using that $r_1^w$ is bounded on $L^2$:

\begin{equation}\label{eq-13h}
\big\| r_1^w \cdot \big( h D_1 \frak{E}_n a \otimes g_{n,h} \big) \big\|_{L^2} \leq C \big\| h D_1 \frak{E}_n a \big\|_{L^2}.
\end{equation}

We then use the integral formula (<ref>) for $\frak{E}_n a$. It gives:

\begin{align}
h D_1 \frak{E}_n a & = (2nh)^{1/2} \cdot \sum_\pm \dfrac{\pm 1}{2\pi h_n^{1/2}} \int_\R \p_{x_1} \varphi(t,x_1,\xi_1) e^{\pm i\varphi(t,x_1,\pm \xi_1) / h_n} \cdot b^\pm_{n,h_n}(t,x_1,\xi_1) b^\pm_{n,h_n}(0,0,\xi_1)^\top \ha(\xi_1) d\xi
\\
& + h \cdot \sum_\pm \dfrac{1}{2\pi h_n^{1/2}} \int_\R e^{\pm i\varphi(t,x_1,\pm \xi_1) / h_n} \cdot D_{x_1} b^\pm_{n,h_n}(t,x_1,\xi_1) b^\pm_{n,h_n}(0,0,\xi_1)^\top \ha(\xi_1) d\xi.
\end{align}

The above integrals are of the form (<ref>). Since they come with coefficients $h^{1/2}$ and $h$, we get

\begin{equation}\label{eq-13g}
h D_1 \frak{E}_n a = O_{L^2} \big( h^{1/2} t^{-1/2} n^{-\infty}\big) .
\end{equation}

This remains true when $n=0$, as can be directly verified from the formula (<ref>). Plugging (<ref>) in (<ref>), and going back to (<ref>), we conclude that (<ref>) holds whenever $r = O(h) \OO(1)$.

3. We now assume that $r = \OO(3)$. Then we can write

r = ∑_α_1+α_2+α_3 = 3 x_2^α_1 ξ_1^α_2 ξ_2^α_1 r_α,

r^w = ∑_α_1+α_2+α_3 = 3 r_α^w ·x_2^α_1 (h D_1)^α_2 (h D_2)^α_3  + ( (2) O(h) )^w + O(h^2)^w,

where we used the expansion for the symbol of the composition of two pseudodifferential operators; see e.g. <cit.>. Again, we can ignore the term $O(h^2)^w$. Symbols $\OO(2) O(h)$ are a fortiori $\OO(1) O(h)$, so therefore we can deal with them using Step 2. We now focus on the first term. Using the relations (<ref>), we can write

\begin{equation}
x_2^{\alpha_1} (h D_1)^{\alpha_2} (h D_2)^{\alpha_3} \big(\frak{E}_n a \otimes g_{n,h}\big) = \sum_{\alpha_2=0}^3  (h D_1)^{\alpha_2} \frak{E}_n a \cdot h^{\frac{3-\alpha_2}{2}} \sum_{j = \alpha_2-3}^{3-\alpha_2} c_{n,\alpha_2,j} g_{n \pm j,h},
\end{equation}

for some coefficients $c_{n,\alpha_2,j}$ that are $O(n^{3/2})$. Following the same path as that leading to (<ref>), we conclude that

\begin{equation}
\big\| r_\alpha^w \cdot x_2^{\alpha_1} (h D_1)^{\alpha_2} (h D_2)^{\alpha_3} \cdot \big( \frak{E}_n a \otimes g_{n,h} \big) \big\|_{L^2} \leq C n^{3/2} \sum_{\alpha_2=0}^3 h^{\frac{3-\alpha_2}{2}} \big\| (h D_1)^{\alpha_2} \frak{E}_n a \big\|_{L^2} \leq C h^{3/2} n^{-\infty} t^{-1/2}.
\end{equation}

This completes the proof.

§ PROOF OF THEOREM A.2

1. Fix $(x_0,\xi_0) \in \Gamma$. According to Theorem <ref>, there exists:

* An operator $\FF$ of the form (<ref>);

* Three smooth functions $\lambda : \R \rightarrow \R^+$, $\mu, s : \R \rightarrow \R$;

* A pseudodifferential operator $R$ with symbol $\OO(3) + \OO(1) O(h) + \OO(h^2)$ near $0$;

such that, with $\Dii$ of the form (<ref>):

\begin{equation}\label{eq-12v}
\Di \FF \  = \FF( \Dii + R ).
\end{equation}

2. Let $\psi_0$ be a wavepacket concentrated at $(x_0,\xi_0)$, see Definition <ref>. Up to a phase $e^{is_h}$,

\begin{equation}
\psi_0(x) = \dfrac{e^{\frac{i}{h} \xi_0(x-x_0)} }{\sqrt{h}} \cdot A\left( \dfrac{x-x_0}{\sqrt{h}} \right), \qquad A \in \SSS(\R^2).
\end{equation}

We define $\rho \in \SSS(\R^2)$ then $\tA \in \SSS(\R^2)$ through the implicit relations:

c(x_0,0) e^i/h ϕ(x_0,0)  ·e^i y,_x^2 ϕ(x_⋆,0)y ρ_0 ( _xξ^2 ϕ(x_⋆,0) y) = A(y),

ρ_0(ξ) = e^i/2 ξ, _ξ^2 ϕ(x_0,0) ξ (ξ),     _0(x) 1√(h) ( x√(h) ).

Then, noting that $\FF \tpsi_0$ is an integral of the form (<ref>), and following Step 3 in the proof of Lemma <ref>, we obtain

_0(x) = c(x_0,0) ·e^i/hϕ(x_0,0) + i/h ξ_0 (x-x_0)√(h) ·e^i/h z,_x^2 ϕ(x_0,0)z ρ_0 ( _xξ^2 ϕ(x_⋆,0) z)  |_z = x-x_0/√(h) + O_L^2(√(h))

= e^i/h ξ_0 (x-x_0)√(h) ·A(x-x_0√(h)) + O_L^2(√(h)) = ψ_0(x) + O_L^2(√(h)).

3. We now define $\tpsi_t(x) = (\frak{E}\tpsi_0)(t,x)$. Thanks to (<ref>), we have:

\begin{equation}
(h D_t + \Di) \big(\psi_t-\FF \tpsi_t\big) = - \FF (h D_t + \Dii + R) \tpsi_t.
\end{equation}

We now apply Duhamel's formula as in <cit.>:

\begin{equation}\label{eq-12b}
\big\| \FF \tpsi_t-\psi_t\big\|_{L^2} \leq \big\| \FF \tpsi_0-\psi_0 \big\|_{L^2} + \dfrac{1}{h} \int_0^t \big\| \FF (h D_\tau + \Dii + R) \tpsi_\tau \big\|_{L^2} d\tau.
\end{equation}

To study the right-hand-side, we decompose $\tA = \sum_{n=0}^\infty \ta_n \otimes G_n$ according to (<ref>). Therefore, using that $\psi_\tau = \frak{E} \tA = \sum_{n=0}^\infty \frak{E}_n \ta_n \otimes G_{n, h}$ and the decomposition (<ref>) of $\Dii$, we deduce that

\begin{equation}
(h D_\tau + \Dii) \tpsi_\tau = (h D_\tau + L) \frak{E}_0 \ta_0 \otimes G_{0,h} + \sum_{n=1}^\infty \sqrt{2nh} (h_n D_\tau + \Dii) \frak{E}_n \ta_n \otimes G_{n,h}.
\end{equation}

In particular, thanks to (<ref>) and (<ref>):

\begin{align}
\big\| (h D_\tau + \Dii) \tpsi_\tau \big\|_{L^2} & \leq \big\| (h D_\tau + L) \frak{E}_0 \ta_0(\tau,\cdot) \big\|_{L^2} + \sum_{n=1}^\infty \sqrt{2nh} \big\| (h_n D_\tau + \Dii) \frak{E}_n \ta_n (\tau,\cdot) \big\|_{L^2}
\\
& \label{eq-13p}
\leq  Ch^{3/2} + C h^{1/2} \tau^{-1/2} \sum_{n=1}^\infty  h_n^2 n^{1/2-\infty} = O(h^{3/2} \tau^{-1/2}).
\end{align}

Likewise, thanks to Lemma <ref>:

\begin{equation}\label{eq-13r}
\big\| R \tpsi_\tau \big\|_{L^2} \leq \sum_{n=0}^\infty \big\| R \big( \frak{E}_n \ta_n \otimes G_{n,h} \big) \big\|_{L^2} = O(\tau^{-1/2} h^{3/2}).
\end{equation}

As a semiclassical Fourier integral operator, $\FF$ is bounded (independently of $h$) on $L^2$. Therefore, plugging the bounds (<ref>) and (<ref>) in (<ref>) and using that $\tau^{-1/2}$ is integrable near $0$, we conclude that uniformly in $t \in (0,T)$ and $h > 0$:

\begin{equation}\label{eq-13s}
\psi_t = \FF \tpsi_t + O_{L^2}(h^{1/2}).
\end{equation}

3. According to (<ref>), the solution $\psi_t$ to $(h D_t + \Di)\psi_t=0$ and the function $\FF \frak{E} \tpsi_0$ nearly coincide. To complete the proof of Theorem <ref>, it remains to understand $\FF \frak{E} \tpsi_0$. We have:

\begin{equation}\label{eq-13v}
\FF \frak{E} \tpsi_0 = \FF \frak{E}_0 \ta_0 \otimes G_{0,h} + \FF \sum_{n=1}^\infty \frak{E}_n \ta_n \otimes G_{n,h}.
\end{equation}

We recall that  $\frak{E}_0 \ta_0 \otimes G_{0,h}$ is a wavepacket concentrated at $(x_t^+,0)$, where $x_t^+ = e^{t\lambda(x_1) \p_{x_1}}(0)$, see (<ref>). Therefore, according to Lemma <ref>, $\FF \frak{E}_0 \ta_0 \otimes G_{0,h}$ is a wavepacket concentrated at $(x_t,\xi_t) = \kappa(x_t^+,0)$. Because of Lemma <ref>, this point satisfies the ODE

\begin{equation}
\dfrac{d(x_t,\xi_t)}{dt} = d\kappa(x_t,\xi_t) \cdot (\dot{x_t^+},0) = d\kappa(x_t,\xi_t) \cdot \lambda(x_t^+) \p_{x_1} = V_\di(x_t,\xi_t).
\end{equation}

That is, $\frak{E}_0 \ta_0 \otimes G_{0,h}(t,\cdot)$ is concentrated along the integral curve of $V_\di$ starting at $(x_0,\xi_0)$.

We now focus on the second term in (<ref>). Fix $t \in (0,T)$ and recall that $h_n = \sqrt{h/2n}$. Thanks to <ref>, we have, uniformly in $h \in (0,1), n \geq 1$:

\begin{align}
\frak{E}_n \ta_n \otimes G_{n,h}(t,y_1)
& =  \sum_{\pm} \left( e^{\pm i \varphi / h_n}  \sqrt{\dfrac{\pm i}{ 2\pi\p_\xi^2 \varphi }} \cdot b^\pm_{n,h_n}  b^\pm_{n,h_n}(0,0,\cdot)^\top \widehat{\ta}_n\right)\big( t,y_1, \pm \Xi(t,y_1) \big) \cdot G_{n,h} (y_2)  \\
& \ \ \ \  + O_{L^2}\big(h^{1/4} n^{-\infty}\big) \label{eq-13d}
\\
&  =  \dfrac{1}{h^{1/4}}\sum_{\pm}  e^{ \pm i \vartheta_{n,\pm}(y_1)/\sqrt{h}} q_\pm(y_1) \cdot G_n \left( \dfrac{y_2}{\sqrt{h}} \right) + O_{L^2}\big(h^{1/4} n^{-\infty}\big),
\end{align}

where we defined (with intentionally omitted time-dependence):

q_±(y_1) (√(±i 2π_ξ^2 φ) ·b^±_n,h_n  b^±_n,h_n(0,0,·)^⊤_n)( t,y_1, ±Ξ(t,y_1) ),

ϑ_n,±(y_1) √(2n) ·φ( t,y_1, ±Ξ(t,y_1) ).

Fourier integral operators are bounded on $L^2$, hence we can forget the remainder term in (<ref>). The function $q$ is smooth, admissible of order $-\infty$, supported in $I_t$ – see the discussion after the proof of Corollary <ref>. Moreover the function $\vartheta_{n,\pm}$ satisfies the estimates (<ref>), see Lemma <ref>. We can then apply Lemma <ref> and obtain that

sup_x ∈^2 | ( ∑_±  e^ ±i ϑ_n,±(y_1)/√(h) q_±(y_1) ·G_n ( y_2√(h) ) )(x)| ≤C n^-∞.

Summing over $n$, we conclude from (<ref>) that

\begin{equation}
\sup_{x \in \R^2} \left| \FF \sum_{n=1}^\infty \frak{E}_n \ta_n \otimes G_{n,h}(t,x) \right| \leq C h^{-1/4}.
\end{equation}

This shows that the second term in (<ref>) is $O_{L^\infty}(h^{-1/4}) + O_{L^2}(h^{1/4})$ for any fixed $t \in (0,T)$, hence completing the proof of (<ref>).

4. To complete the proof of Theorem <ref> it remains to discuss the case $\lambda_\di$ constant and $\Gamma_\di, \pi(\Gamma_\di)$ homeomorphic to $\R$. Fix $t > 0$ and set $L = \lambda_\di \cdot t$.

According to Remark <ref>, we can construct a Fourier integral operator such that (<ref>) holds with a remainder $R$ whose symbol is $\OO(3) + \OO(1) O(h) + \OO(h^2)$ on a neighborhood of $(-L -1 , L +1) \times \{0\}^3$.  Moreover, the estimate (<ref>) holds for all $t > 0$; in particular, $\tpsi_\tau$ js (up to $O_{L^2}(h^{1/4})$ remainder) concentrated in $I_t\times \{0\} = (-L, L) \times \{0\}$. As in Step 2, we deduce that $R \tpsi_\tau = O_{L^2}(\tau^{-1/2} h^{3/2})$. The Duhamel approach written above implies that $\psi_t = \FF \tpsi_t + O_{L^2}(h^{1/2})$.
The rest of the argument applies with no change. This completes the proof.

CHAPTER: APPLICATIONS TO TOPOLOGICAL INSULATORS

In this section we apply our microlocal approach to fundamental models in the field of topological insulators. The simplest one is the domain wall Dirac operator: we find the speed and orientation of propagating modes. The second example analyzes the effect of an external magnetic field. The third model is a natural extension of the latter in curved geometry. In the fourth application, we consider distortions of standard tight-binding models. By comparing our prediction with the second and third examples, we show that the propagation of modes in distorted lattice can be described in terms of pseudo metric and magnetic field.

§ DOMAIN WALL DIRAC OPERATOR

We first consider the domain wall Dirac operator from <ref>:

\begin{equation}\label{eq-13k}
\Di = hD_1 \sigma_1 + hD_2 \sigma_2 + m(x) \sigma_3,
\qquad
\di(x,\xi) = \xi_1 \sigma_1 + \xi_2 \sigma_2 + m(x) \sigma_3,
\end{equation}

where $m \in C^\infty(\R^2,\R)$ is a smooth function that vanishes transversely along its zero set. Such operators emerge as effective Hamiltonians in graphene; see [25, 53, 18, 21] for rigorous derivations starting from continuous, translation-invariant systems.

In this section, we justify that propagating modes for (<ref>) travel along $m^{-1}(0)$ at speed one; and that the estimate (<ref>) holds for all $t > 0$ when $m^{-1}(0)$ is homeomorphic to $\R$. This extends some results from [6].

§.§ Relevant quantities
We first compute the quantities $M_\di$, $\lambda_\di$, $H_\di$ and $V_\di$:

\begin{align}
M_\di(x,\xi) & = \dfrac{1}{2i} \{\di,\di\}(x,\xi) = \p_2 m (x) \sigma_1 - \p_1 m(x) \sigma_2,
\\
\lambda_\di(x,\xi) & = \left(\p_2 m(x)^2 + \p_1 m(x)^2 \right)^{1/4} = \big\| \nabla m(x) \big\|^{1/2},
\\
H_\di(x,\xi) & = \sigma_1 \dd{}{x_1} + \sigma_2 \dd{}{x_2} + \sigma_3 \left( \p_1 m(x) \dd{}{\xi_1} + \p_2 m(x) \dd{}{\xi_2} \right)
\\
V_\di(x,\xi) & = -\dfrac{\Tr(M_\di H_\di)}{2\lambda^2_\di} = \dfrac{1}{\| \nabla m(x) \|} \left( \p_1 m(x) \dd{}{x_2} - \p_2 m(x) \dd{}{x_1} \right) = \dfrac{\nabla m(x)^\perp}{\| \nabla m(x) \|}, \label{eq-0c}
\end{align}

where $u^\perp$ denotes the $\pi/2$ counterclockwise rotation of a vector $u \in \R^2$.

We note that $V_\di$ is a unit vector tangent to $\Gamma_\di = m^{-1}(0) \times \{0\}^2$. In particular, Theorem <ref> predicts the existence of a coherent state traveling (in physical space) along $\Gamma_\di = m^{-1}(0) \times \{0\}^2$ at speed one, in the direction pointing to the left of the positive-topology region (see Figure <ref>). Moreover, its orientation is given by the negative-energy eigenvector of $M_\di$: we find $[-1, e^{i\te(x)}]$ where $\te(x)$ is the angle between $\nabla m(x)$ and the $x_2$ axis.

§.§ Symplectomorphism
We construct explicitly the symplectomorphism of Theorem <ref> (and the associated Fourier integral operator), when $\di$ is the symbol (<ref>), $m^{-1}(0)$ is homeomorphic to $\R$ and $\|\nabla m(x)\| =1$ is constant along $m^{-1}(0)$.  We start with the quadratic form

\begin{equation}\label{eq-10j}
q(x,\xi) = -\det \di(x,\xi) = \xi_1^2 + \xi_2^2 + m(x)^2.
\end{equation}

\begin{equation}\label{eq-10h}
\tau(x) = \nabla m(x)^\perp, \qquad n(x) = \nabla m (x), \qquad \Omega(x) = \big[ \tau(x), n(x) \big].
\end{equation}

We note that $n$ and $\tau$ are normal and tangent to level sets of $m$, in particular to $m^{-1}(0)$; and that $\Omega(x)$ is an orthogonal matrix for every $x$. Fix $y_0 \in m^{-1}(0)$ and define $\Phi(x) = e^{x_2 n} \circ e^{x_1 \tau}(y_0)$. Since $m^{-1}(0)$ is homeomorphic to $\R$, $\Phi$ is a diffeomorphism from a tubular neighborhood of $\R \times \{0\}$ to a tubular neighborhood of $m^{-1}(0)$; we extended to a diffeomorhphism on $\R^2$;  We note that

\begin{equation}
\dd{\Phi}{x_2}(x) = n \circ \Phi(x), \qquad
\dd{\Phi}{x_1}(x_1,0) = \tau \circ \Phi(x_1,0), \qquad \dd{\Phi}{x_1}(x) = \tau \circ \Phi(x) + x_2 u(x),
\end{equation}

for some $u \in C^\infty(\R^2,\R^2)$. From (<ref>) as well as $\tau \cdot \nabla m = 0$, $n \cdot \nabla m = 1$, we deduce that

m ∘ϕ(x) = m ∘ϕ(x_1,0) + ∫_0^x_2 m ∘Φ(x_1,y)x_2 dy =
∫_0^x_1 m ∘Φ(y,0)x_1 dy + ∫_0^x_2 m ∘Φ(x_1,y)x_2 dy

= ∫_0^x_1 (τ·∇m) ∘Φ(x_1,0)  dy + ∫_0^x_2 (n ·∇m) ∘Φ(x_1,y) dy

= x_2 + O(x_2^2).

Likewise, from (<ref>) we obtain

\begin{equation}\label{eq-10i}
(\nabla \Phi^\top)^{-1} =  \left( \Omega \circ \Phi^\top + x_2 [u,0]^\top\right)^{-1} = \Omega \circ \Phi \left( \Id - x_2 [u,0]^\top \Omega \circ \Phi\right) + O(x_2^2).
\end{equation}

Define now

\begin{equation}
\kappa(x) = \big(\Phi(x), (\nabla \Phi(x)^\top)^{-1} \xi \big).
\end{equation}

This is a symplectomorphism. Using (<ref>), (<ref>) and (<ref>), we note that

q ∘κ(x) =  (∇Φ(x)^⊤)^-1 ξ^2 +  m ∘Φ(x)^2

=   ( - x_2 [u(x),0]^⊤Ω(x)) ξ^2 +  x_2^2 + O(x_2^2 ξ^2, x_2 ξ_1^2, x_2 ξ_2^2, x_2^3)

=  ξ_1^2 + ξ_2^2 + x_2^2  - 2 x_2 ξ_1 u(x), Ω(x) ξ + O(x_2^2 ξ^2, x_2 ξ_1^2, x_2 ξ_2^2, x_2^3).

In particular, there are no cubic remainders in $\xi_1 \xi_2^2$ nor $\xi_1 x_2^2$. According to Remark <ref>, $\mu = 0$. Therefore, the coefficient $\nu_t$ defined in (<ref>) vanishes. The dynamical edge state associated to (<ref>) does not experience the weak dispersion discussed below Theorem <ref>.

§.§ Dispersive estimate
Another feature of the symplectomorphism $\kappa$ is that the induced Fourier integral operator is, at least locally, a simple change of coordinates:

\begin{equation}
\FF u(x) = \int_{\R^2} e^{\frac{i}{h}(\Phi(x) - y)\xi } u(y) \dfrac{dy}{(2\pi h)^2} = u \circ \Phi(x).
\end{equation}

Hence, if we assume that $\| \nabla \kappa(x) \|$ is constant along $m^{-1}(0)$ then we are in the setup of <ref>. The dispersive estimate (<ref>) holds for all $t > 0$, see Corollary <ref>. Since $L^\infty$ is invariant under change of coordinates, it transfers back to an estimate on $\Di$. This yields Theorem <ref>(iii).

§ MAGNETIC DIRAC OPERATORS
We couple here the operator (<ref>) with a magnetic field $B = \p_1 A_1 - \p_2 A_1$, $A \in C^\infty_b(\R^2,\R^2)$:

\begin{align}
\Di & = \matrice{m(x) &  hD_1-A_1(x)-i (hD_2-A_2(x)) \\ hD_1-A_1(x) + i  (hD_2-A_2(x)) & - m(x)}  \\
& = \big(hD_1 - A_1(x)\big) \sigma_1 + (hD_2-A_2(x)\big)\sigma_2 +  m(x) \sigma_3, \\
\di(x,\xi) & = \big(\xi_1 - A_1(x)\big) \sigma_1 + (\xi_2-A_2(x)\big)\sigma_2 +  m(x) \sigma_3.\label{eq-10x}
\end{align}

As in <ref>, we first compute the quantities $M_\di$, $\lambda_\di$, $H_\di$ and $V_\di$:

\begin{align}
M_\di(x,\xi) &  = \p_2 m (x) \sigma_1 - \p_1 m(x) \sigma_2 - B(x) \sigma_3 = \matrice{B(x) & \p_2 m(x) + i \p_1 m(x) \\ \p_2 m(x) - i \p_1 m(x) & -B(x)},
\\
\lambda_\di(x,\xi) & = \left( \big\| \nabla m(x) \big\|^2 + B(x)^2 \right)^{1/4},
\\
H_\di(x,\xi) & = \sigma_1\dd{}{x_1}  + \sigma_2 \dd{}{x_2} + \sigma_3 \left( \p_1 m(x) \dd{}{\xi_1} + \p_2 m(x) \dd{}{\xi_2} \right) + O(\p_\xi),
\\
V_\di(x,\xi) & = \dfrac{\p_1 m(x) \p_{x_1} - \p_2 m(x) \p_{x_2}}{\sqrt{ \|\nabla m(x) \big\|2 + B(x)^2 }}  + O(\p_\xi)= \dfrac{\nabla m(x)^\perp}{\sqrt{ \|\nabla m(x) \big\|2 + B(x)^2 }} + O(\p_\xi),
\label{eq-0a}\end{align}

where $O(\p_\xi)$ denotes a vector field generated by $\p_{\xi_1}$ and $\p_{\xi_2}$.

We deduce from (<ref>) the value of the (physical) speed of the propagating mode:

\begin{equation}\label{eq-13m}
\dfrac{\| \nabla m(x) \big\|}{\sqrt{\| \nabla m(x)\|^2 + B(x)^2}}, \qquad x \in m^{-1}(0).
\end{equation}

In particular, (<ref>) is always smaller than $1$: magnetic fields introduced in (<ref>) through minimal coupling may not speed up the propagation of energy.  We also obtain the orientation of the propagating mode: it is the negative-energy eigenvector of (<ref>),

\begin{equation}
\matrice{-\cos(\vp(x)/2) \\ \sin(\vp(x)/2)e^{i\te(x)}}, \qquad \te(x) = \arg\big(\p_2 m(x) - i\p_1 m(x)\big), \qquad \vp(x) = \arg\big(B(x) + i \big \| \nabla m (x) \big\| \big).
\end{equation}

Obtaining these expressions by hand in [7] was a fairly technical procedure, which is now streamlined thanks to our microlocal framework.

§ DIRAC OPERATORS ON SURFACES

In this section, we consider Dirac operators $\Di$ on a curved surface $M$. We find that edge states travel at unit speed (measured with respect to the Riemannian metric). Hence, the curvature does not affect the speed of propagation. We then investigate coupling effect with a magnetic field and derive a formula similar to (<ref>).

§.§ Definitions
Let $M$ be a Riemannian surface. Topological considerations, see e.g. [37], imply that $M$ admits a spin structure. In the present context, this means that there exists a complex vector bundle $\EE \rightarrow M$ of rank $2$, such that $T^*M \otimes \End(\EE) \rightarrow M$ admits a smooth section $\sigma$ with

\begin{equation}\label{eq-11f}
\sigma(x) \xi \cdot \sigma(x)\xi' + \sigma(x) \xi' \cdot \sigma(x)\xi = 2\lr{\xi,\xi'}_{g(x)} \cdot \Id_{\EE_x}, \qquad x \in M, \quad \xi, \xi' \in T_x^*M,
\end{equation}

where $\sigma(x) \xi$ denotes the endomorphism $\sigma(x)(\xi,\cdot)$ of $\EE_x$.

Given $x \in M$ and $v \in T_x$, we let $v^{\perp_g} \in T_x M$ be the unique vector orthogonal to $v$, with same norm, such that $d\vol(v,v^{\perp_g}) > 0$, where $d\vol$ is the Riemannian volume form (in particular, for $f \in C^\infty(M,\R)$, $\nabla_g f^{\perp_g}$ is always tangent to the level sets of $f$). Using the isomorphism between $T_xM$ and $T_x^*M$, we can likewise define $\xi^{\perp_g} \in T_x^*M$ for $\xi \in T^*_xM$. We now set

\begin{equation}\label{eq-11g}
\sigma_3(x) \de \dfrac{1}{i} \sigma(x) \xi \cdot \sigma(x) \xi^{\perp_g}, \qquad |\xi|_{g(x)}^2 = 1.
\end{equation}

We note that the right-hand-side does not depend on $\xi$: indeed, if $\xi' \in T_xM$ has unit norm, then decomposing $\xi' = \cos \te \xi + \sin \te \xi^{\perp_g}$ we obtain

σ(x) ξ' ·σ(x) ξ'^⊥_g = σ(x) ( cosξ+ sinξ^⊥_g ) ·σ(x) ( cosξ^⊥_g - sinξ)

= (cos^2 +sin^2 ) σ(x) ξ·σ(x) ξ^⊥_g + sincos( (σ(x) ξ^⊥_g)^2 - (σ(x) ξ)^2 )
= σ(x) ξ·σ(x) ξ^⊥_g.

We used that $\sigma(x) \xi$ and $\sigma(x) \xi^{\perp_g}$ anticommute, see (<ref>); and that $\xi, \xi^{\perp_g}$ have the same norm.

Given a function $m \in C^\infty(M,\R)$, we define

\begin{equation}\label{eq-11i}
\p(x,\xi) = \sigma(x) \xi + m(x)\sigma_3(x),
\end{equation}

which is a smooth section of $\EE \rightarrow T^*M$. Let $\Di$ be a selfadjoint operator on acting on smooth sections of $\EE \rightarrow M$, with principal symbol $\di(x,\xi)$.

§.§ Speed of propagating state
We compute the quantities $\{\di,\di\}$ and $H_\di$. Let $(x_1,x_2)$ be normal coordinates centered at a point of $m^{-1}(0)$; $(\xi_1,\xi_2)$ be induced local coordinates on the fibers of $T^*M$; and $(e_1,e_2)$ be a local frame on $\EE$. Noting that $\sigma(x) \xi$ is Hermitian and traceless, we may write (in these coordinates):

σ_1(x) σ(x) (1,0) = ∑_k=1^3 h_1k(x) σ_k,     σ_2(x) σ(x) (0,1) = ∑_k=1^3 h_2k(x) σ_k,

for some smooth functions $h_{jk}$ defined near $0$. From (<ref>) and using that we are working in normal coordinates,

∑_k=1^3 h_jk(x) h_ℓk(x) = g^jℓ(x) = δ_jℓ + O(x^2).

Hence, $h_1(0)$ and $h_2(0)$, seen as vectors of $\R^3$, are orthonormal
. Moreover, from (<ref>) we have

σ_3(x) = σ_1(x) σ_2(x)i |g^11(x) g^22(x)|^1/2 = ∑_k,ℓ=1^3 h_2k(x) h_1ℓ(x) σ_k σ_ℓi + O(x^2)

= ∑_k,ℓ,m=1^3  _kℓm h_2k(0) h_1ℓ(0) σ_m + O(x^2) = ∑_m=1^3 ( h_1(0) ×h_2(0))_m σ_m + O(x^2).

Therefore, in the Pauli basis $(\sigma_1, \sigma_2, \sigma_3)$, the matrices $\sigma_1(0), \sigma_2(0), \sigma_3(0)$ have coordinates $h_1(0)$, $h_2(0)$ and $h_1(0) \times h_2(0)$; since $h_1(0)$ and $h_2(0)$ are orthonormal, $\sigma_1(0), \sigma_2(0), \sigma_3(0)$ form a direct orthonormal basis of traceless, Hermitian $2 \times 2$ matrices: their matrix in the Pauli basis is $[h_1,h_2, h_1 \times h_2]$. After conjugating $\di$ with an adequate matrix in $SU(2)$ (which does not affect the value of $\lambda_\di$ or $V_\di$), we can assume that $\sigma_j(0) = \sigma_j$. It follows that (in these coordinates, and up to conjugation):

\begin{align}
\di(x,\xi) & =  \xi_1 \sigma_1 + \sigma_2 \xi_2 + \big(\p_1 m(0) x_1 + \p_2 m(0) x_2\big) \sigma_3 + O(x,\xi)^2,
\\
M_\di(0) & = \p_2 m(0) \sigma_1 -\p_1 m(0) \sigma_2 ,
\\
\label{eq-0g} \lambda_\di(0)^4 & = \p_2 m(0)^2 + \p_1 m(0)^2 =  \| \nabla_g m(0) \|_{g(0)}^2,
\\
H_\di(0) & = \sigma_1 \dd{}{x_1} + \sigma_2 \dd{}{x_2} - \left(\p_1 m(0) \dd{}{\xi_1} + \p_2 m(0) \dd{}{\xi_2}\right) \sigma_3 ,
\\
V_\di(0) & = \dfrac{\p_1m(0)\p_{x_2} - \p_2m(0) \p_{x_1}}{\| \nabla_g m(0) \|_{g(0)}}  = \dfrac{\nabla_g m(0)^{\perp_g}}{\| \nabla_g m(0) \|_{g(0)}},
\end{align}

where we used that $g(0) = dx_1^2 +dx_2^2$ in normal coordinates centered at $0$. Going back to coordinate-free expression, and recalling that the origin of the normal coordinates system was an arbitrary point in $m^{-1}(0)$, we conclude that on $\Gamma = m^{-1}(0) \times \{0\}^2$:

\begin{equation}\label{eq-0z}
V_\di(x,0) = \dfrac{\nabla_g m(x)^{\perp_g}}{\| \nabla_g m (x)\|_g}, \qquad x \in m^{-1}(0).
\end{equation}

The identity (<ref>) implies that even in curved geometry, the speed of dynamical edge states for Dirac operators defined by semiclassical quantizations of (<ref>) (measured with respect to the metric $g$) is independent of the choice of interface function $m$.

We believe that a non-vanishing curvature may lead to a coefficient $\nu_t \neq 0$ in (<ref>), in particular to dispersion of edge states as $t \rightarrow \infty$; we do not investigate this effect further.

§.§ Coupling to magnetic fields
Turning on a magnetic $B \in C^\infty(M,\R)$ in (<ref>) amounts to translating the momentum $\xi$ by a smooth one-form $A$ on $M$ such that $dA = B d\vol_g$; see e.g. [10] for the Schrödinger operator. This produces the symbol

\begin{equation}\label{eq-0f}
\p(x,\xi) = \sigma(x) \big(\xi-A(x)\big) + m(x)\sigma_3(x).
\end{equation}

To compute the speed of a propagating state in curved geometry, with the magnetic field $B$, it suffices again to fix $x \in M$ and work in normal coordinates. Write (in this system of coordinates) $A(x) = A_1(x) dx_1 + A_2(x) dx_2$ and note that $d\vol_g(0) = dx_1 \wedge dx_2$; hence

\begin{equation}
B(0) = \p_1 A_2(0) - \p_2 A_1(0).
\end{equation}

In particular, (<ref>) becomes:

\begin{align}
\di(x,\xi) & =  \sum_{j=1}^2 \big( \xi_j - A_1(0) - \p_{x_1} A_j(0) x_1 - \p_{x_2} A_j(0) x_2 \big)  \sigma_j + \big(\p_1 m(0) x_1 + \p_2 m(0) x_2\big) \sigma_3 + O(x,\xi)^2,
\\
M_\di(0) & = \p_2 m(0) \sigma_1 -\p_1 m(0) \sigma_2 + \big( \p_1 A_2(0)  - \p_2 A_1(0) \big)\sigma_3  = \p_2 m(0) \sigma_1 -\p_1 m(0) \sigma_2 + B(0) \sigma_3,
\\
\lambda_\di(0)^4 & = \p_2 m(0)^2 + \p_1 m(0)^2 + B(0)^2 =  \big\| \nabla_g m(0) \big\|_{g}^2 + B(0)^2,
\\
H_\di(0) & = \sigma_1 \dd{}{x_1} + \sigma_2 \dd{}{x_2}  + O(\p_\xi),
\\
V_\di(0) & = \dfrac{\p_1m(0)\p_{x_2} - \p_2m(0) \p_{x_1}}{\sqrt{\big\| \nabla_g m(0) \big\|_g^2  + B(0)^2}} + O(\p_\xi)  = \dfrac{\nabla_g m(0)^{\perp_g}}{\sqrt{\big\| \nabla_g m(0) \big\|_g^2  + B(0)^2}} + O(\p_\xi).
\end{align}

Varying the origin of the normal coordinate system, we obtain a formula analogous to (<ref>):

\begin{equation}\label{eq-0h}
d\pi(V_\di)(x) = \dfrac{\nabla_g m(x)^{\perp_g}}{\sqrt{\big\| \nabla_g m (x) \big\|_g^2 + B(x)^2}}, \qquad x \in m^{-1}(0).
\end{equation}

§ ANISOTROPIC HALDANE MODELS

In this section, we study symbols of the form

\begin{equation}\label{eq-0b}
\di(x,\xi) = \matrice{m(x) & \overline{\alpha(x)} \cdot \big(\xi-\xi(x)\big) \\ \alpha(x)\cdot \big( \xi-\xi(x)\big) & -m(x)}   = \sum_{j=1}^2 \alpha_j(x) \cdot (\xi-\xi(x)) \sigma_j + m(x)\sigma_3
\end{equation}

where $\alpha = \alpha_1+i\alpha_2$, $\alpha_1, \alpha_2 \in C^\infty_b(\R^2,\R^2)$; $\xi \in C^\infty(\R^2,\R^2)$; and $m \in C^\infty_b(\R^2,\R)$.

We first motivate the origin of (<ref>) as an effective Hamiltonians for distorsions of standard models of condensed matter physics.  We then pull out a metric and a magnetic field such that (<ref>) equals a symbol of the form (<ref>). By analogy with <ref>, this allows us to produce a formula for the speed of propagating modes. At the physical level, our result implies that adiabatic variation in the position of atoms slows down electronic transport along interfaces between topological insulators.

§.§ Tight-binding models
Our starting point is Wallace's Hamiltonian:

\begin{equation}\label{eq-13e}
(H_0\psi)_n = \matrice{\psi_n^B + \psi_{n-v_1}^B + \psi_{n-v_2}^B \\ \psi_n^A + \psi_{n+v_1}^A + \psi_{n+v_2}^A }, \qquad v_1 = -\dfrac{1}{2}\matrice{1 \\ \sqrt{3}}, \qquad v_2 = \dfrac{1}{2}\matrice{1 \\ -\sqrt{3}},
\end{equation}

with wavefunction $\psi \in \ell^2(\Z v_1 \oplus \Z v_2, \C^2)$. This Hamiltonian describes the motion of an electron on a honeycomb lattice, with hopping probabilities equally weighted on the nearest neighbors; see Figure <ref>. Given the periodicity of $H$ with respect to the lattice $\Z v_1 \oplus \Z v_2$, it is natural to take its discrete Fourier transform:

\begin{equation}\label{eq-0n}
H_0(\xi) = \matrice{0 & \w(\xi) \\ \overline{\w(\xi)} & 0}, \qquad \w(\xi) = 1+e^{i\xi v_1} + e^{i\xi v_2}.
\end{equation}

The eigenvalues of $H(\xi)$ are $\pm \big|\w(\xi)\big|$. In particular, they are degenerate precisely when $\xi = \pm \xi_\star \mod 2\pi$, where $\xi_\star = (2\pi/3,-2\pi/3)$. The crossing is conical:

\begin{equation}
\w(\xi_\star + \xi) = \dfrac{\sqrt{3}}{4} (\xi_1 + i \xi_2) + O(\xi^2).
\end{equation}

In particular, the model (<ref>) describes a semimetal.

Small perturbations of semimetals generally transforms them in insulators. In a seminal work [41], Haldane introduced a time-breaking Hermitian perturbation of Wallace's model that was the first model of topological insulator beyond the quantum Hall effect. Specifically:

(H_m ψ)_n  = (H_0 ψ)_n + i m ψ_n+v_1^A - ψ_n-v_1^A  - ψ_n+v_2^A + ψ_n-v_2^A + ψ_n+v_2-v_1^A - ψ_n+v_1-v_2^A

ψ_n+v_1^B - ψ_n-v_1^B  - ψ_n+v_2^B + ψ_n-v_2^B - ψ_n+v_1-v_2^B + ψ_n-v_1+v_2^B,

H_m(ξ) = H_0(ξ) + 2im β(ξ) σ_3,     β(ξ) = sin(ξv_1) - sin(ξv_2) - sin(ξ(v_1-v_2)).

In brief words, the states associated to the valence band (i.e. the negative energy eigenstates of $H_m(\xi)$, seen as functions on a $2$-torus), form a vector bundle with non-trivial topology: its Chern number is $\sgn(m)$. We also comment that $\beta(\xi_\star) = 3\sqrt{3}/2$.

[capbesideposition =right,center]figure[10cm]
A honeycomb structure. An electron on a red site can hop to the nearest blue site (and conversely).

at (0,0) ;

The standard approach to study systems such as (<ref>) near momentum $\xi_\star$ is to look for solutions of $(D_t + H_m) \psi = 0$ that take the form $\psi_n(t)= \Psi(ht,hn) e^{i\xi_\star n}$. This produces the effective equation

\begin{equation}
\left( hD_t + \dfrac{\sqrt{3}}{4}\matrice{m & hD_1-ihD_2 \\ hD_1+ihD_2 & -m} \right) \Psi = 0.
\end{equation}

By slowly changing the sign of $m$ over space, we can generate structures whose topology is globally ill-defined but locally well defined when $m \neq 0$. The set $m^{-1}(0)$ acts as an interface between insulating phases of distinct Chern numbers. The resulting effective equation is:

\begin{equation}\label{eq-0d}
\left( hD_t + \dfrac{\sqrt{3}}{4}\matrice{m(x) & hD_1-ihD_2 \\ hD_1+ihD_2 & -m(x)}\right) \Psi = 0,
\end{equation}

which is precisely that studied in <ref>. According to Theorem <ref>, solutions to (<ref>) split in two parts, one transporting in a direction prescribed by topology (specifically, the positive-topology region lies to the right of the direction of propagation); and one dispersing.

While defects do not affect the insulating character of topological phases, they can influence the speed or profile of conduction along interfaces. To investigate these effects, we introduce anisotropy in the system. This corresponds to a Wallace model with the function

\begin{equation}
\w_a(\xi) = 1-a_1 - a_2 + a_1 e^{i\xi_1} + a_2 e^{i\xi_2}
\end{equation}

instead of $\w(\xi)$ in (<ref>); the parameter $a = (a_1,a_2) \in \R^2$ acounts for anistropic hopping probabilities $\frac{1-a_1-a+2}{3}$,$\frac{1+a_1}{3}$, $\frac{1+a_2}{3}$, see Figure <ref>. For sufficiently small $a$, the equation $\w_a(\xi) = 0$ still has precisely two solutions $\pm (\xi_\star+\xi_a)$ modulo $2\pi$; this means that such perturbations preserve the semimetal character. Since we have

\begin{equation}
\w_a(\xi_\star + \xi) = \p_\xi \w_a(\xi_a) \cdot (\xi-\xi_a) + O(\xi-\xi_a)^2,
\end{equation}

an effective operator for the anisotropic Haldane model near momentum $\xi_\star-\xi_a$ is

\begin{equation}
\matrice{0 &  \ove{\p_\xi \w_a(\xi_a)} \cdot (D_x - \xi_a) \\ \w_a(\xi_a) \cdot (D_x - \xi_a) & 0}.
\end{equation}

[capbesideposition =right,center]figure[10cm]
A distorted honeycomb structure. We drew the original hexagonal structure in the background (light gray).

at (0,0) ;

Adding complex coupling as in [41], and allowing the hopping probabilities and complex coupling to depend on $x$, we end up with an effective Hamiltonian $\Di = \di^w$ with symbol

\begin{equation}\label{eq-0e}
\di(x,\xi) = \matrice{m(x) & \alpha(x) \big(\xi-\xi(x)\big) \\ \alpha(x) \big( \xi-\xi(x)\big) & -m(x)}.
\end{equation}

This Hamiltonian models a honeycomb lattice with microscopic unit cell, submitted to a macroscopic strain.

§.§ Pseudo metric and magnetic field

\begin{equation}
\sigma(x) \xi = (\alpha_j(x) \cdot \xi) \sigma_j.
\end{equation}

In analogy with <ref>, we define a metric $g$ on $\R^2$ by (its action on the cotangent bundle $T^*\R^2$):

\begin{equation}
|\xi|^2_{g(x)}= \sum_{j=1}^2(\alpha_j(x) \cdot \xi)^2 = (\sigma(x) \xi)^2.
\end{equation}

Following (<ref>), we set

\begin{equation}
\sigma_3(x) = \dfrac{1}{i} \sigma(x) \xi \cdot \sigma(x) \xi^{\perp_g}, \qquad |\xi|_{g(x)} = 1,
\end{equation}

which is independent of (adequately normalized) $\xi$. By plugging for $\xi$ a vector in a basis dual to $\alpha_1(x), \alpha_2(x)$, we end up with

\begin{equation}
\sigma_3(x) = \det\big[\alpha_1(x),\alpha_2(x)\big] \sigma_3.
\end{equation}

Hence, the symbol (<ref>) has the form (<ref>) for a domain wall $\tm = m / \det[\alpha_1, \alpha_2]$ and a magnetic potential $A(x) = \xi(x)$. From $dA = B d\vol_g$, we deduce that the effective magnetic field is

\begin{equation}
B(x) = \dfrac{\p_2 \xi_1(x) - \p_1 \xi_2(x)}{\det[\alpha_1(x),\alpha_2(x)]}.
\end{equation}

This implies that to a distorted Haldane model correspond an emerging metric and magnetic field. This relates to the numerical and experimental observation of pseudomagnetism in strained graphene [56, 34]. Based on this analogy and on (<ref>), our microlocal framework produces a direct formula for the (physical) speed of the propagating state:

dπ(V_)(x) = ∇_g (x)^⊥_g√( ∇_g (x)_g^2 + B(x)^2)
= ∇_g m(x)^⊥_g√( ∇_g m (x)_g^2 +  dξ(x) _g^2),     x ∈m^-1(0),

where we used the definition of $B$; and the relation between $m$ and $\tm$ together with $x \in m^{-1}(0)$. This implies that when appropriately measured, microscopic variations in atomic position can only slow down the propagation of edge states.

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