Tunneling effect between radial electric wells in a homogeneous magnetic field

Abstract

We establish a tunneling formula for a Schrödinger operator with symmetric double-well potential and homogeneous magnetic field, in dimension two. Each well is assumed to be radially symmetric and compactly supported. We obtain an asymptotic formula for the difference between the two first eigenvalues of this operator, that is exponentially small in the semiclassical limit.

1 Introduction

This article is devoted to the spectral analysis of electromagnetic Schrödinger operators with symmetries. Without magnetic fields, it is known that symmetries of the potential induce tunneling. This translates into a spectral gap between the two first eigenvalues that becomes exponentially small in the semiclassical limit. This effect was studied in [13, 14] where the spectral gap was estimated (see also [4, 5] and the books [6, 10]). When adding a homogeneous magnetic field to this model, first general results were obtained in [15] (for weak magnetic fields) and [16] (upper bounds on the spectral gap). The problem was recently reconsidered in [7], where some upper and lower bounds on the spectral gap are obtained. These bounds were improved in [11], giving a sharp exponential decay rate.

In this paper, we consider a potential with two symmetric, radial, and compactly supported wells, as in [7, 11]. We prove a sharp estimate on the spectral gap in the presence of a constant magnetic field. As explained in these two references, the spectral gap is given by an integral term measuring the interaction between the wells, the hopping coefficient. However, we suggest here another approach to estimate this coefficient, which is closer to the original spirit of [13]. This approach provides us with a shorter proof and better estimates.

A similar strategy was recently implemented to prove a purely magnetic tunneling formula, between radial magnetic wells [9]. The case of multiple potential wells was also considered in [15] (without magnetic field) and recently adapted to the magnetic case in [12]. In these articles, it is explained how to reduce the problem to many double-well interactions. Therefore, the result we present below should also have applications to that setting.

We consider the following Schrödinger operator acting in the plane $\textbf{R}^2$,

$$\begin{aligned} \mathcal H_h = (-ih\nabla - \textbf{A})^2 + V, \end{aligned}$$

(1.1)

where V is a double-well potential, and $\textbf{A}$ is a vector potential generating a uniform magnetic field of strength $B>0$, i.e., $\nabla \times \textbf{A}= B$. Without loss of generality, we choose the gauge

$$\begin{aligned} A(x,y) = (0,Bx). \end{aligned}$$

(1.2)

The double-well potential is a sum of two disjoint single wells,

$$\begin{aligned} V(x,y) = v(x-L,y) + v(x+L,y), \end{aligned}$$

(1.3)

separated by a distance $2L>0$. On the single well, we make the following assumptions.

Assumption 1

We assume that

• v is smooth, compactly supported, non-positive, and radial.

• v admits a unique minimum $v_0 <0$ reached at 0, and it is non-degenerate.

Let $a>0$ denote the radius of the support of v, i.e. the smallest positive number such that $\textrm{supp}(v) \subset \overline{B}(0,a)$. We exploit the radiality of the single-well problem. We denote by $\varphi$ the ground state of the single-well Hamiltonian,

$$\begin{aligned} \mathcal H_h^{sw} = (-ih\nabla - \textbf{A}^\mathrm{{rad}})^2 + v, \end{aligned}$$

in radial gauge $\textbf{A}^\mathrm{{rad}}(x,y) = \frac{B}{2}(-y,x)$. Then, $\varphi$ is radial, as explained in [11, Section 2.2], and the function $u(|X|) = \varphi (X)$ satisfies the equation

$$\begin{aligned} - h^2 \frac{1}{r} \partial _r (r \partial _r u) + \Big ( \frac{B^2 r^2}{4} + v(r) \Big ) u = \mu _h u \end{aligned}$$

(1.4)

where $\mu _h$ is the eigenvalue of $\varphi$. Hence, the study of the single-well problem is reduced to a radial Schrödinger equation with effective potential $v_B(r) = \frac{B^2 r^2}{4} + v(r)$. As recalled in [11], this is a very standard setting. The decay of $\varphi$ can be optimally measured in terms of the Agmon distance,

$$\begin{aligned} d(r_1,r_2) = \int _{r_1}^{r_2} \sqrt{\frac{B^2 r^2}{4} + v(r) - v_0} \textrm{d}r. \end{aligned}$$

(1.5)

Moreover, we have explicit WKB approximations for $\varphi$ (see Lemma 4). However, the double-well problem is not directly reduced to a Schrödinger equation with potential $v_B$. If that was the case, the spectral gap would have an exponential rate of decay given by $S_0 = 2d(0,L)$, as in the non-magnetic situation [13]. Here, consistently with [11], we prove that the spectral gap is much smaller.

Theorem 1

Let v be a single-well potential satisfying Assumption 1, and let $\mathcal H_h$ be as in (1.1), with V given by (1.3). Also assume that $L > (1+\frac{\sqrt{3}}{2})a$. Then, the spectral gap between the two smallest eigenvalues of $\mathcal H_h$ satisfies

$$\begin{aligned} \lambda _2(h) - \lambda _1(h) = 2C(L,B,v) h^{\frac{1}{2}} e^{-\frac{S}{h} } (1+o(1)) \end{aligned}$$

as $h \rightarrow 0$, for some constant $C(L,B,v) >0$, and with

$$\begin{aligned} S = 2d(0,L) + \int _0^L \Big ( \sqrt{\frac{B^2(2L-r)^2}{4} -v_0} - \sqrt{\frac{B^2r^2}{4} - v_0} \, \Big ) \textrm{d}r. \end{aligned}$$

(1.6)

The constant C(L, B, v) can be computed explicitly, even though we have no simple interpretation, see (4.16).

Theorem 1 is the first result establishing an asymptotic formula for the spectral gap of $\mathcal H_h$. First results in this direction appeared in a general framework in [15], when the magnetic field is weak. Then, first bounds on S were obtained in [7, 16]. Recently, the sharp exponential decay rate (1.6) was found in [11], where it is proven that

$$\begin{aligned} h \ln (\lambda _2(h) - \lambda _1(h)) \sim -S. \end{aligned}$$

Another related problem is the tunneling between purely magnetic wells, when there is no potential, and the confinement is generated by a non-homogeneous magnetic field. This problem was studied in [9] in the case of radial wells, using a similar strategy. Finally, other magnetic tunneling estimates have recently been established in the case of Neumann boundary conditions [3], vanishing magnetic fields [1], or discontinuous magnetic fields [8]. See also [2] in a one-dimensional setting.

Remark 1

The additional integral term in S is a purely magnetic effect, with no analog in the non-magnetic case, thus making this model especially interesting. It is directly related to oscillations of the eigenfunctions. Indeed, the eigenfunctions $\varphi _\ell$ and $\varphi _r$ generated by the left and right well respectively are related by a magnetic translation (see (3.1)),

$$\begin{aligned} \varphi _\ell (x,y) = e^{\frac{i}{h} BL y} \varphi _r(x+2L,y). \end{aligned}$$

(1.7)

Thus, they do not oscillate at the same frequency and the relative oscillation is fast as $h \rightarrow 0$. This rapidly oscillating phase appears in the hopping coefficient, making it smaller than what one could expect (see Sect. 4).

Remark 2

The condition $L > (1+\frac{\sqrt{3}}{2})a$ is a limit of our strategy, which relies entirely on the radiality of the single well. Indeed, the problem centered on the left well is only radial up to distance $2L-a$, and not 2L. This slight difference makes our decay estimates on the eigenfunctions non-optimal (see Lemma 7). To overcome this problem, we would need to understand very precisely the non-radial situation.

Remark 3

We recover the bounds on S stated in [7, 11] as follows. The result [11, Theorem 1.2] can be translated as

$$\begin{aligned} d(0,2L-a) + d(0,a) \le S \le d(0,2L), \end{aligned}$$

(1.8)

which follows from

$$\begin{aligned} S- d(0,2L) = \int _0^L \Big ( \sqrt{\frac{B^2r^2}{4} + v - v_0} - \sqrt{\frac{B^2r^2}{4} - v_0} \, \Big ) \textrm{d}r \le 0, \end{aligned}$$

(1.9)

and

$$\begin{aligned} S- \big ( d(0,2L-a) + d(0,a) \big ) = \int _0^a \Big ( \sqrt{\frac{B^2 (2L-r)^2}{4}-v_0} - \sqrt{\frac{B^2r^2}{4} - v_0} \, \Big ) \textrm{d}r \ge 0. \end{aligned}$$

(1.10)

Moreover, the main result from [7] is

$$\begin{aligned} BL^2 - BLa \le S \le BL^2 + 2 \sqrt{|v_0|} L + \gamma _0, \end{aligned}$$

(1.11)

which follows from (1.8) with $\gamma _0 = \int _0^a \sqrt{v-v_0} \textrm{d}r$.

Remark 4

Using formula (4.16), we can estimate the constant C(L, B, v) in various regimes. For instance,

$$\begin{aligned} C(L,B,v) \underset{B \rightarrow \infty }{\sim }\ \Big ( \frac{B^3 |v_0|}{4\pi } \Big )^{1/2}, \end{aligned}$$

(1.12)

$$\begin{aligned} C(L,B,v) \underset{L\rightarrow \infty }{\sim }\ \sqrt{\pi B L^2} K(L)^2 \exp \Big ( \frac{1}{2} \sqrt{1+ \frac{2v''(0)}{B^2}} \ln \frac{4 |v_0|}{B^2L^2} \Big ), \end{aligned}$$

(1.13)

where K(L) is given in (2.4). For small B, we find

$$\begin{aligned} C(L,B,v) \underset{B \rightarrow 0}{\sim }\ \sqrt{\frac{L}{\pi }} v''(0) |v_0|^{\frac{1}{4}} \exp \Big ( - \int _0^L \frac{v'(s)}{2(v(s)-v_0)} + \frac{1}{s} - \sqrt{\frac{2v''(0)}{v(s)-v(0)}} \textrm{d}s \Big ). \end{aligned}$$

(1.14)

Strategy In Sect. 2, we describe the single well problem. Most of this section is standard since the ground state solves a radial Schrödinger equation without magnetic field. In Sect. 3, we prove that the spectral gap is given by the hopping coefficient. Our proof is inspired by the non-magnetic situation in [6, 10, 13], and is somewhat simpler that the one of [7, 11]. Finally, in Sect. 4, we estimate the hopping coefficient, thus proving Theorem 1.

2 The single-well problem

Let $\varphi$ be the normalized ground state for the single-well problem in radial gauge,

$$\begin{aligned} (-ih\nabla - \textbf{A}^\mathrm{{rad}})^2 \varphi + v \varphi = \mu _h \varphi , \end{aligned}$$

(2.1)

with

$$\begin{aligned} \textbf{A}^\mathrm{{rad}}(x,y) = \frac{B}{2} (-y,x). \end{aligned}$$

(2.2)

We first collect the following basic facts about $\varphi$.

Proposition 2

The ground state energy $\mu _h$ of $(-ih\nabla - \textbf{A}^\mathrm{{rad}})^2$ satisfies

$$\begin{aligned} \mu _h = v_0 + h \sqrt{B^2 + 2 v''(0)} + \mathcal O(h^{2}). \end{aligned}$$

The associated eigenfunction $\varphi$ is radial and real-valued. Moreover, the first excited eigenvalue $\mu _{h,1}$ satisfies

$$\begin{aligned} \mu _{h,1} = v_0 + 3\,h\sqrt{B^2 + 2 v''(0)} + \mathcal O(h^{2}). \end{aligned}$$

A detailed proof of Proposition 2 is given in [11, Section 2.3]. It follows from a harmonic approximation of v near its minimum. Indeed, when v is quadratic, the problem becomes more explicit, and we can prove that the minimizer is radial. The harmonic approximation gives radiality when v has a non-degenerate minimum. The function $\varphi$ being radial, equation (2.1) can be rewritten in polar coordinates, using the notation $\varphi (X) = u(|X|)$,

$$\begin{aligned} -h^2 \frac{1}{r} \partial _r \big ( r \partial _r u \big ) + \frac{B^2r^2}{4}u + vu = \mu _h u. \end{aligned}$$

(2.3)

This is a radial Schrödinger equation, with effective potential $v_B(r) = \frac{B^2r^2}{4} + v(r)$. From this observation, you deduce Proposition 2, by harmonic approximation of $v_B$ near its minimum (see [10, Chapter 2] for instance). Moreover, the ground state $\varphi$ has the following WKB approximation [10, Theorem 2.3.1].

Proposition 3

The ground state $\varphi$ of the single-well problem has the following WKB approximation:

$$\begin{aligned} \Big | e^{\frac{d(0,|X|)}{h}}\varphi (X) - h^{-\frac{1}{2}} K(|X|) \Big | = \mathcal O(h^{\frac{1}{2}}), \end{aligned}$$

uniformly on any compact, where d is the Agmon distance introduced in (1.5), and

$$\begin{aligned} K(r) = \sqrt{\frac{B^2+2v''(0)}{4\pi }} \exp \Big ( - \int _0^{r} \frac{v_B'(s)}{4(v_B(s)-v_0)} + \frac{1}{2s} - \frac{\sqrt{B^2 + 2 v''(0)}}{2\sqrt{v_B(s)-v_0}} \textrm{d}s \Big ) . \end{aligned}$$

(2.4)

As observed in [7, Equation (2.9)], the function $\varphi$ also has an explicit integral formula outside the support of v, since equation (2.3) is related to some special functions.

Lemma 4

The normalized ground state $\varphi$ of the single-well problem, solution of (2.1), is radial. Moreover, it has the following integral formula for $|X| >a$:

$$\begin{aligned} \varphi (X) = C_h \exp \Big ( - \frac{B |X|^2}{4h} \Big ) \int _0^\infty \exp \Big ( - \frac{B|X|^2 t}{2h} \Big ) t^{\alpha -1} (1+t)^{-\alpha } \textrm{d}t, \end{aligned}$$

(2.5)

where $\alpha = \frac{1}{2} - \frac{\mu _h}{2Bh}$, and $C_h$ is a normalization constant. Moreover,

$$\begin{aligned} C_h \sim h^{-1} K(L) \sqrt{\frac{f''(t_L)}{2\pi }} t_L^{1-\nu }(1+t_L)^{\nu } e^{\frac{\tilde{d}(L) - d(0,L)}{h}} \quad \text {as} \quad h \rightarrow 0, \end{aligned}$$

(2.6)

where $\tilde{d}(L) = \int _0^L \sqrt{\frac{B^2r^2}{4} - v_0} \textrm{d}r$, the Agmon distance d(0, L) was defined in (1.5), $t_L$ and $f''(t_L)$ are given in (2.12), and $\nu = \frac{1}{2} - \frac{\sqrt{B^2 + 2v''(0)}}{2B}$.

Proof

Note that $\varphi$ satisfies equation (2.3) which can be solved by special functions outside the support of v (It is related to Kummer functions, see [7]). To estimate $C_h$, we combine the WKB approximation from Proposition 3,

$$\begin{aligned} \varphi (L) = h^{- \frac{1}{2}} K(L) e^{-\frac{d(0,L)}{h}}(1+o(1)), \end{aligned}$$

(2.7)

with an estimate of (2.5) using the Laplace method. Indeed, using that

$$\begin{aligned} \alpha = \frac{|v_0|}{2Bh} + \nu + o(1), \end{aligned}$$

(2.8)

we find

$$\begin{aligned} \varphi (L) = C_h \int _0^\infty \exp \Big ( - \frac{f(t)}{h} \Big ) t^{\nu -1}(1+t)^{-\nu } \textrm{d}t \, \big ( 1+ o(1)\big ), \end{aligned}$$

(2.9)

with

$$\begin{aligned} f(t) = \frac{B L^2}{4}(1+2 t) + \frac{|v_0|}{2B} \ln \Big ( \frac{1+t}{t} \Big ). \end{aligned}$$

(2.10)

The first and second derivatives of f are

$$\begin{aligned} f'(t) = \frac{BL^2}{2} - \frac{|v_0|}{2B} \frac{1}{t(1+t)}, \qquad f''(t) = \frac{|v_0|}{2B} \frac{2t+1}{t^2(1+t)^2}. \end{aligned}$$

(2.11)

In particular, f has a unique critical point $t_L>0$, which is also a global minimum,

$$\begin{aligned} t_L = \frac{1}{2} \Big ( -1 + \sqrt{1 + \frac{4 |v_0|}{B^2 L^2}} \, \Big ), \quad \text {with} \quad f''(t_L) = \frac{B^2L^3}{2|v_0|} \sqrt{B^2 L^2 + 4 |v_0|}. \end{aligned}$$

(2.12)

By the Laplace method, we deduce

$$\begin{aligned} \varphi (L) = C_h \sqrt{\frac{2\pi h}{f''(t_L)}} e^{- \frac{f(t_L)}{h}} t_L^{\nu -1} (1+t_L)^{-\nu } \big ( 1 + o(1) \big ). \end{aligned}$$

(2.13)

Also note that $f(t_L) = \tilde{d}(L)$. We combine this with (2.7) to get the estimate on $C_h$. $\square$

An estimate of $C_h$ is also given in [11] using the same techniques, but the reference point for (2.7) is $r=a$ instead of $r=L$.

3 The double-well problem

In order to study the double-well problem, we follow the strategy of [13] and compute the matrix of $\mathcal H_h$ in a basis $(\varphi _\ell ,\varphi _r)$, where $\varphi _\ell$ (resp. $\varphi _r$) is the ground state generated by the left well (resp. the right well). More precisely, we define $\varphi _\ell$ and $\varphi _r$ as the normalized solutions to

$$\begin{aligned}&(-ih\nabla - \textbf{A})^2 \varphi _\ell + v(x+L,y)\varphi _\ell = \mu _h \varphi _\ell , \end{aligned}$$

(3.1)

$$\begin{aligned}&(-ih\nabla - \textbf{A})^2 \varphi _r + v(x-L,y)\varphi _r = \mu _h \varphi _r, \end{aligned}$$

(3.2)

respectively. Note that $\varphi _\ell$ and $\varphi _r$ are related to the radial solution $\varphi$ of the single-well problem (2.1) as follows. Let us denote by $\textbf{A}^\ell$ and $\textbf{A}^r$ the radial gauges centered at $(-L,0)$ and (L, 0), respectively, namely

$$\begin{aligned} \textbf{A}^\ell (x,y) = \frac{B}{2} (-y, x+L), \qquad \textbf{A}^r(x,y) = \frac{B}{2} (-y,x-L). \end{aligned}$$

We define the two functions $\sigma _\ell$ and $\sigma _r$ by

$$\begin{aligned} \sigma _\ell (x,y) = \frac{B}{2} y (L-x), \qquad \sigma _r(x,y) = - \frac{B}{2} y (L+x). \end{aligned}$$

(3.3)

They satisfy

$$\begin{aligned} \nabla \sigma _\ell = \textbf{A}^\ell - \textbf{A}, \qquad \nabla \sigma _r = \textbf{A}^r - \textbf{A}, \qquad \text {and} \qquad \sigma _\ell - \sigma _r = BL y. \end{aligned}$$

(3.4)

Then, $\varphi _\ell$ and $\varphi _r$ are related to $\varphi$ by a magnetic translation,

$$\begin{aligned} \varphi _\ell (x,y) = e^{-\frac{i \sigma _\ell }{h}} \varphi (x+L,y), \qquad \varphi _r(x,y) = e^{-\frac{i \sigma _r}{h}} \varphi (x-L,y). \end{aligned}$$

(3.5)

The difference between the two smallest eigenvalues of $\mathcal H_h$ can be estimated using $\varphi _\ell$ and $\varphi _r$ through the hopping coefficient, as stated in the following theorem, which can also be found in [7, Section 4], or [12, Section 2.4] in a more general setting.

Theorem 5

The two smallest eigenvalues of $\mathcal H_h$ satisfy

$$\begin{aligned} \lambda _1&= \mu _h - |w_h| + \mathcal O \big ( h^{-3} e^{-\frac{2d(0,2L-a)}{h}} + | \langle \varphi _\ell , \varphi _r \rangle |^2 \big ), \\ \lambda _2&= \mu _h + |w_h| + \mathcal O \big ( h^{-3} e^{-\frac{2d(0,2L-a)}{h}} + | \langle \varphi _\ell , \varphi _r \rangle |^2 \big ), \end{aligned}$$

where

$$\begin{aligned} w_h = \int _{\textbf{R}^2} v(x+L,y) \varphi _\ell (x,y) \overline{\varphi _r(x,y)} \textrm{d}x \textrm{d}y. \end{aligned}$$

(3.6)

The proof of Theorem 5 is a standard application of the Helffer–Sjöstrand strategy (see [13, Theorem 3.9] or [9, 10]). For the reader’s convenience, we recall the main ideas in Sects. 3.1 and 3.2. We show in Sect. 4.2 that the remainder terms are smaller than $w_h$.

3.1 An approximation lemma

First of all, the Agmon estimates give exponential localization of the eigenfunctions of $\mathcal H_h$ inside the wells. This localization is enough to prove that the spectrum of the double-well operator is the superposition of the spectra of the one-well operators, modulo $\mathcal O(h^\infty )$ (as in [10, 13]). The proof of this standard result is omitted.

Lemma 6

There exists $c,h_0>0$ such that for $h \in (0,h_0)$,

$$\begin{aligned} | \lambda _1(h) - \mu _h | = \mathcal O(h^\infty ), \quad | \lambda _2(h) - \mu _h | = \mathcal O(h^\infty ), \end{aligned}$$

and $\lambda _3(h) - \lambda _1(h) \ge ch$.

Let $\Psi _1$, $\Psi _2$ be the two first eigenfunctions of $\mathcal H_h$, and $\Pi$ the spectral projector on $\textrm{Ran}( \Psi _1,\Psi _2)$. The lemma below shows that the single-well states $\varphi _\ell$ and $\varphi _r$ are close to this eigenspace.

Lemma 7

There exists a $C>0$ such that, for h small enough and $j = \ell , r$ we have

$$\begin{aligned} \Vert \varphi _j - \Pi \varphi _j \Vert \le C h^{-\frac{3}{2}} e^{- \frac{d(0,2L-a)}{h}}, \quad \Vert (-ih\nabla - \textbf{A}) (\varphi _j - \Pi \varphi _j) \Vert \le Ch^{-\frac{3}{2}} e^{-\frac{d(0,2L-a)}{h}}. \end{aligned}$$

Proof

We focus on $\varphi _\ell$, the estimates on $\varphi _r$ being identical. Let $\lambda _3$ be the third eigenvalue of $\mathcal H_h$. By definition of $\Pi$, we have the lower bound

$$\begin{aligned} \lambda _3 \Vert (I - \Pi )\varphi _\ell \Vert ^2 \le \langle \mathcal H_h (I-\Pi ) \varphi _\ell , (I-\Pi ) \varphi _\ell \rangle . \end{aligned}$$

(3.7)

On the other hand, since $\mathcal H_h$ and $\Pi$ commute, we can use the eigenvalue equation (3.1) to get

$$\begin{aligned} \langle \mathcal H_h (I-\Pi ) \varphi _\ell , (I-\Pi ) \varphi _\ell \rangle = \mu _h \Vert (I-\Pi ) \varphi _\ell \Vert ^2 + \langle (I-\Pi ) v_r \varphi _\ell , (I-\Pi ) \varphi _\ell \rangle , \end{aligned}$$

(3.8)

with $v_r(x,y) = v(x-L,y)$. Combining (3.7) and (3.8) we find

$$\begin{aligned} \Vert (I - \Pi ) \varphi _\ell \Vert \le (\lambda _3 - \mu _h)^{-1} \Vert v_r \varphi _\ell \Vert . \end{aligned}$$

(3.9)

For the gradient estimate, we start from

$$\begin{aligned} \Vert (-ih\nabla -\textbf{A}) (I-\Pi ) \varphi _\ell \Vert ^2 \le \langle \mathcal H_h (I-\Pi ) \varphi _\ell , (I-\Pi ) \varphi _\ell \rangle + 2 |v_0| \Vert (I-\Pi ) \varphi _\ell \Vert ^2. \end{aligned}$$

(3.10)

We then use (3.8) and (3.9) to deduce

$$\begin{aligned} \Vert (-ih\nabla - \textbf{A}) (I-\Pi ) \varphi _\ell \Vert ^2 \le \big ((\lambda _3 - \mu _h)^{-1} + (\mu _h + 2 |v_0|) (\lambda _3 - \mu _h)^{-2} \big ) \Vert v_r \varphi _\ell \Vert ^2. \end{aligned}$$

(3.11)

With Proposition 3, we finally bound $\Vert v_r \varphi _\ell \Vert$ in (3.9) and (3.11),

$$\begin{aligned} \Vert v_r \varphi _\ell \Vert \le C h^{-\frac{1}{2}}e^{- \frac{d(0,2L-a)}{h}}. \end{aligned}$$

(3.12)

The result follows since $\lambda _3 - \mu _h \ge c h$, by Lemma 6. $\square$

3.2 Proof of Theorem 5

From Lemma 7, we deduce, for $i,j \in \lbrace \ell , r \rbrace$, with $\psi _j = \Pi \varphi _j$,

$$\begin{aligned} \langle \psi _i | \psi _j \rangle&= \langle \varphi _i | \varphi _j \rangle + \mathcal O \big ( h^{-3} e^{- \frac{2d(0,2L-a)}{h}} \big ), \end{aligned}$$

(3.13)

$$\begin{aligned} \langle \psi _i | \mathcal H_h \psi _j \rangle&= \langle \varphi _i | \mathcal H_h \varphi _j \rangle + \mathcal O \big ( h^{-3} e^{-\frac{2d(0,2L-a)}{h}} \big ), \end{aligned}$$

(3.14)

and also

$$\begin{aligned} \langle \varphi _\ell | \mathcal H_h \varphi _\ell \rangle&= \mu _h + \mathcal O \big ( h^{-3} e^{-\frac{2d(0,2L-a)}{h}} \big ), \end{aligned}$$

(3.15)

$$\begin{aligned} \langle \varphi _r | \mathcal H_h \varphi _r \rangle&= \mu _h + \mathcal O \big ( h^{-3} e^{-\frac{2d(0,2L-a)}{h}} \big ), \end{aligned}$$

(3.16)

$$\begin{aligned} \langle \varphi _\ell | \mathcal H_h \varphi _r \rangle&= \mu _h \langle \varphi _\ell | \varphi _r \rangle + w_h. \end{aligned}$$

(3.17)

We now use the orthonormal basis

$$\begin{aligned} \hat{\psi }_\ell = \frac{1}{\Vert \psi _\ell \Vert } \psi _\ell , \qquad \hat{\psi }_r = \frac{\psi _r - \langle \psi _r, \hat{\psi }_\ell \rangle \hat{\psi }_\ell }{\Vert \psi _r - \langle \psi _r, \hat{\psi }_\ell \rangle \hat{\psi }_\ell \Vert }. \end{aligned}$$

(3.18)

It follows from (3.13), (3.14), (3.15), (3.16) and (3.17) that the matrix of $\mathcal H_h$ in the orthonormal basis $(\hat{\psi }_\ell , \hat{\psi }_r)$ is

$$\begin{aligned} \begin{pmatrix} \langle \hat{\psi }_\ell |\mathcal H_h \hat{\psi }_\ell \rangle &{} \langle \hat{\psi }_\ell | \mathcal H_h \hat{\psi }_r \rangle \\ \langle \hat{\psi }_r | \mathcal H_h \hat{\psi }_\ell \rangle &{} \langle \hat{\psi }_r | \mathcal H_h \hat{\psi }_r \rangle \end{pmatrix} = \begin{pmatrix} \mu _h &{} w_h \\ \overline{w}_h &{} \mu _h \end{pmatrix} + \mathcal O \big ( h^{-3} e^{-\frac{2d(0,2L-a)}{h}} + | \langle \varphi _\ell , \varphi _r \rangle |^2 \big ). \end{aligned}$$

(3.19)

We deduce that the two first eigenvalues of $\mathcal H_h$ are

$$\begin{aligned} \lambda _{\pm } = \mu _h \pm | w_h | + \mathcal O \big ( h^{-3} e^{-\frac{2d(0,2L-a)}{h}} + | \langle \varphi _\ell , \varphi _r \rangle |^2 \big ), \end{aligned}$$

(3.20)

and Theorem 5 follows. Note that instead of $\hat{\psi }_j$, one could also use a more symmetric orthonormalization as in [9]. The choice (3.18) is the same as in [7].

4 Estimates on the hopping coefficient $w_h$

We prove here the following estimate on the hopping coefficient, which governs the spectral gap by Theorem 5.

Theorem 8

There exists a constant $C(B,L,v) >0$ such that

$$\begin{aligned} w_h = - C(B,L,v) h^{\frac{1}{2}} e^{- \frac{S}{h}}(1+o(1)) \end{aligned}$$

as $h \rightarrow 0$, with

$$\begin{aligned} S = 2d(0,L) + \int _0^L \Big ( \sqrt{\frac{B^2(2L-r)^2}{4} -v_0} - \sqrt{\frac{B^2r^2}{4} - v_0}\, \Big ) \textrm{d}r. \end{aligned}$$

(4.1)

The constant C(B, L, v) has a long but explicit expression, see (4.16).

Contrary to the approach of [7], we use the representation of $w_h$ as an explicit integral on a line separating the two wells, as in [13, Equation (2.25)]. Compared to the non-magnetic case, the novelty here is that the phase appearing in this integral takes complex values due to the magnetic flux. A similar effect was observed in the purely magnetic situation [9]. However, for constant magnetic fields, the estimate is simpler, since the resulting complex integral is Gaussian. We give the details of the proof in Sect. 4.1.

Our main result, Theorem 1, follows from the reduction to the hopping coefficient Theorem 5, together with Theorem 8. We only need to ensure that the error from Theorem 5 is smaller than $w_h$. We discuss this condition in Sect. 4.2.

4.1 Proof of Theorem 8

Since v is supported in the ball of radius a, and $L>a$, the integral defining $w_h$ can be restricted to the left half-plane, $\Omega _\ell = \lbrace (x,y) | x <0 \rbrace$,

$$\begin{aligned} w_h = \int _{\Omega _\ell } v(x+L,y) \varphi _\ell \overline{\varphi _r} \textrm{d}x \textrm{d}y. \end{aligned}$$

(4.2)

Using Eqs. (3.1) and (3.2) satisfied by $\varphi _\ell$ and $\varphi _r$ on $\Omega _\ell$, we find

$$\begin{aligned} w_h = \int _{\Omega _\ell } \varphi _\ell \cdot \overline{(-ih \nabla - \textbf{A})^2 \varphi _r} \textrm{d}x \textrm{d}y - \int _{\Omega _\ell } (-ih \nabla - \textbf{A})^2 \varphi _\ell \cdot \overline{\varphi _r} \textrm{d}x \textrm{d}y. \end{aligned}$$

(4.3)

A partial integration yields

$$\begin{aligned} w_h = ih \int _{\partial \Omega _\ell } \varphi _\ell \overline{ (-ih\nabla - \textbf{A}) \varphi _r} \cdot \textbf{n}+ \overline{\varphi _r} (-ih \nabla - \textbf{A}) \varphi _\ell \cdot \textbf{n}, \end{aligned}$$

(4.4)

where $\textbf{n}= (1,0)$ is the outward normal to $\Omega _\ell$. Now due to our choice of gauge, $A_1 = 0$ and only remains

$$\begin{aligned} w_h = h^2 \int _{\textbf{R}} \overline{\varphi _r(0,y)} \partial _x \varphi _\ell (0,y) - \varphi _\ell (0,y) \overline{\partial _x \varphi _r(0,y) } \textrm{d}y. \end{aligned}$$

(4.5)

We recall that $\varphi _\ell$ and $\varphi _r$ are related to the radial solution $\varphi$ of the single-well problem by

$$\begin{aligned} \varphi _\ell (x,y) = e^{-\frac{i \sigma _\ell }{h}} \varphi (x+L,y), \qquad \varphi _r(x,y) = e^{-\frac{i \sigma _r}{h}} \varphi (x-L,y). \end{aligned}$$

(4.6)

We now use these relations, together with the integral representation formula in Lemma 2.5, to calculate $w_h$. First of all,

$$\begin{aligned} \varphi _j(0,y) = C_h e^{-\frac{i \sigma _j}{h}} \int _0^\infty e^{- \frac{ B (L^2 + y^2)}{4h} (1+2t)} t^{\alpha -1} (1+t)^{-\alpha } \textrm{d}t, \end{aligned}$$

(4.7)

and

$$\begin{aligned} \partial _x \varphi _\ell (0,y)&= C_h e^{-\frac{i\sigma _\ell }{h}} \int _0^\infty e^{- \frac{ B (L^2 + y^2)}{4h} (1+2t)} \Big (- \frac{i}{h} \partial _x \sigma _\ell - \frac{B L}{2h} (1+2t) \Big ) t^{\alpha -1} (1+t)^{-\alpha } \textrm{d}t,\\ \partial _x \varphi _r(0,y)&= C_h e^{-\frac{i\sigma _r}{h}} \int _0^\infty e^{- \frac{ B (L^2 + y^2)}{4h} (1+2t)} \Big (- \frac{i}{h} \partial _x \sigma _r + \frac{B L}{2h} (1+2t) \Big ) t^{\alpha -1} (1+t)^{-\alpha } \textrm{d}t. \end{aligned}$$

We insert this in (4.5),

$$\begin{aligned} w_h = -h^2 C_h^2&\int _{\textbf{R}\times \textbf{R}_+^2} \Big ( \frac{i}{h} (\partial _x \sigma _r + \partial _x \sigma _\ell ) + \frac{BL}{h} (1+t+s) \Big ) e^{\frac{i}{h} ( \sigma _r - \sigma _\ell )} \nonumber \\&e^{ - \frac{ B (L^2 + y^2)}{2h} (1+t+s) } (ts)^{\alpha -1} (1+t)^{-\alpha }(1+s)^{-\alpha } \textrm{d}y \textrm{d}t \textrm{d}s. \end{aligned}$$

(4.8)

In (4.8), we use $\sigma _r - \sigma _\ell = -BLy$,

$$\begin{aligned} w_h = -h C_h^2 BL \int _{\textbf{R}\times \textbf{R}_+^2} e^{- \frac{B(L^2+y^2)}{2h}(1+t+s) - \frac{iBLy}{h}} \frac{(1+t+s -i y / L) (ts)^{\alpha - 1}}{(1+t)^{\alpha } (1+s)^{\alpha }} \textrm{d}y \textrm{d}t \textrm{d}s. \end{aligned}$$

Here, the y-integral is Gaussian: inside the exponential, we have

$$\begin{aligned} \frac{B(L^2 + y^2)}{2}(1+t+s) + i BL y = \frac{B(1+t+s)}{2} \big ( y + \frac{iL}{1+t+s} \big )^2 + \frac{BL^2}{2} \big ( 1+ t +s + \frac{1}{1+t+s} \big ), \end{aligned}$$

and this complex-centered Gaussian is integrated as

$$\begin{aligned} \int _\textbf{R}\exp \Big (- \frac{B(1+t+s)}{2h} \big ( y +\frac{iL}{1+t+s} \big )^2 \Big ) \textrm{d}y = \sqrt{\frac{2\pi h}{B(1+t+s)}}, \end{aligned}$$

(4.9)

and

$$\begin{aligned} \int _\textbf{R}\frac{-iy}{L}\exp \Big (- \frac{B(1+t+s)}{2h} \big ( y +\frac{iL}{1+t+s} \big )^2 \Big ) \textrm{d}y = \frac{-1}{1+t+s}\sqrt{\frac{2\pi h}{B(1+t+s)}}. \end{aligned}$$

(4.10)

Thus,

$$\begin{aligned} w_h = - h^{\frac{3}{2}} C_h^2 \sqrt{2 \pi B L^2 } \int _{\textbf{R}_+^2}e^{ - \frac{BL^2}{2h} \big ( 1+t+s + \frac{1}{1+t+s} \big ) } \frac{\omega (s,t)(ts)^{\alpha -1}}{(1+t)^\alpha (1+s)^\alpha } \textrm{d}t \textrm{d}s, \end{aligned}$$

(4.11)

with $\omega (s,t) = (1+t+s)^{1/2} - (1+t+s)^{-3/2}\ge 0$. We replace $\alpha = \frac{|v_0|}{2Bh} + \nu + o(1)$ as $h\rightarrow 0$,

$$\begin{aligned} w_h= -h^{\frac{3}{2}} C_h^2 \sqrt{2 \pi B L^2 } \int _{\textbf{R}_+^2} e^{- \frac{g(s,t)}{h}} \frac{\omega (s,t)(ts)^{\nu -1}}{(1+t)^\nu (1+s)^\nu } \textrm{d}t \textrm{d}s \, (1+o(1)), \end{aligned}$$

(4.12)

with

$$\begin{aligned} g(s,t) =\frac{ BL^2}{2} \Big ( 1+t+s + \frac{1}{1+t+s} \Big ) + \frac{|v_0|}{2B} \Big ( \ln \big ( \frac{1+t}{t} \big ) + \ln \big ( \frac{1+s}{s} \big ) \Big ). \end{aligned}$$

(4.13)

The function g has a unique critical point, which is also a global minimum, at $t=s=t_\star$, with

$$\begin{aligned} t_\star = \frac{1}{2} \sqrt{N} - \frac{1}{2} + \frac{1}{2} \sqrt{1+N}, \qquad N = \frac{|v_0|}{B^2L^2}. \end{aligned}$$

(4.14)

Moreover,

$$\begin{aligned} g(t_\star ,t_\star ) = BL^2 \Big ( \sqrt{1+N} + N \ln \big ( \frac{1+\sqrt{1+N}}{\sqrt{N}}\big ) \Big ) = \int _0^{2L} \sqrt{\frac{B^2 r^2}{4}+ |v_0|} \textrm{d}r. \end{aligned}$$

(4.15)

Using the Laplace method, the integral (4.12) can be estimated as

$$\begin{aligned} w_h \sim - h^{5/2} C_h^2 \frac{(2\pi )^{3/2}\sqrt{BL^2}}{|g''(t_\star ,t_\star )|^{\frac{1}{2}}} \frac{\omega (t_\star ,t_\star ) t_\star ^{2\nu -2}}{(1+t_\star )^{2\nu }} \exp \Big ( - \frac{1}{h} \int _0^{2L} \sqrt{\frac{B^2r^2}{4} + |v_0|} \textrm{d}r \Big ), \end{aligned}$$

(4.16)

when $h\rightarrow 0$. We insert the estimate (2.6) on $C_h$, and we find an explicit constant $C(B,L,v) >0$ such that

$$\begin{aligned} w_h \sim - C(B,L,v) h^{\frac{1}{2}} \exp \Big ( - \frac{1}{h} \int _0^{2L} \sqrt{\frac{B^2r^2}{4} + |v_0|} \textrm{d}r + \frac{2}{h} \tilde{d}(L) - \frac{2}{h} d(0,L) \Big ). \end{aligned}$$

(4.17)

We find the exponential rate (4.1) since

$$\begin{aligned} \int _0^{2L}&\sqrt{\frac{B^2r^2}{4} + |v_0|} \textrm{d}r - 2 \tilde{d}(L) = \int _0^L \Big ( \sqrt{\frac{B^2(2L-r)^2}{4} -v_0} - \sqrt{\frac{B^2r^2}{4} - v_0} \, \Big ) \textrm{d}r. \end{aligned}$$

4.2 Condition on the error terms

We prove here that the remainders from Theorem 5 are smaller than $w_h$. We first focus on $\langle \varphi _\ell , \varphi _r \rangle ^2$.

Lemma 9

The following estimate holds

$$\begin{aligned} \langle \varphi _\ell , \varphi _r \rangle = \mathcal O( e^{- \frac{1}{h} d(0,2\,L-a)} + e^{-\frac{S}{h}} ). \end{aligned}$$

Remark 5

The term $\langle \varphi _\ell , \varphi _r \rangle$ was also estimated using different techniques in [12].

Proof

We can first use the WKB approximation, Proposition 3, to estimate the integral at large distances,

$$\begin{aligned} \Big \vert \int _{|x| > L-a} \varphi _\ell (x,y) \overline{\varphi _r(x,y)} \textrm{d}x \textrm{d}y \Big \vert \le C e^{-\frac{1}{h}d(0,2L-a)}, \end{aligned}$$

(4.18)

because $\varphi _\ell$ (resp. $\varphi _r$) is small enough for $x>L-a$ (resp. for $x < -L +a$). It remains to estimate the middle region, for which we can use the explicit formula (2.5) (with (3.5)),

$$\begin{aligned} \int _{|x| < L-a}&\varphi _\ell (x,y) \overline{\varphi _r(x,y)} \textrm{d}x \textrm{d}y \\ {}&= C_h^2 \int _{\textbf{R}^2_+} \int _{\textbf{R}} \int _{-(L-a)}^{L-a} \Big ( e^{-\frac{iBLy}{h} - \frac{B}{4h}(1+2t)((x+L)^2 + y^2) - \frac{B}{4h}(1+2s)((x-L)^2 + y^2)}\\ {}&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{(ts)^{\alpha -1}}{ (1+t)^{\alpha } (1+s)^{\alpha }} \Big ) \textrm{d}x \textrm{d}y \textrm{d}t \textrm{d}s. \end{aligned}$$

Now, this integral can be estimated in the same way as $w_h$. Indeed, the y-integral is Gaussian and can be integrated as in (4.9),

$$\begin{aligned} \int _{|x| < L-a}&\varphi _\ell (x,y) \overline{\varphi _r(x,y)} \textrm{d}x \textrm{d}y \\ {}&= C_h^2 \int _{\textbf{R}^2_+} \int _{-(L-a)}^{L-a} \Big ( e^{ - \frac{B}{4h}(1+2t)(x+L)^2 - \frac{B}{4h}(1+2s)(x-L)^2 - \frac{BL^2}{2h(1+t+s)} } \\ {}&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{(ts)^{\alpha -1}}{ (1+t)^{\alpha } (1+s)^{\alpha }} \Big ) \textrm{d}x \textrm{d}t \textrm{d}s. \end{aligned}$$

The x-integral is also Gaussian, and the quadratic phase is minimal at $x=0$. By the Laplace method, we obtain

$$\begin{aligned} \Big \vert \int _{|x| < L-a}&\varphi _\ell (x,y) \overline{\varphi _r(x,y)} \textrm{d}x \textrm{d}y \Big \vert \\ {}&\le C C_h^2 \int _{\textbf{R}^2_+} \int _{-(L-a)}^{L-a} e^{ - \frac{B}{2h}(1+t+s)L^2 - \frac{BL^2}{2h(1+t+s)} } \frac{(ts)^{\alpha -1}}{ (1+t)^{\alpha } (1+s)^{\alpha }}\textrm{d}t \textrm{d}s. \end{aligned}$$

The right-hand side has the same structure as (4.11), and can be estimated using the Laplace method. We deduce that

$$\begin{aligned} \Big \vert \int _{|x| < L-a} \varphi _\ell (x,y) \overline{\varphi _r(x,y)} \textrm{d}x \textrm{d}y \Big \vert \le C e^{- \frac{S}{h}}. \end{aligned}$$

$\square$

To ensure the error terms from Theorem 5 to be smaller than $w_h$, we only need $S< 2 d(0,2L-a)$. We recall formula (4.1) for S, which can be rewritten as

$$\begin{aligned} S = d(0,2L) + d(0,L) - \tilde{d} (L) = 2d(0,2L) - \tilde{d}(2L). \end{aligned}$$

(4.19)

Thus,

$$\begin{aligned} S-2d(0,2L-a) = 2d(2L-a,2L) - \tilde{d}(2L) = d(2L-a,2L) - \tilde{d}(2L-a). \end{aligned}$$

Hence, the condition $S < 2d(0,2L-a)$ is satisfied as soon as

$$\begin{aligned} d(2L-a,2L) < \tilde{d}(2L-a). \end{aligned}$$

(4.20)

We now show that (4.20) is true as soon as $L> (1+\frac{\sqrt{3}}{2}) a$. First, we bound the left-hand side by

$$\begin{aligned} d(2L-a,2L) \le \int _{2L-a}^{2L} \frac{Br}{2} + \sqrt{|v_0|} \textrm{d}r = \sqrt{|v_0|} a + \frac{Ba}{4}(4L-a), \end{aligned}$$

(4.21)

and the right-hand side

$$\begin{aligned} \tilde{d} (2L-a) \ge \int _0^a \sqrt{| v_0 |} \textrm{d}r + \int _a^{2L-a} \frac{Br}{2} \textrm{d}r = \sqrt{|v_0|} a + BL(L-a). \end{aligned}$$

(4.22)

Thus, a sufficient condition for (4.20) to hold is

$$\begin{aligned} B L(L-a) > B La - \frac{Ba^2}{4}, \end{aligned}$$

which is equivalent to $L > (1+ \frac{\sqrt{3}}{2})a$.