Dispersion for Schrödinger operators on regular trees

Abstract

We prove dispersive estimates for two models: the adjacency matrix on a discrete regular tree, and the Schrödinger equation on a metric regular tree with the same potential on each edge/vertex. The latter model can be thought of as an extension of the case of periodic Schrödinger operators on the real line. We establish a $t^{-3/2}$-decay for both models which is sharp, as we give the first-order asymptotics.

Appendices

Appendix A. Stationary phase result

In this appendix we give an explicit stationary phase estimate. To put things into perspective, we need a stronger statement than the Van der Corput Lemma [30,  Corollary, p. 334] in the sense that we want an asymptotic $\sim$ for the principal term, but we accept a weaker statement than the full asymptotic given in [30,  Proposition 3, p. 334] as we only care about the principal term. Our point is to make the remainder explicit in the phase function and observable, with as few derivatives as possible. This is important for our applications for quantum graphs, where we need to apply this for a series of integrals so we have to ensure the series of errors converge. For this we shall use the explicit version of [25].

Theorem A.1

Let $p\in C^1[a,b]$, $q \in C[a,b]$ and suppose p and q admit a Taylor expansion at $x=a$. Assume $x_0=a$ is the only critical point of p in [a, b], so $p'(a)=0$ and $p'(x)\ne 0$ for $x\in (a,b]$. Assume moreover $p''(a)\ne 0$ and let $\epsilon ={{\,\mathrm{sgn}\,}}p''(a)$. Then

$$\begin{aligned} \int _a^b\mathrm {e}^{\mathrm {i}t p(x)}q(x)\,\mathrm {d}x = \mathrm {e}^{\mathrm {i}t p(a)}\mathrm {e}^{\epsilon \pi \mathrm {i}/4}\sqrt{\frac{\pi }{2\epsilon p''(a)t}}q(a) + \delta _{0,1}(t)-\varepsilon _{0,1}(t), \end{aligned}$$

where

$$\begin{aligned} \delta _{0,1}(t)= & {} \int _a^b\mathrm {e}^{\mathrm {i}tp(x)}\bigg (q(x)-\frac{\epsilon q(a)p'(x)}{\sqrt{2\epsilon p''(a)}\sqrt{\epsilon (p(x)-p(a))}}\bigg )\,\mathrm {d}x, \\ \varepsilon _{0,1}(t)= & {} \mathrm {e}^{\mathrm {i}tp(a)}\mathrm {e}^{\epsilon \pi \mathrm {i}/4}\Gamma \Big (\frac{1}{2},\mathrm {i}t\epsilon (p(a)-\mathrm {i}p(b))\Big )\frac{q(a)}{\sqrt{2\epsilon p''(a) t}} \end{aligned}$$

and $\Gamma (\alpha ,z) = \int _z^\infty \mathrm {e}^{-t}t^{\alpha -1}\,\mathrm {d}t$.

The statement can be greatly generalized : one can replace $p''(a)\ne 0$ by $p^{(k)}(a)\ne 0$, where $p^{(j)}(a)=0$ for all $j<k$. The function q(x) can also have an algebraic singularity $(x-a)^{-\rho }$, $0\le \rho <1$. Finally higher order precision is also available with explicit terms.

Proof

First assume $p'(x)>0$ for all $x\in (a,b]$. We apply [25,  Theorem 1] with $\lambda =1$, $\mu =2$, $m=0$, $n=1$. Since p, q admit Taylor expansions at $x=a$, we have $p(x)=p(a)+\sum _{s=0}^\infty p_s(x-a)^{s+2}$ and $q(x)=\sum _{s=0}^\infty q_s(x-a)^s$ with $p_0 = \frac{p''(a)}{2}$, $q_0=q(a)$. Let $a_0 = \frac{q_0}{2\sqrt{p_0}}$. Noting that $n>m\mu -\lambda$, we take $\nu =1$ and get

$$\begin{aligned} \int _a^b\mathrm {e}^{\mathrm {i}t p(x)}q(x)\,\mathrm {d}x = \mathrm {e}^{\mathrm {i}t p(a)}\mathrm {e}^{\pi \mathrm {i}/4}\Gamma \Big (\frac{1}{2}\Big )\frac{a_0}{t^{1/2}}+\delta _{0,1}(t)-\varepsilon _{0,1}(t), \end{aligned}$$

with $\delta _{0,1}(t)= \int _a^b \mathrm {e}^{\mathrm {i}t p(x)}Q_{0,1}'(x)\,\mathrm {d}x$ and $Q_{0,1}(x) = \int _0^xq(y)\,\mathrm {d}y - \frac{\Gamma (\frac{1}{2})}{\Gamma (\frac{3}{2})}a_0\sqrt{p(x)-p(a)}$, and $\varepsilon _{0,1}(t)=\mathrm {e}^{\mathrm {i}tp(a)}\mathrm {e}^{\pi \mathrm {i}/4}\Gamma (\frac{1}{2},\mathrm {i}tp(a)-\mathrm {i}tp(b))\frac{q_0/2\sqrt{p_0}}{\sqrt{t}}$. The statement follows when ${{\,\mathrm{sgn}\,}}p''(a)>0$. Indeed, expanding $p'(x)= p''(a)(x-a)(1+O(x-a))$, we see that that $p'(x)>0$ iff ${{\,\mathrm{sgn}\,}}p''(a)>0$.

Now assume $p'(x)<0$ for all $x\in (a,b]$, implying ${{\,\mathrm{sgn}\,}}p''(a)<0$. As remarked in [25], the theorem remains true by essentially replacing $\mathrm {i}$ by $-\mathrm {i}$ through most of the proof. More precisely, here p(x) is decreasing so with the notations of [25], one considers the change of variables $v=p(a)-p(x)$ instead. Then $\int _a^b \mathrm {e}^{\mathrm {i}tp(x)}q(x)\,\mathrm {d}x = \mathrm {e}^{\mathrm {i}t p(a)}\int _0^{p(a)-p(b)}\mathrm {e}^{-\mathrm {i}t v}f(v)\,\mathrm {d}v$, with $f(v)=\frac{q(x)}{-p'(x)}$. This shows all $p_s$ get replaced by $-p_s$. Moreover, for (5.4) to hold, we should take $P_j(x)=\left\{ \frac{-1}{p'(x)}\frac{\mathrm {d}}{\mathrm {d}x}\right\} ^j\frac{q(x)}{-p'(x)}$. We take $\beta =p(a)-p(b)$. Lemmas 1,2,3 in [25,  Section 4] continue to hold verbatim if we replace $\mathrm {i}$ by $-\mathrm {i}$ on both hypothesis and conclusion (e.g. now $\lim \nolimits _{\eta \downarrow 0}\int _\beta ^\infty \mathrm {e}^{-\eta v}\mathrm {e}^{-\mathrm {i}tv}v^{\alpha -1}\,\mathrm {d}v = \frac{\mathrm {e}^{-\alpha \pi \mathrm {i}/2}}{x^\alpha }\Gamma (\alpha ,\mathrm {i}t\beta )$). Returning to $\int _0^\beta \mathrm {e}^{-\mathrm {i}tv}f(v)\,\mathrm {d}v$, we see that $\mathrm {i}$ should be replaced by $-\mathrm {i}$ everywhere in (5.8)–(5.11). The same replacement holds for (5.12) and (5.13), except for the terms containing $\varepsilon ,\delta$, i.e. we have $\mathrm {e}^{-\mathrm {i}tp(a)}\varepsilon _{m,n}(t)$ and $\mathrm {e}^{-\mathrm {i}tp(a)}\{\delta _{m,n}(t)-\varepsilon _{m,n}(t)\}$, respectively. (5.14) becomes $\mathrm {e}^{\mathrm {i}tp(a)}(\frac{-\mathrm {i}}{t})^m\int _0^\beta \mathrm {e}^{-\mathrm {i}tv}\phi _n^{(m)}(v)\,\mathrm {d}v$. With our choice of $P_j$, $\phi _n^{(m)}$ has the required form, completing the proof. $\square$

The following corollary is the main tool we use instead of the Van der Corput Lemma, cf. [30,  Corollary, p. 334], to obtain sharp estimates.

Corollary A.2

Under the assumptions of the previous theorem, define

$$\begin{aligned} Q_{1,1}(x) = \frac{q(x)}{\epsilon p'(x)} - \frac{q(a)}{\sqrt{2\epsilon p''(a)}\sqrt{\epsilon (p(x)-p(a))}}. \end{aligned}$$

(A.1)

Then

$$\begin{aligned}&\bigg |\int _a^b\mathrm {e}^{\mathrm {i}tp(x)}q(x)\,\mathrm {d}x - \mathrm {e}^{\mathrm {i}t p(a)}\mathrm {e}^{\epsilon \pi \mathrm {i}/4}\sqrt{\frac{\pi }{2|p''(a)|t}}q(a)\bigg | \\&\quad \le \frac{1}{t}\bigg (| Q_{1,1}(a)| + |Q_{1,1}(b)| + V_{a,b}(Q_{1,1}) + \frac{2|q(a)|}{\sqrt{2|p''(a)|}\sqrt{|p(b)-p(a)|}}\bigg ), \end{aligned}$$

where $V_{a,b}(Q_{1,1}) = \int _a^b|Q_{1,1}'(y)|\,\mathrm {d}y$ is the total variation of $Q_{1,1}$ over [a, b].

Proof

Apply [25,  Eq. (6.3), (6.7)] to the previous theorem. $\square$

Example A.3

As is well-known, for any $\alpha \in \mathbb {R}$, the Fresnel integral $\int _0^\infty \mathrm {e}^{\mathrm {i}t\alpha x^2}\,\mathrm {d}x = \frac{\mathrm {e}^{\epsilon \pi \mathrm {i}/4}}{2}\sqrt{\frac{\pi }{|\alpha |\,t}}$, where $\epsilon = {{\,\mathrm{sgn}\,}}\alpha$. The previous result tells us that if we cutoff at $A>0$, then

$$\begin{aligned} \bigg |\int _0^A\mathrm {e}^{\mathrm {i}t\alpha x^2}-\frac{\mathrm {e}^{\epsilon \pi \mathrm {i}/4}}{2}\sqrt{\frac{\pi }{|\alpha |\,t}}\bigg |\le \frac{1}{A|\alpha |\,t}. \end{aligned}$$

Indeed, here $Q_{1,1}\equiv 0$.

In general, we should compute the limit $Q_{1,1}(a)$ carefully. Say $\epsilon =1$. Then $Q_{1,1}(x) = \frac{q(x)\sqrt{2p''(a)}\sqrt{p(x)-p(a)}-q(a)p'(x)}{p'(x)\sqrt{2p''(a)}\sqrt{p(x)-p(a)}}$. Expanding $p(x)=p(a)+\frac{p''(a)}{2}(x-a)^2+\frac{p'''(a_x)}{6}(x-a)^3$, also $p'(x)=p''(a)(x-a)+\frac{p'''({\tilde{a}}_x)}{2}(x-a)^2$ and $q(x) = q(a)+q'({\hat{a}}_x)(x-a)$, for some $a_x,{\tilde{a}}_x,{\hat{a}}_x\in (a,x)$, the numerator becomes

$$\begin{aligned}&q(x)\bigg [p''(a)(x-a)\sqrt{1+\frac{p'''(a_x)}{3p''(a)}(x-a)}\bigg ]-q(a)p'(x)\\&\quad = \left[ q(a)+q'({\hat{a}}_x)(x-a)\right] \cdot p''(a)(x-a)\cdot \bigg [1+\frac{p'''(a_x)}{6p''(a)}(x-a)+O(x-a)^2\bigg ]\\&\qquad -q(a)\Big [p''(a)(x-a)+\frac{p'''({\tilde{a}}_x)}{2}(x-a)^2\Big ]\\&\quad = q(a)\frac{p'''(a_x)}{6}(x-a)^2+O(x-a)^3+q'({\hat{a}}_x)p''(a)(x-a)^2\\&\qquad -\frac{q(a)p'''({\tilde{a}}_x)}{2}(x-a)^2 \end{aligned}$$

while the denominator is $[p''(a)(x-a)+O(x-a)^2][p''(a)(x-a)\sqrt{1+O(x-a)}]$. Thus,

$$\begin{aligned} Q_{1,1}(a) = \frac{q(a)\frac{p'''(a)}{6}+q'(a)p''(a)-\frac{q(a)p'''(a)}{2}}{p''(a)^2} = \frac{q'(a)}{p''(a)}-\frac{q(a)p'''(a)}{3p''(a)^2}.\qquad \end{aligned}$$

(A.2)

The same calculation shows that in general $Q_{1,1}(a) = \epsilon \big (\frac{q'(a)}{p''(a)}-\frac{q(a)p'''(a)}{3p''(a)^2}\big )$.

We thus have in all cases

$$\begin{aligned}&\bigg |\int _a^b\mathrm {e}^{\mathrm {i}tp(x)}q(x)\,\mathrm {d}x - \mathrm {e}^{\mathrm {i}t p(a)}\mathrm {e}^{\epsilon \pi \mathrm {i}/4}\sqrt{\frac{\pi }{2|p''(a)|t}}q(a)\bigg | \nonumber \\&\quad \le \frac{1}{t}\bigg (\Big |\frac{q'(a)}{p''(a)}-\frac{q(a)p'''(a)}{3p''(a)^2}\Big |\nonumber \\&\qquad + \Big |\frac{q(b)}{p'(b)}\Big | + V_{a,b}(Q_{1,1}) + \frac{3|q(a)|}{\sqrt{2|p''(a)|}\sqrt{|p(b)-p(a)|}}\bigg ). \end{aligned}$$

(A.3)

If the only critical point is at $x=b$ instead, then via the change of variables $y=-x$, ${\tilde{p}}(y)=p(-y)$ and ${\tilde{q}}(y)=q(-y)$, we see that

$$\begin{aligned}&\bigg |\int _a^b\mathrm {e}^{\mathrm {i}tp(x)}q(x)\,\mathrm {d}x - \mathrm {e}^{\mathrm {i}t p(b)}\mathrm {e}^{\epsilon \pi \mathrm {i}/4}\sqrt{\frac{\pi }{2|p''(b)|t}}q(b)\bigg | \nonumber \\&\quad \le \frac{1}{t}\bigg (\Big |\frac{q'(b)}{p''(b)}-\frac{q(b)p'''(b)}{3p''(b)^2}\Big | + \Big |\frac{q(a)}{p'(a)}\Big | + V_{a,b}({\widetilde{Q}}_{1,1}) \nonumber \\&\qquad + \frac{3|q(b)|}{\sqrt{2|p''(b)|}\sqrt{|p(b)-p(a)|}}\bigg ), \end{aligned}$$

(A.4)

where $\epsilon = {{\,\mathrm{sgn}\,}}p''(b)$ and ${\widetilde{Q}}_{1,1}(x) = \frac{q(x)}{\epsilon p'(x)} - \frac{q(b)}{\sqrt{2\epsilon p''(b)}\sqrt{\epsilon (p(x)-p(b))}}$.

Remark A.4

We conclude this appendix by comparing our statement (which is just a streamlined account of [25]) with some classical references.

• (1) The well-known Van der Corput lemma [30,  Corollary p. 334] gives some explicit bound over $\big |\int _a^b\mathrm {e}^{\mathrm {i}t p(x)}q(x)\,\mathrm {d}x \big |$. However, this only yields an upper bound, not an asymptotically equivalent term. Moreover, it requires the additional condition $|p''(x)|\ge 1$ over [a, b].

• (2) The full asymptotics given in [30,  p. 334] or [34,  p. 41] do not have explicit bounds on the remainder. Inspecting the proof of [30], one first restricts the integral to a neighborhood $N_{\epsilon }(a)$ of the critical point a such that $\left| \frac{p'''(x)}{3p''(a)}(x-a)\right| <1$. The remainder integral is estimated using integration by parts. This means one needs to control the size of $N_\epsilon (a)$ and have a lower bound over $p'(x)$ outside $N_\epsilon (a)$. A similar requirement appears in the proofs of [34,  p. 41, p. 45]. When such information is available one can expect to control $V_{a,b}(Q_{1,1})$ in (A.3) by $(b-a)\Vert Q_{1,1}'\Vert _{\infty }$ efficiently; this is in fact what we did in the discussion following (3.34).

• (3) The methods of [30, 34] seem more costly in terms of derivatives. After a change of variables $y=T(x)$, where T depends on the phase function p, the reduced integral (within the neighborhood) becomes $\int _{N_{\epsilon }(a)}\mathrm {e}^{\mathrm {i}tp(x)} q(x)\,\mathrm {d}x = \int _{T(N_\epsilon (a))} \mathrm {e}^{\mathrm {i}t \epsilon y^2} u(y)\,\mathrm {d}y$, where $u(y)=\frac{q(T^{-1}y)}{|T'(T^{-1}y)|}$. It is now necessary to control the derivatives of u. In fact a bound on the error we could extract from [30,  Step 2, p. 335] with $x\eta (x):=u(x)-u(a)$ is $Ct^{-1}|u|_2$, where $|u|_2 = \max (\Vert u'\Vert _\infty ,\Vert u''\Vert _{\infty })$. The first method of [34,  p. 43] is more costly, requiring bounds over $\Vert u^{(k)}\Vert _{\infty }$ for $k\le 4$. However, after involved Taylor–Lagrange expansions, one sees that $\Vert u^{(k)}\Vert _{\infty }$ is controlled by $\displaystyle \max _{\begin{array}{c} j\le k\\ \ell \le k+3 \end{array}} \frac{\Vert q^{(j)}p^{(\ell )}\Vert _{\infty }}{|p''(a)|^2}$. This means that we need to control at least 5 derivatives of p, and 2 derivatives of q.

  The second method in [34,  p. 45] seems more costly for the observable. Taking $m=1$ and controlling the error $I_h^{(1)}(0)$ by taking $N=2$ in [34,  p. 43], one finds it necessary to control all derivatives $\frac{\Vert (g_pq)^{(k)}\Vert _{\infty }}{|p''(a)|^2}$ for $k\le 6$, where $g_p(x) = p(x)-p''(0)x^2/2$, here $a=0$ and $p(0)=0$.

Appendix B. Brief comparison with 1d periodic potentials

We continue here the discussion started in Sect. 1.2.

For transparency, consider first $H=-\Delta$ on $\mathbb {R}$. The Green’s function $G^z_{\mathbb {R}}(x,y)$ for $z\in \mathbb {C}^+$ can be constructed as usual using two semi-$L^2$ ODE solutions. For example take $V_z(x)=\mathrm {e}^{\mathrm {i}\sqrt{z}x}\in L^2[0,\infty )$ and $U_z(x)=\mathrm {e}^{-\mathrm {i}\sqrt{z}x}\in L^2(-\infty ,0]$. Their Wronskian $V_zU_z'-V_z'U_z=-2\mathrm {i}\sqrt{z}$, so $G^z(x,y) = {\left\{ \begin{array}{ll} \frac{\mathrm {e}^{\mathrm {i}\sqrt{z}x}\mathrm {e}^{-\mathrm {i}\sqrt{z}y}}{-2\mathrm {i}\sqrt{z}} &{} \text {if } y\le x,\\ \frac{\mathrm {e}^{\mathrm {i}\sqrt{z}y}\mathrm {e}^{-\mathrm {i}\sqrt{z}x}}{-2\mathrm {i}\sqrt{z}}&{} \text {if } y\ge x. \end{array}\right. }$ Hence, $G^z(x,y) = \frac{\mathrm {e}^{\mathrm {i}\sqrt{z}|x-y|}}{-2\mathrm {i}\sqrt{z}}$ and ${\text {Im}}G^{\lambda +\mathrm {i}0}(x,y)= \frac{\cos \sqrt{\lambda }|x-y|}{2\sqrt{\lambda }}$ for $\lambda \in [0,\infty )=\sigma (-\Delta )$. By the spectral theorem, $\mathrm {e}^{\mathrm {i}tH}(x,y) = \frac{1}{\pi }\int _{\sigma (H)}\mathrm {e}^{\mathrm {i}t\lambda }{\text {Im}}G^{\lambda +\mathrm {i}0}(x,y)\,\mathrm {d}\lambda =\int _0^\infty \frac{\mathrm {e}^{\mathrm {i}t\lambda }\cos \sqrt{\lambda }|x-y|}{2\pi \sqrt{\lambda }}\,\mathrm {d}\lambda = \int _0^\infty \frac{\mathrm {e}^{\mathrm {i}t(\lambda +\frac{\sqrt{\lambda }|x-y|}{t})}+\mathrm {e}^{\mathrm {i}t (\lambda -\frac{\sqrt{\lambda }|x-y|}{t})}}{4\pi \sqrt{\lambda }}\,\mathrm {d}\lambda$. Denote the velocity $v=\frac{|x-y|}{t}$ and consider the changes of variables $\sqrt{\lambda }=k$ and $-\sqrt{\lambda }=k$, respectively. Then $\mathrm {e}^{\mathrm {i}t H}(x,y) = \int _0^\infty \frac{\mathrm {e}^{\mathrm {i}t(k^2+kv)}}{2\pi }\,\mathrm {d}k + \int _{-\infty }^0\frac{\mathrm {e}^{\mathrm {i}t(k^2+kv)}}{2\pi }\,\mathrm {d}k = \frac{1}{2\pi }\int _{-\infty }^\infty \mathrm {e}^{\mathrm {i}t(k^2+kv)}\,\mathrm {d}k$. This is a Fresnel-type integral, it reduces to $\mathrm {e}^{\mathrm {i}tH}(x,y)=\frac{1}{2}\sqrt{\frac{\mathrm {i}}{\pi t}}\mathrm {e}^{\frac{-\mathrm {i}tv^2}{4}}$.

For general periodic Schrödinger operators H on $\mathbb {R}$, one simply replaces $\mathrm {e}^{\pm \mathrm {i}\sqrt{z}}$ by Floquet solutions. The spectrum generally consists of a number of bands $(I_n)_{n\ge 1}$ of purely absolutely continuous spectrum which may be finite or infinite. A corresponding variable k is defined mapping $I_n$ to bands $\Sigma _n\cup (-\Sigma _{n})=:\Sigma (n)$, and one finds that $\mathrm {e}^{\mathrm {i}tH}(x,y) = \sum _n\int _{\Sigma (n)} \mathrm {e}^{\mathrm {i}t(E(k)-kv)} X^+(x,k)X^-(x,k)\,\mathrm {d}k$, here E(k) behaves like $k^2$ away from the band edges and $X^{\pm }$ come from the Floquet solution. See [14, 15] for more details. This integral is now analyzed using the stationary phase method. It was shown in [22,  Corollary 4.4] that for finite bands $\Sigma _n$, $E''(k)$ has a unique zero $k_n\in \Sigma _n$, moreover $E'(k)$ is monotone increasing up to $k_n$, then monotone decreasing. Now consider the phase function $\phi (k) = E(k)-kv$. We have $\phi '(k)=E'(k)-v$ and $\phi ''(k)=E''(k)$. The only possibility that $\phi '(k)=\phi ''(k)=0$, $k\in \Sigma _n$, is if $k=k_n$ and $v=E'(k_n)$, i.e. for a very specific choice of v, hence x, y, t. In this situation the stationary phase method allows to conclude that the speed of dispersion slows down to $t^{-1/3}$ (or slower in principle). This problem does not arise on the infinite band. In all other cases the decay will be $t^{-1/2}$, or even faster when v is very large (in that case $\phi '(k)$ does not vanish). See [15] for details when the number of bands is finite. The paper [14] considers the case of infinite number of bands, but only provides upper bounds, as it relies on the Van der Corput lemma; it can be an interesting question to test for sharpness by providing asymptotic equivalents as in [15].

Back to our case of quantum trees, the idea of constructing the resolvent kernel from two semi-$L^2$ functions works again, see [11, 17]. The Floquet functions $\mathrm {e}^{\pm \mathrm {i}k x}X^{\pm }(x,k)$ are replaced by $(\mu ^-(\lambda )^m)V^+_\lambda (x)$ and $(\mu ^-(\lambda ))^mU^-_{\lambda }(x)$, where m is the distance of x from a fixed edge $b_0$ (think of $b_0=[0,1]$ in $\mathbb {R}$) and $V^+_\lambda ,U_\lambda ^-$ are fixed functions repeated on all edges (i.e. can be regarded as periodic). The main difference is that the multiplicative factor $\mu ^-(\lambda )^m$ decays exponentially in m, in fact $|\mu ^-(\lambda )^m|=q^{-m/2}$, in contrast to $|\mathrm {e}^{\mathrm {i}k m}|=1$ in case of $\mathbb {R}$, and the $\lambda$-variations of ${\text {Im}}G^{\lambda }_{\mathbf {T}_q}(x,y)$ also decay exponentially with d(x, y). This is in contrast to $\left| \frac{\mathrm {d}^j}{\mathrm {d}k^j}\mathrm {e}^{\mathrm {i}km}\right| = m^j$ which grows with the distance. These differences make it more reasonable to consider the phase function as $\phi (k)= E(k)$ and keep the analog of $\mathrm {e}^{\mathrm {i}k(x-y)}$ in the observable part; the aforementioned control over its modulus and derivatives allows for a good control using the stationary phase method. This is the qualitative reason why we observe a fixed speed of dispersion $t^{-3/2}$ independently of the potentials W and $\alpha$ we put on the edges/vertices.