Landau–Zener formula in a “non-adiabatic” regime for avoided crossings

Abstract

We study a two-level transition probability for a finite number of avoided crossings with a small interaction. Landau–Zener formula, which gives the transition probability for one avoided crossing as $e^{-\pi \frac{\varepsilon ^{2}}{h}}$, implies that the parameter h and the interaction $\varepsilon$ play an opposite role when both tend to 0. The exact WKB method produces a generalization of that formula under the optimal regime $\scriptstyle {\frac{h}{\varepsilon ^2}}$ tends to 0. In this paper, we investigate the case $\scriptstyle {\frac{\varepsilon ^2}{h}}$ tends to 0, called “non-adiabatic” regime. This is done by reducing the associated Hamiltonian to a microlocal branching model which gives us the asymptotic expansions of the local transfer matrices.

Appendices

Exact WKB approach

This appendix is devoted to a quick review of the exact WKB method. In particular, we fix the notations used in the paper.

1.1 Construction of the exact WKB solutions

In this subsection, we recall the construction of exact WKB solutions specific to our situation with the parameter $\varepsilon$ fixed for the moment. This construction was initiated by Gérard-Grigis (see [12]) and developed to a first order $2\times 2$ system by Fujiié-Lasser-Nédélec (see [9]).

Set $\psi (t;h):= \frac{1}{2}\begin{pmatrix}1&{}i\\ i&{}1\end{pmatrix}\phi (t;h)$, where $\phi (t;h)$ is the solution of (1). Then the original Eq. (1) can be reduced to the following first order $2\times 2$ system:

$$\begin{aligned} \frac{h}{i} \frac{d}{dt} \phi (t;h) = \begin{pmatrix} 0 &{} \alpha (t)\\ -\beta (t) &{} 0 \end{pmatrix} \phi (t;h), \end{aligned}$$

(45)

where $\alpha (t) = -iV(t) - \varepsilon$ and $\beta (t) = -iV(t) + \varepsilon$. One sees that the Eq. (45) is a natural extension of the Schrödinger equation by taking $\alpha (t)=1$ and $\beta (t) = V(t)-E$.

We treat this equation on a simply connected domain ${\mathcal S}\subset \mathbb {C}$ given in (H1) and define for any fixed point $a\in {{\mathcal {S}}}$

$$\begin{aligned} z_a(t) = \int _a^t \sqrt{\alpha (s)\beta (s)}\,ds = i \int _a^t \sqrt{V(s)^2 + \varepsilon ^2}\, ds, \end{aligned}$$

where the branch of the integrand $\sqrt{V(t)^2 + \varepsilon ^2}$ is taken $\varepsilon$ at vanishing points of V(t) i.e., $t= t_k$ for $k = 1,\ldots ,n$. One sees that $z_a(t)$ satisfies so-called eikonal Eq. (45). Notice that for any $a, {\tilde{a}} \in {{\mathcal {S}}}$ one has

$$\begin{aligned} z_a(t) = z_{{\tilde{a}}}(t) + \int _a^{{\tilde{a}}} \sqrt{\alpha (s)\beta (s)} ds. \end{aligned}$$

(46)

We denote by $\varLambda$ a set of turning points which are zeros of $\alpha (t)\beta (t)$, and by $\widetilde{{\mathcal {S}}}$ the simply connected domain ${{\mathcal {S}}}\setminus \varLambda$. Remark that the mapping $z_a$ is bijective from $\widetilde{{\mathcal {S}}}$ to $z_a(\widetilde{{\mathcal {S}}})$. From (H1), we can find a suitable small constant c and define the strip domain ${\mathcal S}_\mathrm{bdd} =\{ t\in {{\mathcal {S}}}\,; \ |\text {Im}\, t| < c \}$ such that ${{\mathcal {S}}}_\mathrm{bdd} \cap \varLambda = \{ \zeta _k, \overline{\zeta _k}\, ;\, k= 1,\ldots , n \}$ by the assumption (H3) and the Rouché theorem. We make a branch cut from each $\zeta _k$ (resp. $\overline{\zeta _k}$) in the direction parallel to the imaginary axis with the positive (resp. negative) imaginary part (see Fig. 4). Note that under this choice of the branch cut the whole real axis is included in the corresponding simply connected subdomain of $\widetilde{{\mathcal {S}}}_\mathrm{bdd}:=\widetilde{{\mathcal {S}}}\cap {{\mathcal {S}}}_\mathrm{bdd}$. This fact permits us to know that $\pm \text {Re}\,z_a(t)$ increase as $\pm \text {Im}\,t$ decrease.

In this context, we can consider the solution of (45), $\phi (t;h)$, as a function of the variable z by setting $\phi _\pm (t;h) = e^{\pm \frac{z}{h}} M_\pm (z) w_\pm (z;h)$, with

$$\begin{aligned} M_\pm (z) = \begin{pmatrix} K(z)^{-1} &{}\quad K(z)^{-1}\\ \mp i K(z) &{}\quad \pm i K(z) \end{pmatrix}, \quad K(z(t)) = \left( \frac{\beta (t)}{\alpha (t)}\right) ^{\frac{1}{4}} = \left( \frac{-iV(t) + \varepsilon }{-iV(t) -\varepsilon }\right) ^{\frac{1}{4}}. \end{aligned}$$

Notice that K(z(t)) is independent of the base point a involved in the definition of the function $z(t)=z_a(t)$. The branch of K(z(t)) is taken $e^{-i\frac{\pi }{4}}$ at each $t=t_k$ with the branch cut along a positive real axis on $\mathbb {C}_z$. Here the vector-valued function $w_\pm (z;h)$ are determined as solutions of

$$\begin{aligned} \frac{d}{dz} w_\pm (z;h) = \begin{pmatrix} 0 &{}\quad \frac{K'(z)}{K(z)}\\ \frac{K'(z)}{K(z)} &{}\quad \mp \frac{2}{h} \end{pmatrix} w_\pm (z;h). \end{aligned}$$

Moreover, by identity (46) and the following equality

$$\begin{aligned} \frac{\frac{d}{dz}K(z(t))}{K(z(t))} = \frac{\alpha (t)\beta '(t) - \alpha '(t)\beta (t)}{4(\alpha (t)\beta (t))^{3/2}}\,, \end{aligned}$$

(47)

we see that $\frac{\frac{d}{dz}K(z(t))}{K(z(t))}$ and $w_\pm (z(t);h)$ are independent of a. The above equality (47) implies that when $\zeta$ is a simple turning point, the function $\frac{K'(z)}{K(z)}$ has a simple pole at $z=z(\zeta )$.

Generally, even if the vector-valued symbols $w_\pm (z;h)$ are developed with respect to h small enough, the series do not converge. The essential idea of [12] (see also [9]) is to introduce a resummation by using the following integral recurrence system on $\mathbb {C}_z$. More precisely, for any $b\in \widetilde{{\mathcal {S}}}_\mathrm{bdd}$, the vector-valued functions $w_\pm (z;h)=w_{\pm }(z,z(b);h)$ are of the form:

$$\begin{aligned} w_{\pm }(z,z(b);h) = \sum _{k\ge 0} w_{\pm ,k}(z,z(b);h), = \sum _{k\ge 0}\left( \begin{matrix} w_{\pm ,2k}(z,z(b);h)\\ w_{\pm ,2k-1}(z,z(b);h) \end{matrix}\right) , \end{aligned}$$

(48)

where the sequences $\big \{w_{\pm ,k}(z,z(b);h)\big \}_{k\in \mathbb {N}}$ are defined by

$$\begin{aligned} \left\{ \begin{aligned} w_{\pm ,0}(z,z(b);h)&\equiv \, 1, \quad w_{\pm ,-1}(z,z(b);h) \equiv \, 0,\\ \ w_{\pm ,2k+1}(z,z(b);h)&=\int _{z(b)}^z e^{\pm \frac{2}{h}(\zeta -z)}\frac{K'(\zeta )}{K(\zeta )} w_{\pm ,2k}(\zeta ,z(b);h)\,d\zeta&(k\ge 0),\\ w_{\pm ,2k}(z,z(b);h)&=\int _{z(b)}^z \frac{K'(\zeta )}{K(\zeta )} w_{\pm ,2k-1}(\zeta ,z(b);h)\,d\zeta&(k\ge 1). \end{aligned} \right. \end{aligned}$$

Thanks to the above resummation, the vector-valued symbol expansions (48) converge absolutely and uniformly in a neighborhood of z(b) for $b\in \widetilde{{\mathcal {S}}}_\mathrm{bdd}$ (see, for example, [9, Lemma 3.2]). Hence, for any fixed $(a,b)\in {{\mathcal {S}}}_\mathrm{bdd}\times \widetilde{{\mathcal {S}}}_\mathrm{bdd}$, we can define the exact WKB solutions of type ± as follows:

$$\begin{aligned} \psi _{\pm }(t,a,b;h) = \frac{1}{2}\begin{pmatrix}1&{}\quad i\\ i&{}\quad 1\end{pmatrix} e^{\pm \frac{z_a(t)}{h}} M_{\pm }(z(t))w_{\pm }(z(t),z(b);h), \end{aligned}$$

(49)

which are linearly independent exact solutions of (1). Notice that $a\in {{\mathcal {S}}}_\mathrm{bdd}$ is the base point of the phase and $b\in \widetilde{{\mathcal {S}}}_\mathrm{bdd}$ is that of the symbol.

We conclude this subsection by recalling some results concerning the exact WKB solutions given by (49). In fact, the exact WKB method is based on two properties, which are the Wronskian formula between the exact WKB solutions of type ± and the asymptotic expansion with respect to h of the symbol.

Lemma 6

([31, Proposition 2.2.2]) The Wronskian between any exact WKB solutions of type ± with the same base point of the phase satisfies:

$$\begin{aligned} {{\mathcal {W}}}[\psi _+(t,a,b_+;h), \psi _-(t,a,b_-;h)] = 2i \sum _{k\ge 0} w_{+,2k}(z(b_-),z(b_+);h), \end{aligned}$$

(50)

where $a\in {{\mathcal {S}}}_\mathrm{bdd}$ and $b_{\pm }\in \widetilde{{\mathcal {S}}}_\mathrm{bdd}$. Here the Wronskian between $\mathbb {C}^2$-valued functions $\psi _1$ and $\psi _2$ is defined by ${{\mathcal {W}}}[\psi _1,\psi _2] := \mathrm{det}\left( \psi _1 \psi _2\right)$.

The proof of this lemma is based on a direct computation and the independence of the Wronskian with respect to the variable t thanks to the trace-free matrix in (45). The prefactor 2i is exactly the $\det M_\pm$.

To state the next result, we introduce canonical curves of type ± in $\widetilde{{\mathcal {S}}}_\mathrm{bdd}$ from a fixed point b to t along which $\pm \text {Re}\,z_a(t)$ increase strictly, for a fixed $a\in {{\mathcal {S}}}_\mathrm{bdd}$. The advantage of the integral recurrence system is to give not only an absolutely convergence but also $\mathbb {C}^2$-valued asymptotic sequences with respect to h uniformly away from turning points. More precisely,

Lemma 7

([31, Proposition 2.3.1]) If there exist canonical curves of type ± from $b_\pm$ to t denoted by $\gamma _\pm$, then the vector-valued symbols have the following asymptotic expansions:

$$\begin{aligned} w_{\pm }(z(t),z(b_\pm );h) = \begin{pmatrix} 1\\ 0 \end{pmatrix} \left( 1+ {{\mathcal {O}}}\left( \frac{h}{\mathrm{dist}(\gamma _\pm ;\varLambda ) }\right) \right) \end{aligned}$$

(51)

as h tends to 0, where $\mathrm{dist}(\gamma _\pm ;\varLambda )$ stands for $\underset{t\in \gamma _\pm , \zeta \in \varLambda }{\inf }|z_{\zeta }(t)|$.

This lemma can be proved by an integration by parts thanks to the exponential decaying along the canonical curve.

Combining Lemmas 6 and 7, we obtain the asymptotic expansion of the Wronskian:

Lemma 8

([31, Proposition 2.4.1]) If there exists a canonical curve of type $+$ from $b_+$ to $b_-$ denoted by $\gamma$, the Wronskian between any exact WKB solutions of type ± with the same base point of the phase has the following asymptotic expansion,

$$\begin{aligned} {{\mathcal {W}}}[\psi _+(t,a,b_+;h), \psi _-(t,a,b_-;h)] = 2i + {{\mathcal {O}}}\left( \frac{h}{\mathrm{dist}(\gamma ;\varLambda ) }\right) \end{aligned}$$

(52)

as h tends to 0, where $\mathrm{dist}(\gamma ;\varLambda )=\underset{t\in \gamma , \zeta \in \varLambda }{\inf }|z_{\zeta }(t)|$.

Remark 5

(“Adiabatic” v.s. “Non-Adiabatic”) In order to derive the exponential decay of the transition probability (see Proposition 2), we apply the above Wronskian formula to the local transfer matrix given by (56)), but we must find a canonical curve $\gamma$ passing between the two turning points $\zeta$ and $\overline{\zeta }$ which accumulate to the same crossing point as $\varepsilon$ tends to 0. In this case, one sees that $\mathrm{dist}(\gamma ;\varLambda )$ is of order ${{\mathcal {O}}}(\varepsilon ^2)$. Hence the WKB method works only under the regime ${{\scriptstyle \frac{h}{\varepsilon ^2}}} \rightarrow 0$, called “adiabatic” regime.

When ${\scriptstyle \frac{\varepsilon ^2}{h}} \rightarrow 0$, the above lemma is obsolete at the crossing point. But we can control the behavior of the error in (52) far from this point, for example outside an $\mathcal {O}(\sqrt{h})$-neighborhood of it. We say this regime “non-adiabatic”.

1.2 Representation of the scattering matrix

In this subsection we give the proof of Proposition 1. When we construct exact WKB solutions globally for a sake of expressing the scattering matrix, it is difficult to deal with various turning points, so that we treat only two turning points near each vanishing point of V(t), without loss of generality. We put $\scriptstyle d_1 = \frac{t_1 - t_2}{2}$, $\scriptstyle d_n = \frac{t_{n-1} - t_n}{2}$ and $d_k =\frac{1}{2} \max \{ t_k - t_{k+1},\ t_{k-1} - t_k \}$ for $k= 2,\ldots , n-1$. Let ${{\mathcal {S}}}_k \subset {{\mathcal {S}}}_\mathrm{bdd}$ be a simply connected small box in ${{\mathcal {S}}}_\mathrm{bdd}$ including only one vanishing point $t_k$, given by

$$\begin{aligned} {{\mathcal {S}}}_k := \{ t\in {{\mathcal {S}}}_\mathrm{bdd}\ ; \ |\text {Re}\,t - t_k|< d_k +\rho \}\quad (k=1,2,\ldots , n), \end{aligned}$$

(53)

where $\rho$ is a suitable small constant. We see that ${\mathcal S}_k \cap {{\mathcal {S}}}_{k+1} \ne \emptyset$ for each $k= 1,\ldots , n-1$ and we put there a symbol base point

$$\begin{aligned} \delta _{k,k+1} := \frac{t_k + t_{k+1}}{2} + i c_{k} \in {{\mathcal {S}}}_k \cap {{\mathcal {S}}}_{k+1} \end{aligned}$$

and its complex conjugate (see Fig. 4). Here $0<c_k<c$ where the constant c is involved in the definition of ${\mathcal S}_\mathrm{bdd}$.

Fig. 4

Picture of ${{\mathcal {S}}}_k$

In each ${{\mathcal {S}}}_k$ $(k=1,\ldots , n)$, we introduce the intermediate WKB solutions, which consist of the bases in ${\mathcal {S}}_k$,

$$\begin{aligned} \psi _+(t,\zeta _k,\delta _{k-1,k};h)= & {} \exp \left[ +\frac{z_{\zeta _k}(t)}{h}\right] M_+(z(t))w_+(z(t),z(\delta _{k-1,k});h) := \psi _{+,k}^r,\nonumber \\ \psi _-(t,\overline{\zeta _k},\overline{\delta _{k-1,k}};h)= & {} \exp \left[ -\frac{z_{\overline{\zeta _k}}(t)}{h}\right] M_-(z(t))w_-(z(t),z(\overline{\delta _{k-1,k}});h) := \psi _{-,k}^r,\nonumber \\ \psi _+(t,\zeta _k,\delta _{k,k+1};h)= & {} \exp \left[ +\frac{z_{\zeta _k}(t)}{h}\right] M_+(z(t))w_+(z(t),z(\delta _{k,k+1});h) := \psi _{+,k}^l,\nonumber \\ \psi _- (t,\overline{\zeta _k},\overline{\delta _{k,k+1}};h)= & {} \exp \left[ -\frac{z_{\overline{\zeta _k}}(t)}{h}\right] M_-(z(t))w_-(z(t),z(\overline{\delta _{k,k+1}});h) := \psi _{-,k}^l.\nonumber \\ \end{aligned}$$

(54)

Remark that each exact WKB solution has a valid asymptotic expansion for h small enough in the direction from its symbol base point toward the vanishing point $t_k$ in ${{\mathcal {S}}}_k$ thanks to Lemma 7.

As mentioned in Sect. 2.1, there exist two kind of the transfer matrices. One of them is a change of bases with respect to the base points of the symbol function denoted by $T_k(\varepsilon ,h)$, that is, it transfers from right side to left side over the crossing point in ${{\mathcal {S}}}_k$. The other is one with respect to the base points of the phase function denoted by $T_{k,k+1}(\varepsilon ,h)$, that is, it transfers on the intersection between ${{\mathcal {S}}}_k$ and ${{\mathcal {S}}}_{k+1}$. In fact, they are written by

$$\begin{aligned} \left( \psi _{+,k}^l \ \psi _{-,k}^l \right) = \left( \psi _{+,k}^r\ \psi _{-,k}^r\right) T_{k},\quad \left( \psi _{+,k+1}^r \ \psi _{-,k+1}^r \right) = \left( \psi _{+,k}^l\ \psi _{-,k}^l\right) T_{k,k+1}. \end{aligned}$$

(55)

The former transfer matrix $T_k(\varepsilon ,h)$ is our main target, which is given by

$$\begin{aligned} T_k(\varepsilon ,h) = \frac{1}{{{\mathcal {W}}}[\psi _{+,k}^r, \psi _{-,k}^r]} \begin{pmatrix} {{\mathcal {W}}}[\psi _{+,k}^l, \psi _{-,k}^r] &{}\quad {{\mathcal {W}}}[\psi _{-,k}^l, \psi _{-,k}^r]\\ {{\mathcal {W}}}[\psi _{+,k}^r, \psi _{+,k}^l] &{}\quad {{\mathcal {W}}}[\psi _{+,k}^r, \psi _{-,k}^l] \end{pmatrix}. \end{aligned}$$

(56)

The asymptotic behaviors of these Wronskians can be computed by Lemma 8 under the regime ${\scriptstyle \frac{h}{\varepsilon ^2}} \rightarrow 0$ (see Lemma 5). On the other hand, the asymptotic behaviors of them under the regime ${\scriptstyle \frac{\varepsilon ^2}{h}} \rightarrow 0$ must be investigated more carefully (see Remark 5).

From (46), the latter transfer matrix $T_{k,k+1}(\varepsilon ,h)$ is the diagonal one, whose diagonal elements are complex conjugate each other. Put

$$\begin{aligned} a_k(\varepsilon ,h) = e^{\frac{i}{2h}\left( A_k(\varepsilon ) - A_{k+1}(\varepsilon ) + R_{k}(\varepsilon )\right) } \end{aligned}$$

(57)

for $k = 1,2, \ldots , n-1$, where $A_k(\varepsilon )$ (resp. $R_k(\varepsilon )$) is given by (4) (resp. 6). Then, for $k = 1,2, \ldots , n-1$,

$$\begin{aligned} T_{k,k+1}(\varepsilon ,h)&= \left( \begin{array}{cc} a_k(\varepsilon ,h) &{}\quad 0\\ 0 &{}\quad \overline{a_k(\varepsilon ,h)} \end{array} \right) . \end{aligned}$$

(58)

In addition, the transfer matrices between Jost solutions and exact WKB solutions at $\pm \infty$ must be treated separately. For a fixed $R>0$ we denote an unbounded simply connected domain by ${{\mathcal {S}}}_R = {{\mathcal {S}}}\cap \{ t\in \mathbb {C}\, ;\ |\mathrm{Re}\, t | > R\}$. From (H2), we can find a suitable constant $R>0$ such that ${{\mathcal {S}}}_R \cap \varLambda = \emptyset$. Recall that $\varLambda$ is the set of turning points. In ${{\mathcal {S}}}_R$ we can construct the exact WKB solutions of (45) corresponding to Jost solutions as

$$\begin{aligned} \phi _\pm ^\star (t;h) = e^{\pm \frac{z_{\star }(t)}{h}} M_\pm (z(t)) w_\pm ^\star (z(t);h), \end{aligned}$$

where the index $\star \in \{r,l\}$ stands for a direction corresponding to either $+\infty$ or $-\infty$, and a modified phase function $z_{\star }(t)$ is given by

$$\begin{aligned} z_{\star }(t)&= i \int _{\pm \infty }^t \left( \sqrt{V^2(s) + \varepsilon ^2} - \lambda _\star \right) ds + i\lambda _\star \, t, \end{aligned}$$

with $\lambda _\star = \sqrt{E_\star ^2 + \varepsilon ^2}$ and a modified symbol functions $w_\pm ^{\star }(z(t))$ are given by the same integral recurrence system described in Sect. A.1, but the integral paths taken from $\infty e^{\pm i\theta _1}$ to $t\in {{\mathcal {S}}}_R$ for $\theta _1\in (0,\theta _0)$ with $\theta _0$ given by (H1). Remark that the modified phase functions are convergent thanks to (H2) and the resummation based on the modified integral paths works also thanks to (H1).

Lemma 9

([31, Proposition 3.1.1]) We obtain the relation between Jost solutions and the corresponding WKB solutions:

$$\begin{aligned}&J_+^\star (t) = -\frac{1}{2}\begin{pmatrix}1&{}i\\ i&{}1\end{pmatrix} \phi _+^\star (t), \quad J_-^\star (t) = -\frac{i}{2}\begin{pmatrix}1&{}i\\ i&{}1\end{pmatrix} \phi _-^\star (t),\qquad (\star \in \{r,l\}). \end{aligned}$$

One sees that the computations of the asymptotic behaviors of $M_\pm (z(t))$ are essentially same as in Sect. 2.4 and about those of $w_\pm ^{\star }(z(t);h)$ one can consult with [26] (see also [13, Lemma 3.2]), in fact $w_\pm ^{\star }(z(t);h) \rightarrow { }^t (1, 0)$ as $\mathrm{Re}\, t \rightarrow \pm \infty$. From this lemma, when we denote by $T_r(\varepsilon ,h)$ (resp. $T_l(\varepsilon ,h)$) the transfer matrices from ${{\mathcal {S}}}_R$ to ${{\mathcal {S}}}_1$ (resp. ${{\mathcal {S}}}_n$) as

$$\begin{aligned} (J_+^r\ J_-^r) = (\psi _{+,1}^r \ \psi _{-,1}^r)T_r,\quad (J_+^l\ J_-^l) = (\psi _{+,n}^l \ \psi _{-,n}^l)T_l, \end{aligned}$$

and do the action integrals corresponding to $\pm \infty$ by

$$\begin{aligned} A_{\star }(\varepsilon )&= 2\int _{t_\star }^{\pm \infty } {\left( \sqrt{V(t)^2 + \varepsilon ^2} - \sqrt{E_\star ^2+\varepsilon ^2}\right) }\,dt, \end{aligned}$$

(59)

with $t_r=t_1$ and $t_l = t_n$, the transfer matrices $T_r(\varepsilon ,h)$ and $T_l(\varepsilon ,h)$ are diagonal. Actually, they are given by

$$\begin{aligned} T_{\star }(\varepsilon ,h)&= \left( \begin{array}{cc} -a_\star (\varepsilon ,h) &{}\quad 0\\ 0 &{}\quad \overline{ia_\star (\varepsilon ,h)} \end{array} \right) \Bigl (1+f_\star (h)\Bigr ), \end{aligned}$$

(60)

where each error $f_\star (h)$ $(\star \in \{r,l\})$ is ${\mathcal O}(h)$ as h tends to 0 uniformly with respect to small $\varepsilon$ and

$$\begin{aligned} a_r (\varepsilon ,h) = e^{\frac{i}{2h}\left( A_1(\varepsilon )-A_{r}(\varepsilon )+2\lambda _rt_1\right) },\qquad a_l(\varepsilon ,h) = e^{\frac{i}{2h}\left( A_n(\varepsilon )-A_{l}(\varepsilon )+2\lambda _lt_n\right) }. \end{aligned}$$

(61)

Summing up, by using all kinds of the transfer matrices, we have a representation of the scattering matrix as we state in Proposition 1.

Remark 6

The asymptotic behaviors of the scattering matrix $S(\varepsilon ,h)$ are essentially given by the asymptotic expansions of the local transfer matrices $(T_k)_{1\le k \le n}$. So, the shape of the function V(t) near its vanishing points is crucial. The assumption (H3) implies that we have the same geometrical configuration in each ${{\mathcal {S}}}_k$ for $k= 1,2,\ldots , n$. Hence, without loss of generality, we may assume that $t_k = 0$ and $V(0) = 1$, that is, $V(t) = t + {{\mathcal {O}}}(t^2)$ in an h-independent neighborhood of 0.

1.3 Proof of Proposition 4

For the proof of Proposition 4, it is enough to compute the leading term of the exact WKB solutions $\widetilde{\psi }_{\pm }^\star$ with $\star \in \{r,l\}$. Remembering that V(t) is bounded and real-valued in $I^\star (h)$, where $I^\star (h)$ is given by (22), we first study $K(z(t)) = (\kappa (t))^{\frac{1}{4}}$ and see that

$$\begin{aligned} \kappa (t) = \frac{-iV(t) + \varepsilon }{-iV(t) - \varepsilon }&= \frac{V(t)^2 - \varepsilon ^2}{V(t)^2 + \varepsilon ^2} + i \frac{2\varepsilon V(t)}{V(t)^2 + \varepsilon ^2}\quad \mathrm{and} \quad |\kappa (t)| = 1. \end{aligned}$$

Recall that, under our branch of K, the function $\kappa (t)$ tends to $e^{-\pi i}$ as t goes to 0. We express $\kappa (t)$ and K(z(t)) as follows:

$$\begin{aligned} \kappa (t)&= \left\{ \begin{aligned}&e^{i(\theta (t;\varepsilon ) - 2\pi )}&\text {for} \ t>0,\\&e^{-i\theta (t;\varepsilon )}&\text {for} \ t<0, \end{aligned} \right. \quad \mathrm{and}\quad K(z(t)) = \left\{ \begin{aligned} -i e^{i\frac{\theta (t;\varepsilon )}{4}}&\quad \text {for} \ t>0,\\ e^{-i\frac{\theta (t;\varepsilon )}{4}}&\quad \text {for} \ t<0, \end{aligned} \right. \end{aligned}$$

where $\theta (t;\varepsilon )$ satisfies $\tan \theta (t;\varepsilon ) = \frac{2\varepsilon |V(t)|}{|V(t)|^2-\varepsilon ^2}$ for $0< \theta (t;\varepsilon ) <\frac{\pi }{2}$. In fact, we know

$$\begin{aligned} \theta (t;\varepsilon ) = \tan ^{-1} \frac{2\varepsilon |V(t)|}{|V(t)|^2-\varepsilon ^2} = \frac{2\varepsilon }{|V(t)|} + {{\mathcal {O}}}\left( \left( \frac{\varepsilon }{\sqrt{h}}\right) ^3\right) . \end{aligned}$$

Hence we have

$$\begin{aligned}&K(z(t))^{-1} + K(z(t))\\&\quad = \left\{ \begin{aligned}&2\sin \frac{\theta (t;\varepsilon )}{4} = \frac{\varepsilon }{t}(1 + {{\mathcal {O}}}(\sqrt{h})) + {{\mathcal {O}}}\left( \left( \frac{\varepsilon }{ \sqrt{h}}\right) ^{3}\right)&\qquad \text {for} \ t>0,\\&2\cos \frac{\theta (t;\varepsilon )}{4} = 2 + {{\mathcal {O}}}\left( \frac{\varepsilon ^2}{ h}\right)&\qquad \text {for} \ t<0, \end{aligned} \right. \\&K(z(t))^{-1} - K(z(t)) \\&\quad =\left\{ \begin{aligned}&2i\cos \frac{\theta (t;\varepsilon )}{4} = 2i + {{\mathcal {O}}}\left( \frac{\varepsilon ^2}{h}\right)&\quad \text {for} \ t>0,\\&2i\sin \frac{\theta (t;\varepsilon )}{4} = -i\frac{\varepsilon }{t}(1 + {{\mathcal {O}}}(\sqrt{h})) + {{\mathcal {O}}}\left( \left( \frac{\varepsilon }{\sqrt{h}}\right) ^{3}\right)&\quad \text {for} \ t<0, \end{aligned} \right. \end{aligned}$$

as $\varepsilon$ and h go to 0 and $\frac{\varepsilon ^2}{h}$ tends to 0 uniformly in $I^\star (h)$. These last asymptotic expansions give us the behaviors of the leading terms involved in the vector-valued symbols of the exact WKB solutions.

Next, we give the asymptotic behaviors of the phase functions for $t \in I^r(h)$, by the same way we have them in $I^l(h)$.

We decompose the phase functions as follows, involving the crossing point 0,

$$\begin{aligned} \int _{\zeta _\pm }^t \sqrt{V(s)^2+\varepsilon ^2}\,ds = \int _0^t \sqrt{V(s)^2+\varepsilon ^2}\,ds - \int _0^{\zeta _\pm } \sqrt{V(s)^2+\varepsilon ^2}\,ds. \end{aligned}$$

(62)

The second integral of (62) is the action integral $A(\zeta _\pm )$, which is ${{\mathcal {O}}}(\varepsilon ^2)$. The first of (62) is real-valued and moreover can be decomposed with some constant $c>0$ as

$$\begin{aligned} \int _0^t \sqrt{V(s)^2+\varepsilon ^2}\,ds = \int _0^{c \varepsilon } \sqrt{V(s)^2+\varepsilon ^2}\,ds + \int _{c \varepsilon }^t \sqrt{V(s)^2+\varepsilon ^2}\,ds. \end{aligned}$$

(63)

The first integral of (63) is also ${{\mathcal {O}}}(\varepsilon ^2)$. We put

$$\begin{aligned} F(s; \varepsilon )&:= \sqrt{V(s)^2 + \varepsilon ^2} - V(s) - \frac{\varepsilon ^2}{2V(s)} = \frac{-\varepsilon ^4}{2V(s)(\sqrt{V(s)^2 + \varepsilon ^2}+V(s))^2},\\ G(s)&:= \frac{1}{V(s)} - \frac{1}{s} = \frac{-V_1(s)}{s(1+V_1(s))}, \end{aligned}$$

where $V_1$ is a holomorphic function satisfying $V(s) = s(1+V_1(s))$ such that $V_1(s) = {{\mathcal {O}}}(s)$. Then we can estimate, by using the fact that $|1+ V_1(s)|^{-1} \le 2$ along the integral path on $I^r(h)$ for h small enough, the integrals of the absolute value of these functions as

$$\begin{aligned} \int _{c\varepsilon }^t |F(s; \varepsilon )| ds&= {{\mathcal {O}}}(\varepsilon ^2),\qquad \int _{c\varepsilon }^t |G(s)| ds = {{\mathcal {O}}}(\sqrt{h}). \end{aligned}$$

These estimates imply that

$$\begin{aligned}&\int _{c\varepsilon }^t \sqrt{V(s)^2 + \varepsilon ^2} ds\\&\quad = \int _{c\varepsilon }^t V(s)ds + \frac{\varepsilon ^2}{2} \int _{c\varepsilon }^t \frac{ds}{V(s)} + {{\mathcal {O}}}(\varepsilon ^2)\\&\quad = \left( \int _{0}^t V(s)ds + {{\mathcal {O}}}(\varepsilon ^2)\right) + \frac{\varepsilon ^2}{2} \left( \int _{c\varepsilon }^t \frac{ds}{s} + {{\mathcal {O}}}(\sqrt{h})\right) + {{\mathcal {O}}}(\varepsilon ^2)\\&\quad = \int _{0}^t V(s)ds + \frac{\varepsilon ^2}{2}\log t - \frac{\varepsilon ^2}{2}\log \varepsilon + {{\mathcal {O}}}(\varepsilon ^2). \end{aligned}$$

Hence we have

$$\begin{aligned} e^{\textstyle \frac{i}{h}\int _{\zeta _+}^t\sqrt{V(s)^2 + \varepsilon ^2}ds}&= e^{\textstyle \frac{i}{h}\int _{0}^t V(s)ds} t^{\frac{i\varepsilon ^2}{2h}} e^{-\frac{i\varepsilon ^2}{2h}\log \varepsilon } \left( 1 + {{\mathcal {O}}}\left( \frac{\varepsilon ^2}{h} \right) \right) . \end{aligned}$$

Finally, in the case where $(\varepsilon ,h)$ goes to (0, 0) and $\frac{\varepsilon ^2}{h}$ tends to 0, we combine the asymptotic behaviors of each part in the intervals $I^\star (h)$, and then we obtain Proposition 4.

Branching-model and its applications

1.1 Solutions of the branching model

The branching model:

$$\begin{aligned} Q\phi (y) = \left( \begin{matrix} y &{}\quad \frac{\mu }{\sqrt{2}}\\ \frac{\mu }{\sqrt{2}} &{}\quad \frac{\mu }{i}\frac{d}{dy} \end{matrix} \right) \phi (y) = 0 \end{aligned}$$

(64)

with a small parameter $\mu >0$ has two solutions of the forms:

$$\begin{aligned} \begin{aligned} \phi ^{\vdash }(y)&= Y(y)y^{\frac{i}{2}\mu }\left( \begin{matrix} -\frac{\mu }{\sqrt{2}y} \\ 1 \end{matrix} \right) ,\\ \phi ^{\dashv }(y)&= Y(-y)|y|^{\frac{i}{2}\mu }\left( \begin{matrix} -\frac{\mu }{\sqrt{2}y} \\ 1 \end{matrix} \right) , \end{aligned} \end{aligned}$$

(65)

where Y(y) is the Heaviside function. The properties of these distributions can be found in [11]. Notice that this Eq. (64) is treated in Sect. 2.2. Here we give properties of the solutions (65).

The differential operator Q commutes with the operator $\left( \begin{matrix} 0&{}\quad 1\\ 1&{}\quad 0 \end{matrix} \right) {{\mathcal {C}}} {{\mathcal {F}}}_\mu$, where ${{\mathcal {C}}}$ is a complex conjugate operator, that is ${{\mathcal {C}}}\phi (y) = \overline{\phi (y)}$, and ${{\mathcal {F}}}_\mu$ is a semi-classical Fourier transform:

$$\begin{aligned} {{\mathcal {F}}}_\mu [u](\xi ) = \frac{1}{\sqrt{2\pi \mu }} \int _{\mathbb {R}} e^{-\frac{i}{\mu }x\xi } u(x) dx. \end{aligned}$$

(66)

Then the functions $\phi ^{\bot }(y)$ and $\phi ^{\top }(y)$ given by

$$\begin{aligned} \phi ^{\bot }(y)&= \left( \begin{matrix} 0&{}\quad 1\\ 1&{}\quad 0 \end{matrix} \right) {{\mathcal {C}}} {{\mathcal {F}}}_\mu \phi ^{\vdash }(y), \qquad \phi ^{\top }(y) = \left( \begin{matrix} 0&{}\quad 1\\ 1&{}\quad 0 \end{matrix} \right) {{\mathcal {C}}} {{\mathcal {F}}}_\mu \phi ^{\dashv }(y) \end{aligned}$$

(67)

are also solutions of (64). Computing $\phi ^{\bot }(y)$, $\phi ^{\top }(y)$ by using the property ${{\mathcal {C}}} {\mathcal F}_\mu = {{\mathcal {F}}}_\mu ^{-1} {{\mathcal {C}}}$, we obtain the relation between the pairs $(\phi ^{\vdash }, \phi ^{\dashv })$ and $(\phi ^{\bot }, \phi ^{\top })$.

Proposition 6

Let R be a $2\times 2$ matrix such that $(\phi ^\bot \phi ^\top ) = (\phi ^\vdash \phi ^\dashv )R$. Then R is of the form: $R = \left( \begin{matrix} p &{} -q\\ q &{} -p \end{matrix} \right)$ with

$$\begin{aligned} p = \gamma e^{\frac{\pi \mu }{4}}, \qquad q = \gamma e^{-\frac{\pi \mu }{4}},\qquad \gamma = \frac{1}{i\sqrt{\pi \mu }} \mu ^{-\frac{i\mu }{2}} \varGamma \left( 1-\frac{i\mu }{2}\right) . \end{aligned}$$

(68)

Proof of Proposition 6

The direct computations of $\phi ^{\bot }(y)$ and $\phi ^{\top }(y)$ under the definition (67) give us the entries of $\phi ^{\bot }(y)$ and $\phi ^{\top }(y)$ expressed by those of $\phi ^{\vdash }(y) = \, ^t(\phi _1^{\vdash }, \phi _2^{\vdash })$ and $\phi ^{\dashv }(y) = \, ^t(\phi _1^{\dashv }, \phi _2^{\dashv })$ as

$$\begin{aligned} \left( \begin{matrix} \phi _1^{\star }\\ \phi _2^{\star } \end{matrix} \right) = \left( \begin{matrix} 0&{}1\\ 1&{}0 \end{matrix} \right) \left( \begin{matrix} {{\mathcal {C}}}{{\mathcal {F}}}_\mu [\phi _1^{*}]\\[5pt] {{\mathcal {C}}}{{\mathcal {F}}}_\mu [\phi _2^{*}] \end{matrix} \right) = \left( \begin{matrix} {{\mathcal {C}}}{{\mathcal {F}}}_\mu [\phi _2^{*}]\\[5pt] {{\mathcal {C}}}{{\mathcal {F}}}_\mu [\phi _1^{*}] \end{matrix} \right) , \end{aligned}$$

where the pair of the indexes $(*,\star ) \in \{ (\vdash , \bot ), (\dashv , \top ) \}$. For example, the first entries of $\phi ^{\bot }(y)$ are expressed by

$$\begin{aligned} \phi _1^{\bot }(y) = {{\mathcal {C}}}{{\mathcal {F}}}_\mu [\phi _2^{\vdash }](y) = {{\mathcal {F}}}_\mu ^{-1}[{{\mathcal {C}}}\phi _2^{\vdash }](y) = \frac{1}{\sqrt{2\pi \mu }}\int _{\mathbb {R}} e^{\frac{i}{\mu }y\eta } Y(\eta ) \eta ^{-\frac{i}{2}\mu } d\eta . \end{aligned}$$

We demonstrate the computation concerning only $\phi ^\bot _1$.

$$\begin{aligned} \phi ^\bot _1 (y)&= \frac{1}{\sqrt{2\pi \mu }}\int _{\mathbb {R}} e^{\frac{i}{\mu }y \eta } Y(\eta ) |\eta |^{-\frac{i}{2}\mu } d\eta = \frac{1}{\sqrt{2\pi \mu }}\int _{0}^{\infty } e^{\frac{i}{\mu }y \eta } \eta ^{-\frac{i}{2}\mu } d\eta . \end{aligned}$$

In order to reduce this integral to the Gamma function, we treat it separately as a positive part and a negative part with respect to y.

$$\begin{aligned} \phi _1^\bot (y)&= \frac{Y(y)}{\sqrt{2\pi \mu }}\int _{0}^{\infty } e^{\frac{i}{\mu }y \eta } \eta ^{-\frac{i}{2}\mu } d\eta + \frac{Y(-y)}{\sqrt{2\pi \mu }}\int _{0}^{\infty } e^{\frac{i}{\mu }y \eta } \eta ^{-\frac{i}{2}\mu } d\eta . \end{aligned}$$

For the first integral, by the change of variable $\frac{i}{\mu }y\eta = -z$ and by the Cauchy integral theorem, we have

$$\begin{aligned} \frac{1}{\sqrt{2\pi \mu }}\int _{0}^{\infty } e^{\frac{i}{\mu }y \eta } \eta ^{-\frac{i}{2}\mu } d\eta&= \frac{i}{\sqrt{2\pi }} \mu ^{\frac{1}{2}-\frac{i}{2}\mu } e^{\frac{\pi }{4}\mu } \varGamma \left( 1-\frac{i}{2}\mu \right) y^{\frac{i}{2}\mu -1}. \end{aligned}$$

Recalling the form of the first entry of $\phi ^\vdash$, we see

$$\begin{aligned} \frac{1}{\sqrt{2\pi \mu }}\int _{0}^{\infty } e^{\frac{i}{\mu }y \eta } \eta ^{-\frac{i}{2}\mu } d\eta&= \frac{-i}{\sqrt{\pi }\mu } \mu ^{\frac{1}{2}-\frac{i}{2}\mu } e^{\frac{\pi }{4}\mu } \varGamma \left( 1-\frac{i}{2}\mu \right) \phi _1^\vdash (y). \end{aligned}$$

Similarly we compute the second one with the change of variable $\frac{i}{\mu }(-y)\eta = z$ as

$$\begin{aligned} \frac{1}{\sqrt{2\pi \mu }}\int _{0}^{\infty } e^{-\frac{i}{\mu }(-y) \eta } \eta ^{-\frac{i}{2}\mu } d\eta&= \frac{i}{\sqrt{\pi }\mu } \mu ^{\frac{1}{2}-\frac{i}{2}\mu } e^{-\frac{\pi }{4}\mu } \varGamma \left( 1-\frac{i}{2}\mu \right) \phi _1^\dashv (y). \end{aligned}$$

Therefore we obtain

$$\begin{aligned} \phi _1^\bot (y)&= \frac{1}{i\sqrt{\pi \mu }} \mu ^{-\frac{i}{2}\mu } \varGamma \left( 1-\frac{i}{2}\mu \right) \left( e^{\frac{\pi }{4}\mu } \phi _1^\vdash (y) + e^{-\frac{\pi }{4}\mu } \phi _1^\dashv (y) \right) . \end{aligned}$$

By similar computations, we have the followings:

$$\begin{aligned} \phi _2^\bot (y)&= \frac{1}{i\sqrt{\pi \mu }} \mu ^{-\frac{i}{2}\mu } \varGamma \left( 1-\frac{i}{2}\mu \right) \left( e^{\frac{\pi }{4}\mu } \phi _2^\vdash (y) + e^{-\frac{\pi }{4}\mu } \phi _2^\dashv (y) \right) ,\\ \phi _1^\top (y)&= \frac{1}{i\sqrt{\pi \mu }} \mu ^{-\frac{i}{2}\mu } \varGamma \left( 1-\frac{i}{2}\mu \right) \left( -e^{-\frac{\pi }{4}\mu } \phi _1^\vdash (y) - e^{\frac{\pi }{4}\mu } \phi _1^\dashv (y) \right) ,\\ \phi _2^\top (y)&= \frac{1}{i\sqrt{\pi \mu }} \mu ^{-\frac{i}{2}\mu } \varGamma \left( 1-\frac{i}{2}\mu \right) \left( -e^{-\frac{\pi }{4}\mu } \phi _2^\vdash (y) - e^{\frac{\pi }{4}\mu } \phi _2^\dashv (y) \right) . \end{aligned}$$

Hence we get the relation betweens $(\phi ^\vdash , \phi ^\dashv )$ and $(\phi ^\bot , \phi ^\top )$.

$$\begin{aligned} \left( \phi ^\bot \phi ^\top \right) = \frac{1}{i\sqrt{\pi \mu }} \mu ^{-\frac{i}{2}\mu } \varGamma \left( 1-\frac{i}{2}\mu \right) \left( \phi ^\vdash \phi ^\dashv \right) \left( \begin{matrix} {\displaystyle e^{\frac{\pi }{4}\mu } } &{} {\displaystyle -e^{-\frac{\pi }{4}\mu } } \\ {\displaystyle e^{-\frac{\pi }{4}\mu } } &{} {\displaystyle -e^{\frac{\pi }{4}\mu } } \end{matrix} \right) . \end{aligned}$$

$\square$

From the reflection property of the Gamma function:

$$\begin{aligned} \varGamma (z)\varGamma (1-z) = \frac{\pi }{\sin \pi z} = \frac{2\pi i}{e^{i\pi z} - e^{-i\pi z}}\qquad (z\in \mathbb {C}\setminus \mathbb {Z}) \end{aligned}$$

and thanks to $\frac{q}{p}\in \mathbb {R}$, we get the properties of the constants $\gamma$, p and q as follows:

$$\begin{aligned} |\gamma |^2 = \left( e^{\frac{\pi }{2}\mu } - e^{-\frac{\pi }{2}\mu }\right) ^{-1},\quad |p|^2 - |q|^2 = 1, \quad \frac{p^2 - q^2}{p} = \frac{1}{\bar{p}}. \end{aligned}$$

(69)

1.2 Asymptotic expansions of pull-back solutions of the branching model

In this subsection, we give the asymptotic behaviors of the images of Fourier integral operator $U_{\frac{\pi }{4}}$ of the solutions of the branching model $Q\phi = 0$, which are studied in Appendix “B.1”. We put $I_c(\mu ) := \{ x\in \mathbb {R};\ c \sqrt{\mu } \le |x| \le 2c \sqrt{\mu } \}$, for some constant $c>0$, where the notation $I_c(\mu )$ is introduced in (18).

Proposition 7

There exists $\mu _0 >0$ small enough such that for any $\mu \in (0, \mu _0]$ we obtain

$$\begin{aligned} U_{\frac{\pi }{4}} [\phi ^\vdash ](x)&= e^{-\frac{\pi }{8}i} 2^{\frac{1}{4}} e^{\frac{ix^2}{2\mu }} x^{\frac{i}{2}\mu } \begin{pmatrix} -\frac{\mu }{2x} \\ 1 \end{pmatrix} \left( 1 + {{\mathcal {E}}}^\vdash (x; \mu ) \right)&(x\in I_c(\mu )\cap \mathbb {R}_+),\\ U_{\frac{\pi }{4}} [\phi ^\dashv ](x)&= e^{-\frac{\pi }{8}i} 2^{\frac{1}{4}} e^{\frac{ix^2}{2\mu }} (-x)^{\frac{i}{2}\mu } \begin{pmatrix} -\frac{\mu }{2x} \\ 1 \end{pmatrix} \left( 1 + {{\mathcal {E}}}^\dashv (x; \mu ) \right)&(x\in I_c(\mu )\cap \mathbb {R}_-),\\ U_{\frac{\pi }{4}} [\phi ^\bot ](x)&= e^{\frac{\pi }{8}i} 2^{\frac{1}{4}} e^{-\frac{ix^2}{2\mu }} (-x)^{-\frac{i}{2}\mu } \begin{pmatrix} 1 \\ \frac{\mu }{2x} \end{pmatrix} \left( 1 + {{\mathcal {E}}}^\bot (x; \mu ) \right)&(x\in I_c(\mu )\cap \mathbb {R}_-),\\ U_{\frac{\pi }{4}} [\phi ^\top ](x)&= e^{\frac{\pi }{8}i} 2^{\frac{1}{4}} e^{-\frac{ix^2}{2\mu }} x^{-\frac{i}{2}\mu } \begin{pmatrix} 1 \\ \frac{\mu }{2x} \end{pmatrix} \left( 1 + {{\mathcal {E}}}^\top (x; \mu ) \right)&(x\in I_c(\mu )\cap \mathbb {R}_+), \end{aligned}$$

where each error ${{\mathcal {E}}}^{*} (x;\mu )$ is a function satisfying ${{\mathcal {E}}}^{*} (x;\mu ) = {{\mathcal {O}}}(\mu ^2 |x|^{-2}) = {{\mathcal {O}}}(\mu )$ uniformly on $I_c(\mu )$ with $*\in \{ \vdash , \dashv , \bot , \top \}$.

Proof of Proposition 7

The proof is the direct calculation by using the stationary phase method. We show the calculation of the asymptotic behavior of $U_{\frac{\pi }{4}}[\phi ^\vdash ](x)$ when $\mu$ tends to 0 for $x \in I_c(\mu )\cap \mathbb {R}_+$. From the definition of the Fourier integral operator $U_{\frac{\pi }{4}}$, we compute

$$\begin{aligned} U_{\frac{\pi }{4}}[\phi _2^\vdash (y)](x)&= \frac{e^{\frac{\pi i}{8}}2^{\frac{1}{4}}}{\sqrt{2\pi \mu }} e^{-\frac{ix^2}{2\mu }} \int _{0}^{\infty } e^{\frac{i}{\mu }f(x,y)} y^{\frac{i\mu }{2}} dy, \end{aligned}$$

where the phase function f(x, y) is of the form: $f(x,y) = \sqrt{2}xy - \frac{y^2}{2}$. The first and second derivatives of f(x, y) with respect to y are $f_y (x,y) = \sqrt{2}x - y$ and $f_{yy} (x,y) = -1$. One sees that the stationary point $y = \sqrt{2}x = {{\mathcal {O}}}(\sqrt{\mu })$ lies on the finite integral path for $x \in I_c(\mu )\cap \mathbb {R}_+$. Applying the stationary phase method to $U_{\frac{\pi }{4}}[\phi _2^\vdash ](x)$, we obtain

$$\begin{aligned} U_{\frac{\pi }{4}}[\phi _2^\vdash ](x)&= e^{-\frac{\pi i}{8}} 2^{\frac{1}{4}} e^{-\frac{ix^2}{2\mu }} e^{\frac{i}{\mu } f(x,\sqrt{2}x)} \left( y^{\frac{i}{2}\mu } + \mu \partial _y^2 (y^{\frac{i}{2}\mu })\Bigr |_{y=\sqrt{2}x} + {{\mathcal {O}}}(\mu ^2) \right) \\&= e^{-\frac{\pi i}{8}} 2^{\frac{1}{4} + \frac{i}{4}\mu } e^{\frac{ix^2}{2h}} x^{\frac{i}{2}\mu } (1+ C \mu ^2 x^{-2} + {{\mathcal {O}}}(\mu ^2)) \end{aligned}$$

as $\mu$ tends to 0 with some constant C. In fact, the second term $\mu ^2 x^{-2} = {{\mathcal {O}}}(\mu )$ uniformly on $I_c(\mu )\cap \mathbb {R}_+$. Moreover we can compute $U_{\frac{\pi }{4}}[\phi _1^\vdash ](x)$ as $-2^{-\frac{1}{2}}\mu \, U_{\frac{\pi }{4}}[ y^{-1}\phi _2^\vdash (y)](x)$ similarly thanks to the fact that the phase function is the same. Hence we obtain

$$\begin{aligned} U_{\frac{\pi }{4}}[\phi ^\vdash ](x) = e^{-\frac{\pi i}{8}} 2^{\frac{1}{4} + \frac{i}{4}\mu } e^{\frac{ix^2}{2h}} x^{\frac{i}{2}\mu } \begin{pmatrix} {\displaystyle - \frac{\mu }{2 x}}\\ 1 \end{pmatrix} \left( 1 + {{\mathcal {O}}}(\mu )\right) \end{aligned}$$

as $\mu$ tends to 0 uniformly on $I_c(\mu )\cap \mathbb {R}_+$.

On the other hand, the calculation of $U_{\frac{\pi }{4}}[\phi ^\dashv ](x)$ for $x \in I_c(\mu )\cap \mathbb {R}_-$, we see that the stationary point $y= \sqrt{2}x$ also lies on the finite integral path. Similarly we obtain

$$\begin{aligned} U_{\frac{\pi }{4}}[\phi _2^\dashv ](x)&= e^{-\frac{\pi i}{8}} 2^{\frac{1}{4} + \frac{i}{4}\mu } e^{\frac{ix^2}{2\mu }} (-x)^{\frac{i}{2}\mu } \left( 1+ {{\mathcal {O}}}(\mu )\right) \end{aligned}$$

as $\mu$ tends to 0 uniformly on $I_c(\mu )\cap \mathbb {R}_-$. From the fact that $U_{\frac{\pi }{4}}[\phi _1^\dashv ](x) = 2^{-\frac{1}{2}} \mu \, U_{\frac{\pi }{4}}[(-y)^{-1}\phi _2^\dashv (y)](x)$, we have

$$\begin{aligned} U_{\frac{\pi }{4}}[\phi ^\dashv ](x)&= e^{-\frac{\pi i}{8}} 2^{\frac{1}{4} + \frac{i}{4}\mu } e^{\frac{ix^2}{2\mu }} (-x)^{\frac{i}{2}\mu } \begin{pmatrix} {\displaystyle - \frac{\mu }{2 x}}\\ 1 \end{pmatrix} \left( 1+ {{\mathcal {O}}}(\mu )\right) . \end{aligned}$$

Applying the following lemma to the computation of $U_{\frac{\pi }{4}}[\phi ^\bot ](x)$ and $U_{\frac{\pi }{4}}[\phi ^\top ](x)$, we complete the proof of Proposition 7. $\square$

Lemma 10

The following relations holds.

$$\begin{aligned} U_{\frac{\pi }{4}}[\phi ^{\bot }] = \begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix}{{\mathcal {C}}} U_{\frac{\pi }{4}}[\phi ^\dashv ],\qquad U_{\frac{\pi }{4}}[\phi ^{\top }] = \begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix}{{\mathcal {C}}} U_{\frac{\pi }{4}}[\phi ^\vdash ], \end{aligned}$$

where ${{\mathcal {C}}}$ is a complex conjugate operator, that is ${{\mathcal {C}}}\phi (y) = \overline{\phi (y)}$.

Proof of Lemma 10

Recalling the Fourier integral operators and their canonical transformations on the phase space, we obtain the following identities:

$$\begin{aligned}&{{\mathcal {F}}}_\mu ^2\phi ^{\vdash } = \begin{pmatrix}-1&{}\quad 0\\ 0&{}\quad 1\end{pmatrix}\phi ^{\dashv },&{{\mathcal {F}}}_\mu ^2\phi ^{\dashv } = \begin{pmatrix}-1&{}\quad 0\\ 0&{}\quad 1\end{pmatrix}\phi ^{\vdash },\\&U_{\frac{\pi }{4}}{{\mathcal {C}}} = {{\mathcal {C}}} U_{\frac{\pi }{4}}^{-1},&U_{\frac{\pi }{4}}^{-1} {\mathcal {F}}_\mu ^{-1} = U_{\frac{\pi }{4}}, \end{aligned}$$

where ${{\mathcal {F}}}_\mu$ is a semi-classical Fourier transform defined by (66). Lemma 10 can be obtained by the following computation:

$$\begin{aligned} U_{\frac{\pi }{4}}[\phi ^{\bot }]&= U_{\frac{\pi }{4}} \begin{pmatrix}0&{}1\\ 1&{}0\end{pmatrix}{{\mathcal {C}}} {\mathcal {F}}_\mu [\phi ^{\vdash }] = \begin{pmatrix}0&{}1\\ 1&{}0\end{pmatrix} U_{\frac{\pi }{4}} {{\mathcal {C}}} {{\mathcal {F}}}_\mu ^{-1} {{\mathcal {F}}}_\mu ^2[\phi ^{\vdash }] \\&= \begin{pmatrix}0&{}1\\ 1&{}0\end{pmatrix} U_{\frac{\pi }{4}} {\mathcal {C}} {{\mathcal {F}}}_\mu ^{-1} \left[ \begin{pmatrix}-1&{}0\\ 0&{}1\end{pmatrix}\phi ^{\dashv }\right] = \begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix}{{\mathcal {C}}} U_{\frac{\pi }{4}}^{-1} {{\mathcal {F}}}_\mu ^{-1}[\phi ^{\dashv }] \\&= \begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix}{{\mathcal {C}}} U_{\frac{\pi }{4}}[\phi ^{\dashv }], \end{aligned}$$

and by the similar one concerning $U_{\frac{\pi }{4}}[\phi ^{\top }]$. $\square$

A kind of Neumann’s lemma

We introduce an iteration scheme for a $2\times 2$ system which reduces a function in the off-diagonal to a constant.

Lemma 11

Let J be an interval on ${\mathbb R}$ and $\delta$ be a parameter. Let $\psi (z)$ be a solution of

$$\begin{aligned} \begin{pmatrix} D_z + z &{}\quad \delta g(z;h)\\ \delta g(z;h) &{}\quad D_z -z \end{pmatrix} \psi (z) = 0, \end{aligned}$$

(70)

where the map $g: z \mapsto g(z;h)$ is $C^\infty (J;{\mathbb R})$ with $g(0;h) =1$ and bounded on J uniformly with respect to $h \in (0, h_0]$ with all its derivatives. There exist $\delta _0 >0$ small enough such that for any $\delta \in (0, \delta _0]$, we can find a $C^\infty$-matrix $M(z; \delta , h)$ given by

$$\begin{aligned} M(z; \delta , h) = \text {Id} + \sum _{k\ge 1} M_k(z;h) \delta ^k, \end{aligned}$$

where $M_k(z;h)$ is bounded on J uniformly with respect to $h \in (0, h_0]$ together with all its derivatives, and then $M(z; \delta , h) \psi (z)$ is a solution of

$$\begin{aligned} \begin{pmatrix} D_z + z &{}\quad \delta \\[7pt] \delta &{}\quad D_z - z \end{pmatrix} \Psi (z) =0,\quad z\in J. \end{aligned}$$

(71)

Remark 7

This kind of lemma is performed in [5] and [9] for the study of a $2\times 2$ system but depending only on one parameter. This lemma allows us to reduce our microlocal model to a solvable one’s. Such a technique of reducing a symbol to a special form is well-known as a Birkhoff normal from. See for example [29] in the case of non-degenerate potential wells. And see also [4, Section 5.4] for the symplectic case. The more geometric general settings were studied in [3].

Proof of Lemma 11

We start by constructing a $2\times 2$ matrix $M(z;\delta ,h)$ such that $M(z;\delta ,h)\psi$ satisfies (71) if $\psi$ is a solution of (70). We put ${\tilde{g}}(z;h) := g(z;h) -1$ and rewrite (70) and (71) as

$$\begin{aligned} \left( D_z \mathrm{Id} + z B + \delta N \right) \psi (z)&= -\delta \tilde{g}(z;h) N \psi (z), \end{aligned}$$

(72)

$$\begin{aligned} \left( D_z \mathrm{Id} + z B + \delta N\right) M(z; \delta ,h)\psi (z)&= 0, \end{aligned}$$

(73)

where B and N are $2\times 2$ constant matrices of the forms

$$\begin{aligned} B = \begin{pmatrix} 1 &{}\quad 0\\ 0 &{}\quad -1\end{pmatrix},\qquad N = \begin{pmatrix} 0 &{}\quad 1\\ 1 &{}\quad 0\end{pmatrix}. \end{aligned}$$

We look for a $C^\infty$-matrix $M(z;\delta ,h)$ having the next form:

$$\begin{aligned} M(z;\delta ,h) = \mathrm{Id} + M_+(z;\delta ,h), \end{aligned}$$

where $M_+(z;\delta ,h) = \underset{k\ge 1}{\sum } \delta ^k M_k(z;h)$ with

$$\begin{aligned} M_k(z;h) = \begin{pmatrix} a_k(z;h) &{} b_k(z;h)\\[7pt] -\overline{b_k(z;h)} &{} \overline{a_k(z;h)} \end{pmatrix}\quad \text {for}\ k\ge 1. \end{aligned}$$

We compute the left-hand side of (73) by using the system (72).

$$\begin{aligned} \left( D_z \mathrm{Id} + z B +\delta N\right)&(\mathrm{Id}+ M_+)\psi \nonumber \\&= \left( D_z M_+ + z[B, M_+] + \delta [N, M_+] - \delta {\tilde{g}} MN \right) \psi . \end{aligned}$$

(74)

From the computation (74), we determine M satisfying

$$\begin{aligned}&\delta \left( D_z M_1 + z[B, M_1] - {\tilde{g}} N \right) \\&\quad +\,\sum _{k\ge 2} \delta ^k \left( D_z M_k + z [B, M_k] + [N, M_{k-1}] - {\tilde{g}} M_{k-1} N \right) = 0. \end{aligned}$$

Note that, concerning the matrices B, N and $M_k$, the followings are useful.

$$\begin{aligned} \begin{aligned} \left[ B,\begin{pmatrix} a_k &{} b_k\\ -\bar{b}_k &{} \bar{a}_k\end{pmatrix} \right]&= 2\begin{pmatrix} 0 &{} b_k\\ \bar{b}_k &{} 0\end{pmatrix}, \\ N \begin{pmatrix} a_k &{} b_k\\ -\bar{b}_k &{} \bar{a}_k \end{pmatrix}&= \begin{pmatrix} -\bar{b}_k &{} \bar{a}_k\\ a_k &{} b_k\end{pmatrix},\\ \begin{pmatrix} a_k &{} b_k\\ -\bar{b}_k &{} \bar{a}_k\end{pmatrix} N&= \begin{pmatrix} b_k &{} a_k\\ \bar{a}_k &{} -\bar{b}_k\end{pmatrix}. \end{aligned} \end{aligned}$$

(75)

In the case $k=1$, one sees from (75) that the recurrence system is

$$\begin{aligned} -i\begin{pmatrix}a_1'(z;h)&{}b_1'(z;h)\\[7pt]-\overline{b_1'(z;h)}&{}\overline{a_1'(z;h)}\end{pmatrix} + 2z \begin{pmatrix}0&{}b_1(z;h)\\[7pt]\overline{b_1(z;h)}&{}0\end{pmatrix} = {\tilde{g}}(z;h) \begin{pmatrix}0&{}1\\[7pt]1&{}0\end{pmatrix}. \end{aligned}$$

We remark that two equations in the off-diagonal entries are the same each other by taking their complex conjugates thanks to the assumption that g(z; h) is real. From the diagonal entry, we can choose $a_1(z;h)$ as some constant independent of h. The off-diagonal entry

$$\begin{aligned} \begin{aligned} -i b_1'(z;h) + 2z b_1(z;h) = {\tilde{g}}(z;h) \end{aligned} \end{aligned}$$

is a first order differential equation and one sees that this equation can be solved as

$$\begin{aligned} b_1(z;h) = i e^{-iz^2}\int _0^z e^{is^2}{\tilde{g}}(s;h) ds \end{aligned}$$

with a choice of $b_1(0;h) = 0$. Note that $b_1(z;h) = {\mathcal O}(1)$ on $I \times (0, h_0]$ form the boundedness of g(z; h).

In the case $k\ge 2$, the recurrence system is given by

$$\begin{aligned}&-i\begin{pmatrix} a_k'&{}b_k'\\[7pt]-\overline{b_k'}&{}\overline{a_k'} \end{pmatrix} + 2z \begin{pmatrix}0&{}b_k\\[7pt]\overline{b_k}&{}0 \end{pmatrix} \\&\quad + \begin{pmatrix} -( b_{k-1} + \overline{b_{k-1}}) &{} \overline{a_{k-1}} - a_{k-1}\\[7pt] a_{k-1}-\overline{a_{k-1}} &{} b_{k-1} + \overline{b_{k-1}} \end{pmatrix} = {\tilde{g}} \begin{pmatrix}b_{k-1}&{}a_{k-1}\\[7pt]\overline{a_{k-1}}&{}-\overline{b_{k-1}}\end{pmatrix}. \end{aligned}$$

We also notice that each two equations in the diagonal and off-diagonal entries are the same each other by taking their complex conjugates. One can solve the diagonal entry with an initial condition $a_k(0;h)=0$ as

$$\begin{aligned} a_k(z;h) = i\int _0^z \left( g(s;h) b_{k-1}(s;h) + \overline{b_{k-1}(s;h)} \right) ds \end{aligned}$$

and the off-diagonal entry with $b_k(0;h)=0$ as

$$\begin{aligned} b_k(z;h) = ie^{-iz^2}\int _0^z e^{is^2} \left( g(s;h) a_{k-1}(s;h) - \overline{a_{k-1}(s;h)} \right) ds. \end{aligned}$$

Hence we can construct recursively each entry of $M_k(z;h)$ for all $k\in {\mathbb N}$ and we see that $a_k(z;h)$ and $b_k(z;h)$ are ${{\mathcal {O}}}(1)$ on $J \times (0, h_0]$. The way to construct $a_k(z;h)$ and $b_k(z;h)$ as before and the assumption of a smoothness of g(z; h) imply that

$$\begin{aligned} \Vert M_k(z;h) \Vert \le \left( 2(||g||_{\infty } +1) \sup _{z\in J} |z| \right) ^{2k-1}. \end{aligned}$$

Hence we set $\delta _0 = \left( 2(||g||_{\infty } +1)\, \underset{z\in J}{\sup } \, |z| \right) ^{-2}$ and then for any $\delta < \delta _0$ the constructed matrix $M(z;\delta , h)$ is $C^\infty$ on z and bounded on J uniformly on $(0, h_0]$ together with its all derivatives. $\square$

Algebraic computation

In order to compute the product of the transfer matrices appearing in (9) , we prepare algebraic lemmas. First we introduce four matrices which consist of $M_2(\mathbb {C})$.

$$\begin{aligned} D_1 = \begin{pmatrix} 1&{}0\\ 0&{}0 \end{pmatrix},\quad D_2 = \begin{pmatrix} 0&{}0\\ 0&{}1 \end{pmatrix},\quad N_1 = \begin{pmatrix} 0&{}0\\ 1&{}0 \end{pmatrix},\quad N_2 = \begin{pmatrix} 0&{}1\\ 0&{}0 \end{pmatrix}, \end{aligned}$$

(76)

whose subscript corresponds to a non-zero column. One sees that the set $\{ D_1, D_2, N_1, N_2\}$ is closed under the usual product $M_2(\mathbb {C})$ as follows:

$$\begin{aligned} \begin{aligned} D_1^2&= D_1,&D_2^2 = D_2,&D_1D_2 = D_2D_1 = O,\\ N_1N_2&= D_2,&N_2N_1 = D_1,&N_1^2 = N_2^2 = O,\\ D_1N_2&= N_2,&D_2N_1 = N_1,&D_1N_1 = D_2N_2 = O,\\ N_1D_1&= N_1,&N_2D_2 = N_2,&N_1D_2 = N_2D_1 =O. \end{aligned} \end{aligned}$$

(77)

We put

$$\begin{aligned} T_{k,k+1} = a_k D_1 + \overline{a_k}D_2,\qquad T_k = b_k D_1 + \overline{b_k} D_2 + c_k N_1 + c_k N_2, \end{aligned}$$

where $a_k, b_k, c_k \in \mathbb {C}$ and set ${{\mathcal {T}}}_k := T_k T_{k,k+1}$ for $k=0,1,2,\ldots$ with $T_0 := \mathrm{Id}$. Notice that these matrices appear in the representation of the scattering matrix. From the above properties, we know

$$\begin{aligned} {{\mathcal {T}}}_k = a_kb_k D_1 + \overline{a_kb_k} D_2 + a_kc_k N_1 + \overline{a_k}c_k N_2. \end{aligned}$$

(78)

Thanks to these decompositions, we can understand the algebraic properties of the entries of the product of ${{\mathcal {T}}}_k$, which corresponds to the scattering matrix $S(\varepsilon ,h)$ as in Proposition 1. We focus on the term which includes just one factor $b_j$ or $\overline{b_j}$ in the coefficients of these matrices, for the reason why under our setting of this paper such term contributes the principal and sub principal terms of the the scattering matrix (see Lemma 12) and the transition probability (see Lemma 13).

Definition 3

We define the sequences $\{\sigma _n(\varvec{a},\varvec{c}) \}$ and $\{ \tau _{n}(\varvec{a},\varvec{b},\varvec{c}) \}$ for $n \in \mathbb {N}$ by

$$\begin{aligned}&\left\{ \begin{aligned} \sigma _1(\varvec{a},\varvec{c})&= a_0\overline{a_1} c_1,\\ \sigma _{n}(\varvec{a}, \varvec{c})&= \sigma _{n-1}(\varvec{a},\varvec{c}) \left( {{\mathcal {C}}}^{(n)}a_{n}\right) c_{n}, \end{aligned} \right. \end{aligned}$$

(79)

$$\begin{aligned}&\left\{ \begin{aligned} \tau _1(\varvec{a},\varvec{b},\varvec{c})&= a_0a_1 b_1,\\ \tau _{n}(\varvec{a},\varvec{b},\varvec{c})&= \tau _{n-1}(\varvec{a},\varvec{b},\varvec{c}) \left( {{\mathcal {C}}}^{(n-1)}a_{n}\right) c_{n} + \sigma _{n-1}(\varvec{a},\varvec{c}) \left( {\mathcal C}^{(n-1)}a_{n}b_{n}\right) , \end{aligned} \right. \end{aligned}$$

(80)

where the notation ${{\mathcal {C}}}^{(n)}$ stands for the n-th iterated composition ${{\mathcal {C}}}\circ {{\mathcal {C}}} \circ \cdots \circ {{\mathcal {C}}}$ with ${{\mathcal {C}}}$ the operator of taking its complex conjugate as in Lemma 10.

From the above definition (79) and (80), we see that $\{\sigma _n(\varvec{a},\varvec{c}) \}$ (resp. $\{ \tau _{n}(\varvec{a},\varvec{b},\varvec{c}) \}$) is determined by $(a_0, a_1, \ldots , a_n)\in \mathbb {C}^{n+1}$ and $(c_1,\ldots , c_n)\in \mathbb {C}^n$ (resp. $(a_0, a_1, \ldots , a_n)\in \mathbb {C}^{n+1}$, $(b_1,\ldots , b_n)\in \mathbb {C}^n$ and $(c_1,\ldots , c_n)\in \mathbb {C}^n$ and their complex conjugates depending on the index n. Omitting the dependence on n for simplicity, we denote these vectors by $\varvec{a}\in \mathbb {C}^{n+1}$ and $\varvec{b}, \varvec{c}\in \mathbb {C}^n$.

Lemma 12

Let k be a fixed positive integer and $a_j, b_j, c_j$ complex numbers for $j = 1, \ldots , k$ such that $|b_j| \ll 1$ for each j. We denote $\max \{ |b_1|, \dots , |b_k| \}$ by b. There exist two kinds of functions $f_k(\varvec{a},\varvec{b},\varvec{c})$ and $g_k(\varvec{a},\varvec{b},\varvec{c})$ satisfying ${{\mathcal {O}}}(b^2)$ such that

$$\begin{aligned} \begin{aligned} \prod _{j=1}^{2k-1} {{\mathcal {T}}}_j&= \left( \tau _{2k-1}(\varvec{a},\varvec{b},\varvec{c}) + {{\mathcal {O}}}(b^3)\right) D_1 + \left( \overline{\tau _{2k-1}(\varvec{a},\varvec{b},\bar{\varvec{c}} )} + {{\mathcal {O}}}(b^3)\right) D_2 \\&\quad + \left( \overline{\sigma _{2k-1}(\varvec{a},\overline{\varvec{c}})} + g_{2k-1}(\varvec{a},\varvec{b},\varvec{c}) +{{\mathcal {O}}}(b^3) \right) N_1\\&\quad + \left( \sigma _{2k-1}(\varvec{a},\varvec{c}) + f_{2k-1}(\varvec{a},\varvec{b},\varvec{c}) + {{\mathcal {O}}}(b^3) \right) N_2,\\ \prod _{j=1}^{2k} {{\mathcal {T}}}_j&= \left( \sigma _{2k}(\varvec{a},\varvec{c}) + f_{2k}(\varvec{a},\varvec{b},\varvec{c}) + {{\mathcal {O}}}(b^3) \right) D_1\\&\ + \left( \overline{\sigma _{2k}(\varvec{a},\overline{\varvec{c}})} + g_{2k}(\varvec{a},\varvec{b},\varvec{c}) +{{\mathcal {O}}}(b^3) \right) D_2\\&\ + \left( \overline{\tau _{2k}(\varvec{a},\varvec{b},\overline{\varvec{c}})} +{{\mathcal {O}}}(b^3) \right) N_1 + \left( \tau _{2k}(\varvec{a},\varvec{b},\varvec{c}) + {{\mathcal {O}}}(b^3) \right) N_2. \end{aligned} \end{aligned}$$

(81)

The proof of this lemma is just algebraic computation based on the properties (77) and the definition of $\{\sigma _n(\varvec{a},\varvec{c}) \}$ and $\{ \tau _{n}(\varvec{a},\varvec{b},\varvec{c}) \}$, so it can be omitted.

The final step for the proof of Theorem 1 (see Sect. 3.2) requires the computation of $|\tau _n(\varepsilon ,h)|^2$. From the definition of $\{ \sigma _{n}(\varvec{a},\varvec{b},\varvec{c}) \}$ given by (79), we see for any $n\in \mathbb {N}$,

$$\begin{aligned} \sigma _n = \left( \prod _{l=0}^n {{\mathcal {C}}}^{(l)} a_l \right) \left( \prod _{l=1}^n c_l \right) ,\qquad |\sigma _n|^2 = \left( \prod _{l=0}^n | a_l |^2 \right) \left( \prod _{l=1}^n |c_l |^2 \right) . \end{aligned}$$

(82)

By (82) and from the definition of $\{\tau _{n}(\varvec{a},\varvec{b},\varvec{c}) \}$ (see 80), we have

$$\begin{aligned} \tau _n = \left( \prod _{l=1}^n c_l \right) \sum _{k=1}^{n} \left[ \left( \prod _{l=0}^{k-1} {{\mathcal {C}}}^{(l)} a_l \right) \left( \prod _{l=k-1}^{n-1} {{\mathcal {C}}}^{(l)} a_{l+1} \right) \left( {{\mathcal {C}}}^{(k-1)} b_{k} \right) \frac{1}{c_{k}} \right] . \end{aligned}$$

(83)

In order to compute the transition probability, the form of $|\tau _n|^2$ is required and given by the following lemma:

Lemma 13

$$\begin{aligned} \begin{aligned} |\tau _n|^2&= \left( \prod _{l=0}^n |a_l|^2 \right) \left( \prod _{l=1}^n |c_l|^2 \right) \sum _{k=1}^n \left| \frac{b_k}{c_k} \right| ^2 \\&\quad + 2 \left( \prod _{l=1}^n |c_l|^2 \right) \mathrm{Re}\, \sum _{k=2}^{n} \Biggl [ \left( \prod _{l=k}^n |a_l|^2 \right) \left( {{\mathcal {C}}}^{(k)} b_k\right) \left( {{\mathcal {C}}} \frac{1}{c_k} \right) \\&\quad \quad \sum _{j=1}^{k-1} \left( \prod _{l=0}^{j-1} |a_l|^2 \right) \!\! \left( \prod _{l=j-1}^{k-2} \left( {{\mathcal {C}}}^{(l)} a_{l+1}^2 \right) \right) \!\! \left( {{\mathcal {C}}}^{(j-1)} b_{j}\right) \frac{1}{c_{j}} \Biggr ]. \end{aligned} \end{aligned}$$

(84)

The proof of this lemma follows directly from (83) and from the useful fact:

$$\begin{aligned} \left( \sum _{k=1}^n \tau _k \right) \left( \sum _{k=1}^n \overline{\tau _k} \right) = \sum _{k=1}^n |\tau _k|^2 + 2 \mathrm{Re}\, \sum _{k=2}^n \overline{\tau _k} \sum _{j=1}^{k-1} \tau _j. \end{aligned}$$