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On the Asymptotic Dynamics of 2-D Magnetic Quantum Systems

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Abstract

In this work, we provide results on the long-time localization in space (dynamical localization) of certain two-dimensional magnetic quantum systems. The underlying Hamiltonian may have the form \(H=H_0+W\), where \(H_0\) is rotationally symmetric and has dense point spectrum and W is a perturbation that breaks the rotational symmetry. In the latter case, we also give estimates for the growth of the angular momentum operator in time.

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Acknowledgements

D.H. and S.V. gratefully acknowledge the funding by Deutsche Forschungsgemeinschft (DFG) through project ID 258734477–SFB 1173. E.S. has been partially funded by Fondecyt (Chile) project # 118–0355.

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Correspondence to Edgardo Stockmeyer.

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Communicated by Alain Joye.

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Appendices

Appendix A: The Magnetic Schrödinger Operator

Here, we want to review the form definition of the magnetic Schrödinger operator \(H_0\). We have to be a little bit careful, since we want to be able to handle rotationally symmetric, but possibly singular, magnetic fields B. Recall that we choose the vector potential A in the Poincaré gauge given by (4).

Lemma A.1

If the magnetic field \(B:\mathbb {R}^2 \rightarrow \mathbb {R}\) is rotationally symmetric and locally square integrable, then the function

$$\begin{aligned} \mathbb {R}^2\ni x\mapsto \Phi (|x|)/|x|= \frac{1}{|x|}\int _0^{|x|} B(s) s \mathrm{d}s \end{aligned}$$

is locally square integrable. In particular, the magnetic vector potential A given by (4) is in \( L^2_\mathrm{loc} (\mathbb {R}^2) \).

Proof

This is a simple consequence of Jensen’s inequality. Since \(\frac{2}{r^2}\int _0^r s \mathrm{d}s =1\), Jensen’s inequality shows

$$\begin{aligned} \left( \frac{2\Phi (r)}{r^2}\right) ^2=\left( \frac{2}{r^2} \int _0^r B(s) s\mathrm{d}s\right) ^2\leqslant \frac{2}{r^2} \int _0^r B(s)^2 s\mathrm{d}s \end{aligned}$$

for all \(r>0\). Thus

(76)
(77)

for all \(R>0\). By the definition (4) we have \(|A(x)| =\frac{\Phi (|x|)}{|x|}\), so also the magnetic vector potential in the Poincaré gauge A is locally square integrable. \(\square \)

Denote by \(p = - i \nabla \) the usual momentum operator. We will need a representation of the magnetic Schrödinger operator \((p-A)^2\), when A is in the Poincaré gauge and the magnetic field is rotationally symmetric. This is well known, but we want to include singular magnetic fields, so we have to be a bit careful.

Lemma A.2

The quadratic form \(q_0\) of the free magnetic Schrödinger operator \((p-A)^2\) is given by

$$\begin{aligned} q_0(\varphi ,\varphi )&= \left\langle (p-A)\varphi ,(p-A)\varphi \right\rangle = \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle + \left\langle {\frac{1}{r}(\Phi -L)\varphi }{\frac{1}{r}(\Phi -L)\varphi }\right\rangle \end{aligned}$$
(78)

for all \(\varphi \in \mathcal {D}(q_0)\). Here, \(L= x_1 p_2 -x_2 p_1 \) is the generator of rotations, i.e. the angular momentum operator, in \(L^2(\mathbb {R}^2)\), \(r=|x|\), and the radial derivative is given by \(\partial _r= \frac{x}{|x|}\cdot \nabla \). In particular, \(\mathcal {D}(q_0)= \mathcal {D}(\partial _r)\cap \mathcal {D}(\frac{1}{r}(\Phi -L))\).

Before we show this, we collect one more result, which is needed

Lemma A.3

The quadratic form corresponding to the kinetic energy \(p^2\) in dimension \(d\geqslant 2\) is given by

$$\begin{aligned} \left\langle p\varphi ,p\varphi \right\rangle = \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle + \sum _{1\leqslant j<k\leqslant d} \left\langle {\frac{1}{r}L_{j,k}\varphi }{\frac{1}{r}L_{j,k}\varphi }\right\rangle , \end{aligned}$$
(79)

for all \(\varphi \in H^1(\mathbb {R}^d)\), the form domain of \(p^2\), where \(r=|x|\), \(\partial _r=\frac{x}{|x|}\cdot \nabla \) is the radial derivative on \(\mathbb {R}^d\) and \(L_{j,k}= x_jp_k-x_kp_j\), \(1\leqslant j<k\leqslant d\) are the angular momentum generators.

In particular, \(H^1(\mathbb {R}^d)= \mathcal {D}(\partial _r)\cap \cap _{1\leqslant j<k\leqslant d}\mathcal {D}(\frac{1}{r}L_{j,k})\).

Proof of Lemma A.2

We assume that \(\varphi \in \mathcal {C}_0^\infty (\mathbb {R}^2)\), by density. Then, since A is locally square integrable,

$$\begin{aligned} \left\langle \varphi ,H_0\varphi \right\rangle&= \left\langle (p-A)\varphi ,(p-A)\varphi \right\rangle , \\&= \left\langle p\varphi ,p\varphi \right\rangle -2 \mathrm {Re}\left\langle A_1\varphi ,p_1\varphi \right\rangle -2 \mathrm {Re}\left\langle A_2\varphi ,p_2\varphi \right\rangle +\left\langle A\varphi ,A\varphi \right\rangle . \end{aligned}$$

Using the explicit form of vector potential in the gauge (4), one also sees that

$$\begin{aligned} \left\langle A_1\varphi ,p_1\varphi \right\rangle = - \left\langle {x_2\frac{\Phi }{r^2}\varphi }{p_1\varphi }\right\rangle = - \left\langle {\frac{\Phi }{r}\varphi }{\frac{1}{r}x_2p_1\varphi }\right\rangle \end{aligned}$$

and similarly for

$$\begin{aligned} \left\langle A_2\varphi ,p_2\varphi \right\rangle = \left\langle {\frac{\Phi }{r}\varphi }{\frac{1}{r}x_1p_2\varphi }\right\rangle , \end{aligned}$$

where all terms are well defined, since Lemma A.1 shows that \(\frac{\phi (|x|)}{|x|}\) is locally square integrable over \(\mathbb {R}^2\). Thus,

$$\begin{aligned} \left\langle A_1\varphi ,p_1\varphi \right\rangle + \left\langle A_2\varphi ,p_2\varphi \right\rangle = \left\langle {\frac{\Phi }{r}\varphi }{\frac{1}{r}(x_1p_2 -x_2p_1)\varphi }\right\rangle = \left\langle {\frac{\Phi }{r}\varphi }{\frac{1}{r}L\varphi }\right\rangle \end{aligned}$$

with \(L= x_1p_2 -x_2p_1\). Since also

$$\begin{aligned} \left\langle A\varphi ,A\varphi \right\rangle = \left\langle {\frac{\Phi }{r}\varphi }{\frac{\Phi }{r}\varphi }\right\rangle , \end{aligned}$$

this yields

$$\begin{aligned} \left\langle (p-A)\varphi ,(p-A)\varphi \right\rangle&= \left\langle p\varphi ,p\varphi \right\rangle + 2\mathrm {Re}\left\langle {\frac{\Phi }{r}\varphi }{\frac{1}{r}L\varphi }\right\rangle + \left\langle A\varphi ,A\varphi \right\rangle , \\&= \left\langle p\varphi ,p\varphi \right\rangle + 2\left\langle {\frac{\Phi }{r}\varphi }{\frac{1}{r}L\varphi }\right\rangle + \left\langle {\frac{\Phi }{r}\varphi }{\frac{\Phi }{r}\varphi }\right\rangle , \end{aligned}$$

since the angular momentum commutes with rotationally symmetric functions, so \(\left\langle \frac{\Phi }{r}\varphi ,\frac{1}{r}L\varphi \right\rangle \) is real. Moreover, by Lemma A.3, and again using that L commutes with multiplication by rotationally symmetric functions, this gives

$$\begin{aligned} \left\langle (p-A)\varphi ,(p-A)\varphi \right\rangle&= \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle + \left\langle {\frac{1}{r}L\varphi }{\frac{1}{r}L\varphi }\right\rangle + 2 \left\langle {\frac{\Phi }{r}\varphi }{\frac{1}{r}L\varphi }\right\rangle + \left\langle {\frac{\Phi }{r}\varphi }{\frac{\Phi }{r}\varphi }\right\rangle , \\&= \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle + \left\langle {\frac{1}{r}(\Phi -L)\varphi }{\frac{1}{r}(\Phi -L)\varphi }\right\rangle , \end{aligned}$$

which proves (78) when \(\varphi \in \mathcal {C}^\infty _0(\mathbb {R}^2)\). Since \(\mathcal {C}^\infty _0(\mathbb {R}^2)\) is dense in the domain of \(q_0\) [18] and all terms on the right-hand side of (78) are non-negative, a standard density argument shows that the domain of \(\mathcal {D}(q_0)\) is equal to the intersection of \(\mathcal {D}(\partial _r)\) and \(\mathcal {D}(\tfrac{1}{r}(\Phi -L))\). This proves Lemma A.2. \(\square \)

Proof of Lemma A.3

We can use the same density argument as above to see that it is enough to assume that \(\varphi \in \mathcal {C}_0^\infty (\mathbb {R}^d)\). Then,

$$\begin{aligned} \sum _{1\leqslant j<k\leqslant d} \left\langle L_{j,k}\varphi ,L_{j,k}\varphi \right\rangle = - \sum _{1\leqslant j<k\leqslant d} \left\langle \varphi ,(x_j\partial _k-x_k\partial _j)^2\varphi \right\rangle . \end{aligned}$$

Moreover,

$$\begin{aligned}&\sum _{1\leqslant j<k\leqslant d} (x_j\partial _k-x_k\partial _j)^2 = \frac{1}{2}\sum _{ j\ne k } (x_j\partial _k-x_k\partial _j)^2, \\&\quad = \frac{1}{2}\sum _{ j\ne k } (x_j\partial _k x_j\partial _k - x_j\partial _k x_k\partial _j- x_k\partial _j x_j\partial _k + x_k\partial _j x_k\partial _j)\, , \\&\quad = \frac{1}{2}\sum _{ 1\leqslant j, k \leqslant d} (x_j^2\partial _k^2 + x_k^2\partial _j^2 - \partial _k x_kx_j\partial _j -x_j\partial _j - \partial _j x_j x_k\partial _k -x_k\partial _k) \\&\qquad - \sum _{ j } (x_j^2\partial _j^2 - \partial _j x_j^2\partial _j - x_j\partial _j),\\&\quad = |x|^2 \Delta -(\nabla \cdot x)(x\cdot \nabla ) + 2x\cdot \nabla = |x|^2 \Delta -(x\cdot \nabla )^2 -(d- 2)x\cdot \nabla . \end{aligned}$$

Thus, one obtains that

$$\begin{aligned} p^2= -\Delta = -\frac{1}{|x|^2}(x\cdot \nabla )^2 -\frac{(d- 2)}{|x|^2}x\cdot \nabla + \frac{1}{|x|^2}\sum _{1\leqslant j<k\leqslant d} L_{j,k}^2\, , \end{aligned}$$

that is, as quadratic forms

$$\begin{aligned} \left\langle p\varphi ,p\varphi \right\rangle = -\left\langle {\varphi }{\frac{1}{|x|^2}(x\cdot \nabla )^2\varphi }\right\rangle -\left\langle {\varphi }{\frac{d-2}{|x|^2}(x\cdot \nabla )\varphi }\right\rangle + \sum _{1\leqslant j<k\leqslant d}\left\langle {\varphi }{\frac{1}{|x|^2}L_{j,k}^2\varphi }\right\rangle \end{aligned}$$
(80)

at least when \(\varphi \in \mathcal {C}_0^\infty (\mathbb {R}^d)\). For such \(\varphi \) we set

$$\begin{aligned} \psi (r,\omega ) := \varphi (r\omega ), \end{aligned}$$

when \(r\geqslant 0\) and \(|\omega |=1\). That is, \(\varphi (x)= \psi (|x|, x/|x|)\). Then, clearly, \(\partial _r\psi (r,\omega )= \omega \cdot \nabla \varphi (r\omega )\), so

$$\begin{aligned} x\cdot \varphi (x) = r\partial _r\psi (r, \omega ) \end{aligned}$$
(81)

with \(r=|x|\) and \(\omega =x/|x|\in S^{d-1}\). Then,

$$\begin{aligned} -\left\langle {\varphi }{\frac{1}{|x|^2}(x\cdot \nabla )^2\varphi }\right\rangle = - \int _{S^{d-1}} {\mathrm{d}}\omega \int _0^\infty {\mathrm{d}}r r^{d-1} \, \overline{\psi (r,\omega )} r^{-2}(r\partial _r)^2 \psi (r,\omega ). \end{aligned}$$
(82)

Now for fixed \(\omega \), we have

$$\begin{aligned}&-\int _0^\infty \mathrm{d}r r^{d-1} \overline{\psi (r,\omega )} r^{-2}(r\partial _r)^2 \psi (r,\omega ) \\&\quad = - \left[ r^{d-1} \, \overline{\psi (r,\omega )} \partial _r\psi (r,\omega ) \right] _{r=0}^{r=\infty } + \int _0^\infty \mathrm{d}r \overline{(\partial _r(r^{d-2} \, \psi (r,\omega )))} (r\partial _r) \psi (r,\omega ),\\&\quad = (d-2)\int _0^\infty \mathrm{d}r r^{d-3}\overline{ \psi (r,\omega ))} \, r\partial _r \psi (r,\omega ) + \int _0^\infty \mathrm{d}r r^{d-1}|\partial _r\psi (r,\omega ))|^2. \end{aligned}$$

The first term in the second line above vanishes, since \(\varphi \) has compact support and \(d\geqslant 2\). Thus,

$$\begin{aligned} -\left\langle {\varphi }{\frac{1}{|x|^2}(x\cdot \nabla )^2\varphi }\right\rangle = (d-2)\left\langle {\varphi }{\frac{1}{|x|^2}(x\cdot \nabla )\varphi }\right\rangle + \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle \end{aligned}$$

with \(\partial _r= \frac{x}{|x|}\cdot \nabla \). Using this in (80) shows

$$\begin{aligned} \left\langle p\varphi ,p\varphi \right\rangle&= \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle + \sum _{1\leqslant j<k\leqslant d}\left\langle {\varphi }{\frac{1}{|x|^2}L_{j,k}^2\varphi }\right\rangle ,\\&= \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle + \sum _{1\leqslant j<k\leqslant d}\left\langle {\frac{1}{|x|}L_{j,k}\varphi }{\frac{1}{|x|}L_{j,k}\varphi }\right\rangle , \end{aligned}$$

since \(L_{j,k}\) commutes with multiplication with radial functions. This proves (79), at least when \(\varphi \in \mathcal {C}_0^\infty (\mathbb {R}^d)\). On the other hand, since all the terms on the right-hand side of (79) are positive and \(\mathcal {C}_0^\infty (\mathbb {R}^d)\) is dense in the Sobolev space \(H^1(\mathbb {R}^d)\), the form domain of \(p^2\), it is an easy exercise to show that then \(\partial _r\varphi , \frac{1}{|x|}L_{j,k}\varphi \in L^2(\mathbb {R}^d)\) for all \(\varphi \in H^1(\mathbb {R}^d)\) and (79) holds for all \(\varphi \in H^1(\mathbb {R}^d)\)\(\square \)

To work in polar coordinates, we identify \(L^2(\mathbb {R}^2)\) with the Hilbert space \(\mathcal {H}=L^2(\mathbb {R}^+ \times \mathbb {S}^1, r \text {d}r\, \text {d}\theta )\) with the scalar product

$$\begin{aligned} \left\langle f,g\right\rangle _{\mathcal {H}}\equiv \left\langle f,g\right\rangle =\int _{\mathbb {R}^+\times [0,2\pi )} \overline{f(r,\theta )} g(r,\theta )\, r\text {d}r \text {d}\theta . \end{aligned}$$

By means of the unitary map \(\mathcal {U}:L^2(\mathbb {R}^2)\rightarrow \mathcal {H}\) introduced in (6), the angular momentum component \(L_{1,2}\) takes the form \(J:= -i\frac{\partial }{\partial \theta } = \mathcal {U}L_{1,2}\mathcal {U}^* \). In these coordinates we have

Proposition A.4

Let \(\varphi \) be in the domain of the quadratic form \(q_0\) corresponding to \((P-A)^2\), and expand \(\widetilde{\varphi }=\mathcal {U}\varphi \) as in (99). Then,

$$\begin{aligned} q_0(\varphi ,\varphi ) = \sum _{j\in \mathbb {Z}} \left( \left\langle \partial _r\varphi _j,\partial _r\varphi _j \right\rangle _{L^2(\mathbb {R}_+,rdr)} + \left\langle {\varphi _j}{\frac{1}{r^2}(\Phi (r)-j)^2\varphi _j}\right\rangle _{L^2(\mathbb {R}_+,rdr)} \right) . \end{aligned}$$
(83)

So, the eigenspaces corresponding to \(P_j\), introduced in (9), are invariant subspaces for the unperturbed magnetic Schrödinger operator with a rotationally symmetric magnetic field, when the magnetic vector potential is in the Poincaré gauge (4).

Because of the above identity, it is convenient to recall the definition of the effective potential, namely,

$$\begin{aligned} V_j(r):= \frac{1}{r^2} (\Phi (r)-j)^2. \end{aligned}$$
(84)

By polarization, Proposition A.4 shows that when \(\varphi ,\psi \) are in the domain of the form \(q_0\) corresponding to \((p-A)^2\) and \(\widetilde{\varphi }=\mathcal {U}\varphi , \widetilde{\psi }=\mathcal {U}\psi \) are expanded as in (99) then

$$\begin{aligned} q_0(\varphi ,\psi ) = \sum _{j\in \mathbb {Z}} \Big ( \left\langle \partial _r\varphi _j,\partial _r\psi _j \right\rangle _{L^2(\mathbb {R}_+,rdr)} + \left\langle \varphi _j,V_j\psi _j \right\rangle _{L^2(\mathbb {R}_+,rdr)} \Big ). \end{aligned}$$
(85)

We need one more result, concerning the form boundedness of potentials W satisfying Condition 1 with respect to the radial kinetic energy.

Lemma A.5

Assume that v is a rotationally symmetric potential which is form bounded with respect to \(p^2\), that is, for any \(0<\varepsilon \) there exists \(C(\varepsilon )<\infty \) with

$$\begin{aligned} |\left\langle \varphi ,v\varphi \right\rangle |\leqslant \varepsilon \left\| \nabla \varphi \right\| ^2 +C(\varepsilon )\left\| \varphi \right\| ^2 \end{aligned}$$
(86)

for all \(\varphi \in \mathcal {D}(p)\). Then, also

$$\begin{aligned} |\left\langle \varphi ,v\varphi \right\rangle |\leqslant \varepsilon \left\| \partial _r\varphi \right\| ^2 +C(\varepsilon )\left\| \varphi \right\| ^2 \end{aligned}$$
(87)

for all \(\varphi \in \mathcal {D}(\partial _r)\), where \(\partial _r=\tfrac{x}{|x|}\cdot \nabla \) is the radial derivative.

Proof

We expand \(\widetilde{\varphi }= \mathcal {U}\varphi = \sum _{j\in \mathbb {Z}}\varphi _j e_j\), where \(\varphi _j\) are purely radial functions and \(e_j\) are the basis of complex exponentials. Then, for a radial potential v we have

$$\begin{aligned} \left\langle \varphi ,v\varphi \right\rangle&= \sum _{j\in \mathbb {Z}} \left\langle P_j\varphi ,v\varphi \right\rangle = \sum _{j\in \mathbb {Z}} \left\langle P_j^2\varphi ,v\varphi \right\rangle = \sum _{j\in \mathbb {Z}} \left\langle P_j\varphi ,vP_j\varphi \right\rangle \\&= \sum _{j\in \mathbb {Z}} \left\langle \varphi _j,v\varphi _j \right\rangle _{L^2(\mathbb {R}_+,rdr )}, \end{aligned}$$

with the angular momentum projections \(P_j\). Lifting each \(\varphi _j\) back to \(L^2(\mathbb {R}^2)\), by considering it to be constant in the angular coordinate, i.e. identifying it as the function \(\mathbb {R}^2\ni x\mapsto \varphi _j(|x|)\), we see have by assumption (86)

$$\begin{aligned} \left| \left\langle \varphi ,v\varphi \right\rangle \right|&\leqslant \varepsilon \sum _{j\in \mathbb {Z}} \left\langle \nabla \varphi _j,\nabla \varphi _j \right\rangle + C(\varepsilon )\sum _{j\in \mathbb {Z}} \left\langle \varphi _j,\varphi _j \right\rangle . \end{aligned}$$

Now, since \(\varphi _j\) lifted back to \(\mathbb {R}^2\) is radial, we have

$$\begin{aligned} \left\langle \nabla \varphi _j,\nabla \varphi _j \right\rangle = \left\langle \partial _r\varphi _j,\partial _r\varphi _j \right\rangle = \left\langle \partial _rP_j\varphi ,\partial _rP_j\varphi \right\rangle = \left\langle P_j\partial _r\varphi ,\partial _r\varphi \right\rangle \end{aligned}$$

since each angular momentum projection \(P_j\) commutes with the radial part of the kinetic energy. We also have

$$\begin{aligned} \left\langle \varphi _j,\varphi _j \right\rangle = \left\langle P_j\varphi ,P_j\varphi \right\rangle = \left\langle P_j\varphi ,\varphi \right\rangle \end{aligned}$$

so combining the above yields

$$\begin{aligned} \left| \left\langle \varphi ,v\varphi \right\rangle \right|&\leqslant \varepsilon \sum _{j\in \mathbb {Z}} \left\langle P_j\partial _r\varphi ,\partial _r\varphi \right\rangle + C(\varepsilon ) \sum _{j\in \mathbb {Z}} \left\langle P_j\varphi ,\varphi \right\rangle = \varepsilon \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle + C(\varepsilon ) \left\langle \varphi ,\varphi \right\rangle \end{aligned}$$

which proves the claim. \(\square \)

Remark A.6

The above result also shows that any radial potential v which is form bounded with respect to the nonmagnetic kinetic energy is also form bounded with respect to the magnetic kinetic energy with a rotationally symmetric magnetic field, with the same constants, since by Lemma A.5 we have

(88)
(89)

since the effective potential \(V=(V_j)_{j\in \mathbb {Z}}\geqslant 0\).

Appendix B: The Exponentially Twisted Magnetic Quadratic Forms

The sequences of weights \(F = (F_j)_{j\in \mathbb {Z}}\) we have interest in, are bounded and have derivatives that can be dominated by the magnetic quadratic form \(q_0\). This turns to be sufficient to prove invariance of the form domain.

In what follows, we denote by \(\Vert \cdot \Vert _{q_0}^2 = q_0[\cdot ] + \Vert \cdot \Vert ^2\) the norm induced by \(q_0\) on \(\mathcal {D}(q_0)\).

Lemma B.1

Let \(G = (G_j)_{j\in \mathbb {Z}} \subset PC_{bd}^1 (\mathbb {R}^+, \mathbb {R})\) be a sequence of functions for which there exists \(c>0\) such that

$$\begin{aligned} \sum _{j\in \mathbb {Z}} \left\langle \varphi _j , |G_j ' |^2 \varphi _j \right\rangle \leqslant c\, \Vert \varphi \Vert _{q_0}^2 \, \quad \varphi \in \mathcal {D}(q_0). \end{aligned}$$
(90)

Then, \(e^{G} : \mathcal {D}(q_0) \rightarrow \mathcal {D}(q_0) \) is a continuous bijection with respect to \(\Vert \cdot \Vert _{q_0}\).

Proof

Recall that \(e^G\) is bounded in \(L^2(\mathbb {R}^2)\) and \(\mathcal {D}(q_0 ) = \mathcal {D}(\partial _r) \cap \mathcal {D}(r^{-1}(\Phi - L))\) because of Lemma A.2. Let \(\varphi \in C_0^\infty (\mathbb {R}^2)\) and compute that

$$\begin{aligned} \Vert \partial _r(e^G \varphi ) \Vert ^2&\leqslant 2 \Vert e^G \Vert ^2 ( \Vert G' \varphi \Vert ^2 + \Vert \partial _r \varphi \Vert ^2 ) \leqslant 2 \Vert e^G\Vert ^2 ( c \, q_0 [\varphi ] + c \Vert \varphi \Vert ^2 + \Vert \partial _r \varphi \Vert ^2 ) . \end{aligned}$$
(91)

Since \(C_0^\infty (\mathbb {R}^2) \subset \mathcal {D}(q_0 ) \subset \mathcal {D}(\partial _r )\), the right-hand side above is finite and we get \(\partial _r(e^G \varphi )\in L^2(\mathbb {R}^2)\). By a similar argument, \(r^{-1}(\Phi - L )e^G \varphi \in L^2(\mathbb {R}^2)\) as well. Therefore, we conclude that \(e^G \varphi \in \mathcal {D}(q_0)\). Moreover, we compute for \(\varphi \in C_0^\infty (\mathbb {R}^2)\)

(92)
(93)
(94)

Now, recalling that \(C_0^\infty (\mathbb {R}^2)\) is a form core for \(H_0\), for a given \(\varphi \in \mathcal {D}(q_0)\) we find a sequence \((\varphi _n)_{n\in \mathbb {N}}\subset C_0^\infty (\mathbb {R}^2)\) converging to \(\varphi \) with respect to the norm \(\Vert \cdot \Vert _{q_0} \). In view of the above estimate, we have that \((e^G\varphi _n)_{n\in \mathbb {N}}\) is a Cauchy sequence in \(\Vert \cdot \Vert _{q_0}\), and so a limit \(\psi \in \mathcal {D}(q_0)\) exists. Since \(\varphi _n \rightarrow \varphi \) in \(L^2(\mathbb {R}^2)\), continuity of \(e^G\) then ensures that \(e^G \varphi _n \rightarrow e^G \varphi \) in \(L^2(\mathbb {R}^2)\). Uniqueness of the limits then gives us that \(e^G \varphi = \psi \), and we are able to conclude that \(e^G \mathcal {D}(q_0 )\subset \mathcal {D}(q_0)\). The same analysis holds for \(e^{-G}\) as well; we conclude that \(e^G \mathcal {D}(q_0) = \mathcal {D}(q_0)\). Continuity now follows from (94). \(\square \)

For sequences \(F = (F_j)_{j\in \mathbb {Z}}\in PC_{bd}^1(\mathbb {R}^+,\mathbb {R})\) satisfying the bound (90), we define the exponentially twisted sesquilinear form

$$\begin{aligned} q_F (\psi , \varphi ) := q_0(e^F \psi , e^{-F} \varphi ), \quad \psi , \varphi \in \mathcal {D}(q_0). \end{aligned}$$
(95)

In view of Lemma B.1, \(q_F\) is well defined. Moreover, it is easily verified that

$$\begin{aligned} q_F (\psi , \varphi ) = \left\langle \psi , e^{F}H_0 e^{-F}\varphi \right\rangle , \quad \psi \in \mathcal {D}(q_0), \varphi \in e^F \mathcal {D}(H_0). \end{aligned}$$
(96)

Its explicit representation will be recorded here.

Proposition B.2

For the sesquilinear form \(q_F\) defined above, we have for any \(\varphi ,\psi \in \mathcal {D}(q_0)\)

$$\begin{aligned} q_F (\psi ,\varphi ) = q_0 (\psi ,\varphi )+ \sum _{j\in \mathbb {Z}} \left( \left\langle F_j'\psi _j,\partial _r\varphi _j \right\rangle - \left\langle \partial _r\psi _j,F_j'\varphi _j \right\rangle - \left\langle F_j'\psi _j,F_j'\varphi _j \right\rangle \right) . \end{aligned}$$
(97)

In particular, for any \(\varphi \in \mathcal {D}(q_0)\)

$$\begin{aligned} \mathrm {Re}\, q_F [\varphi ] = \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle + \left\langle \varphi ,(V-(F')^2)\varphi \right\rangle . \end{aligned}$$
(98)

Proof

A straightforward calculation starting form (85). \(\square \)

To control a perturbation W which is not rotationally symmetric, we recall that the Fourier transformation of the angular variable is given through the unitary operator

$$\begin{aligned} \mathcal {F}:\mathcal {H}\longrightarrow \bigoplus _{j\in \mathbb {Z}} L^2(\mathbb {R}^+,\text {d}r) \end{aligned}$$
(99)

acting as the closure of the map

$$\begin{aligned} \psi \longmapsto (\mathcal {F}\psi )_j\equiv {\psi }_j:= \left( \frac{1}{\sqrt{2\pi }}\int _{0}^{2\pi } \psi (\cdot ,\theta )e^{-ij\theta }\mathrm{d}\theta \right) _{j\in \mathbb {Z}}, \end{aligned}$$

initially defined on \(\mathcal {U}\mathcal {C}^\infty _0(\mathbb {R}^2)\).

It is easy to check that, for any \(j\in \mathbb {Z}\) and \(\psi \in \mathcal {U}\mathcal {C}^\infty _0(\mathbb {R}^2)\),

$$\begin{aligned}{}[P_j \psi ](r,\theta ) = {\psi }_j(r) e_j(\theta ),\quad r>0, \theta \in [0,2\pi ), \end{aligned}$$

with \(e_j(\theta )=e^{ij\theta }/\sqrt{2\pi }\), since \(P_j = \mathbf {1}\otimes |e_j {\rangle }{ \langle } e_j|\) on \(L^2(\mathbb {R}^+)\otimes L^2(\mathbb {S}^1)\simeq \mathcal {H}\). Moreover, we have

$$\begin{aligned} \langle P_j\varphi ,W P_k\psi \rangle _{\mathcal {H}}=\langle {\varphi }_j, \widehat{W}(\cdot , j-k) {\psi }_k \rangle _{L^2(\mathbb {R}^+)}. \end{aligned}$$
(100)

Lemma B.3

Assume that W satisfies Condition 1 for some \(a>0\), \(0<\zeta \leqslant 1\). Let \(F= (F_j)_{j\in \mathbb {Z}} \in PC_{bd}^1(\mathbb {R}^+,\mathbb {R})\) satisfy (42). Then, for any \(\varphi \in \mathcal {C}^\infty _0(\mathbb {R}^2)\),

$$\begin{aligned} \left| \langle \varphi ,e^{F} W e^{-F}\varphi \rangle \right| \leqslant \xi (a,\zeta )\langle \varphi ,v \,\varphi \rangle . \end{aligned}$$
(101)

Moreover, for any \(\varphi \in C^\infty _0(\mathbb {R}^2)\),

$$\begin{aligned} \left| \langle \varphi , W \varphi \rangle \right| \leqslant \xi (2a,\zeta )\langle \varphi ,v \,\varphi \rangle . \end{aligned}$$
(102)

Here, v is defined through Condition 1 and \(\xi (a,\zeta ):=\sum _{k\in \mathbb {Z}} e^{ - \frac{a}{2}|k|^\zeta }\).

Proof

We estimate using (11) for any \(\varphi \in C^\infty _0(\mathbb {R}^2)\)

$$\begin{aligned} \left| \langle \varphi ,e^{F} W e^{-F}\varphi \rangle \right|&\leqslant \sum _{j,k\in \mathbb {Z}} \left| \left\langle e^{F_j}P_j\varphi ,W e^{-F_k}P_k\varphi \right\rangle \right| \\&=\sum _{j,k\in \mathbb {Z}} \left| \left\langle e^{F_j} {\varphi }_j,\widehat{W}(\cdot , j-k) e^{-F_k} {\varphi }_k \right\rangle _{L^2(\mathbb {R}^+)} \right| \\&\leqslant \sum _{j,k\in \mathbb {Z}} e^{-a\left| j-k \right| ^\zeta }\left\langle e^{F_j} \left| {\varphi }_j \right| ,b \,e^{-F_k}\left| {\varphi }_k \right| \right\rangle _{L^2(\mathbb {R}^+)}\\&\leqslant \sum _{j,k\in \mathbb {Z}} e^{-a\left| j-k \right| ^\zeta /2}\left\langle \left| {\varphi }_j \right| ,b \,\left| {\varphi }_k \right| \right\rangle _{L^2(\mathbb {R}^+)}\\&\leqslant \sum _{j,k\in \mathbb {Z}} e^{-a\left| j-k \right| ^\zeta /2} \left\| b^{1/2} {\varphi }_j \right\| _{L^2(\mathbb {R}^+)} \left\| b^{1/2} {\varphi }_k \right\| _{L^2(\mathbb {R}^+)} \end{aligned}$$

where in the last two inequalities we use (42) and Cauchy–Schwarz inequality for the scalar product, respectively. We can estimate the last sums applying Young’s inequality for convolutions to get

$$\begin{aligned} \left| \langle \varphi ,e^{F} W e^{-F}\varphi \rangle \right| \leqslant \left( \sum _{k\in \mathbb {Z}} e^{-a\left| k \right| ^\zeta /2}\right) \left( \sum _{j\in \mathbb {Z}} \left\| b^{1/2} { {\varphi }_j} \right\| _{L^2(\mathbb {R}^+)}^2\right) =\xi (a,\zeta )\langle \varphi ,v \,\varphi \rangle . \end{aligned}$$

This proves (101). In the case, \(F=0\), we clearly obtain the same estimate as above with a/2 replaced by a. This concludes the proof of the lemma. \(\square \)

Since v is infinitesimally \(q_0\)-bounded, as shown in “Appendix A”, we conclude that so is \(\left\langle \psi ,e^F W e^{-F} \varphi \right\rangle \). In particular, its form domain contains \(\mathcal {D}(q_0)\) and the perturbed sesquilinear form

$$\begin{aligned} q_{F,W}(\psi ,\varphi ) = q_F(\psi , \varphi ) + \left\langle \psi , e^F W e^{-F}\varphi \right\rangle ,\quad \psi ,\, \varphi \in \mathcal {D}(q_0), \end{aligned}$$
(103)

is well defined and satisfies

$$\begin{aligned} q_{F,W}(\psi ,\varphi ) = \left\langle \psi , e^F H e^{-F}\varphi \right\rangle , \quad \psi \in \mathcal {D}(q_0), \, \varphi \in e^F \mathcal {D}(H_0). \end{aligned}$$
(104)

We write here the explicit form of its real part, which will be used in the proof of Lemma 3.3.

Proposition B.4

Assume that W satisfies Condition 1 for some \(a>0\), \(0<\zeta \leqslant 1\). Let \(F= (F_j)_{j\in \mathbb {Z}} \in PC_{bd}^1(\mathbb {R}^+,\mathbb {R})\) satisfy (42) and (90). Then, the quadratic form \(q_{F,W}\) satisfies

$$\begin{aligned} \mathrm {Re}\, q_{F,W} [\varphi ] = \left\langle \partial _r\varphi ,\partial _r\varphi \right\rangle + \left\langle \varphi ,(V-(F')^2)\varphi \right\rangle + \mathrm {Re}\left\langle \varphi ,e^FWe^{-F}\varphi \right\rangle \end{aligned}$$
(105)

as quadratic forms on \(\mathcal {D}(q_0)\).

Proof

Follows directly from (97). \(\square \)

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Cárdenas, E., Hundertmark, D., Stockmeyer, E. et al. On the Asymptotic Dynamics of 2-D Magnetic Quantum Systems. Ann. Henri Poincaré 22, 415–445 (2021). https://doi.org/10.1007/s00023-020-01012-1

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