Quantum Hamiltonians with Weak Random Abstract Perturbation. II. Localization in the Expanded Spectrum

Abstract

We consider multi-dimensional Schrödinger operators with a weak random perturbation distributed in the cells of some periodic lattice. In every cell the perturbation is described by the translate of a fixed abstract operator depending on a random variable. The random variables, indexed by the lattice, are assumed to be independent and identically distributed according to an absolutely continuous probability density. A small global coupling constant tunes the strength of the perturbation. We treat analogous random Hamiltonians defined on multi-dimensional layers, as well. For such models we determine the location of the almost sure spectrum and its dependence on the global coupling constant. In this paper we concentrate on the case that the spectrum expands when the perturbation is switched on. Furthermore, we derive a Wegner estimate and an initial length scale estimate, which together with Combes–Thomas estimate allow to invoke the multi-scale analysis proof of localization. We specify an energy region, including the bottom of the almost sure spectrum, which exhibits spectral and dynamical localization. Due to our treatment of general, abstract perturbations our results apply at once to many interesting examples both known and new.

Appendices

Appendix A. Random Magnetic Field with Non-zero Electric Potential

Remark A.1

We show here that for a random magnetic field as in Sect. 4.4, an arbitrary measurable and bounded $V_0$ and no random electric potential (i.e. $W_1 = W_2 = 0$ in the notation of Sect. 4.4) we have $\Lambda _1 = 0$ and $\Lambda _2 \geqslant 0$, i.e. we are neither in (Case I) nor (Case II). For that purpose, we recall parts of the calculation in [2, Sec. 3.3] and study the effect of adding a non-zero background potential $V_0$. Here we also assume that the considered magnetic field is non-trivial in the sense that it can not be removed by an appropriate gauge transformation. As it is known, this is equivalent to assuming that the magnetic potential A is not a gradient of some scalar function.

Recall that $\Lambda _0$ is the smallest eigenvalue of the operator

$$\begin{aligned} -\frac{d^2}{dx_{n+1}^2}+V_0\quad \text {on}\quad (0,d) \end{aligned}$$

subject to Dirichlet or Neumann boundary condition and $\Psi _0=\Psi _0(x_{n+1})$ is the associated positive eigenfunction, extended to $\square$ by $\Psi _0(x',x_{n+1})=\Psi _0(x_{n+1})$, and normalized appropriately. We then have

$$\begin{aligned} \Lambda _1 = ({\mathcal {L}}_1\Psi _0,\Psi _0)_{L^2(\square )}, \qquad \Lambda _2 = ({\mathcal {L}}_2\Psi _0,\Psi _0)_{L^2(\square )} + (\Psi _1,{\mathcal {L}}_1\Psi _0)_{L^2(\square )}, \end{aligned}$$

where $\Psi _1$ is the unique solution to

$$\begin{aligned} ({\mathcal {H}}^0_\square -\Lambda _0)\Psi _1=-{\mathcal {L}}_1\Psi _0 + \Lambda _1\Psi _0, \qquad (\Psi _1,\Psi _0)_{L^2(\square )}=0. \end{aligned}$$

As in Sect. 4.4, the operators ${\mathcal {L}}_1$ and ${\mathcal {L}}_2$ can be written as

$$\begin{aligned} {\mathcal {L}}_1 = \mathrm {i}[\nabla \cdot A + A \cdot \nabla ] = 2 \mathrm {i}A \cdot \nabla + \mathrm {i}{\text {div}}A, \quad {\mathcal {L}}_2 = |A |^2. \end{aligned}$$

(A.1)

We observe that the formula for ${\mathcal {L}}_1$ implies in particular that the function $\Psi _1$ is pure imaginary.

Since $\Lambda _0$ is a ground state of a 1-dimensional Schrödinger equation, it is non-degenerate and the corresponding eigenfunction $\Psi _0$ is up to a phase real-valued (real part and imaginary part are linearly dependent because else they would yield two linearly independent ground states). Thus, without loss of generality, we assume that $\Psi _0$ is real-valued and since A is also real-valued, we can calculate by employing integration by parts:

$$\begin{aligned} \Lambda _1 = \int \limits _{\square } \mathrm {i}\Psi _0 \left[ \nabla \cdot A + A \cdot \nabla \right] \Psi _0 \mathrm {d}x = \mathrm {i}\int \limits _{\square } \big ((A \Psi _0) \cdot \nabla \Psi _0 - \nabla \Psi _0 \cdot (A \Psi _0)\big ) \mathrm {d}x=0.\nonumber \\ \end{aligned}$$

(A.2)

Hence, also for non-zero $V_0$, random perturbations consisting purely of magnetic fields always imply $\Lambda _1 = 0$. Thus, we are not in (Case I)

Worse, we also have $\Lambda _2 > 0$, such that we are not even in (Case II). To see this, let us first note that

$$\begin{aligned}&\Lambda _2 = ( {\mathcal {L}}_2 \Psi _0, \Psi _0)_{L^2(\square )} + (\Psi _1, {\mathcal {L}}_1 \Psi _0)_{L^2(\square )}\\&= \int _\square |A |^2 \cdot |\Psi _0 |^2 - 2 \mathrm {i}\Psi _1 A \cdot \nabla \Psi _0 - \mathrm {i}\Psi _0 \Psi _1 {\text {div}} A \mathrm {d}x. \end{aligned}$$

Now, to see that $\Lambda _2 > 0$ it suffices to establish

$$\begin{aligned} \Lambda _2 = \int _{\square } |\Psi _0 |^2 \left|A + \mathrm {i}\nabla \frac{\Psi _1}{\Psi _0} \right|^2 \mathrm {d}x. \end{aligned}$$

(A.3)

Indeed, the right hand side is obviously non-negative and the sum $A + \mathrm {i}\nabla \frac{\Psi _1}{\Psi _0}$ cannot vanish since otherwise A would be the gradient of a real-valued scalar function because $\Psi _1$ is purely imaginary.

Identity (A.3) is proved by calculations which are explicitly performed in [2, Section 3.3] using in particular that

$$\begin{aligned} \Psi _0 \nabla \frac{\Psi _1}{\Psi _0} = \nabla \Psi _1 - \frac{\Psi _1}{\Psi _0} \nabla \Psi _0 \end{aligned}$$

Note that since $\Psi _0$ is a ground state, it is bounded away from zero by Harnack’s inequality and we can divide by $\Psi _0$ without any trouble.

Appendix B. Chasing the Wegner Estimate: An Example

Remark B.1

In [25], operators with random magnetic fields are studied. In particular, Sect. 6 of [25] treats random operators of the form

$$\begin{aligned} {\mathcal {H}}^\varepsilon (\omega ) = \big (\mathrm {i}\nabla + A_0 + A^\varepsilon (\omega )\big )^2 + V_0,\qquad A^\varepsilon (\omega ):=\varepsilon \sum \limits _{k\in \Gamma } \omega _k A(x'-k,x_n), \end{aligned}$$

(B.1)

with a random magnetic field $A^\varepsilon (\omega )$, a deterministic magnetic field $A_0$ and a deterministic electric potential $V_0$. We assume that $A_0:\overline{\Pi } \rightarrow \mathbb {R}^{n+1}$, $A:\overline{\Pi } \rightarrow \mathbb {R}^{n+1}$ and $V_0:\overline{\Pi }\rightarrow \mathbb {R}$ are $\square$-periodic, the potential A vanishes on the boundary of $\square$, and all these functions are twice continuously differentiable.

Note that in this case, the zero disorder limit of the random operator is the magnetic Schrödinger operator $(\mathrm {i}\nabla + A_0)^2 + V_0$. This is a more general situation than considered in the main body of this paper, where the corresponding limit operator is $- \Delta + V_0$. In [25, Theorem 6.1.(a)], a Wegner estimate is proved, however only in an energy region strictly below the infimum of the spectrum of the unperturbed operator and at small disorder.

In such a situation it is crucial to investigate whether at small disorder there is any spectrum in the region where the Wegner estimate can be proven. Else it would concern the resolvent set and would be a trivial statement. Unfortunately from the discussion in Appendix A it follows that at least in the special case where $A_0 = 0$, we have $\Lambda _1 = 0$ and $\Lambda _2 > 0$. Thus, in this special case, at small disorder there is no spectrum at all below the infimum of the unperturbed operator.

Furthermore, even if the spectrum expanded below the infimum of the deterministic operator, another issue would arise:

More precisely, Theorem 6.1.(a) of [25] states the following: Fix parameters $E_0 < \min \Sigma _0:=\min \sigma ({\mathcal {H}}^0)$ and $\eta \in (0, \eta _{\sup })$, where $\eta _{\sup } = {{\,\mathrm{dist}\,}}(E_0, \Sigma _0)/2$ and ${\mathcal {H}}^0=(i \nabla + A_0)^2 + V_0$ denotes the unperturbed operator. Then there exists $\varepsilon _0 > 0$ (depending on $E_0$ and $\eta$) such that for all disorder strengths²$\varepsilon \leqslant \varepsilon _0$, we have a Wegner estimate in $[E_0 - \eta , E_0 + \eta ]$. Clearly, since the spectrum expands continuously with the disorder strength and since $\eta < {{\,\mathrm{dist}\,}}(E_0, \Sigma _0)/2$, the region of the Wegner estimate $[E_0 - \eta , E_0 + \eta ]$ will contain no spectrum for small $\varepsilon$. Whether there is a parameter $\varepsilon \in (0, \varepsilon _0]$ such that region where the Wegner estimate holds contains any spectrum at all depends on the rate of expansion of the spectrum with respect to $\varepsilon$ and on the interplay of $\eta$, $E_0$, and $\varepsilon$.

In fact, in the proof of [25, Theorem 6.1 (a)] (text between Formulas (6.13) and (6.14)) once $E_0 < \min \Sigma _0$ is chosen, the disorder strength $\varepsilon$ must satisfy

$$\begin{aligned} \varepsilon ^2 \leqslant \varepsilon _0^2 := \frac{1 - 2 \eta / {{\,\mathrm{dist}\,}}(E_0, \Sigma _0)}{2 \Vert R_0(E_0)^{\frac{1}{2}} {\mathcal {L}}_2 R_0(E_0)^{\frac{1}{2}} \Vert } = \frac{ \eta _{\sup } - \eta }{ 2 \eta _{\sup } \Vert R_0(E_0)^{\frac{1}{2}} {\mathcal {L}}_2 R_0(E_0)^{\frac{1}{2}} \Vert } \end{aligned}$$

(B.2)

where $R_0(E_0) = ({\mathcal {H}}^0 - E_0)^{-1}$. If we choose $\eta$ close to $\eta _{\sup }$, we have $\varepsilon _0 \sim 0$, such that $\sigma ({\mathcal {H}}^\varepsilon (\omega ))$ does not intersect $[E_0 - \eta , E_0 + \eta ]$, cf. Fig. 1

Fig. 1

The maximal possible expansion of the spectrum as a function of $\eta$ and the area where the Wegner estimate holds

This shows that it is more natural to choose the interval $[E_0 - \eta , E_0 + \eta ]$ depending on the disorder $\varepsilon$, in particular $E_0=E_0(\varepsilon )$ and $\eta = \eta (\varepsilon )$. To make sure that this interval intersects $\Sigma _\varepsilon$ we need to have $\min \Sigma <E_0(\varepsilon )+\eta (\varepsilon )$. Possibly adding a constant to ${\mathcal {H}}^0$, we can assume w.l.o.g $\min \Sigma _0=0$. Since $\eta (\varepsilon ) < \eta _{\sup }= |E_0(\varepsilon )|/2$ we have $E_0(\varepsilon )+\eta (\varepsilon ) \leqslant E_0(\varepsilon )/2$.

Let us first consider the case that $\min \Sigma _\varepsilon$ decreases linearly for small $\varepsilon >0$, i.e. there is an $c_{\ell }>0$ such that $\min \Sigma _\varepsilon \leqslant -c_{\ell } \varepsilon$ (in analogy to Case (I) in the main body of the paper). In order to ensure $E_0(\varepsilon )+\eta (\varepsilon ) \in \Sigma _\varepsilon$, we choose $E_0(\varepsilon ) := -c_{\ell }\varepsilon$ and $\eta (\varepsilon ) = \frac{c_{\ell }}{4}\varepsilon$. Then

$$\begin{aligned} \eta _{\sup } = {{\,\mathrm{dist}\,}}(E_0(\varepsilon ), 0)/2=\frac{c_{\ell }}{2}\varepsilon > \eta (\varepsilon ) \end{aligned}$$

and, indeed, $E_0(\varepsilon )+\eta (\varepsilon ) = -\frac{3c_{\ell }}{4}\varepsilon > -c_{\ell }\varepsilon \geqslant \min \Sigma _\varepsilon$. Hence for this choice of $\eta (\varepsilon )$, in the light of (B.2), [25, Theorem 6.1 (a)] allows disorder strengths

$$\begin{aligned} \varepsilon ^2 \leqslant \frac{1}{4 \Vert R_0(E_0(\varepsilon ))^{\frac{1}{2}} {\mathcal {L}}_2 R_0(E_0(\varepsilon ))^{\frac{1}{2}} \Vert }. \end{aligned}$$

Let us bound the denominator, using representation (A.1)

$$\begin{aligned} \Vert R_0(E_0(\varepsilon ))^{\frac{1}{2}} {\mathcal {L}}_2 R_0(E_0(\varepsilon ))^{\frac{1}{2}} \Vert \leqslant \Vert R_0(E_0(\varepsilon ))^{\frac{1}{2}}\Vert ^2 \Vert {\mathcal {L}}_2 \Vert =|E_0(\varepsilon )|^{-1} \Vert |A|^2 \Vert _\infty =\frac{\sup |A|^2 }{c_{\ell }\varepsilon } \end{aligned}$$

Thus Ineq. (B.2) is satisfied if $\varepsilon ^2 \leqslant c_{\ell } \varepsilon /( 4\sup |A|^2)$. This is true for sufficiently small $\varepsilon$.

Let us turn to the case that $\min \Sigma _\varepsilon$ decreases quadratically for small $\varepsilon >0$, i.e. there is an $c_{q}>0$ such that $\min \Sigma _\varepsilon \leqslant -c_{q} \varepsilon ^2$ (in analogy to Case (II) in the main body of the paper).

In order to ensure $E_0(\varepsilon )+\eta (\varepsilon ) \in \Sigma _\varepsilon$, we choose $E_0(\varepsilon ) := -c_{q}\varepsilon ^2$ and $\eta (\varepsilon ) = \frac{c_{q}}{4}\varepsilon ^2$. Then

$$\begin{aligned} \eta _{\sup } = {{\,\mathrm{dist}\,}}(E_0(\varepsilon ), 0)/2=\frac{c_{q}}{2}\varepsilon ^2> \eta (\varepsilon ) \end{aligned}$$

and, indeed, $E_0(\varepsilon )+\eta (\varepsilon ) = -\frac{3c_{q}}{4}\varepsilon ^2 > -c_{q}\varepsilon ^2\geqslant \min \Sigma _\varepsilon$. Hence for this choice of $\eta (\varepsilon )$, in the light of (B.2), [25, Theorem 6.1 (a)] allows disorder strengths

$$\begin{aligned} \varepsilon ^2 \leqslant \frac{1}{4 \Vert R_0(E_0(\varepsilon ))^{\frac{1}{2}} {\mathcal {L}}_2 R_0(E_0(\varepsilon ))^{\frac{1}{2}} \Vert }. \end{aligned}$$

Let us bound the denominator, using representation (A.1)

$$\begin{aligned} \Vert R_0(E_0(\varepsilon ))^{\frac{1}{2}} {\mathcal {L}}_2 R_0(E_0(\varepsilon ))^{\frac{1}{2}} \Vert \leqslant \Vert R_0(E_0(\varepsilon ))^{\frac{1}{2}}\Vert ^2 \Vert {\mathcal {L}}_2 \Vert =|E_0(\varepsilon )|^{-1} \Vert |A|^2 \Vert _\infty =\frac{\sup |A|^2 }{c_{q}\varepsilon ^2} \end{aligned}$$

Thus Ineq. (B.2) is satisfied if $\varepsilon ^2 \leqslant c_{q}\varepsilon ^2 /( 4\sup |A|^2)$, i. e. $4\sup |A|^2 \leqslant c_{q}$ To decide whether this condition holds, one needs to provide a lower bound on the expansion coefficient $c_{q}$.

In any case, merely assuming small disorder is not sufficient to ensure that [25, Theorem 6.1 (a)] is a non-trivial statement. Additional arguments or assumptions are required. To elucidate this further we want to exhibit situations where indeed we have linear expansion, i.e. $\min \Sigma _\varepsilon \leqslant -c_{\ell } \varepsilon$ with $c_{\ell } >0$.

Remark B.2

The operator (B.1) does not fit the assumptions of the present paper. Indeed, for $\varepsilon =0$ the unperturbed operator becomes ${\mathcal {H}}^0=(\mathrm {i}\nabla +A_0)^2$, possibly with $A_0\ne 0$. Nevertheless, the issue on how the spectrum expands can be analyzed in this situation, as well, by applying the general results of [4]. In this context Theorem 2.3 is replaced by Theorem 2.1 in [4] still ensuring that there exist an almost sure spectral set $\Sigma _\varepsilon$. To fit the setting of [4] we introduce some notation and assumptions. We begin with the Floquet-Bloch expansion for the unperturbed operator, namely, we consider the operator ${\mathcal {H}}^0_{per}:=(\mathrm {i}\nabla +A_0)^2$ on the periodicity cell $\square$ subject to $\theta$-quasiperiodic boundary conditions on the (lateral) boundaries $\gamma$ and denote its lowest eigenvalue by $E_0(\theta )$. We assume further that there exists a unique quasimomentum $\theta _0$ such that $E_0(\theta _0)=\min _\theta E_0(\theta )=\Lambda _0 :=\min \Sigma _\varepsilon$, that $\Lambda _0$ is a simple eigenvalue of ${\mathcal {H}}^0$ on $\square$ with $\theta _0$-periodic boundary conditions on $\gamma$ and $\Psi _0$ is the associated eigenfunction normalized in $L^2(\square )$. Continuous dependence of the Floquet eigenvalues on the quasimomentum implies that there is closed ball U around $\theta _0$ such that $\min _{\theta \in U}{{\,\mathrm{dist}\,}}(E_0(\theta ), \sigma ({\mathcal {H}}^0)\setminus \{E_0(\theta )\})>0$. Such a scenario indeed occurs for electromagnetic Schrödinger operators, see e.g. [20, 41]. Since here we focus on the effects of magnetic vector potentials we will assume in the following $V_0\equiv 0$. Assume also that the function $\varpi :=\frac{\Psi _0}{|\Psi _0|}$ belongs to $C^2(\overline{\square })$.

Then one can prove along the lines of [4] that the bottom of the spectrum of the operator ${\mathcal {H}}^\varepsilon (\omega )$ satisfies the asymptotic formula

$$\begin{aligned} \inf \sigma \big ({\mathcal {H}}^\varepsilon (\omega )\big )\leqslant \Lambda _0 + \varepsilon \min \{b \Lambda _1, \Lambda _1\} +O(\varepsilon ^2), \end{aligned}$$

(B.3)

where the constant $\Lambda _1$ is determined by a formula similar to (A.2):

$$\begin{aligned} \Lambda _1 = \int \limits _{\square } \overline{\Psi _0} \left[ (\mathrm {i}\nabla +A_0) \cdot A + A \cdot (\mathrm {i}\nabla +A_0) \right] \Psi _0 \mathrm {d}x. \end{aligned}$$

(B.4)

We also note that if the function $\frac{1}{\Psi _0}\frac{\partial \Psi _0}{\partial \nu }$ belongs to $C^1(\partial \square )$ then according Theorem 2.3 in [4], the inequality in (B.3) can be replaced by the identity.

Integrating by parts in (B.4) and taking into consideration that A vanishes on the boundary of $\square$, we obtain:

$$\begin{aligned}&\Lambda _1 = \int \limits _{\square } \Big (\overline{\Psi _0 A} \cdot (\mathrm {i}\nabla +A_0) + \Psi _0 A \cdot \overline{(\mathrm {i}\nabla +A_0)\Psi _0}\Big ) \mathrm {d}x\nonumber \\&=2{{\,\mathrm{Re}\,}}\big (\Psi _0 A,(\mathrm {i}\nabla +A_0)\Psi _0 \big )_{L^2(\square )}. \end{aligned}$$

(B.5)

Thanks to the assumed smoothness of $A_0$, A and $V_0$ and by the Schauder estimates, we have $\Psi _0\in C^2(\overline{\square })$. Then we can rewrite formula (B.5) as

$$\begin{aligned} \Lambda _1=2{{\,\mathrm{Re}\,}}\big (A,\overline{\Psi _0}(\mathrm {i}\nabla +A_0)\Psi _0 \big )_{L^2(\square )}. \end{aligned}$$

(B.6)

This representation for $\Lambda _1$ shows that once $A_0$ is fixed, the vector function $\overline{\Psi _0}(\mathrm {i}\nabla +A_0)\Psi _0$ is fixed as well, and we can vary the potential A to achieve $\Lambda _1\ne 0$. Let us prove this fact rigorously.

We are going to show that there exists at least one potential A such the constant $\Lambda _1$ defined by (B.6) is non-zero. We argue by contradiction assuming that $\Lambda _1$ vanishes for all A. Then formula (B.6) implies that

$$\begin{aligned} 0\equiv {{\,\mathrm{Re}\,}}\overline{\Psi _0}(\mathrm {i}\nabla +A_0)\Psi _0=-{{\,\mathrm{Im}\,}}\overline{\Psi _0}\nabla \Psi _0+ A_0|\Psi _0|^2=0\quad \text {in}\quad \square . \end{aligned}$$

Employing the definition of the function $\varpi$, it is straightforward to check that the above identity is rewritten equivalently as

$$\begin{aligned} A_0=-\mathrm {i}\overline{\varpi }\nabla \varpi . \end{aligned}$$

The right hand side in the above formula is a real-valued vector function. Indeed, since $\varpi \overline{\varpi }\equiv 1$, we easily confirm that

$$\begin{aligned} \overline{\varpi }\nabla \varpi + \overline{\overline{\varpi }\nabla \varpi }=\overline{\varpi }\nabla \varpi + \varpi \nabla \overline{\varpi }=\nabla \varpi \overline{\varpi }\equiv 0 \end{aligned}$$

and hence, $\overline{\varpi }\nabla \varpi$ is a pure imaginary vector function.

On $L^2(\square )$, we introduce a unitary transformation by the formula ${\mathcal {U}}u:=\varpi u$, and ${\mathcal {U}}^{-1}u=\overline{\varpi } u$. It is straightforward to check that the operator ${\mathcal {H}}^0_{per}$ is unitary equivalent to

$$\begin{aligned} {\mathcal {U}}^{-1}{\mathcal {H}}^0_{per}{\mathcal {U}}=\big (\mathrm {i}\nabla + \mathrm {i}\overline{\varpi }\nabla \varpi +A_0\big )^2=-\Delta , \end{aligned}$$

where the differential expressions in the right hand side are treated as operators in $L^2(\square )$ subject to the same boundary conditions as ${\mathcal {H}}^0_{per}$. This means that $\Lambda _1$ can vanish simultaneously for all A only in the above discussed case $A_0=0$. Once $A_0$ describes a non-trivial magnetic field, there exists at least one potential A, for which $\Lambda _1$ is non-zero. Once such potential A is found and fixed, we can consider small variation of A of the form $A+\delta A$ and in view of the continuity of $\Lambda _1$ in A, we conclude immediately that $\Lambda _1$ is also non-zero for $A+\delta A$ provide $\delta A$ is small enough. Hence, there exist infinitely many examples of A, for which $\Lambda _1$ is non-zero.

Moreover, given some $A_0$, A, for which $\Lambda _1$ is non-zero, we easily see that replacing A by $-A$, we change the sign of $\Lambda _1$. This means that in general, $\Lambda _1$ is non-zero and can be both negative and positive. Hence, returning back to formula (B.3), we conclude that in general, the term of order $O(\varepsilon )$ in this formula is non-zero and can be both positive and negative.

Let us discuss the assumption $\frac{\Psi _0}{|\Psi _0|}=\varpi \in C^2(\overline{\square })$. This is surely true once we know that $\Psi _0$ does not vanish in $\square$ and behaves “well enough” at the boundaries of $\square$. Apriori, we failed trying to prove this fact for general $A_0$, but the set of potentials $A_0$ obeying this assumption is non-empty. Indeed, we can start from $A_0\equiv 0$, then, as it was discussed in Appendix A, the ground state $\Psi _0$ is real and sign, hence in this case the function $\varpi$ is identically constant: $\varpi \equiv 1$. Now, consider a (sufficiently) small non-trivial magnetic potential $A_0$. Its ground state will be a small perturbation of the one for $A_0\equiv 0$, i. e. a (sufficiently) small variation of the function $\varpi \equiv 1$ The perturbed ground state will obviously still belong to $C^2(\overline{\square })$.