## 1 Introduction

Let $$\Omega$$ be an open subset of $$\mathbb{R }^d$$ such that $$\Omega ^\mathsf{c }=\mathbb{R }^d\setminus \Omega$$ is of finite measure. In [2] Berezin proved that the Dirichlet Laplacian operator $$-\Delta _{\Omega ^\mathsf c }^{\mathscr{D}}$$ on $$\Omega ^\mathsf c$$ satisfies the inequality

\begin{aligned} \mathrm {tr} (-\Delta _{\Omega ^\mathsf c }^{\mathscr{D}}-\lambda )_-^\gamma \le (2\pi )^{-d}|\Omega ^{c}|\int _\mathbb{R ^d}(|p|^2-\lambda )_-^\gamma \, \mathrm {d} p =L_{\gamma ,d}^\mathrm{cl }|\Omega ^{c}|\lambda ^{\gamma +\frac{d}{2}} \end{aligned}
(1)

for all $$\lambda \ge 0,\gamma \ge 1$$. Here and below, the measure of a set $$S\subset \mathbb{R }^d$$ is denoted by $$|S|$$ and $$x_-=\frac{1}{2} (|x|- x)$$ is the negative part of a variable, a function or a self-adjoint operator. The so-called Lieb–Thirring constant $$L_{\gamma ,d}^\mathrm{cl }$$ can be computed to be

\begin{aligned} L_{\gamma ,d}^\mathrm{cl }=(4\pi )^{-\frac{d}{2}}\frac{\Gamma (\gamma +1)}{\Gamma \left( \gamma +1+\frac{d}{2}\right) } \end{aligned}

and is sharp, which follows from an asymptotic result by Weyl [17]. For the inequality to hold it is essential that the Laplace operator is considered on the set $$\Omega ^\mathsf{c }$$ of finite volume. This guarantees that $$H$$ only has discrete spectrum consisting of eigenvalues converging to infinity, showing that the left-hand-side of (1) exists.

In [5] a similar result to the Berezin inequality (1) has been established for the Dirichlet Laplace operator on the set $$\Omega$$ of infinite measure. To this end one introduces the orthogonal projection $$P_\Omega :L^2(\mathbb{R }^d)\rightarrow L^2(\Omega )$$, i.e. the multiplication with the characteristic function $$\chi _\Omega$$. The operator $$P_\Omega (-\Delta )P_\Omega$$ corresponds to the Laplacian on the set $$\Omega$$ with Dirichlet boundary conditions. Since the continuous spectrum of $$P_\Omega (-\Delta )P_\Omega$$ contains the positive real axis, the operator $$(P_\Omega (-\Delta )P_\Omega -\lambda )_-$$ is not trace-class on $$L^2(\mathbb{R }^d)$$. However, it can be compared to a suitable operator to achieve similar results to (1). The authors of [5] considered the difference $$(-\Delta -\lambda )_- -(P_\Omega (-\Delta )P_\Omega -\lambda )_-$$ and proved that

\begin{aligned} \mathrm {tr} \left( (-\Delta -\lambda )_- -(P_\Omega (-\Delta )P_\Omega -\lambda )_-\right) \ge L_{1,d}^\mathrm{cl }|\mathbb R ^d\setminus \Omega |\lambda ^{1+\frac{d}{2}}, \end{aligned}
(2)

which can be seen as an analogue of the Berezin inequality for perturbations of the continuous spectrum of the Laplace operator. Bounds on traces for these types of problems are a fairly recent research area and we point to [5] for a generalisation of Lieb–Thirring inequalities to this setting.

In our paper we aim to find an analogous inequality to (2) for the magnetic operator $$H_B=(-\mathrm i \nabla +A(x))^2$$. Similar to the case of the Laplacian, problems stem from the fact that $$(P_\Omega H_B P_\Omega -\lambda )_-$$ is not trace-class. Thus we consider the difference $$(H_B-\lambda )^\gamma _- -(P_\Omega H_B P_\Omega -\lambda )^\gamma _-$$ and establish lower bounds on the trace of this operator. We also prove a similar inequality for the sub-Laplacian $$L$$ on the first Heisenberg group $$\mathbb H ^1$$.A key observation for our results is that, for any self-adjoint operator $$H$$, a formal computation involving the Berezin–Lieb inequality for convex functions (see [1] and [14]) yields the result

\begin{aligned} \mathrm {tr} \left( (H-\lambda )^\gamma _- -(P_\Omega H P_\Omega -\lambda )^\gamma _-\right) \ge \mathrm {tr} \left( (H-\lambda )^\gamma _--P_\Omega (H-\lambda )^\gamma _-P_\Omega \right) . \end{aligned}
(3)

It is the object of this work to give correct mathematical meaning to this observation and to explicitly calculate the right-hand-side for the two special choices of $$H$$.

The Berezin inequality (1) on domains of finite measure has inspired a number of authors and is related to the Li–Yau inequality [13]. In their paper the authors showed that the sum over the first $$k$$ eigenvalues $$\lambda _1,\dots ,\lambda _k$$ of $$-\Delta _{\Omega ^\mathsf{c }}^{\mathscr{D}}$$ can be bounded from below as

\begin{aligned} \sum _{j=1}^k\lambda _j\ge \frac{d}{d+2}\left( L_{0,d}^\mathrm{cl }|\Omega ^\mathsf{c }|\right) ^{-\frac{2}{d}}k^{\frac{d+2}{d}}. \end{aligned}

This was later proven to be a corollary of (1) via the Legendre transformation, see [11]. In [9] comparable inequalities were established for various classes of differential and pseudo-differential operators including $$((-\Delta )^\alpha )_ {\Omega ^\mathsf c }^\mathcal D$$ with $$\alpha >0$$. A similar inequality to (1) can be found for Schrödinger operators with magnetic fields in the case $$d=2$$. The operator $$H_{B,\Omega ^\mathsf{c }}^{\mathscr{D}}:=(-\mathrm i \nabla +A(x))^2$$ on $$L^2(\Omega ^\mathsf{c })$$ with Dirichlet boundary conditions and arbitrary vector field $$A$$ satisfies

\begin{aligned} \mathrm {tr} (H_{B,\Omega ^\mathsf{c }}^\mathscr{D}-\lambda )^\gamma _-\le L_{\gamma ,2}^\mathrm{cl }\lambda ^{\gamma +1}|\Omega ^\mathsf{c }| \end{aligned}
(4)

for all $$\gamma \ge \frac{3}{2}$$, which follows from a result by Laptev and Weidl in [12] (see also [6]). In [4] this was generalised to $$\gamma \ge 1$$ under the restriction that the magnetic field $$B=\, \mathrm {d} A$$ is constant. In this case the upper bound in (4) can be improved by allowing it to depend on $$B$$

\begin{aligned} \mathrm {tr} (H_{B,\Omega ^\mathsf{c }}^\mathscr{D}-\lambda )^\gamma _- \le |\Omega ^\mathsf{c }|\frac{B}{2\pi }\sum _{k=0}^{\infty } \left( (2k+1)B-\lambda \right) ^\gamma _- \end{aligned}
(5)

as shown in [6]. In their paper the authors also proved that, under the assumption that $$\Omega ^\mathsf{c }$$ is a tiling domain, this inequality also holds if $$0\le \gamma <1$$, where it is sharp. For $$\gamma =1$$ the right-hand-side of (5) can be adapted to magnetic operators with additional external potentials $$V$$, see [7]. The Berezin inequality was furthermore extended to the sub-Laplacian $$L$$ on the Heisenberg group $$\mathbb H ^1$$. In [8] (see also [16]), it was proven that the Dirichlet realisation $$L^\mathscr{D}_{\Omega ^\mathsf{c }}$$ of $$L$$ on a domain $$\Omega ^\mathsf{c }\subset \mathbb H ^1$$ of finite measure satisfies

\begin{aligned} \mathrm {tr} (L^\mathscr{D}_{\Omega ^\mathsf c }-\lambda )^\gamma _-\le |\Omega ^\mathsf{c }|\frac{1}{16}\frac{1}{(\gamma +1)(\gamma +2)}\lambda ^{\gamma +2}. \end{aligned}
(6)

In our paper we obtain lower bounds on the traces of the differences $$(H_B-\lambda )^\gamma _- -(P_\Omega H_B P_\Omega -\lambda )^\gamma _-$$ and $$(L-\lambda )^\gamma _- -(P_\Omega L P_\Omega -\lambda )^\gamma _-$$ which are of the same form as the upper bounds in (5) and (6), respectively.

The paper is organised as follows. In Sect. 2 we discuss (3) in the general setting of $$H$$ being a self-adjoint operator on $$L^2(\mathbb{R }^d)$$. We then state our main results for the magnetic operator $$H_B$$ with constant magnetic field and the sub-Laplacian $$L$$ in Theorems 2 and 3, respectively. The complete proofs of these results are given in the subsequent sections.

## 2 Statement of the main results

Let $$H$$ be a self-adjoint operator on the space $$L^2(\mathbb{R }^d)$$ and let $$\varphi :\mathbb R \rightarrow \mathbb{R }$$ be a convex function such that $$\varphi (H)-\varphi (P_\Omega H P_\Omega )$$ and $$\varphi (H)-P_\Omega \varphi (H)P_\Omega$$ are both trace-class. Under these assumptions a generalisation of the Berezin–Lieb inequality as derived in [10] states that

\begin{aligned} \mathrm {tr} \left( P_\Omega \varphi (H)P_\Omega -\varphi (P_\Omega H P_\Omega )\right) \ge 0. \end{aligned}

As a consequence, we obtain the inequality

\begin{aligned} \mathrm {tr} \left( \varphi (H)-\varphi (P_\Omega H P_\Omega )\right) \ge \mathrm {tr} \left( \varphi (H)-P_\Omega \varphi (H)P_\Omega \right) \end{aligned}
(7)

by making use of the additivity of the trace. We now simplify the right-hand-side of (7) as follows. To shorten notation denote the trace-class operator $$Q:=\varphi (H)-P_\Omega \varphi (H)P_\Omega$$ and let $$P_{\Omega ^\mathsf{c }}=\mathbb I -P_\Omega$$ be the complementary projection of $$P_\Omega$$. Clearly $$Q$$ can be written as the sum of four operators corresponding to the decomposition of $$L^2(\mathbb{R }^d)$$ into $$\mathrm {ran} P_\Omega$$ and $$\mathrm {ran} P_{\Omega ^\mathsf{c }}$$, i.e.

\begin{aligned} Q=P_\Omega Q P_\Omega + P_\Omega Q P_{\Omega ^\mathsf{c }}+ P_{\Omega ^\mathsf{c }} Q P_\Omega + P_{\Omega ^\mathsf{c }} Q P_{\Omega ^\mathsf{c }} . \end{aligned}

In [15][Theorem VI.25] it is shown that, if $$T$$ is trace-class and $$S$$ is bounded, then $$\mathrm {tr} (ST)=\mathrm {tr} (TS)$$. As a result $$\mathrm {tr} (P_\Omega Q P_{\Omega ^\mathsf{c }})=0$$ as well as $$\mathrm {tr} (P_{\Omega ^\mathsf{c }} Q P_\Omega )=0$$. Thus the trace of $$Q$$ consists only of the diagonal terms

\begin{aligned} \mathrm {tr} \left( \varphi (H)-P_\Omega \varphi (H)P_\Omega \right) =\mathrm {tr} (P_\Omega Q P_\Omega + P_{\Omega ^\mathsf c } Q P_{\Omega ^\mathsf{c }}) =\mathrm {tr} \left( P_{\Omega ^\mathsf c } \varphi (H) P_{\Omega ^\mathsf c } \right) \!. \end{aligned}

These results are summarised in the following theorem.

### Theorem 1

Let $$H$$ be a self-adjoint operator on $$L^2(\mathbb{R }^d)$$ and $$\varphi :\mathbb R \rightarrow \mathbb R$$ a convex function such that $$\varphi (H)-\varphi (P_\Omega H P_\Omega )$$ and $$\varphi (H)-P_\Omega \varphi (H)P_\Omega$$ are both trace-class. Then the Berezin–Lieb type inequality

\begin{aligned} \mathrm {tr} \left( \varphi (H)-\varphi (P_\Omega H P_\Omega )\right) \ge \mathrm {tr} \left( \varphi (H)-P_\Omega \varphi (H)P_\Omega \right) =\mathrm {tr} \left( P_{\Omega ^\mathsf c } \varphi (H) P_{\Omega ^\mathsf c } \right) \end{aligned}

holds.

While this result is true for arbitrary self-adjoint operators $$H$$, we shall now apply it to two special choices of $$H$$ to obtain the main results of this work. First, consider Schrödinger operators with magnetic fields. Let the magnetic field $$B(x)$$ be a two-form on $$\mathbb R ^d$$ and the magnetic vector potential $$A(x)$$ a one-form satisfying $$B(x)=\, \mathrm {d} A(x)$$. We shall restrict ourselves to the case $$d=2$$ and in the remainder of this work, we furthermore assume that $$B$$ is constant and positive. Consider the magnetic operator $$H_B=(-\mathrm i \nabla +A(x))^2$$, which is defined as the closure of the form

\begin{aligned} (\psi ,H_B\psi ):=\int _\mathbb{R ^2}\left| \left( -\mathrm i \nabla +A(x)\right) \psi (x)\right| ^2\, \mathrm {d} x \end{aligned}

on $$\mathscr{C}_c^\infty (\mathbb R ^2)$$, the set of smooth functions with compact support. The obtained operator is found to be self-adjoint and we can state the first main result.

### Theorem 2

Assume $$d=2,\lambda \ge 0,B>0,\gamma \ge 1$$ and let $$\Omega$$ be an open subset of $$\mathbb R ^2$$ such that $$\mathbb R ^2\setminus \Omega$$ has finite measure. Then the inequality

\begin{aligned} \mathrm {tr} \left( (H_B-\lambda )^\gamma _-- (P_\Omega H_B P_\Omega -\lambda )^\gamma _-\right)&\ge \mathrm {tr} \left( (H_B-\lambda )^\gamma _- -P_\Omega (H_B-\lambda )^\gamma _- P_\Omega \right) \nonumber \\&= \mathrm {tr} \left( P_{\Omega ^\mathsf c }(H_B-\lambda )^\gamma _-P_{\Omega ^\mathsf c }\right) \end{aligned}
(8)

holds and the right-hand-side can be calculated explicitly as

\begin{aligned} \mathrm {tr} \left( P_{\Omega ^\mathsf c }(H_B-\lambda )^\gamma _-P_{\Omega ^\mathsf c }\right) =|\mathbb R ^2\setminus \Omega |\frac{B}{2\pi }\sum _{k=0}^{\infty } \left( (2k+1)B-\lambda \right) ^\gamma _-. \end{aligned}
(9)

The proof of Theorem 2 is provided in Sect. 3. The lower bound (9) coincides with the upper bound (5) for the magnetic operator on the set $$\Omega ^\mathsf c$$ of finite volume. In essence the proof is the same.

Similar results can also be obtained on the first Heisenberg group $$\mathbb H ^1$$. Here, $$\mathbb H ^1$$ is considered to be the three-dimensional space $$\mathbb R ^3$$ equipped with the non-commutative multiplication

\begin{aligned} (x_1,x_2,x_3)\circ (y_1,y_2,y_3)=\left( x_1+y_1,x_2+y_2,x_3+y_3-\frac{1}{2}(x_1y_2-x_2y_1)\right) \end{aligned}

for $$(x_1,x_2,x_3),(y_1,y_2,y_3)\in \mathbb R ^3$$. On the Heisenberg group, we introduce the two left-invariant vector fields

\begin{aligned} X_1=\frac{\partial }{\partial x_1}+\frac{1}{2} x_2 \frac{\partial }{\partial x_3},\quad X_2=\frac{\partial }{\partial x_2}-\frac{1}{2} x_1 \frac{\partial }{\partial x_3}. \end{aligned}

Using these definitions, we consider the quadratic form

\begin{aligned} \ell (\psi )=\int \limits _\mathbb{R ^3}\left( |X_1\psi |^2+|X_2\psi |^2\right) \, \mathrm {d} x_1\, \mathrm {d} x_2\, \mathrm {d} x_3 \end{aligned}

on $$\mathcal C _c^\infty (\mathbb R ^3)$$ and note that the closure of this form gives the self-adjoint sub-Laplacian $$L=-X_1^2-X_2^2$$ on $$\mathbb H ^1$$. For a detailed background we refer to the literature, e.g. [3]. The sub-Laplacian $$L$$ is found to satisfy the following analogue of the Berezin inequality.

### Theorem 3

Assume $$\lambda \ge 0,\gamma \ge 1$$ and let $$\Omega$$ be an open subset of $$\mathbb R ^3$$ such that $$\mathbb R ^3\setminus \Omega$$ has finite measure. Then the inequality

\begin{aligned} \mathrm {tr} \left( (L-\lambda )^\gamma _-- (P_\Omega L P_\Omega -\lambda )^\gamma _-\right)&\ge \mathrm {tr} \left( (L-\lambda )^\gamma _- -P_\Omega (L-\lambda )^\gamma _- P_\Omega \right) \nonumber \\&= \mathrm {tr} \left( P_{\Omega ^\mathsf c }(L-\lambda )^\gamma _-P_{\Omega ^\mathsf c }\right) \end{aligned}
(10)

holds for the sub-Laplacian $$L$$ on $$\mathbb H ^1$$ and the right-hand-side can be calculated explicitly as

\begin{aligned} \mathrm {tr} \left( P_{\Omega ^\mathsf c }(L-\lambda )^\gamma _-P_{\Omega ^\mathsf c }\right) =|\mathbb R ^3\setminus \Omega |\frac{1}{16}\frac{1}{(\gamma +1)(\gamma +2)}\lambda ^{\gamma +2}. \end{aligned}
(11)

Similarly to the previous application, the lower bound (11) coincides with the upper bound in the case of the Heisenberg sub-Laplacian being defined on the domain $$\Omega ^\mathsf c$$ of finite measure with Dirichlet boundary conditions, see (6). The proof of Theorem 3 is basically the same as in the case of finite measure [8] and can be found in Sect. 4. Note that this result can easily be generalised to the $$N$$-th Heisenberg group $$\mathbb H ^N$$.

### Remark 1

Using Theorem 1 we can also reproduce the results of Frank, Lewin, Lieb and Seiringer [5] and show that

\begin{aligned} \mathrm {tr} \left( (-\Delta -\lambda )^\gamma _- -(P_\Omega (-\Delta )P_\Omega -\lambda )^\gamma _-\right) \ge L_{\gamma ,d}^\mathrm{cl }|\mathbb R ^d\setminus \Omega |\lambda ^{\gamma +\frac{d}{2}} \end{aligned}

for the Laplacian on a set $$\Omega \subset \mathbb R ^d$$ with complement of finite measure.

## 3 The proof of Theorem 2

Let $$\varphi _{\lambda ,\gamma }:\mathbb R \rightarrow \mathbb R$$ be the convex function defined as

\begin{aligned} \varphi _{\lambda ,\gamma }(t)=(t-\lambda )^\gamma _-= \left\{ \begin{array}{ll} (\lambda -t)^\gamma , &{}\quad t\le \lambda \\ 0, &{}\quad t>\lambda . \end{array} \right. \end{aligned}
(12)

Applying Theorem 1 to this function and the operator $$H_B$$ yields (8) and it only remains to prove (9). This can be done in complete analogy to calculations by Frank, Loss and Weidl [6]. The spectrum of $$H_B$$ is entirely discrete and can be calculated to be $$(2k+1)B$$ for $$k\in \mathbb N \cup \{0\}$$. The projection onto the $$k$$-th Landau level is denoted by $$\Pi _{B,k}$$. The spectral theorem implies that the operator $$\varphi _{\lambda ,\gamma }(H_B)$$ can then be written as

\begin{aligned} \varphi _{\lambda ,\gamma }(H_B)=\sum _{k=0}^\infty \varphi _{\lambda ,\gamma }\left( (2k+1)B\right) \Pi _{B,k}. \end{aligned}

We multiply this identity from both sides with the projection $$P_{\Omega ^\mathsf c }$$ and consider the trace of the obtained expression, that is

\begin{aligned} \mathrm {tr} \left( P_{\Omega ^\mathsf c }\varphi _{\lambda ,\gamma }(H_B)P_{\Omega ^\mathsf c }\right) =\sum _{k=0}^\infty \varphi _{\lambda ,\gamma }\left( (2k+1)B\right) \mathrm {tr} (P_{\Omega ^\mathsf{c }}\Pi _{B,k}). \end{aligned}
(13)

To explicitly calculate the summands on the right-hand-side of (13), we observe that by the cyclicity of the trace

\begin{aligned} \mathrm {tr} (P_{\Omega ^\mathsf c }\Pi _{B,k})=\mathrm {tr} (P_{\Omega ^\mathsf c }\Pi _{B,k}\Pi _{B,k}P_{\Omega ^\mathsf c })=\left\| P_{\Omega ^\mathsf c }\Pi _{B,k} \right\| _{\sigma _2}^2, \end{aligned}
(14)

where $$\left\| \cdot \right\| _{\sigma _2}$$ denotes the Hilbert–Schmidt norm. This norm can be calculated explicitly by using the integral kernel of the operator $$P_{\Omega ^\mathsf c }\Pi _{B,k}$$. Let $$\Pi _{B,k}(x,y)$$ be the integral kernel of $$\Pi _{B,k}$$ such that $$\Pi _{B,k} \psi (x)=\int _\mathbb{R ^2} \Pi _{B,k}(x,y)\psi (y)\, \mathrm {d} y$$. The integral kernel of the composition $$P_{\Omega ^\mathsf c }\Pi _{B,k}$$ is then given by

\begin{aligned} (P_{\Omega ^\mathsf c }\Pi _{B,k})(x,y)=\chi _{\Omega ^\mathsf c }(x)\Pi _{B,k}(x,y). \end{aligned}

We can calculate the Hilbert-Schmidt norm on the right-hand-side of (14) by double integration of the square of the modulus of this integral kernel, that is

\begin{aligned} \mathrm {tr} (P_{\Omega ^\mathsf c }\Pi _{B,k} ) =\left\| P_{\Omega ^\mathsf c }\Pi _{B,k}\right\| _{\sigma _2}^2 =\int _\mathbb{R ^2}\int _\mathbb{R ^2} |\Pi _{B,k}(x,y)|^2\chi _{\Omega ^\mathsf c }(x)\, \mathrm {d} y \, \mathrm {d} x. \end{aligned}
(15)

To explicitly solve this integral, we point out some important properties of the function $$\Pi _{B,k}(x,y)$$. As the orthogonal projection $$\Pi _{B,k}$$ is self-adjoint it must hold that $$\Pi _{B,k}(x,y)=\overline{\Pi _{B,k}(y,x)}$$. By evaluating $$\Pi _{B,k}$$ at the delta distribution $$\delta (x-x_0)$$ and using the defining property of projections, $$\Pi _{B,k}=\Pi _{B,k}\Pi _{B,k}$$, it can easily be concluded that

\begin{aligned} \int _\mathbb{R ^2}|\Pi _{B,k}(x_0,y)|^2\, \mathrm {d} y =\Pi _{B,k}(x_0,x_0). \end{aligned}
(16)

It is furthermore a remarkable fact that the diagonal of the integral kernel of $$\Pi _{B,k}$$ is given by the constant $$\Pi _{B,k}(x,x)=\frac{B}{2\pi }$$ for all $$k\in \mathbb N \cup \{0\}$$. Using these properties, identity (15) can be continued as

\begin{aligned} \mathrm {tr} (P_{\Omega ^\mathsf c }\Pi _{B,k} ) =\int _\mathbb{R ^2}\Pi _{B,k}(x,x) \chi _{\Omega ^\mathsf c }(x)\, \mathrm {d} x=\frac{B}{2\pi }|\mathbb R ^2\setminus \Omega |. \end{aligned}

Inserting this equation back into (13) yields the final result

\begin{aligned} \mathrm {tr} \left( P_{\Omega ^\mathsf c }\varphi _{\lambda ,\gamma }(H_B)P_{\Omega ^\mathsf c }\right) =|\mathbb R ^2\setminus \Omega |\frac{B}{2\pi }\sum _{k=0}^{\infty } \varphi _{\lambda ,\gamma }\left( (2k+1)B\right) \end{aligned}

which finishes the proof.

## 4 The proof of Theorem 3

Let the convex function $$\varphi _{\lambda ,\gamma }:\mathbb R \rightarrow \mathbb R$$ be defined as in (12). Theorem 1 applied to $$L$$ yields (10) and it only remains to show (11) which can be proven following calculations by Hansson and Laptev [8]. Firstly, we introduce the Fourier transformation $$\mathscr{F}_{x_3}$$ with respect to the variable $$x_3$$,

\begin{aligned} \mathscr{F}_{x_3} \psi (x_1,x_2,x_3)=(2\pi )^{-\frac{1}{2}}\int \limits _{-\infty }^{+\infty }\mathrm e ^{-\mathrm i x_3 y_3} \psi (x_1,x_2,y_3)\, \mathrm {d} y_3. \end{aligned}

A simple calculation shows that the Heisenberg sub-Laplacian $$L$$ satisfies the identity

\begin{aligned} \mathscr{F}_{x_3} L \mathscr{F}_{x_3}^*=\left( \mathrm i \frac{\partial }{\partial x_1}-\frac{1}{2}x_2x_3\right) ^2+\left( \mathrm i \frac{\partial }{\partial x_2}+\frac{1}{2} x_1x_3\right) ^2. \end{aligned}
(17)

For fixed $$x_3$$, the right-hand-side of (17) can be identified with a two-dimensional Schrödinger operator with vector field $$A=\frac{x_3}{2}(- x_2,x_1)$$. The corresponding magnetic field $$B=\, \mathrm {d} A$$ is constant and can be calculated to be $$B=x_3$$. The eigenvalues of this magnetic operator $$H_{x_3}$$ are $$(2k+1)|x_3|$$ for $$k\in \mathbb N \cup \{0\}$$. Note that they depend on the variable $$x_3$$. Similar to the previous section we shall use the Landau projections $$\Pi _{x_3,k}$$ to prove (11). With respect to the variable $$x_3$$, the operator $$\mathscr{F}_{x_3} L \mathscr{F}_{x_3}^*$$ simply acts as a multiplication operator and consequently the spectral theorem allows us to write

\begin{aligned} \varphi _{\lambda ,\gamma }(\mathscr{F}_{x_3} L \mathscr{F}_{x_3}^*)=\sum _{k=0}^\infty \varphi _{\lambda ,\gamma }\left( (2k+1)|x_3|\right) \widehat{\Pi }_{x_3,k} \end{aligned}
(18)

where we have used the tensor product $$\widehat{\Pi }_{x_3,k}=\Pi _{x_3,k}\otimes \mathbb I _{L^2(\mathbb R )}$$ with $$\mathbb I _{L^2(\mathbb R )}$$ denoting the identity on $$L^2(\mathbb R )$$. For convenience we introduce the notation $$\mu _k(x_3)=\varphi _{\lambda ,\gamma }\left( (2k+1)|x_3|\right)$$ swallowing the dependence on $$\lambda$$ and $$\gamma$$ for the moment. As a consequence of (18) we obtain the identity

\begin{aligned} \mathrm {tr} \left( P_{\Omega ^\mathsf c }\varphi _{\lambda ,\gamma }(L)P_{\Omega ^\mathsf c }\right) =\sum _{k=0}^\infty \mathrm {tr} \left( P_{\Omega ^\mathsf c }\mathscr{F}_{x_3}^*\mu _k(x_3)\widehat{\Pi }_{x_3,k}\mathscr{F}_{x_3}P_{\Omega ^\mathsf{c }}\right) . \end{aligned}
(19)

This result can be compared to the analogous equation in the case of a magnetic operator (13). While in this setting the magnetic field was a given constant, we now consider a magnetic field that changes with the variable $$x_3$$. In addition, the Fourier transformation $$\mathscr{F}_{x_3}$$ has to be dealt with. The summands on the right-hand-side of (19) can be written as Hilbert–Schmidt norms of certain operators, that is

\begin{aligned} \mathrm {tr} \left( P_{\Omega ^\mathsf{c }}\mathscr{F}_{x_3}^*\mu _k(x_3)\widehat{\Pi }_{x_3,k}\mathscr{F}_{x_3}P_{\Omega ^\mathsf c }\right) =\left\| P_{\Omega ^\mathsf c }\mathscr{F}_{x_3}^*\mu _k(x_3)^{\frac{1}{2}}\widehat{\Pi }_{x_3,k}\right\| _{\sigma _2}^2 \end{aligned}

for every $$k\in \mathbb N \cup \{0\}$$. Here we have used that the multiplication operator $$\mu _k(x_3)$$ and the projection $$\Pi _{x_3,k}$$ commute. The investigation of these Hilbert–Schmidt norms requires us to calculate the integral kernels of the operators involved. Let $$x=(x_1,x_2,x_3)$$ and $$y=(y_1,y_2,y_3)$$ be two vectors in $$\mathbb R ^3$$. The integral kernel of $$P_{\Omega ^\mathsf c }\mathscr{F}_{x_3}^*\mu _k(x_3)^{\frac{1}{2}}\widehat{\Pi }_{x_3,k}$$ can then be computed to be $$\chi _{\Omega ^\mathsf c }(x)\frac{1}{\sqrt{2\pi }}\mathrm e ^\mathrm{i x_3y_3}\Pi _{y_3,k}(x_1,x_2,y_1,y_2)\mu _k(y_3)^{\frac{1}{2}}$$. As a consequence we obtain the identity

\begin{aligned}&\left\| P_{\Omega ^\mathsf c }\mathscr{F}_{x_3}^*\mu _k(x_3)^{\frac{1}{2}}\widehat{\Pi }_{x_3,k}\right\| _{\sigma _2}^2\\ \nonumber&=\dfrac{1}{2\pi }\displaystyle \int _\mathbb{R ^3}\int _\mathbb{R ^3}\chi _{\Omega ^\mathsf c }(x)|\Pi _{y_3,k}(x_1,x_2,y_1,y_2)|^2\mu _k(y_3)\, \mathrm {d} y\, \mathrm {d} x\\ \nonumber&=\dfrac{1}{2\pi }\displaystyle \int _\mathbb{R }\!\mu _k(y_3)\int _\mathbb{R ^3}\chi _{\Omega ^\mathsf c }(x)\!\int _\mathbb{R ^2}|\Pi _{y_3,k}(x_1,x_2,y_1,y_2)|^2\, \mathrm {d} y_1\, \mathrm {d} y_2\, \mathrm {d} x \, \mathrm {d} y_3. \end{aligned}

To calculate this integral, we recall (16) and stress again that the diagonal of the integral kernel $$\Pi _{y_3,k}(x_1,x_2,y_1,y_2)$$ is is known to be the constant $$\frac{|y_3|}{2\pi }$$. This results in

\begin{aligned} \left\| P_{\Omega ^\mathsf c }\mathscr{F}_{x_3}^*\mu _k(x_3)^{\frac{1}{2}}\widehat{\Pi }_{x_3,k}\right\| _{\sigma _2}^2&= \dfrac{1}{2\pi }\int _\mathbb{R }\mu _k(y_3)\int _\mathbb{R ^3}\chi _{\Omega ^\mathsf c }(x)\frac{|y_3|}{2\pi }\, \mathrm {d} x\, \mathrm {d} y_3\\ \nonumber&= \dfrac{1}{2\pi ^2}|\mathbb R ^3\setminus \Omega | \int _{0}^{+\infty }\left( (2k+1)y_3-\lambda \right) ^\gamma _-y_3\, \mathrm {d} y_3 \end{aligned}

where we have used the definition of $$\mu _k(y_3)$$ to obtain the last equality. We insert this identity back into (19) and substitute $$p=(2k+1)y_3$$ to conclude that

\begin{aligned} \mathrm {tr} \left( P_{\Omega ^\mathsf c }\varphi _{\lambda ,\gamma }(L)P_{\Omega ^\mathsf c }\right)&= \dfrac{1}{2\pi ^2}|\mathbb R ^3\setminus \Omega |\sum _{k=0}^\infty \frac{1}{(2k+1)^2}\int _{0}^{+\infty }\left( p-\lambda \right) ^\gamma _{-} p\, \mathrm {d} p\\&= \dfrac{1}{16}|\mathbb R ^3\setminus \Omega |\int _{0}^{+\infty }\left( p-\lambda \right) ^\gamma _{-} p\, \mathrm {d} p. \end{aligned}

Here, the last equality follows from the well-known fact that $$\sum _{k=0}^\infty \frac{1}{(2k+1)^2}=\frac{\pi ^2}{8}$$. The remaining integral can be easily calculated using partial integration

\begin{aligned} \int _{0}^{+\infty }\left( p-\lambda \right) ^\gamma _{-}, p\, \mathrm {d} p=\frac{\lambda ^{\gamma +2}}{(\gamma +1)(\gamma +2)} \end{aligned}

and this yields the desired result.