Spectral Properties of Relativistic Quantum Waveguides

Abstract

We make a spectral analysis of the massive Dirac operator in a tubular neighbourhood of an unbounded planar curve, subject to infinite mass boundary conditions. Under general assumptions on the curvature, we locate the essential spectrum and derive an effective Hamiltonian on the base curve which approximates the original operator in the thin-strip limit. We also investigate the existence of bound states in the non-relativistic limit and give a geometric quantitative condition for the bound states to exist.

Appendix A. Proof of some technical results

In this section, we collect the proofs of some technical results stated in the paper, in order to simplify the overall presentation.

Proof of Proposition 10 and Corollary 11

The multiplication operators by $\sigma _1$ and $\sigma _3$ are bounded and self-adjoint in $L^2\big ((-1,1),{\mathbb {C}}^2\big )$ thus ${\mathcal {T}}(k,m)$ is self-adjoint if and only if ${\mathcal {T}}_0$ is self-adjoint. An integration by parts easily yields that ${\mathcal {T}}_0$ is symmetric and by definition, one has

$$\begin{aligned}&\mathrm {dom}\left( {\mathcal {T}}_0^*\right) = \Big \{ u \in L^2\big ((-1,1),{\mathbb {C}}^2\big ) :\exists \ w \in L^2\big ((-1,1),{\mathbb {C}}^2\big ) \text { such that }\\&\quad \forall \ v \in \mathrm {dom}\left( {\mathcal {T}}_0\right) , \ \langle u,{\mathcal {T}}_0 v\rangle _{L^2((-1,1),{\mathbb {C}}^2)} = \langle w,v\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\Big \}. \end{aligned}$$

For every $v\in {\mathcal {D}}:= C_0^\infty \big ((-1,1),{\mathbb {C}}^2\big )$ and $u \in \mathrm {dom}\left( {\mathcal {T}}_0^*\right)$, there holds

$$\begin{aligned} \langle {\mathcal {T}}_0^*u,v\rangle _{L^2((-1,1),{\mathbb {C}}^2)} = \langle u,{\mathcal {T}}_0 v\rangle _{L^2((-1,1),{\mathbb {C}}^2)}&= \langle u, -i \sigma _2 v'\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\\&= \langle u, i \overline{\sigma _2v'}\rangle _{{\mathcal {D}}',{\mathcal {D}}} \\&= \langle -i \sigma _2 u', {\overline{v}}\rangle _{{\mathcal {D}}',{\mathcal {D}}}\\&= \langle {\mathcal {T}}_0^*u,{\overline{v}}\rangle _{{\mathcal {D}}',{\mathcal {D}}}, \end{aligned}$$

where $\langle \cdot ,\cdot \rangle _{{\mathcal {D}}',{\mathcal {D}}}$ is the duality bracket of distributions. In particular, we know that ${\mathcal {T}}_0^*u = -i \sigma _2 u' \in L^2\big ((-1,1),{\mathbb {C}}^2\big )$ thus we get $u\in H^1\big ((-1,1),{\mathbb {C}}^2\big )$. Moreover, if $v \in \mathrm {dom}\left( {\mathcal {T}}_0\right)$, there holds

$$\begin{aligned} \langle {\mathcal {T}}_0^*u,v\rangle _{L^2((-1,1),{\mathbb {C}}^2)}&= \langle -i \sigma _2 u',v\rangle _{L^2((-1,1),{\mathbb {C}}^2)} \\&= \langle u,-i \sigma _2 v'\rangle _{L^2((-1,1),{\mathbb {C}}^2)} + \Big [\langle -i \sigma _2 u,v\rangle _{{\mathbb {C}}^2}\Big ]_{-1}^1\\&= \langle u,{\mathcal {T}}_0 v\rangle _{L^2((-1,1),{\mathbb {C}}^2)} - u_2(1)\overline{v_1}(1) + u_1(1)\overline{v_2}(1)\\ {}&\qquad + u_2(-1)\overline{v_1}(-1) - u_1(-1)\overline{v_2}(-1). \end{aligned}$$

Since $v\in \mathrm {dom}\left( {\mathcal {T}}_0\right)$, we obtain

$$\begin{aligned} 0 = - (u_2(1) + u_1(1))\overline{v_1}(1) + (u_2(-1) -u_1(-1))\overline{v_1}(-1). \end{aligned}$$

This holds for any $v\in \mathrm {dom}\left( {\mathcal {T}}_0\right)$, so that $u_2(\pm 1) = \mp u_1(\pm 1)$ and $v \in \mathrm {dom}\left( {\mathcal {T}}_0\right)$. In particular ${\mathcal {T}}_0^* = {\mathcal {T}}_0$. Observe that, by the closed graph theorem, $\mathrm {dom}\left( {\mathcal {T}}(k,m)\right)$ is continuously embedded in $H^1\big ((-1,1),{\mathbb {C}}^2\big )$ which itself is compactly embedded in $L^2\big ((-1,1),{\mathbb {C}}^2\big )$. Thus, ${\mathcal {T}}(k,m)$ has compact resolvent.

Let us prove Point (i) by picking $u\in \mathrm {dom}\left( {\mathcal {T}}(k,m)\right)$ and considering

$$\begin{aligned} \begin{aligned} \Vert {\mathcal {T}}(k,m)u\Vert ^2&=\Vert u'\Vert _{L^2((-1,1),{\mathbb {C}}^2)}^2 + (m^2+k^2) \Vert u\Vert _{L^2((-1,1),{\mathbb {C}}^2)}^2 \\&\quad + 2 mk \mathfrak {R}\big (\langle \sigma _3 u,\sigma _1 u\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\big ) \\&\quad + 2m \mathfrak {R}\big (\langle -i\sigma _2 u',\sigma _3 u\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\big )\\&\quad + 2k \mathfrak {R}\big (\langle -i\sigma _2 u',\sigma _1 u\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\big ). \end{aligned} \end{aligned}$$

(69)

We rewrite (69), arguing as follows. Using the anti-commutation rules of Pauli matrices and the boundary condition, we get

$$\begin{aligned} 2 \mathfrak {R}\big (\langle \sigma _3 u,\sigma _1 u\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\big ) = 2 \mathfrak {R}\big (\langle -i\sigma _2 u',\sigma _1 u\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\big ) = 0 \end{aligned}$$

and

$$\begin{aligned} 2 \mathfrak {R}\big (\langle -i\sigma _2 u',\sigma _3 u\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\big ) = \Vert u(1)\Vert _{{\mathbb {C}}^2}^2 + \Vert u(-1)\Vert _{{\mathbb {C}}^2}^2. \end{aligned}$$

(70)

In particular, we obtain

$$\begin{aligned} \Vert {\mathcal {T}}(k,m) u\Vert _{L^2((-1,1),{\mathbb {C}}^2)}^2&= \Vert u'\Vert _{L^2((-1,1),{\mathbb {C}}^2)}^2 + (m^2+k^2) \Vert u\Vert _{L^2((-1,1),{\mathbb {C}}^2)}^2 \\ {}&\qquad + m (\Vert u(1)\Vert _{{\mathbb {C}}^2}^2 + \Vert u(-1)\Vert _{{\mathbb {C}}^2}^2)\\ {}&\ge (m^2+k^2) \Vert u\Vert _{L^2((-1,1),{\mathbb {C}}^2)}^2. \end{aligned}$$

Hence, by the min–max principle (see Proposition 6), if $\lambda \in Sp({\mathcal {T}}(k,m))$, we get $|\lambda |\ge \sqrt{m^2+k^2}$. Moreover, the last inequality is strict. Indeed, if u is an eigenfunction of ${\mathcal {T}}(k,m)$ associated with an eigenvalue $\lambda$ such that $|\lambda | = \sqrt{m^2 +k^2}$ we necessarily get that u is a constant ${\mathbb {C}}^2$-valued function on $(-1,1)$ satisfying the boundary conditions given in (15). It is a contradiction because it implies that $u = 0$ identically. Hence, ${{\,\mathrm{Sp}\,}}({\mathcal {T}}(k,m)) \cap [-\sqrt{m^2+k^2},\sqrt{m^2+k^2}] = \emptyset$ and Point (i) is proved.

Now, let $\lambda \in {{\,\mathrm{Sp}\,}}({\mathcal {T}}(k,m))$ and pick an associated eigenfunction $u = (u_1,u_2)^\top \in \mathrm {dom}\left( {\mathcal {T}}(k,m)\right)$. There holds

$$\begin{aligned} \left\{ \begin{array}{rcl} mu_1+ku_2-u'_2 &{}=&{} \lambda u_1\,,\\ ku_1+u'_1-mu_2&{}= &{} \lambda u_2\,. \end{array}\right. \end{aligned}$$

(71)

The second equation gives $(m+\lambda )u_2=ku_1+u'_1$ and multiplying the first line by $(\lambda +m)$, we get

$$\begin{aligned} -u''_1=Eu_1\,,\qquad E:=\lambda ^2-(m^2+k^2)\,. \end{aligned}$$

Recall that $m\ge 0$ and that by Point (i), we have $E>0$ for all $k\in {\mathbb {R}}$. Thus, we find

$$\begin{aligned} u_1(t)=\alpha \cos \big (\sqrt{E}(t+1)\big ) +\beta \sin \big (\sqrt{E}(t+1)\big ), \end{aligned}$$

for some constants $\alpha ,\beta \in {\mathbb {C}}$ and as $m+\lambda \ne 0$ we get

$$\begin{aligned}&u_2(t)=\frac{1}{\lambda +m}\cos \big (\sqrt{E}(t+1)\big )\big (k\alpha +\sqrt{E}\beta \big )\\&\quad +\frac{1}{\lambda +m}\sin \big (\sqrt{E}(t+1)\big )\big (k\beta -\sqrt{E}\alpha \big ). \end{aligned}$$

The boundary condition at $t=-1$ gives

$$\begin{aligned} (m+\lambda -k)\alpha -\sqrt{E}\beta = 0\,. \end{aligned}$$

The boundary condition at $t=1$ gives

$$\begin{aligned}&\big ((m+\lambda +k)\cos (2\sqrt{E}) - \sqrt{E}\sin (2\sqrt{E})\big ) \alpha + \big ((m+\lambda +k)\sin (2\sqrt{E}) +\\&\quad \sqrt{E}\cos (2\sqrt{E})\big )\beta = 0 . \end{aligned}$$

To obtain a nonzero eigenfunction u, there has to hold

$$\begin{aligned} 0 = \begin{vmatrix} m+\lambda -k&-\sqrt{E}\\ (m+\lambda +k)\cos (2\sqrt{E}) - \sqrt{E}\sin (2\sqrt{E})&(m+\lambda +k)\sin (2\sqrt{E}) + \sqrt{E}\cos (2\sqrt{E}) \end{vmatrix}. \end{aligned}$$

Computing the determinant, we are left with the implicit equation

$$\begin{aligned} m \sin (2\sqrt{E}) + \sqrt{E}\cos (2\sqrt{E}) = 0. \end{aligned}$$

(72)

In particular, it yields that the spectrum of ${\mathcal {T}}(k,m)$ is symmetric with respect to the origin and we remark that when $m = k = 0$, we necessarily have $\sqrt{E} = |\lambda |= p \frac{\pi }{4}$ (with $p\in {\mathbb {N}}$), and that in this case, a normalized eigenfunction associated with $\lambda = \pm \frac{\pi }{4}$ is given by

$$\begin{aligned} u_k^\pm (t) = \frac{1}{2}\cos \left( k\frac{\pi }{4}(t+1)\right) \begin{pmatrix}1\\ 1\end{pmatrix} \pm \frac{1}{2} \sin \left( k\frac{\pi }{4}(t+1)\right) \begin{pmatrix}1\\ -1\end{pmatrix}, \end{aligned}$$

which proves Corollary 11.

Remark that for $m > 0$, a solution E to (72) verifies $\cos (2\sqrt{E})\ne 0$, and we obtain

$$\begin{aligned} \tan (2\sqrt{E}) + \frac{\sqrt{E}}{m} = 0. \end{aligned}$$

(73)

Now, for $p \in {\mathbb {N}}_0 = {\mathbb {N}}\cup \{0\}$, define the line segments $I_0 := [0,\frac{\pi }{2})$ and $I_{p+1} = ((2p+1)\frac{\pi }{2}, (2p+3)\frac{\pi }{2})$

$$\begin{aligned} g_p : I_p \rightarrow {\mathbb {R}},\quad g_p(x) = \tan (2x) +\frac{x}{m}. \end{aligned}$$

(74)

Remark that $g_p'(x) >0$, and in particular, the only solution to $g_0(x) = 0$ is $x=0$. For all $p \ge 1$, we have

$$\begin{aligned} \lim _{x\rightarrow (2p-1)\frac{\pi }{2}^+} g_p(x) = -\infty ,\quad g_p(p\pi ) = p\frac{\pi }{m}>0. \end{aligned}$$

In particular, for all $p\ge 1$, there is a unique solution $x_p \in I_p$ to $g_p(x) =0$. Moreover, it satisfies $x_p \in \big ((2p-1)\frac{\pi }{2},p\pi \big )$. Hence, for $p\ge 1$, $E_p(m)$ is defined as the unique solution E to $g_p(2\sqrt{E}) = 0$. In particular, $E_p(m) \in ((2p-1)^2 \frac{\pi ^2}{16},p^2\frac{\pi ^2}{4})$ which proves Points (ii) and (iii).

Now, we prove (iv). Guided by (72), we define the $C^\infty$ function

$$\begin{aligned} F :\left\{ \begin{array}{lcl} {\mathbb {R}}\times {\mathbb {R}} &{}\rightarrow &{} {\mathbb {R}}\\ (\mu ,m) &{} \mapsto &{} 2m\sin (\mu ) + \mu \cos (\mu ) \end{array}\right. . \end{aligned}$$

One remarks that $F(\frac{\pi }{2},0) = 0$ and $\partial _\mu F(\frac{\pi }{2},0) = \frac{\pi }{2}$. Hence, by the implicit function theorem, there exists $\delta _1,\delta _2 > 0$ and a $C^\infty$ function $\mu : (-\delta _1,\delta _1) \rightarrow (\frac{\pi }{2}-\delta _2,\frac{\pi }{2}+\delta _2)$ verifying $\mu (0) = \frac{\pi }{2}$ and such that for all $|m|<\delta _1$, there holds $F(\mu (m),m) = 0$. Moreover, when $m\rightarrow 0$, there holds

$$\begin{aligned} \mu (m) = \mu (0) + \mu '(0)m + {\mathcal {O}}(m^2) = \frac{\pi }{2} + \frac{4}{\pi }m + {\mathcal {O}}(m^2). \end{aligned}$$

Necessarily, for $m > 0$ sufficiently small, there holds $E_1(m) = \frac{1}{4}\mu (m)^2$. Hence, when $m\rightarrow 0$, there holds

$$\begin{aligned} E_1(m) = \frac{\pi ^2}{16} + m + {\mathcal {O}}(m^2), \end{aligned}$$

which is precisely Point (iv).

Finally, we prove (v). Once again, guided by (72), we define the $C^\infty$ function

$$\begin{aligned} G :\left\{ \begin{array}{lcl} {\mathbb {R}}\times {\mathbb {R}} &{}\rightarrow &{} {\mathbb {R}}\\ (\mu ,\nu ) &{} \mapsto &{} 2\sin (\mu ) + \mu \nu \cos (\mu ) \end{array}\right. . \end{aligned}$$

One remarks that $G(\pi ,0) = 0$ and $\partial _\mu G(\pi ,0) = -2$. Hence, by the implicit function theorem, there exists $\delta _1,\delta _2 > 0$ and a $C^\infty$ function $\mu : (-\delta _1,\delta _1) \rightarrow (\pi -\delta _2,\pi +\delta _2)$ verifying $\mu (0) = \pi$ and such that for all $|\nu |<\delta _1$ there holds $G(\mu (\nu ),\nu ) = 0$. Moreover, when $\nu \rightarrow 0$, there holds

$$\begin{aligned} \mu (\nu ) = \mu (0) + \mu '(0)\nu + {\mathcal {O}}(\nu ^2) = \pi -\frac{\pi }{2} \nu + {\mathcal {O}}(\nu ^2). \end{aligned}$$

Necessarily, for $m > 0$ sufficiently large, there holds $E_1(m) = \frac{1}{4}\mu (m^{-1})^2$. Hence, when $m\rightarrow +\infty$, there holds

$$\begin{aligned} E_1(m) = \frac{\pi ^2}{4} -\frac{\pi ^2}{4m} + O(m^{-2}), \end{aligned}$$

which gives (v). $\square$

Proof of Lemma 28

Let $u \in \mathrm {dom}\left( {\mathcal {E}}_0(\varepsilon ,m)\right)$ and remark that there holds

$$\begin{aligned} \Vert {\mathcal {C}}(\varepsilon ,m) u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2&= \Vert (-i\sigma _1)\partial _s u + m \sigma _3 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2\nonumber \\&\quad + \frac{1}{\varepsilon ^2}\underset{:= A}{\underbrace{\Vert (-i\sigma _2)\partial _t u - \frac{\pi }{4}(P^+ - P^-)u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2}}\nonumber \\&\quad + \frac{1}{\varepsilon }\underset{:=B}{\underbrace{2\mathfrak {R}(\langle (-i\sigma _1)\partial _s u, (-i\sigma _2)\partial _t u - \frac{\pi }{4}(P^+ - P^-)u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)})}} \nonumber \\&\quad + \frac{m}{\varepsilon }\underset{:= C}{\underbrace{2\mathfrak {R}(\langle \sigma _3 u, (-i\sigma _2)\partial _t u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)})}}\nonumber \\&\quad - \frac{m\pi }{4\varepsilon }\underset{:=D}{\underbrace{2\mathfrak {R}(\langle \sigma _3 u, (P^+ - P^-)u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)})}}. \end{aligned}$$

(75)

Now, we deal with each term appearing on the right-hand side of (75). For further use, for all $k\ge 1$, we set $f_k^\pm := \langle u,u_k^\pm \rangle _{L^2((-1,1),{\mathbb {C}}^2)}$ and recall that $\Pi _k$ denotes the projector defined in (25). In particular, for all $k\ge 1$, there holds

$$\begin{aligned} \Vert \Pi _k u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 = \int _{{\mathbb {R}}}\Big (|f_k^+(s)|^2 + |f_k^-(s)|^2\Big )\mathrm{d}s. \end{aligned}$$

Step 1. In this step, we analyze the term A appearing in (75). We remark that

$$\begin{aligned} (-i\sigma _2\partial _t -\frac{\pi }{4}(P^+-P^-))u = \sum _{k\ge 2} \frac{(k-1)\pi }{4}(f_k^+ u_k^+ - f_k^-u_k^-). \end{aligned}$$

(76)

In particular, it gives

$$\begin{aligned} A = \frac{\pi ^2}{16}\sum _{k\ge 2}(k-1)^2 \Vert \Pi _k u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2. \end{aligned}$$

(77)

Step 2. A straightforward computation gives

$$\begin{aligned} -i\sigma _1\partial _s u = \sum _{k\ge 1} -i (f^-_k)' u_k^+ - i (f^+_k)'u_k^-. \end{aligned}$$

In particular, using (76), there holds

$$\begin{aligned}&\langle -i\sigma _1\partial _s u, \big (-i\sigma _2\partial _t - \frac{\pi }{4}(P^+-P^-)\big ) u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)} \nonumber \\&\quad = \frac{\pi }{4}\sum _{k\ge 2}(k-1)\big (-i\int _{\mathbb {R}}(f^-_k)'(s)\overline{f_k^+(s)} \mathrm{d}s + i\int _{\mathbb {R}}(f^+_k)'(s)\overline{f_k^-(s)} \mathrm{d}s\big ). \qquad \end{aligned}$$

(78)

Integrating by parts, we find

$$\begin{aligned}&\overline{-i\int _{\mathbb {R}}(f^-_k)'(s)\overline{f_k^+(s)} ds + i\int _{\mathbb {R}}(f^+_k)' \overline{f_k^-(s)} \mathrm{d}s} \\&\quad = i \int _{\mathbb {R}}\overline{(f^-_k)'(s)}f_k^+(s) \mathrm{d}s - i \int _{\mathbb {R}}\overline{(f^+_k)'(s)} f_k^-(s) \mathrm{d}s \\&\quad =-\left( -i\int _{\mathbb {R}}(f^-_k)'(s)\overline{f_k^+(s)} ds + i\int _{\mathbb {R}}(f^+_k)'(s)\overline{f_k^-(s)} \mathrm{d}s\right) \,, \end{aligned}$$

and then using (78), we get

$$\begin{aligned}&\langle -i\sigma _1\partial _s u, \big (-i\sigma _2\partial _t - \frac{\pi }{4}(P^+-P^-)\big ) u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)} \\&\quad = -\langle \big (-i\sigma _2\partial _t - \frac{\pi }{4}(P^+-P^-)\big ) u,-i\sigma _1\partial _s u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)}. \end{aligned}$$

In particular, we obtain

$$\begin{aligned} B = 2\mathfrak {R}(\langle -i\sigma _1\partial _s u, \big (-i\sigma _2\partial _t - \frac{\pi }{4}(P^+-P^-)\big ) u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)}) = 0. \end{aligned}$$

(79)

Step 3. In this step, we deal with the term C. Integrating by parts as in (70), we obtain:

$$\begin{aligned} C = \int _{\mathbb {R}}\vert u(s,1)\vert ^2 + \vert u(s,-1)\vert ^2 \mathrm{d}s. \end{aligned}$$

(80)

Step 4 It remains to deal with the term D. To do so, we remark that:

$$\begin{aligned} \langle \sigma _3 u, (P^+-P^-)u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)}&= \underset{:= \alpha }{\underbrace{\langle \Pi _1\sigma _3\Pi _1 u, (P^+-P^-)u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)}}}\nonumber \\&\quad + \underset{:= \beta }{\underbrace{\langle \Pi _1^\perp \sigma _3\Pi _1^\perp u, (P^+-P^-)u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)}}} \nonumber \\&\quad +\underset{:=\gamma }{\underbrace{\langle \Pi _1\sigma _3\Pi _1^\perp u, (P^+-P^-)u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)}}}\nonumber \\&\quad + \underset{:= \delta }{\underbrace{\langle \Pi _1^\perp \sigma _3\Pi _1 u, (P^+-P^-)u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)}}}. \end{aligned}$$

(81)

Now, in each of the next substep, we deal with the terms appearing on the right-hand side of (81).

Substep 4.1 Remark that there holds

$$\begin{aligned} \alpha&= \langle f_1^+\sigma _3 u_1^+ + f_1^-\sigma _3 u_1^-, f_1^+u_1^+ - f_1^- u_1^-\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)}\\ {}&= \langle \sigma _3 u_1^+,u_1^+\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\Vert f_1^+\Vert _{L^2({\mathbb {R}})}^2 - \langle \sigma _3 u_1^-,u_1^-\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\Vert f_1^-\Vert _{L^2({\mathbb {R}})}\\ {}&\qquad - \langle \sigma _3 u_1^+,u_1^-\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\langle f_1^+,f_1^-\rangle _{L^2({\mathbb {R}})} \\&\quad + \langle \sigma _3 u_1^-,u_1^+\rangle _{L^2((-1,1),{\mathbb {C}}^2)}\langle f_1^-,f_1^+\rangle _{L^2({\mathbb {R}})}. \end{aligned}$$

Thanks to (31), we get

$$\begin{aligned} \alpha = \frac{2}{\pi }\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2. \end{aligned}$$

(82)

Substep 4.2 We handle the term $\beta$ by obtaining the following upper-bound thanks to the Cauchy–Schwarz inequality:

$$\begin{aligned} |\beta | = |\langle \Pi _1^\perp \sigma _3\Pi _1^\perp u, (P^+-P^-)u\rangle _{L^2(\mathsf {Str},{\mathbb {C}}^2)}| \le \Vert \Pi _1^\perp u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2. \end{aligned}$$

(83)

Substep 4.3 Now, let us focus on the two off-diagonal terms $\gamma$ and $\delta$. A direct computation shows that

$$\begin{aligned} \langle \sigma _3 u^-_k,u^+_1 \rangle _{{\mathbb {C}}^2}=-\langle \sigma _3 u^+_k,u^-_1 \rangle _{{\mathbb {C}}^2}\,,\quad \langle \sigma _3 u^-_k,u^-_1 \rangle _{{\mathbb {C}}^2}=-\langle \sigma _3 u^+_k,u^+_1 \rangle _{{\mathbb {C}}^2}\,. \end{aligned}$$

Then, we get

$$\begin{aligned} \Pi _1\sigma _3\Pi _1^\perp u = \left( \sum _{k\ge 2}a_k f_k^+ - b_k f_k^-\right) u_1^+ + \left( \sum _{k\ge 2}b_k f_k^+ - a_k f_k^-\right) u_1^-, \end{aligned}$$

where we have set for $k\ge 2$

$$\begin{aligned} a_k&:= \langle \sigma _3 u_k^+,u_1^+\rangle _{L^2((-1,1),{\mathbb {C}}^2)} = \frac{4}{\pi }\frac{\sin ^2(\frac{\pi }{4}(k+1))}{(k+1)},\nonumber \\ b_k&:= \langle \sigma _3 u_k^+,u_1^-\rangle _{L^2((-1,1),{\mathbb {C}}^2)} = \frac{4}{\pi }\frac{\sin ^2(\frac{\pi }{4}(k-1))}{(k-1)}. \end{aligned}$$

(84)

Thus, we find

$$\begin{aligned} \gamma = \sum _{k\ge 2} \int _{\mathbb {R}}\langle (a_k - \sigma _1 b_k)\begin{pmatrix}f_k^+\\ f_k^-\end{pmatrix},\begin{pmatrix}f_1^+\\ f_1^-\end{pmatrix}\rangle _{{\mathbb {C}}^2}\mathrm{d}s. \end{aligned}$$

(85)

A similar computation gives

$$\begin{aligned} \delta = \sum _{k\ge 2} \int _{\mathbb {R}}\langle \begin{pmatrix}f_1^+\\ f_1^-\end{pmatrix},(a_k + \sigma _1 b_k)\begin{pmatrix}f_k^+\\ f_k^-\end{pmatrix}\rangle _{{\mathbb {C}}^2}\mathrm{d}s. \end{aligned}$$

(86)

In particular, using (85) and (86), we get

$$\begin{aligned} \gamma + \delta&=2\mathfrak {R}\Big (\sum _{k\ge 2}a_k\int _{\mathbb {R}}\langle \begin{pmatrix}f_1^+\\ f_1^-\end{pmatrix},\begin{pmatrix}f_k^+\\ f_k^-\end{pmatrix}\rangle _{{\mathbb {C}}^2}\mathrm{d}s\Big )\nonumber \\&\quad + 2i\mathfrak {I}\Big (\sum _{k\ge 2}b_k \int _{\mathbb {R}}\langle \begin{pmatrix}f_1^+\\ f_1^-\end{pmatrix},\sigma _1\begin{pmatrix}f_k^+\\ f_k^-\end{pmatrix}\rangle _{{\mathbb {C}}^2}\mathrm{d}s\Big ). \end{aligned}$$

(87)

Using (81), (82) and (87), we obtain

$$\begin{aligned} D = \frac{4}{\pi }\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 + 2 \mathfrak {R}(\beta ) + 4 \mathfrak {R}\Big (\sum _{k\ge 2}a_k\int _{\mathbb {R}}\langle \begin{pmatrix}f_1^+\\ f_1^-\end{pmatrix},\begin{pmatrix}f_k^+\\ f_k^-\end{pmatrix}\rangle _{{\mathbb {C}}^2}\mathrm{d}s\Big ). \end{aligned}$$

In particular, using the Cauchy–Schwartz inequality, we get

$$\begin{aligned} D \le \frac{4}{\pi }\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 + 2 |\beta | + 4\sum _{k\ge 2}\Big (|a_k|\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}\Vert \Pi _ku\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}\Big ).\nonumber \\ \end{aligned}$$

(88)

Now, let us fix $c>0$ to be chosen later. For all $a,b\in {\mathbb {R}}$ and $\varepsilon >0$, we recall the elementary inequality $ab \le \frac{c\varepsilon }{2} a^2 + \frac{1}{2c\varepsilon }b^2$ that we use to get for all $k\ge 2$:

$$\begin{aligned} |a_k|\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}\Vert \Pi _ku\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)} \le \frac{c\varepsilon }{2}a_k^2 \Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 + \frac{1}{2c\varepsilon }\Vert \Pi _k u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2. \end{aligned}$$

Then, summing up for $k\ge 2$, we get

$$\begin{aligned} \sum _{k\ge 2}\Big (|a_k|\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}\Vert \Pi _ku\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}\Big )&\le \frac{1}{2}c\varepsilon S\Vert \Pi _1u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 \nonumber \\&\quad + \frac{1}{2c\varepsilon }\Vert \Pi _1^\perp u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2, \end{aligned}$$

(89)

where we have set $S = \sum _{k\ge 2} a_k^2 < +\infty$ because $a_k^2 = {\mathcal {O}}(k^{-2})$ when $k\rightarrow +\infty$ by (84). Taking into account (83) and (89), (88) gives

$$\begin{aligned} D \le (\frac{4}{\pi }+2cS\varepsilon )\Vert \Pi _1u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 + 2(1+\frac{1}{c\varepsilon }) \Vert \Pi _1^\perp u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2. \end{aligned}$$

(90)

Step 5. In this step, we conclude the proof. Using (77), (79) and (80), (75) becomes

$$\begin{aligned}&\Vert {\mathcal {C}}(\varepsilon ,m) u\Vert ^2_{L^2(\mathsf {Str},{\mathbb {C}}^2)} \\&\quad = \Vert (- i\sigma _1 \partial _s + m\sigma _3)u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 + \frac{\pi ^2}{16\varepsilon ^2}\sum _{k\ge 2}(k-1)^2\Vert \Pi _k u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2\\&\qquad + \frac{m}{\varepsilon }\int _{{\mathbb {R}}} \Big (\vert u(s,1)\vert ^2 + \vert u(s,-1)\vert ^2\Big )\mathrm{d}s - \frac{m\pi }{4\varepsilon } D\\&\quad = \Vert (- i\sigma _1 \partial _s + m\sigma _3)u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 + \frac{\pi ^2}{16\varepsilon ^2}\sum _{k\ge 2}(k-1)^2\Vert \Pi _k u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2\\&\qquad +\frac{1}{\varepsilon ^2}\int _{\mathbb {R}}(\mathfrak {\tau }_{m\varepsilon }(u)(s)-\mathfrak {\tau }_0(u)(s))\mathrm{d}s -\frac{m\pi }{4\varepsilon } D\\&\quad = \Vert (- i\sigma _1 \partial _s + m\sigma _3)u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 + \frac{\pi ^2}{16\varepsilon ^2}\sum _{k\ge 2}(k-1)^2\Vert \Pi _k u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2\\&\qquad - \frac{\pi ^2}{16\varepsilon ^2}\Vert \Pi _1 u\Vert ^2_{L^2(\mathsf {Str},{\mathbb {C}}^2)}+\frac{1}{\varepsilon ^2}\int _{{\mathbb {R}}}(\mathfrak {\tau }_{m\varepsilon }(u)(s)-\mathfrak {\tau }_0(\Pi _1^\perp u)(s))\mathrm{d}s -\frac{m\pi }{4\varepsilon } D, \end{aligned}$$

where the quadratic forms $\mathfrak {\tau }_{\varepsilon m}$ and $\mathfrak {\tau }_0$ are defined in (43). Notice that in the above formula, we used the fact that

$$\begin{aligned}&\mathfrak {\tau }_0(u)(s)=\mathfrak {\tau }_0(\Pi _1 u)(s)+\mathfrak {\tau }_0(\Pi ^\perp _1 u)(s)\\&\quad =\frac{\pi ^2}{16}\Vert (\Pi _1 u)(s)\Vert ^2_{L^2(-1,1,{\mathbb {C}}^2)}+\mathfrak {\tau }_0(\Pi ^\perp _1 u)(s)\,,\qquad s\in {\mathbb {R}}\, \end{aligned}$$

and

$$\begin{aligned} \Vert \Pi _1 u\Vert ^2_{L^2(\mathsf {Str},{\mathbb {C}}^2)}=\int _{{\mathbb {R}}}\Vert (\Pi _1 u)(s)\Vert ^2_{L^2((-1,1),{\mathbb {C}}^2)}\mathrm{d}s\,. \end{aligned}$$

Using Lemma (26), this last inequality becomes

$$\begin{aligned} \Vert {\mathcal {C}}(\varepsilon ,m) u\Vert ^2_{L^2(\mathsf {Str},{\mathbb {C}}^2)} \ge \&\Vert (- i\sigma _1 \partial _s + m\sigma _3)u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 + \frac{\pi ^2}{16\varepsilon ^2}\Vert \Pi _1^\perp u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2\\ {}&+ \frac{1}{\varepsilon ^2}\Big (E_1(m\varepsilon ) - \frac{\pi ^2}{16}\Big )\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 -\frac{m\pi }{4\varepsilon } D \end{aligned}$$

and (90) yields

$$\begin{aligned} \Vert {\mathcal {C}}(\varepsilon ,m) u\Vert ^2_{L^2(\mathsf {Str},{\mathbb {C}}^2)}&\ge \Vert (- i\sigma _1 \partial _s + m\sigma _3)u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 \nonumber \\&\quad + \frac{1}{\varepsilon ^2}\Big (\frac{\pi ^2}{16} - \frac{m\pi }{2c} - \frac{m\pi }{2}\varepsilon \Big )\Vert \Pi _1^\perp u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2\nonumber \\&\quad + \frac{1}{\varepsilon ^2}\Big (E_1(m\varepsilon ) - \frac{\pi ^2}{16} - m\varepsilon - \frac{m\pi cS}{2}\varepsilon ^2\Big )\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2. \end{aligned}$$

(91)

Now, we choose $c > \frac{8m}{\pi }$ and remark that there exists $\varepsilon _1> 0$ such that for all $\varepsilon \in (0,\varepsilon _1)$, there holds

$$\begin{aligned} \frac{\pi ^2}{16} - \frac{m\pi }{2c} - \frac{m\pi }{2}\varepsilon > 0. \end{aligned}$$

(92)

Moreover, thanks to (iv) of Proposition 10, there exists $\varepsilon _2$ and $K>0$ such that for all $\varepsilon \in (0,\varepsilon _2)$

$$\begin{aligned} E_1(m\varepsilon ) - \frac{\pi ^2}{16} - m\varepsilon - \frac{m\pi cS}{2}\varepsilon ^2 > -K \varepsilon ^2. \end{aligned}$$

(93)

Setting $\varepsilon _0 := \min (\varepsilon _1,\varepsilon _2)$ and taking into account (92) and (93) in (91) we obtain that for all $\varepsilon \in (0,\varepsilon _0)$, there holds

$$\begin{aligned} K\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 + \Vert {\mathcal {C}}(\varepsilon ,m)u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 \ge \Vert (-i\sigma _1\partial _s + m\sigma _3)u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2. \end{aligned}$$

The proof of Lemma 28 is completed remarking that $\Vert \Pi _1 u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2 \le \Vert u\Vert _{L^2(\mathsf {Str},{\mathbb {C}}^2)}^2$. $\square$