Lower bounds on the spectral gap of one-dimensional Schrödinger operators

Abstract

We establish an explicit lower bound on the spectral gap of one-dimensional Schrödinger operators with non-negative bounded potentials and subject to Neumann boundary conditions. In addition, for a smaller class of potentials, we provide an improved lower bound which holds on large intervals.

1 Introduction

In this paper, one aim is to derive a lower bound on the spectral gap for (deterministic) Schrödinger operators defined on an interval of length $L > 0$, subject to Neumann boundary conditions and with non-negative potentials in $L^{\infty }({\mathbb {R}})$. Our bound will be valid for all values of $L > 0$ and will depend on the underlying potential in an explicit way. In addition to that, for certain compactly supported and bounded potentials, we shall also provide an improved lower bound to the spectral gap which holds on large intervals. As a matter of fact, for the potentials considered, we are able to prove that the spectral gap cannot close faster than $\sim L^{-4}$.

The spectral gap is defined as the distance between the first two eigenvalues and constitutes a classical quantity in the spectral theory of operators. For example, in [1, 2, 9] and more recently in [3, 4], the spectral gap of the Laplacian (subject to certain self-adjoint boundary conditions) on a fixed interval was compared with the spectral gap of the Laplacian plus some additional potential on the same interval. It turns out that the spectral gap may increase or decrease, depending on specific properties of the potential considered; for example, as shown in [9], the spectral gap for the (Dirichlet or Neumann) Laplacian always increases given the added potential is convex. Unfortunately, since a generic potential fails, e.g., to be convex, it seems rather difficult to control the spectral gap for a Schrödinger operator on a fixed interval. Surprisingly, as discovered in a recent paper [8], it turns out that the asymptotic behaviour of the spectral gap can nevertheless be studied for a rather general class of potentials. A main finding of [8] was that the spectral gap converges to zero strictly faster than the spectral gap of the free (Dirichlet-) Laplacian for potentials that decay fast enough at infinity. This holds, in particular, for bounded potentials of compact support. Furthermore, in [8], the authors also put forward the conjecture that the spectral gap cannot close faster than $\sim L^{-3}$ for bounded compactly supported potentials and this is part of what motivates the present paper. At this point, let us also refer to [6] where lower bounds on the spectral gaps between consecutive negative eigenvalues for one-dimensional Schrödinger operators have been studied.

The paper is organized as follows: In Section 2, we describe the setting. In Section 3, we derive a Harnack-type inequality which then allows us to establish an explicit lower bound to the spectral gap that holds for a large class of potentials. Finally, in Section 4, we restrict ourselves to a smaller class of potentials and derive an improved lower bound on large intervals.

2 The model

On the interval $\Lambda _L=(-L/2,+L/2)$, we consider the Schrö-dinger operator

$$\begin{aligned} h_L=-\frac{\mathrm {d}^2}{\mathrm {d}x^2}+v \end{aligned}$$

with real, non-negative potentials $v \in L^{\infty }({\mathbb {R}})$ and Neumann boundary conditions at the endpoints $x=\pm L/2$. Standard operator theory tells us that $h_L$ is self-adjoint with purely discrete spectrum. We denote its eigenvalues as $\lambda ^{v}_0(L)\le \lambda ^{v}_1(L) \le \cdots$; the normalized ground-state eigenfunction shall be denoted as $\varphi ^{v,L}_0 \in L^2(\Lambda _L)$. We also recall that $\varphi ^{v,L}_0$ is a positive function. We set $k^v_0(L):=\sqrt{\lambda ^{v}_0(L)}$.

The main object of interest in this paper is the spectral gap

$$\begin{aligned} \Gamma _v(L) := \lambda ^v_1(L)- \lambda ^v_0(L). \end{aligned}$$

(2.1)

Since the ground state is non-degenerate [10], one has $\Gamma _v(L) > 0$ for every value $L > 0$. Furthermore, for potentials that decay fast enough at infinity (i.e. at least quadratically), $\Gamma _v(L)$ converges to zero like $\sim L^{-2}$ as $L \rightarrow \infty$; this holds, in particular, for potentials v of compact support. Note that this follows from an immediate generalization of [8, Theorem 2.1] to the case of Neumann boundary conditions.

3 A general lower bound

In a first step, we derive a lower bound to the infimum of the ground-state eigenfunction as well as a Harnack-type inequality. This inequality will then be the key ingredient to establish our lower bound to the spectral gap.

Lemma 3.1

Assume $v \in L^\infty ({\mathbb {R}})$ and $v \ge 0$. Then

$$\begin{aligned} \inf _{x \in \Lambda _L}\varphi ^{v,L}_0(x) \ge \frac{\mathrm {e}^{-2L\Vert v\Vert _{L^1(\Lambda _L)}}}{\sqrt{L}} \end{aligned}$$

(3.1)

holds for all $L > 0$. Furthermore, for all $L > 0$, one has

$$\begin{aligned} \inf _{x \in \Lambda _L}\varphi ^{v,L}_0(x) \ge \mathrm {e}^{-2L\Vert v\Vert _{L^1(\Lambda _L)}} \cdot \sup _{x \in \Lambda _L}\varphi ^{v,L}_0(x). \end{aligned}$$

(3.2)

.

Proof

We follow the strategy outlined in [5] and, in particular, the proof of [5, Theorem 1.2]: The eigenvalue equation for the ground state reads

$$\begin{aligned} -(\varphi ^{v,L}_0)^{\prime \prime }(x)+v(x)\varphi ^{v,L}_0(x)=\lambda ^{v}_0(L)\varphi ^{v,L}_0(x). \end{aligned}$$

One then defines the auxiliary function, for $x \in \Lambda _L$, by

$$\begin{aligned} w^v_L(x):=\frac{(\varphi ^{v,L}_0)^{\prime }(x)}{\varphi ^{v,L}_0(x)}+\int \limits _{-\frac{L}{2}}^{x}q^v_L(t) \ \mathrm {d}t \end{aligned}$$

where $q^v_L(x):=\lambda ^v_0(L)-v(x)$. The eigenvalue equation then implies that $w^v_L$ is monotonically decreasing and therefore, for all $x \in \Lambda _L$,

$$\begin{aligned} w^v_L\left( +\frac{L}{2}\right) \le w^v_L(x) \le w^v_L\left( -\frac{L}{2}\right) . \end{aligned}$$

Using the fact that the $\varphi ^{v,L}_0$ satisfies Neumann boundary conditions at $\pm L/2$, one has for all $x \in \Lambda _L$,

$$\begin{aligned} \int \limits _{-\frac{L}{2}}^{+\frac{L}{2}}q^v_L(t) \ \mathrm {d}t \le w^v_L(x) \le 0 \end{aligned}$$

and hence

$$\begin{aligned} \int \limits _{x}^{+\frac{L}{2}}q^{v}_L(t) \ \mathrm {d}t \le \frac{(\varphi ^{v,L}_0)^{\prime }(x)}{\varphi ^{v,L}_0(x)} \le -\int \limits _{-\frac{L}{2}}^{x}q^v_L(t) \ \mathrm {d}t . \end{aligned}$$

This immediately gives, for all $x \in \Lambda _L$,

$$\begin{aligned} \left| \frac{(\varphi ^{v,L}_0)^{\prime }(x)}{\varphi ^{v,L}_0(x)} \right| \le \Vert q^v_L\Vert _{L^1(\Lambda _L)}. \end{aligned}$$

(3.3)

Since, by the minmax-principle (using $1/\sqrt{L}$ as a test function in the Rayleigh quotient),

$$\begin{aligned} \lambda ^v_0(L) \le \frac{\Vert v\Vert _{L^1(\Lambda _L)}}{L}, \end{aligned}$$

(3.4)

one concludes $\Vert q\Vert _{L^1(\Lambda _L)} \le 2\Vert v\Vert _{L^1(\Lambda _L)}$.

In a next step, one introduces the function

$$\begin{aligned} h^v_L(t):= \ln \left( \varphi ^{v,L}_0\left( t(x-y)+y\right) \right) \end{aligned}$$

for $0 \le t \le 1$ and $x,y \in \Lambda _L$. This implies

$$\begin{aligned} (h^v_L)^{\prime }(t)=(x-y)\frac{(\varphi ^{v,L}_0)^{\prime }\left( t(x-y)+y\right) }{\varphi ^{v,L}_0\left( t(x-y)+y\right) }. \end{aligned}$$

Then, starting from the identity

$$\begin{aligned} \ln \left( \frac{\varphi ^{v,L}_0(x)}{\varphi ^{v,L}_0(y)}\right) =h^v_L(1)-h^v_L(0)=\int _0^1 (h^v_L)^{\prime }(t)\ \mathrm {d}t, \end{aligned}$$

a simple estimate using (3.3) shows that, for $x,y \in \Lambda _L$,

$$\begin{aligned} \frac{\varphi ^{v,L}_0(x)}{\varphi ^{v,L}_0(y)} \le \mathrm {e}^{2L\Vert v\Vert _{L^1(\Lambda _L)}}. \end{aligned}$$

(3.5)

Eq. (3.5) then immediately implies (3.2). Finally, since $\varphi ^{v,L}_0$ has $L^2$-norm one, one concludes

$$\begin{aligned} \sup _{x \in \Lambda _L} \varphi ^{v,L}_0(x) \ge \frac{1}{\sqrt{L}} \end{aligned}$$

which implies (3.1). $\square$

Remark 3.2

Eq.(3.1) in Lemma 3.1 is sharp: choosing $v \equiv 0$ yields the lower bound $\frac{1}{\sqrt{L}}$. On the other hand, for the zero-potential, the ground state is given by $\varphi ^{v=0,L}_0(x)=\frac{1}{\sqrt{L}}$.

Using [7, Theorem 1.4] in combination with Lemma 3.1 then yields the following result.

Theorem 3.3

(Lower bound spectral gap I). Assume $v \in L^\infty ({\mathbb {R}})$ and $v \ge 0$. Then

$$\begin{aligned}\begin{aligned} \Gamma _v(L) \ge \mathrm {e}^{-4L\Vert v\Vert _{L^1(\Lambda _L)}} \cdot \frac{\pi ^2}{L^2} \end{aligned} \end{aligned}$$

holds for all $L > 0$.

Proof

The strategy is to compare the spectral gap of the operator $h_L$ with the spectral gap of the Neumann Laplacian (i.e., setting $v \equiv 0$) which is given by $\pi ^2/L^2$. Such a comparison result has been provided in [7, Theorem 1.4]: more explicitly, taking into account that the normalized ground-state eigenfunction for the Neumann Laplacian is the constant function $1/\sqrt{L}$, one has

$$\begin{aligned}\begin{aligned} \Gamma _v(L)&\ge \left( \frac{\inf _{x \in \Lambda _L} \varphi ^{v,L}_0(x)}{\sup _{x \in \Lambda _L} \varphi ^{v,L}_0(x)}\right) ^2 \cdot \frac{\pi ^2}{L^2}. \end{aligned} \end{aligned}$$

The result then readily follows with Lemma 3.1. $\square$

Remark 3.4

Theorem 3.3 establishes a lower bound which is (for non-zero potentials) at least exponentially small in the interval length. Hence, this bound is still far away from a lower bound as established, for example, in [8, Proposition 2.9] for a symmetric step-potential. In this case, the lower bound reads $\alpha L^{-3}$ for some constant $\alpha > 0$ and all $L > 0$ large enough. However, due to the fact that a Harnack-type inequality has been used in the proof of Theorem 3.3, an exponential factor seems expectable.

Also, for the zero-potential $v \equiv 0$, the lower bound in Theorem 3.3 is sharp.

4 An improved bound for certain potentials

In this section, we shall study a smaller class of potentials for which we are going to derive another lower bound to the spectral gap which holds on large intervals. More explicitly, we shall make the following additional assumptions on the non-negative potential $v \in L^{\infty }({\mathbb {R}})$:

• i) $v(x) \ne 0 \quad \Leftrightarrow \quad x \in (-b,+b)$ for some $0< b < \infty$,

• ii) $v(x)=v(-x)$       (symmetry),

• iii) v is strictly monotonically increasing on $[-b,-\varepsilon ]$ for some $0< \varepsilon < b$ and such that $\inf _{x \in [-\varepsilon ,0]} v(x) > \gamma$ for some $\gamma > 0$ (alternatively, it is enough to assume $\inf _{x \in [-b,0]} v(x) > \gamma$ for some $\gamma > 0$),

• iv) $b^2\Vert v\Vert _{L^\infty ({\mathbb {R}})} < 1/2$       (smallness condition).

In a first result, we derive a lower bound to $\Gamma _v(L)$ which only depends on the asymptotic behaviour of $k^v_0(L)$. The important aspect here is that asymptotic behaviour of the gap is reduced to studying the asymptotic behaviour of the ground state eigenvalue which is usually more accessible.

Lemma 4.1

Consider $h_L$ with a potential $v \in L^{\infty }({\mathbb {R}})$, $v \ge 0$, that also fulfils conditions i)-iv). Then

$$\begin{aligned} \Gamma _v(L) \ge (1-2b^2\Vert v\Vert _{L^\infty ({\mathbb {R}})})^2 \cdot \frac{\pi ^2}{L^2}\cdot \cos ^2\left( k^v_0(L)(L/2-b)\right) \end{aligned}$$

holds for all $L > 0$ large enough.

Proof

First note that, for L large enough, the assumptions on the potential guarantee that $\varphi ^{v,L}_0$ attains its maximum at the boundary of $(-L/2,+L/2)$ and the minimum at $x=0$: To see this, we first note that $\varphi ^{v,L}_0 \in C^1(\overline{\Lambda _L})$ since $h_L$ is self-adjoint on a domain contained in $H^2(\Lambda _L)$ which itself is a consequence of v being bounded. Furthermore, assumption ii) implies that the ground state is a symmetric function and hence its derivative vanishes at $x=0$. Also, the eigenvalue equation $-(\varphi ^{v,L}_0)^{\prime \prime }(x)+v(x)\varphi ^{v,L}_0(x)=\lambda ^{v}_0(L)\varphi ^{v,L}_0(x)$ together with the fact that $\lambda ^{v}_0(L) \rightarrow 0$ as $L \rightarrow \infty$ (see (3.4) or the remark below (2.1)) and iii) imply that the second derivative of $\varphi ^{v,L}_0$ changes sign exactly once in $(-L/2,0)$ for all L large enough. Since $(\varphi ^{v,L}_0)^{\prime }(-L/2)=0$ and $(\varphi ^{v,L}_0)^{\prime \prime }$ is negative in a neighbourhood around $x=-L/2$, the maximum is assumed at $x=-L/2$ and the minimum at $x=0$. In addition, we conclude that $(\varphi ^{v,L}_0)^{\prime }(x) \le 0$ for $x \in [-L/2,0]$.

Now, taking assumption i) into account, the restriction of the ground state eigenfunction $\varphi ^{v,L}_0$ to $(-L/2,-b)$ is given by

$$\begin{aligned} (\varphi ^{v,L}_0|_{(-L/2,-b)})(x)=A \cos (k^v_0(L) (x+L/2)) \end{aligned}$$

with $A=\varphi ^{v,L}_0(-L/2)=\sup _{x \in \Lambda _L}\varphi ^{v,L}_0(x) > 0$. On the other hand, using the eigenvalue equation and the symmetry of v with respect to $x=0$ (assumption ii)), we obtain

$$\begin{aligned} (\varphi ^{v,L}_0)^{\prime }(x)=\int \limits _{x}^{0}(\lambda ^{v}_0(L)-v(x))\varphi ^{v,L}_0(x)\ \mathrm {d}x , \quad x \in [-b,0], \end{aligned}$$

(4.1)

and this yields, for $x \in [-b,0]$,

$$\begin{aligned}\begin{aligned} |(\varphi ^{v,L}_0)^{\prime }(x)|&\le b\Vert \lambda ^{v}_0(L)-v\Vert _{L^\infty ({\mathbb {R}})}\cdot \sup _{x \in [-b,0)}\varphi ^{v,L}_0(x)\ \\&= b\Vert \lambda ^{v}_0(L)-v\Vert _{L^\infty ({\mathbb {R}})}\cdot A\cos \left( k^v_0(L)(L/2-b)\right) \ \\&\le 2Ab\Vert v\Vert _{L^\infty ({\mathbb {R}})}\cdot \cos \left( k^v_0(L)(L/2-b)\right) \ \end{aligned} \end{aligned}$$

for all $L > 0$ large enough. Hence,

$$\begin{aligned} \inf _{x \in \Lambda _L}\varphi ^{v,L}_0(x) \ge (1-2b^2\Vert v\Vert _{L^\infty ({\mathbb {R}})})\cdot A\cos \left( k^v_0(L)(L/2-b)\right) \ \end{aligned}$$

which is a useful bound assuming iv).

Finally, as in the proof of Theorem 3.3, [7, Theorem 1.4] yields (taking into account that the ground state eigenfunction of the Neumann Laplacian with zero potential is the constant function $1/\sqrt{L}$)

$$\begin{aligned} \Gamma _v(L) \ge \left( \frac{\inf _{x \in \Lambda _L}\varphi ^{v,L}_0(x)}{\sup _{x \in \Lambda _L}\varphi ^{v,L}_0(x)}\right) ^2 \cdot \frac{\pi ^2}{L^2} , \end{aligned}$$

and hence the statement follows readily. $\square$

In a next step, we need to investigate the asymptotic behaviour of $k^v_0(L)$ as $L \rightarrow \infty$.

Proposition 4.2

Consider $h_L$ with a potential $v \in L^{\infty }({\mathbb {R}})$, $v \ge 0$, that also fulfils conditions i)-iv). Then

$$\begin{aligned} \lim _{L \rightarrow \infty }\left| \frac{\pi }{2}-k^v_0(L)L\left( \frac{1}{2}-\frac{b}{L}\right) \right| =0. \end{aligned}$$

(4.2)

Furthermore, there exists a constant $\delta > 0$ such that

$$\begin{aligned} \frac{\delta }{L} \le \left( \frac{\pi }{2}-k^v_0(L)L\left( \frac{1}{2}-\frac{b}{L}\right) \right) \end{aligned}$$

(4.3)

for all $L > 0$ large enough.

Proof

For the proof, we introduce the characteristic function $\mathbbm {1}_A(\cdot )$ of a measurable set $A \subset {\mathbb {R}}$. The idea is to compare $k^v_0(L)$ with the square root of the lowest eigenvalue for the Laplacian with a step-potential ${\tilde{v}}(x):={\tilde{v}} \cdot \mathbbm {1}_{[-c,+c]}(x)$ with suitable ${\tilde{v}}, c > 0$. The square root of such an eigenvalue shall be denoted by ${\tilde{k}}_0(L)$ (for notational simplicity, we do not state the dependence of ${\tilde{k}}_0(L)$ on ${\tilde{v}}$ and c explicitly in the following).

As shown in the appendix, the quantitation condition (meaning the relation determining ${\tilde{k}}_0(L)$) for the step-potential ${\tilde{v}}$ with Neumann boundary conditions and large enough $L>0$ reads

$$\begin{aligned} \frac{M_0(L)}{{\tilde{\omega }}_0(L)}\tanh (M_0(L)l_2(L))=\tan ({\tilde{\omega }}_0(L)l_1(L)) \end{aligned}$$

(4.4)

where ${\tilde{\omega }}_0(L):={\tilde{k}}_0(L)L$, $M_0(L):=\sqrt{L^2{\tilde{v}}-{\tilde{\omega }}^2_0(L)}$, $l_1(L):=\frac{1}{2}-\frac{c}{L}$, and $l_2(L):=\frac{c}{L}$.

Now, to prove (4.2), we pick two step potentials, one with $c:=\varepsilon$ and ${\tilde{v}}:=\gamma$ and the other one with $c:=b$ and ${\tilde{v}}:=\Vert v\Vert _{L^\infty ({\mathbb {R}})}$. Let ${\tilde{k}}^{(1)}_0(L)$ denote the square root of the lowest eigenvalue for the first step-potential and ${\tilde{k}}^{(2)}_0(L)$ the square root of the lowest eigenvalue for the second step-potential. Consequently, by an operator-bracketing, we have

$$\begin{aligned} {\tilde{k}}^{(1)}_0(L)L \le k^{v}_0(L)L \le {\tilde{k}}^{(2)}_0(L)L \end{aligned}$$

and the statement readily follows from the fact that both, the left-hand as well as the right-hand side, converge to $\pi$ as $L \rightarrow \infty$. This, on the other hand, follows from the quantization condition (4.4).

On the other hand, in order to prove (4.3), we again choose the step-potential with $c:=b$ and ${\tilde{v}}=\Vert v\Vert _{L^\infty ({\mathbb {R}})}$. Hence, ${\tilde{k}}^{(2)}_0(L) \ge k^v_0(L)$ and it is therefore enough to prove (4.3) with ${\tilde{k}}^{(2)}_0(L)$ instead; note there that, by (4.4),

$$\begin{aligned} \frac{\pi }{2}-{\tilde{k}}^{(2)}_0(L)L\left( \frac{1}{2}-\frac{b}{L}\right) > 0. \end{aligned}$$

Now, for ${\tilde{\omega }}_0(L)l_1(L) < \pi /2$ close enough to $\pi /2$ (hence for $L > 0$ large enough), one obtains

$$\begin{aligned} \begin{aligned} \tan ({\tilde{\omega }}_0(L)l_1(L))&\ge \frac{1}{2\cos ({\tilde{\omega }}_0(L)l_1(L))}\ \\&\ge \frac{1}{2(-{\tilde{\omega }}_0(L)l_1(L)+\pi /2)}. \end{aligned} \end{aligned}$$

(4.5)

Furthermore, setting $A(L):=M_0(L)l_1(L)\tanh (M_0(L)l_2(L))$, (4.4) and (4.5) imply

$$\begin{aligned} \begin{aligned} {\tilde{\omega }}_0(L)l_1(L)&\le \frac{\pi A(L)}{(1+2A(L))}\ \\&=\frac{\pi }{2 (1+\frac{1}{2A(L)})}\ \end{aligned} \end{aligned}$$

(4.6)

for L large enough. This now yields, for some constant $\delta > 0$ and all L large enough,

$$\begin{aligned}\begin{aligned} \left( \frac{\pi }{2}-{\tilde{\omega }}_0(L)l_1(L)\right)&\ge \frac{\pi }{2}\left( \frac{1}{2A(L)(1+\frac{1}{2A(L)})}\right) \ \\&\ge \frac{\pi }{8A(L)}\ \\&\ge \frac{\delta }{L}. \end{aligned} \end{aligned}$$

$\square$

Combining Lemma 4.1 and Proposition 4.2 then yields the following statement.

Theorem 4.3

(Lower bound spectral gap II). Consider $h_L$ with a potential $v \in L^{\infty }({\mathbb {R}})$, $v \ge 0$, that also fulfils conditions i)-iv). Then there exists a constant $\beta > 0$ such that

$$\begin{aligned} \Gamma _v(L) \ge \frac{\beta }{L^4} \end{aligned}$$

for all $L>0$ large enough.

Proof

We establish a lower bound to

$$\begin{aligned} \cos ^2\left( k^v_0(L)(L/2-b)\right) =\sin ^2\left( \frac{\pi }{2}+k^v_0(L)(L/2-b)\right) \end{aligned}$$

for large $L > 0$. Since $|\sin (x)| \ge |-\frac{1}{2}x+\frac{\pi }{2}|$ in a small neighbourhood around $x=\pi$, Proposition 4.2 yields

$$\begin{aligned}\begin{aligned} \left| \sin \left( \frac{\pi }{2}+k^v_0(L)(L/2-b)\right) \right|&\ge \left| -\frac{1}{2}\left( \frac{\pi }{2}+k^v_0(L)(L/2-b)\right) + \frac{\pi }{2}\right| \ \\&= \left| \frac{1}{2}\left( \pi -\frac{\pi }{2}-k^v_0(L)(L/2-b)\right) \right| \ \end{aligned} \end{aligned}$$

for all large enough $L > 0$. Taking the square, we get, for L large enough,

$$\begin{aligned} \sin ^2 \left( \frac{\pi }{2}+k^v_0(L)(L/2-b)\right) \ge \frac{1}{4} \left( \frac{\pi }{2}-k^v_0(L)L\left( \frac{1}{2}-\frac{b}{L}\right) \right) ^2. \end{aligned}$$

The statement then follows with Proposition 4.2 and Lemma 4.1. $\square$

Appendix

In this appendix, we shall derive relation (4.4). Recall that $\mathbbm {1}_A(\cdot )$ denotes the characteristic function of a measurable set $A \subset {\mathbb {R}}$. We first remark that the Schrödinger operator

$$\begin{aligned}\begin{aligned} h_L=-\frac{\mathrm {d}^2}{\mathrm {d}x^2}+{\tilde{v}} \cdot \mathbbm {1}_{[-c,+c]}(x) \end{aligned} \end{aligned}$$

with ${\tilde{v}},c > 0$ and defined on $L^2(\Lambda _L)$ is unitarily equivalent to the operator $L^{-2}{\tilde{h}}_L$ with

$$\begin{aligned} {\tilde{h}}_L=-\frac{\mathrm {d}^2}{\mathrm {d}x^2}+L^2{\tilde{v}} \cdot \mathbbm {1}_{[-c/L,+c/L]}(x) \end{aligned}$$

(A.1)

defined on $L^2(\Lambda _{1})$ (for a proof, we refer to [8, Proposition 2.4]). Hence, given ${\tilde{\omega }}_0(L)$ denotes the square root of the lowest eigenvalue of ${\tilde{h}}_L$, one has ${\tilde{\omega }}_0(L)={\tilde{k}}_0(L)L$ where ${\tilde{k}}_0(L)$ is the square root of the lowest eigenvalue of $h_L$. Furthermore, we set $M_0(L):=\sqrt{L^2{\tilde{v}}-{\tilde{\omega }}^2_0(L)}$, $l_1(L):=\frac{1}{2}-\frac{c}{L}$, and $l_2(L):=\frac{c}{L}$.

Equation (4.4) now follows from suitable matching conditions when constructing the ground state eigenfunction of ${\tilde{h}}_L$ (for large $L > 0$). More explicitly, we identify the interval $(-1/2,-c/L)$ with $I_1:=(0,l_1(L))$ and the interval $(-c/L,0)$ with $I_2:=(0,l_2(L))$. Note that, since the ground state eigenfunction of $h_L$ and ${\tilde{h}}_L$ are symmetric with respect to $x=0$, it is sufficient to construct the ground state on $(-L/2,0)$ or $(-1/2,0)$, respectively.

On the interval $I_1$, the potential is zero and hence we make the ansatz

$$\begin{aligned} {\tilde{\varphi }}_{1,L}(x):=A\cos ({\tilde{\omega }}_0(L)x)\ \end{aligned}$$

for some non-zero $A \in {\mathbb {R}}$. On $I_2$, the potential is constant equal to $L^2{\tilde{v}}$ and this yields, for $L > 0$ large enough (i.e. one requires that $L^2{\tilde{v}}-{\tilde{\omega }}^2_0(L) > 0$), the ansatz

$$\begin{aligned} {\tilde{\varphi }}_{2,L}(x):=B \sinh (M_0(L)x)+C\cosh (M_0(L)x) \end{aligned}$$

for some non-zero $B,C \in {\mathbb {R}}$ to be determined. Since the ground state eigenfunction is symmetric and differentiable, we obtain the matching conditions

$$\begin{aligned}\begin{aligned} {\tilde{\varphi }}_{1,L}(l_1(L))&={\tilde{\varphi }}_{2,L}(0),\ \\ {\tilde{\varphi }}^{\prime }_{1,L}(l_1(L))&={\tilde{\varphi }}^{\prime }_{2,L}(0), \\ {\tilde{\varphi }}^{\prime }_{2,L}(l_2(L))&=0. \end{aligned} \end{aligned}$$

The first condition implies

$$\begin{aligned} \cos ({\tilde{\omega }}_0(L)l_1(L))=\frac{C}{A}, \end{aligned}$$

the second condition gives

$$\begin{aligned} \sin ({\tilde{\omega }}_0(L)l_1(L))=-\frac{BM_0(L)}{A{\tilde{\omega }}_0(L)}, \end{aligned}$$

and the third condition yields

$$\begin{aligned} \tanh (M_0(L)l_2(L))=-\frac{B}{C}. \end{aligned}$$

Combining them, we eventually arrive at (4.4).