Location of eigenvalues of three-dimensional non-self-adjoint Dirac operators

Abstract

We prove the absence of eigenvalues of the three-dimensional Dirac operator with non-Hermitian potentials in unbounded regions of the complex plane under smallness conditions on the potentials in Lebesgue spaces. Our sufficient conditions are quantitative and easily checkable.

1 Introduction

Let us consider a relativistic quantum particle of spin $\frac{1}{2}$ and mass $m \ge 0$ in $\mathbb {R}^3$, subject to an external electric field described by a potential V. The dynamics is governed by the Dirac Hamiltonian

$$\begin{aligned} H_{V} := - i \; \alpha \cdot \nabla + m \, \alpha _4 + V \end{aligned}$$

(1)

acting in the Hilbert space of spinors $\mathcal {H}:= L^2(\mathbb {R}^3;\mathbb {C}^4)$. Here $\alpha := (\alpha _1,\alpha _2,\alpha _3)$ with $\alpha _\mu$ being the usual $4 \times 4$ Hermitian Dirac matrices satisfying the anticommutation rules

$$\begin{aligned} \alpha _\mu \alpha _\nu + \alpha _\nu \alpha _\mu = 2 \delta _{\mu \nu } I_{\mathbb {C}^4} \end{aligned}$$

(2)

for $\mu ,\nu \in \{1,\dots ,4\}$ and the dot denotes the scalar product in $\mathbb {R}^3$. Motivated by a growing interest in non-self-adjoint operators in quantum mechanics (cf.  [1] for a mathematical overview), we proceed in a greater generality by allowing V to be a possibly non-Hermitian $4 \times 4$ matrix in (1).

In the traditional self-adjoint case (i.e. V is a real scalar multiple of the diagonal matrix), the literature on spectral properties of $H_V$ is enormous and we limit ourselves to quoting the classical Thaller’s monograph [17]. For V being matrix-valued and possibly non-Hermitian, a systematic study of the spectrum of $H_V$ was initiated by the pioneering work of Cuenin, Laptev and Tretter [5] in the one-dimensional setting and followed by [2, 6, 8]. Some spectral aspects in the present three-dimensional situation are covered by [3, 7, 15].

In the field-free case (i.e.$V=0$), it is well known that $\sigma (H_{0}) = (-\,\infty ,-\,m] \cup [+\,m,+\,\infty )$ and that the spectrum is purely continuous. That is, the residual spectrum is empty and there are no eigenvalues. The objective of this paper is to derive quantitative smallness conditions on the potential V, which guarantee that the spectrum of $H_{V}$ remains purely continuous, at least in certain regions of the complex plane. Denoting by |V(x)| the operator norm of the matrix $V(x) \in \mathbb {C}^{4 \times 4}$ for almost every fixed $x \in \mathbb {R}^3$, the smallness is measured through Lebesgue norms of the real-valued function |V|.

We present two types of results in this paper. The first reads as follows:

Theorem 1

Assume $|V| \in L^3(\mathbb {R}^3)$. If

$$\begin{aligned} C \,f(\lambda ,m) \,\Vert |V|\Vert _{L^3(\mathbb {R}^3)} < 1 \end{aligned}$$

(3)

with

$$\begin{aligned} C := \left( \frac{\pi }{2}\right) ^{1/3} \sqrt{1+e^{-1}+2e^{-2}} \qquad \text{ and } \qquad f(\lambda ,m) := \sqrt{ 1 + \frac{(\mathfrak {R}\lambda )^2}{\big (\mathfrak {R}\sqrt{m^2-\lambda ^2}\big )^2} } \,, \end{aligned}$$

then $\lambda \not \in \sigma _\mathrm {p}(H_V)$.

The hypothesis (3) is essentially a smallness requirement about V. In view of Kato’s general smoothness theory [12] (see also [14, Sec. XIII.7]), it is not surprising that a result of this type should hold for weakly coupled potentials. For the Dirac operator in all dimensions, a similar result has been obtained by Cuenin [3, Thm. 6.1.b] (see also [4, Thm. 1.4.6]). The strength of our result lies in that the condition is explicit and easy to check in applications. Notice that $C \approx 1.5$ and that $f(\lambda ,m)$ is finite if, and only if, $\lambda \not \in (-\,\infty ,-\,m]\cup [+\,m,+\,\infty )$. Consequently, the condition (3) is obeyed for such $\lambda$ whenever the norm $\Vert |V|\Vert _{L^3(\mathbb {R}^3)}$ is sufficiently small. Let us also remark that $f(\lambda ,m) \sim |\lambda |/|\mathfrak {I}(\lambda )|$ as $|\mathfrak {I}(\lambda )| \rightarrow +\infty$ and in fact $f(\lambda ,0)=|\lambda |/|\mathfrak {I}(\lambda )|$ in the massless case. We illustrate the dependence of $f(\lambda ,m)$ on $\lambda$ in Fig. 1. We leave as an open problem whether the asymptotic double cone is optimal.

Remark 1

An analogous result holds (with an unspecified constant) provided that $L^{3}(\mathbb {R}^3)$ is replaced by the Lorentz space $L^{3,\infty }(\mathbb {R}^3)$. Since this space in particular contains the Coulomb potential $V_Z(x) := -Z/|x|$, which creates discrete eigenvalues in the gap $(-\,m,+\,m)$ whenever $Z>0$, cf.  [17, Sec. 7.4], we see that the presence of a $\lambda$-dependent function $f(\lambda ,m)$ diverging as $\lambda \rightarrow \pm \, m$ is in fact unavoidable in (3).

Fig. 1

Graph of the surface $\lambda \mapsto f(\lambda ,m)$ and its contour plot for $m=5$

To get a more uniform control over the spectrum and exclude possible eigenvalues embedded in the essential spectrum $(-\,\infty ,-\,m]\cup [+\,m,+\,\infty )$, we have to strengthen the condition about the potential V.

Theorem 2

Assume $|V| \in L^3(\mathbb {R}^3) \cap L^{3/2}(\mathbb {R}^3)$. If

$$\begin{aligned} C \, \Vert |V|\Vert _{L^3(\mathbb {R}^3)} + C' \, |\mathfrak {R}\lambda | \, \Vert |V|\Vert _{L^{3/2}(\mathbb {R}^{3})} < 1 \,, \end{aligned}$$

(4)

where C is the same constant as in Theorem 1 and

$$\begin{aligned} C' := \frac{2^{17/6}}{3\pi ^{2/3}} \,, \end{aligned}$$

then $\lambda \not \in \sigma _\mathrm {p}(H_V)$.

In this case, given a potential V with sufficiently small norm $\Vert |V|\Vert _{L^3(\mathbb {R}^3)}$, the hypothesis (4) excludes the existence of eigenvalues in thin tubular neighbourhoods of the imaginary axis, with the thinness determined by the norm $\Vert |V|\Vert _{L^{3/2}(\mathbb {R}^3)}$. Notice that $C' \approx 1.1$ and that the condition (4) is m-independent.

Of course, Theorems 1 and 2 can be combined to disprove the existence of eigenvalues in a union of the unbounded regions of the complex plane initially covered by the theorems separately. From this perspective, our theorems are an improvement upon Cuenin [3, Thm. 2.3], who disproves the existence of eigenvalues in compact regions only. Moreover, our results are much more explicit and quantitative. On the other hand, our method seems to be restricted to the three-dimensional situation, while the results of [3] are stated in all dimensions greater than or equal to two.

Remark 2

We note that V obeys (3) (respectively, (4)) if, and only if, the adjoint $V^*$ does. Noticing additionally that $f({\bar{\lambda }},m)=f(\lambda ,m)$, we see that any $\lambda \in \mathbb {C}$ satisfying (3) or (4) is not in the residual spectrum of $H_V$ either. Consequently, $\lambda$ is either in the continuous spectrum or in the resolvent set of $H_V$. Finally, let us notice that $\sigma _\mathrm {ess}(H_V) = \sigma _\mathrm {ess}(H_0) = (-\,\infty ,-\,m]\cup [+\,m,+\,\infty )$ under the conditions stated in Theorems 1 and 2.

Our approach to establish Theorems 1 and 2 is based on two ingredients. The main technique is the Birman–Schwinger principle stating that if $\lambda$ is in the point spectrum of the differential operator $H_V$, then $-1$ is in the spectrum of the (formal) integral operator

$$\begin{aligned} W^{1/2} \, (H_{0}-\lambda )^{-1} \, U \, W^{1/2} \,, \end{aligned}$$

(5)

where $V=UW$ is the polar decomposition of V (i.e. W is the absolute value of V and U is unitary). In this paper, we make an effort to justify the principle under the minimal regularity assumption that V is in a sense relatively form-bounded with respect to $H_0$. Moreover, we rigorously cover the embedded eigenvalues $\lambda \in (-\,\infty ,-\,m]\cup [+\,m,+\,\infty )$. Once the Birman–Schwinger principle is established, it is enough to show that the norm of (5) is less than one in order to establish Theorems 1 and 2. To this aim, we use the second ingredient based on a careful estimate of the resolvent integral kernel of $H_0$. Putting these two together, we end up with the left-hand sides of the conditions (3) and (4) being precisely upper bounds to the norm of (5) obtained additionally with help of Sobolev-type inequalities.

The same approach has been recently employed in [9, Sect. 2] to disprove the existence of eigenvalues of Schrödinger operators with form-subordinated complex potentials (cf.  [9, Thm. 1]). As an additional ingredient in comparison with [9, Sect. 2], here we also use the Hardy–Littlewood–Sobolev inequality [13, Thm. 4.3] to deal with the resolvent integral kernel of $H_0$ (see [10, Thm. 2] for a similar trick in the case of Schrödinger operators). In the present case of Dirac operator, however, the method does not seem to lead to uniform results in $\lambda$.

The rest of the paper consists of a precise definition of $H_V$ as a pseudo-Friedrichs extension in Sect. 2, the analysis of the free resolvent in Sect. 3, the justification of the Birman–Schwinger principle in Sect. 4 and its application to the proofs of Theorems 1 and 2 in Sect. 5.

2 The pseudo-Friedrichs extension

We consider $H_{V}$ as a perturbation of $H_{0}$, the latter being the operator that acts as (1) with $V=0$ and has the domain

$$\begin{aligned} \mathsf {D}(H_{0}) := \left\{ \psi \in \mathcal {H}: \nabla \psi \in \mathcal {H}^3 \right\} . \end{aligned}$$

It is well known that $H_0$ is self-adjoint and that $C_0^\infty (\mathbb {R}^3;\mathbb {C}^4)$ is a core of $H_0$.

Notice that $H_0^2 = (-\Delta +m^2) I_{\mathbb {C}^4}$, where $-\Delta +m^2$ is the self-adjoint Schrödinger operator in $L^2(\mathbb {R}^3)$ with the usual domain $H^2(\mathbb {R}^3)$. The absolute value of $H_0$ thus equals $(H_0^2)^{1/2} = \sqrt{-\Delta +m^2} I_{\mathbb {C}^4}$, which is again a self-adjoint operator when considered on the domain $H^1(\mathbb {R}^3;\mathbb {C}^4)$. The form domain of $\sqrt{-\Delta +m^2}$ equals the fractional Sobolev space $H^{1/2}(\mathbb {R}^3)$, cf.  [13, Sect. 7.11]. Notice that $C_0^\infty (\mathbb {R}^3)$ is dense in $H^{1/2}(\mathbb {R}^3)$, cf.  [13, Sect. 7.14]. Clearly, $\sqrt{-\Delta +m^2} \ge \sqrt{-\Delta }$.

We always assume that $|V| \in L^2_\mathrm {loc}(\mathbb {R}^3)$ and that V is relatively form-bounded with respect to the massless $H_0$ in the following sense: There exist numbers $a \in (0,1)$ and $b\in \mathbb {R}$ such that, for all $\psi \in C_0^\infty (\mathbb {R}^3)$,

$$\begin{aligned} \int _{\mathbb {R}^3} |V(x)| |\psi (x)|^2 \, \mathrm {d}x \le a \int _{\mathbb {R}^3} |\root 4 \of {-\Delta }\,\psi (x)|^2 \, \mathrm {d}x + b \int _{\mathbb {R}^3} |\psi (x)|^2 \, \mathrm {d}x \,. \end{aligned}$$

(6)

We recall that |V(x)| denotes the operator norm of the matrix V(x) in $\mathbb {C}^4$. Similarly, $|\psi (x)|$ denotes the norm of the vector $\psi (x) \in \mathbb {C}^4$. The double norm $\Vert \cdot \Vert _\mathcal {X}$ is reserved for functional spaces $\mathcal {X}$. For instance, $\Vert \psi \Vert _{\mathcal {H}} = \Vert |\psi |\Vert _{L^2(\mathbb {R})}$. The symbol $(\cdot ,\cdot )_\mathcal {X}$ stands for an inner product in a Hilbert space $\mathcal {X}$.

By the pseudo-Friedrichs extension [11, Thm. VI.3.11]¹ (see also [18] for more recent developments), there exists a unique closed extension $H_V$ of the operator sum $H_0+V$, where V is understood as the multiplication operator by the matrix V in $\mathcal {H}$ with initial domain $C_0^\infty (\mathbb {R}^3;\mathbb {C}^4)$. More specifically, the operator $H_V$ satisfies

$$\begin{aligned} (\phi ,H_V\psi )_{\mathcal {H}} = \big (G_0^{1/2}\phi ,H_0 G_0^{-1} G_0^{1/2}\psi \big )_{\mathcal {H}} + \int _{\mathbb {R}^3} \phi (x)^* V(x) \psi (x) \, \mathrm {d}x \end{aligned}$$

(7)

for all $\psi \in \mathsf {D}(H_V) \subset H^{1/2}(\mathbb {R}^3;\mathbb {C}^4)$ and $\phi \in H^{1/2}(\mathbb {R}^3;\mathbb {C}^4)$. Here $G_0 := (H_0^2)^{1/2} + (a^{-1} b + \delta ) I_{\mathbb {C}^4}$ with any positive $\delta$ (which does not influence the definition of $H_V$). Notice that $H_0$ and $G_0$ commute (in the usual sense for unbounded operators), so the first term on the right-hand side of (7) equals $(\phi ,H_0\psi )_{\mathcal {H}}$ if $\psi \in H^1(\mathbb {R}^3;\mathbb {C}^4)$ Consequently, $H_V$ realises in a sense the form sum of $H_0$ and V. We have $\mathsf {D}({H_V}^{*}) \subset H^{1/2}(\mathbb {R}^3;\mathbb {C}^4)$ and $i\eta \not \in \sigma (H_V)$ for all real $\eta$ with sufficiently large $|\eta |$.

Remark 3

By Kato’s inequality $\sqrt{-\Delta } \ge (2/\pi ) |x|^{-1}$ valid in the sense of quadratic forms in $L^2(\mathbb {R}^3)$, cf.  [11, Rem. V.5.12], we see that the potentials V satisfying the pointwise inequality

$$\begin{aligned} |V(x)| \le a \, \frac{2}{\pi }\frac{1}{|x|} + b \end{aligned}$$

for almost every $x \in \mathbb {R}^3$ obey the hypothesis (6).

Another sufficient condition is given by the following proposition. (More optimal results can be obtained by using the Lorentz spaces.)

Proposition 1

Let $|V| = v_1 + v_2$ with $v_1 \in L^3(\mathbb {R}^3)$ and $v_2 \in L^\infty (\mathbb {R}^3)$ and assume that

$$\begin{aligned} \Vert v_1\Vert _{L^3 (\mathbb {R}^3)} < (2\pi ^2)^{1/3} \,. \end{aligned}$$

Then (6) holds true.

Proof

For every $\psi \in C_0^\infty (\mathbb {R}^3)$, we have

$$\begin{aligned} \big \Vert v_1^{1/2}\psi \big \Vert _{L^2(\mathbb {R}^3)}^2 \le \Vert v_1\Vert _{L^3(\mathbb {R}^3)} \, \big \Vert \psi \big \Vert _{L^3(\mathbb {R}^3)}^{2} \le (2\pi ^2)^{-1/3} \, \Vert v_1\Vert _{L^3(\mathbb {R}^3)} \, \big \Vert \root 4 \of {-\Delta }\,\psi \big \Vert _{L^2(\mathbb {R}^3)}^{2} \,, \end{aligned}$$

where the first estimate follows by Hölder inequality and the second inequality quantifies the Sobolev-type embedding ${\dot{H}}^{1/2}(\mathbb {R}^3) \hookrightarrow L^3(\mathbb {R}^3)$, cf.  [13, Thm. 8.4]. At the same time, $\Vert v_2^{1/2}\psi \Vert _{L^2(\mathbb {R}^3)}^2 \le b \, \Vert \psi \Vert _{L^2(\mathbb {R}^3)}^2$ with some non-negative constant b. $\square$

Remark 4

Since (6) allows V to be merely defined through a quadratic form, one can in principle introduce $H_V$ under the weaker hypothesis $|V| \in L^1_\mathrm {loc}(\mathbb {R}^3)$.

3 The free resolvent

Using the well-known trick

$$\begin{aligned} (H_0-z)^{-1} = (H_0+z) (H_0^2-z^2)^{-1} \end{aligned}$$

and the knowledge of the integral kernel of the free Schrödinger operator $-\Delta + m^2$, the integral kernel of the free resolvent $(H_{0}-z)^{-1}$ can be written down explicitly:

$$\begin{aligned}&(H_0-z)^{-1}(x,x') \nonumber \\&\quad = \frac{e^{-\sqrt{m^2-z^2}\;|x-x'|}}{4\pi \,|x-x'|} \left( \frac{i \alpha \cdot (x-x')}{|x-x'|^2} + \sqrt{m^2-z^2} \ \frac{i \alpha \cdot (x-x')}{|x-x'|} + m \alpha _4 + z \right) \qquad \quad \end{aligned}$$

(8)

for every $z \not \in \sigma (H_0)$. Here and in the sequel, we choose the principal branch of the square root. From now on, we also usually suppress writing the identity operators in the formulae; therefore, we simply write z instead of $z I_\mathcal {H}$ (respectively, $z I_{\mathbb {C}^4}$) on the left-hand (respectively, right-hand) side of (8) and elsewhere.

Given any matrix $M \in \mathbb {C}^{4 \times 4}$, we use the notation $|M|_\mathrm {HS}:=\sqrt{\mathop {\mathrm {tr}}\nolimits (M^*M)}$ for the Hilbert–Schmidt (or Frobenius) norm of M. Recall that $|M| \le |M|_\mathrm {HS}$.

Lemma 1

(Hilbert–Schmidt norm) For almost every $x,x' \in \mathbb {R}^3$, one has

$$\begin{aligned}&\big |(H_0-z)^{-1}(x,x')\big |_\mathrm {HS}^2 \\&\quad = 4 \, \frac{e^{-2\,\mathfrak {R}\sqrt{m^2-z^2}\,|x-x'|}}{(4\pi )^2 |x-x'|^4} \left( 1+2\, \mathfrak {R}\sqrt{m^2-z^2} \, |x-x'|\right. \\&\left. \qquad +\,2 \left[ \big (\mathfrak {R}\sqrt{m^2-z^2}\big )^2 + (\mathfrak {R}z)^2\right] |x-x'|^2\right) . \end{aligned}$$

Proof

Writing

$$\begin{aligned} \big |(H_0-z)^{-1}(x,x')\big |_\mathrm {HS}^2 = \frac{e^{-2\,\mathfrak {R}\sqrt{m^2-z^2}\,|x-x'|}}{(4\pi )^2 |x-x'|^6} \, \mathop {\mathrm {tr}}\nolimits (A^*A) \,, \end{aligned}$$

where

$$\begin{aligned} A := a \cdot \alpha + a_4 \alpha _4 + a_0 I_{\mathbb {C}^4} \end{aligned}$$

(9)

with

$$\begin{aligned} a := i \left( 1+\sqrt{m^2-z^2}\,|x-x'|\right) (x-x') , \quad a_4 := m \, |x-x'|^2 \,, \qquad a_0 := z \, |x-x'|^2 \,, \end{aligned}$$

the proof reduces to straightforward manipulations with the Dirac matrices. Using the anticommutation relations (2), we immediately arrive at

$$\begin{aligned} A^* A = (|a|^2 + a_4^2 + |a_0|^2) I_{\mathbb {C}^4} + B \,, \end{aligned}$$

where

$$\begin{aligned} B := b \cdot \alpha + b_4 \, \alpha _4 + c \cdot i\alpha \alpha _4 \end{aligned}$$

with

$$\begin{aligned} \begin{aligned} b :=&~2 \mathfrak {R}({\overline{a}} a_0) (x-x') \\ =&~2 \left[ \mathfrak {I}z - \big (\mathfrak {I}z \, \mathfrak {R}\sqrt{m^2-z^2} + \mathfrak {R}z \ \mathfrak {I}\sqrt{m^2-z^2} \big ) |x-x'| \right] |x-x'|^2 (x-x') \,, \\ b_4 :=&~2 a_4 \mathfrak {R}(a_0) = 2 m \, \mathfrak {R}z \, |x-x'|^4 \,, \\ c :=&-2 \mathfrak {I}(a) a_4 (x-x') = -2 m \big ( 1 + \mathfrak {R}\sqrt{m^2-z^2} \, |x-x'| \big ) |x-x'|^2 (x-x') \,. \end{aligned} \end{aligned}$$

Since $\mathop {\mathrm {tr}}\nolimits (\alpha _\mu )=0$ for $\mu \in \{1,\dots ,4\}$ as well as $\mathop {\mathrm {tr}}\nolimits (\alpha _4\alpha _k)=0$ for $k \in \{1,\dots ,3\}$ and $\mathop {\mathrm {tr}}\nolimits (I_{\mathbb {C}^4})=4$, we see that also the matrix B is traceless. Consequently, using in addition the explicit expressions for $a_1$, $a_2$ and $a_3$, we obtain

$$\begin{aligned} \begin{aligned} \frac{1}{4}\mathop {\mathrm {tr}}\nolimits (A^*A)&= (|z|^2+m^2) |x-x'|^4 + \left| 1+\sqrt{m^2-z^2}\,|x-x'|\right| ^2 |x-x'|^2 \\&= \left( |z|^2+m^2+|m^2-z^2|\right) |x-x'|^4 + + 2 \mathfrak {R}\sqrt{m^2-z^2}\,|x-x'|^3 + |x-x'|^2 \,. \end{aligned} \end{aligned}$$

It remains to use the identity $|z|^2+m^2+|m^2-z^2| = 2 \left[ \big (\mathfrak {R}\sqrt{m^2-z^2}\big )^2+(\mathfrak {R}z)^2\right]$.

$\square$

The present paper extensively uses the following explicit bound on the Hilbert–Schmidt norm of the free resolvent.

Lemma 2

For almost every $x,x' \in \mathbb {R}^3$, one has

$$\begin{aligned} \big |(H_0-z)^{-1}(x,x')\big |_\mathrm {HS}^2 \le c_1^2 \, \frac{1}{|x-x'|^4} + c_2^2 \, (\mathfrak {R}z)^2 \, \frac{e^{-2\,\mathfrak {R}\sqrt{m^2-z^2}\,|x-x'|}}{|x-x'|^2} \,, \end{aligned}$$

where

$$\begin{aligned} c_1 := \frac{\sqrt{1+e^{-1}+2e^{-2}}}{2\pi } \,, \qquad c_2 := \frac{\sqrt{2}}{2\pi } \,. \end{aligned}$$

Proof

The estimate follows from Lemma 1 by using the elementary bounds $r e^{-r} \le e^{-1}$ and $r^2 e^{-r} \le 4 e^{-2}$ valid for every $r \ge 0$. $\square$

4 The Birman–Schwinger principle

For almost every $x \in \mathbb {R}^3$, let us introduce the matrix $W(x) := (V(x)^* V(x))^{1/2}$, the absolute value of V(x). Using the polar decomposition for matrices, we have

$$\begin{aligned} V(x) = U(x) W(x) = U(x) W(x)^{1/2} W(x)^{1/2} \,, \end{aligned}$$

where U(x) is a unitary matrix on $\mathbb {C}^4$. Notice that $|V(x)| = |W(x)| = |W(x)^{1/2}|^2$, where we recall that $|\cdot |$ stands for the operator norm of a matrix.

Given any $z\not \in \sigma (H_0)$, we introduce the Birman–Schwinger operator

$$\begin{aligned} K_z := \big [W^{1/2}G_0^{-1/2}\big ] \big [G_0(H_{0}-z)^{-1}\big ] \big [(UW^{1/2})^*G_0^{-1/2}\big ]^* \,, \end{aligned}$$

(10)

where $G_0$ is the shifted absolute value of $H_0$ introduced in (7). With an abuse of notation, we denote by the same symbols $W^{1/2}$ and U the maximal multiplication operators in $\mathcal {H}$ generated by the matrix-valued function $x \mapsto W(x)^{1/2}$ and $x \mapsto U(x)$, respectively. It follows from (6) that, for every $\psi \in \mathcal {H}$,

$$\begin{aligned} \Vert W^{1/2}\psi \Vert _\mathcal {H}^2 \le a \, \Vert G_0^{1/2}\psi \Vert _\mathcal {H}^2 \qquad \text{ and }\qquad \Vert (UW^{1/2})^*\psi \Vert _\mathcal {H}^2 \le a \, \Vert G_0^{1/2}\psi \Vert _\mathcal {H}^2 \,. \end{aligned}$$

Consequently, $\Vert W^{1/2}G_0^{-1/2}\Vert _{\mathcal {H}\rightarrow \mathcal {H}}^2 \le a$ and $\Vert (UW^{1/2})^*G_0^{-1/2}\Vert _{\mathcal {H}\rightarrow \mathcal {H}}^2 \le a$. Hence, $K_z$ is a well-defined bounded operator on $\mathcal {H}$ (as a composition of three bounded operators) and one has the rough bound

$$\begin{aligned} \Vert K_z\Vert _{\mathcal {H}\rightarrow \mathcal {H}} \le a \, \Vert G_0(H_{0}-z)^{-1}\Vert _{\mathcal {H}\rightarrow \mathcal {H}} \le a \, \sup _{\xi \in (-\infty ,-m)\cup (+m,\infty )} \frac{|\xi |+a^{-1}b+\delta }{|\xi -z|} \,. \end{aligned}$$

(11)

Using the commutativity of $H_0$ and $G_0$, the operator $K_z$ admits the familiar form (5) provided that V is bounded. But we insist working under the minimal regularity assumption (6) in this section.

By $\varphi \in L_0^2(\mathbb {R}^3;\mathbb {C}^4)$ in the lemma below we mean $\varphi \in L^2(\mathbb {R}^3;\mathbb {C}^4)$ and that $\mathop {\mathrm {supp}}\nolimits \varphi$ is compact.

Lemma 3

(Birman–Schwinger principle) Assume (6). Let $H_V\psi =\lambda \psi$ with some $\lambda \in \mathbb {C}$ and $\psi \in \mathsf {D}(H_V)$, and set $\phi := W^{1/2} \psi$. One has $\phi \in \mathcal {H}$ and $\phi \not =0$ if $\psi \not =0$.

• (i) If $\lambda \not \in \sigma (H_0)$, then $K_{\lambda } \phi = - \phi$.

• (ii) If $\lambda \in \sigma (H_0)$, then $\displaystyle \lim _{\varepsilon \rightarrow 0^\pm } (\varphi ,K_{\lambda + i\varepsilon } \phi )_\mathcal {H}= - (\varphi ,\phi )_\mathcal {H}$ for every $\varphi \in L_0^2(\mathbb {R}^3;\mathbb {C}^4)$.

Proof

By (6), $\phi \in \mathcal {H}$ if $\psi \in \mathsf {D}(H_V) \subset H^{1/2}(\mathbb {R}^3;\mathbb {C}^4)$. From (7) we deduce that $H_V\psi =H_0\psi$ provided that $\phi =0$. But then $\lambda$ is an eigenvalue of $H_0$ (which is impossible) unless $\psi =0$.

If $\lambda \not \in \sigma (H_0)$, then

$$\begin{aligned} \begin{aligned} (\varphi ,K_\lambda \phi )_\mathcal {H}&= \big ([W^{1/2}G_0^{-1/2}]^*\varphi , [G_0(H_{0}-\lambda )^{-1}] [(UW^{1/2})^*G_0^{-1/2}]^*\phi \big )_\mathcal {H}\\&= \big ( [UW^{1/2})^*G_0^{-1/2}] [G_0(H_{0}-\lambda )^{-1}]^* [W^{1/2}G_0^{-1/2}]^* \varphi , W^{1/2}\psi \big )_\mathcal {H}\\&= \int _{\mathbb {R}^3} \eta (x)^* V(x) \psi (x) \, \mathrm {d}x \end{aligned} \end{aligned}$$

for every $\varphi \in \mathcal {H}$ with $\eta := G_0^{-1/2}[G_0(H_{0}-\lambda )^{-1}]^* [W^{1/2}G_0^{-1/2}]^*\varphi \in H^{1/2}(\mathbb {R}^3;\mathbb {C}^4) \,.$ Using (7), it follows that

$$\begin{aligned} \begin{aligned} (\varphi ,K_\lambda \phi )_\mathcal {H}&= (\eta ,H_V\psi )_\mathcal {H}- \big (G_0^{1/2}\eta ,H_0 G_0^{-1} G_0^{1/2}\psi \big )_{\mathcal {H}} \\&= \lambda (\eta ,\psi )_\mathcal {H}- \big (G_0^{1/2}\eta ,H_0 G_0^{-1} G_0^{1/2}\psi \big )_{\mathcal {H}} \\&= \lambda \big (G_0^{1/2}\eta ,G_0^{-1}G_0^{1/2}\psi \big )_\mathcal {H}- \big (G_0^{1/2}\eta ,H_0 G_0^{-1} G_0^{1/2}\psi \big )_{\mathcal {H}} \\&= - \big (G_0^{1/2}\eta ,(H_0-\lambda ) G_0^{-1} G_0^{1/2}\psi \big )_{\mathcal {H}} \\&= - \big ([W^{1/2}G_0^{-1/2}]^*\varphi , G_0(H_{0}-\lambda )^{-1}(H_0-\lambda ) G_0^{-1} G_0^{1/2}\psi \big )_{\mathcal {H}} \\&= - \big ([W^{1/2}G_0^{-1/2}]^*\varphi , G_0^{1/2}\psi \big )_{\mathcal {H}} \\&= - \big (\varphi ,W^{1/2}\psi \big )_{\mathcal {H}} = - (\varphi ,\phi )_{\mathcal {H}} \end{aligned} \end{aligned}$$

for every $\varphi \in \mathcal {H}$. This proves that $K_\lambda \phi =-\phi$ and therefore (i).

If $\lambda \in \sigma (H_0)$, then there exists $\varepsilon _0 > 0$ such that $\lambda +\varepsilon \not \in \sigma (H_0)$ for all real $\varepsilon$ satisfying $0< |\varepsilon | < \varepsilon _0$. Now let us assume that $\varphi \in L_0^2(\mathbb {R}^3;\mathbb {C}^4)$. As above, we have

$$\begin{aligned} \begin{aligned} (\varphi ,K_{\lambda +i\varepsilon }\phi )_\mathcal {H}&= - \big ([W^{1/2}G_0^{-1/2}]^*\varphi , G_0(H_{0}-\lambda -i\varepsilon )^{-1}(H_0-\lambda ) G_0^{-1} G_0^{1/2}\psi \big )_{\mathcal {H}} \\&= - (\varphi ,\phi )_{\mathcal {H}} - i\varepsilon \, \big ([W^{1/2}G_0^{-1/2}]^*\varphi , G_0(H_{0}-\lambda -i\varepsilon )^{-1} G_0^{-1} G_0^{1/2}\psi \big )_{\mathcal {H}} \\&= - (\varphi ,\phi )_{\mathcal {H}} - i\varepsilon \, \big (\varphi , W^{1/2}(H_{0}-\lambda -i\varepsilon )^{-1}\psi \big )_{\mathcal {H}} \end{aligned} \end{aligned}$$

for every $\varphi \in \mathcal {H}$. Let us show that the last inner product vanishes as $\varepsilon \rightarrow 0$. Using Lemma 2, we have

$$\begin{aligned} \big |\big (\varphi , W^{1/2}(H_{0}-\lambda -i\varepsilon )^{-1}\psi \big )_{\mathcal {H}}\big | \le c_1 \, I_1 + c_2 \, |\lambda | \, I_2(\varepsilon ) \,, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} I_1&:= \int _{\mathbb {R}^3\times \mathbb {R}^3} \frac{|\varphi (x)|\,|W(x)|^{1/2}\,|\psi (x')|}{|x-x'|^2} \, \mathrm {d}x \, \mathrm {d}x' \,, \\ I_2(\varepsilon )&:= \int _{\mathbb {R}^3\times \mathbb {R}^3} |\varphi (x)|\,|W(x)|^{1/2} \, \frac{e^{-\,\mathfrak {R}\sqrt{m^2-(\lambda +i\varepsilon )^2}\,|x-x'|}}{|x-x'|} \, |\psi (x')| \, \mathrm {d}x \, \mathrm {d}x' \,. \end{aligned} \end{aligned}$$

The first integral is estimated as follows:

$$\begin{aligned} \begin{aligned} I_1&\le (2\pi )^{2/3} \Vert |\varphi ||W|^{1/2}\Vert _{L^1(\mathbb {R}^3)} \Vert |\psi |\Vert _{L^3(\mathbb {R}^3)} \\&\le (2\pi )^{2/3} \Vert |\varphi ||W|^{1/2}\Vert _{L^1(\mathbb {R}^3)} \, (2\pi ^2)^{-1/6} \big \Vert \root 4 \of {-\Delta }\,|\psi |\big \Vert _{L^2(\mathbb {R}^3)} \\&\le (2\pi )^{2/3} \Vert |\varphi |\Vert _{L^2(\mathbb {R}^3)} \, \Vert \chi _\Omega |V|\Vert _{L^1(\mathbb {R}^3)}^{1/2} \, (2\pi ^2)^{-1/6} \big \Vert \root 4 \of {-\Delta }\,|\psi |\big \Vert _{L^2(\mathbb {R}^3)} \end{aligned} \end{aligned}$$

with $\Omega := \mathop {\mathrm {supp}}\nolimits \varphi$. Here, the first bound is due to the Hardy–Littlewood–Sobolev inequality [13, Thm. 4.3], the second estimate is a Sobolev-type inequality [13, Thm. 8.4] quantifying the embedding ${\dot{H}}^{1/2}(\mathbb {R}^3) \hookrightarrow L^3(\mathbb {R}^3)$ and the last bound is the Schwarz inequality. Notice that the integrals on the last line are all finite, in particular because $\psi \in H^{1/2}(\mathbb {R}^3;\mathbb {C}^4)$ and $|V| \in L_\mathrm {loc}^1(\mathbb {R}^3)$. To estimate the second integral, we proceed as in [9, proof of Lem. 2]:

$$\begin{aligned} I_2(\varepsilon ) \le \Vert |\varphi | \Vert _{L^2(\mathbb {R}^3)} \Vert M_\varepsilon \Vert _{L^2(\mathbb {R}^3) \rightarrow L^2(\mathbb {R}^3)} \Vert |\psi | \Vert _{L^2(\mathbb {R}^3)} \,, \end{aligned}$$

where $M_\varepsilon$ is the integral operator with kernel

$$\begin{aligned} M_\varepsilon (x,x') := \chi _{\Omega }(x) \, |W(x)|^{1/2} \, \frac{e^{-\,\mathfrak {R}\sqrt{m^2-(\lambda +i\varepsilon )^2}\,|x-x'|}}{|x-x'|} \,. \end{aligned}$$

Using the Hilbert–Schmidt norm, we have

$$\begin{aligned} \begin{aligned} \Vert M_\varepsilon \Vert _{L^2(\mathbb {R}^3) \rightarrow L^2(\mathbb {R}^3)}^2 \le \Vert M_\varepsilon \Vert _\mathrm {HS}^2&= \int _{\Omega \times \mathbb {R}^3} |V(x)| \, \frac{e^{-\,2\mathfrak {R}\sqrt{m^2-(\lambda +i\varepsilon )^2}\,|x-x'|}}{|x-x'|^2} \, \mathrm {d}x \, \mathrm {d}x' \\&= \frac{2\pi }{\mathfrak {R}\sqrt{m^2-(\lambda +i\varepsilon )^2}} \int _\Omega |V(x)| \, \mathrm {d}x \,, \end{aligned} \end{aligned}$$

where the integral is again finite because of the local integrability of V. It remains to realise that

$$ \begin{aligned} \mathfrak {R}\sqrt{m^2-(\lambda +i\varepsilon )^2} \sim {\left\{ \begin{array}{ll} |\varepsilon |^{1/2} &{}\quad \text{ if } \,\, \lambda ^2=m^2 \ \& \ m\not =0 \,, \\ |\varepsilon | &{}\quad \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Hence, $\varepsilon I_2(\varepsilon ) = O(|\varepsilon |^{1/2})$ as $\varepsilon \rightarrow 0$. Putting the estimates together, we have proved

$$\begin{aligned} \begin{aligned}&\frac{|(\varphi ,K_{\lambda +i\varepsilon }\phi +\phi )_\mathcal {H}|}{\Vert \varphi \Vert _{\mathcal {H}}}\\&\quad \le \varepsilon \, \Vert \chi _\Omega |V|\Vert _{L^1(\mathbb {R}^3)}^{1/2} \, \Vert |\psi |\Vert _{H^{1/2}(\mathbb {R}^3)} \left( c_1 2^{1/2} \pi ^{1/3} + \frac{c_2 |\lambda | (2\pi )^{1/2}}{\big (\mathfrak {R}\sqrt{m^2-(\lambda +i\varepsilon )^2}\big )^{1/2}} \right) \xrightarrow [\varepsilon \rightarrow 0]{} 0 \end{aligned} \end{aligned}$$

for every given $\varphi \in L_0^2(\mathbb {R}^3;\mathbb {C}^4)$. This shows (ii) and concludes the proof of the lemma. $\square$

The preceding lemma is a precise statement of one side of the Birman–Schwinger principle under the minimal regularity assumption (6). It says that if $\lambda$ is an eigenvalue of $H_V$, then $-1$ is an eigenvalue of an integral equation related to $K_\lambda$. If $\lambda \not \in \sigma (H_0)$ the converse implication also holds, but it is not generally true if $\lambda \in \sigma (H_0)$, cf.  [16, Sec. III.2], and it is not needed for the purpose of this paper. In fact, we exclusively use the following corollary of Lemma 3.

Corollary 1

Assume (6) and let $\lambda \in \sigma _\mathrm {p}(H_V)$.

• (i) If $\lambda \not \in \sigma (H_0)$, then $\Vert K_{\lambda }\Vert \ge 1$.

• (ii) If $\lambda \in \sigma (H_0)$, then $\displaystyle \liminf _{\varepsilon \rightarrow 0^\pm } \Vert K_{\lambda +i\varepsilon }\Vert _{\mathcal {H}\rightarrow \mathcal {H}} \ge 1$.

Proof

Let $\lambda \in \sigma _\mathrm {p}(H_V)$, let $\psi$ be a corresponding eigenfunction and set $\phi = W^{1/2}\psi \not = 0$.

If $\lambda \not \in \sigma (H_0)$, then the statement (i) of Lemma 3 implies

$$\begin{aligned} \Vert \phi \Vert _\mathcal {H}^2 \, \Vert K_\lambda \Vert _{\mathcal {H}\rightarrow \mathcal {H}} \ge |(\phi ,K_\lambda \phi )_\mathcal {H}| = \Vert \phi \Vert _\mathcal {H}^2 \,, \end{aligned}$$

from which the claim (i) immediately follows.

If $\lambda \in \sigma (H_0)$, we set $\phi _n := \xi _n \phi$ for every positive n, where $\xi _n(x):=\xi (x/n)$ and $\xi \in C_0^\infty (\mathbb {R}^3)$ is a usual cut-off function satisfying $\xi (x)=1$ for $|x| \le 1$ and $\xi (x)=0$ for $|x| \ge 2$. As above, we write

$$\begin{aligned} \Vert \phi _n\Vert _\mathcal {H}\Vert \phi \Vert _\mathcal {H}\, \Vert K_{\lambda +i\varepsilon }\Vert _{\mathcal {H}\rightarrow \mathcal {H}} \ge |(\phi _n,K_{\lambda +i\varepsilon }\phi )_\mathcal {H}| \,, \end{aligned}$$

Taking the limit $\varepsilon \rightarrow 0^\pm$, the statement (ii) of Lemma 3 implies

$$\begin{aligned} \Vert \phi _n\Vert _\mathcal {H}\Vert \phi \Vert _\mathcal {H}\, \liminf _{\varepsilon \rightarrow 0^\pm } \Vert K_{\lambda +i\varepsilon }\Vert _{\mathcal {H}\rightarrow \mathcal {H}} \ge |(\phi _n,\phi )_\mathcal {H}| \,. \end{aligned}$$

The desired claim (ii) then follows by taking the limit $n\rightarrow \infty$. $\square$

To disprove the existence of eigenvalues in a complex region, it is thus enough to show that the norm of the Birman–Schwinger operator is strictly less than one there.

5 Proofs

To implement the above idea, we first strengthen our hypotheses about V and establish a more precise estimate on the norm of $K_z$ as compared to (11).

Lemma 4

Let $|V| \in L^3(\mathbb {R}^3)$. For every $z \not \in \sigma (H_0)$,

$$\begin{aligned} \Vert K_z\Vert _{\mathcal {H}\rightarrow \mathcal {H}} \le \left( \frac{\pi }{2}\right) ^{1/3} \sqrt{1+e^{-1}+e^{-2}} \, \sqrt{ 1 + \frac{(\mathfrak {R}z)^2}{\big (\mathfrak {R}\sqrt{m^2-z^2}\big )^2} } \ \Vert |V|\Vert _{L^3(\mathbb {R}^3)} \,. \end{aligned}$$

Proof

By the elementary bound $r^2 e^{-r} \le 4 e^{-2}$ valid for every $r \ge 0$, we get from Lemma 2 the estimate

$$\begin{aligned} \big |(H_0-z)^{-1}(x,x')\big |_\mathrm {HS}^2 \le \left( c_1^2 + {\tilde{c}}_2^2 \frac{(\mathfrak {R}z)^2}{\big (\mathfrak {R}\sqrt{m^2-z^2}\big )^2} \right) \frac{1}{|x-x'|^4} \end{aligned}$$

for almost every $x,x' \in \mathbb {R}^3$ with ${\tilde{c}}_2 := c_2 e^{-2} = \sqrt{2e^{-2}}/(2\pi )$. Consequently, for every $\phi ,\psi \in \mathcal {H}$,

$$\begin{aligned} \begin{aligned}&|(\phi ,K_z\psi )_{\mathcal {H}}| \\&\quad \le \sqrt{ c_1^2 + {\tilde{c}}_2^2 \frac{(\mathfrak {R}z)^2}{\big (\mathfrak {R}\sqrt{m^2-z^2}\big )^2} } \int _{\mathbb {R}^3 \times \mathbb {R}^3} \frac{|V(x)|^{1/2} \, |\phi (x)| \, |V(x')|^{1/2} \, |\psi (x')| }{|x-x'|^2} \, \mathrm {d}x \, \mathrm {d}x' \\&\quad \le \sqrt{ c_1^2 + {\tilde{c}}_2^2 \frac{(\mathfrak {R}z)^2}{\big (\mathfrak {R}\sqrt{m^2-z^2}\big )^2} } \ 2^{2/3}\pi ^{4/3} \, \Vert |V|\Vert _{L^3(\mathbb {R}^3)} \, \Vert \phi \Vert _\mathcal {H}\Vert \psi \Vert _\mathcal {H}\,, \end{aligned} \end{aligned}$$

where the second estimate is due to the Hardy–Littlewood–Sobolev inequality with optimal constants [13, Thm. 4.3] together with the Hölder inequality. We eventually use ${\tilde{c}}_2 \le c_1$, just to make the final result look more elegant. $\square$

Further strengthening hypothesis about V, we can also get a bound which is independent of the mass m and has a different behaviour in z.

Lemma 5

Let $|V| \in L^3(\mathbb {R}^3) \cap L^{3/2}(\mathbb {R}^3)$. For every $z \not \in \sigma (H_0)$,

$$\begin{aligned} \Vert K_z\Vert _{\mathcal {H}\rightarrow \mathcal {H}} \le \left( \frac{\pi }{2}\right) ^{1/3} \sqrt{1+e^{-1}+e^{-2}} \, \Vert |V|\Vert _{L^3(\mathbb {R}^3)} + \frac{2^{17/6}}{3\pi ^{2/3}} \, |\mathfrak {R}z| \, \Vert |V|\Vert _{L^{3/2}(\mathbb {R}^{3})} \,. \end{aligned}$$

Proof

For every $\phi ,\psi \in \mathcal {H}$, we have

$$\begin{aligned} \begin{aligned}&|(\phi ,K_z\psi )_{\mathcal {H}}| \\&\quad \le \int _{\mathbb {R}^3 \times \mathbb {R}^3} |V(x)|^{1/2} \, |\phi (x)| \left( \frac{c_1}{|x-x'|^2} + \frac{c_2 \, |\mathfrak {R}z|}{|x-x'|} \right) |V(x')|^{1/2} \, |\psi (x')| \, \mathrm {d}x \, \mathrm {d}x' \\&\quad \le \left( c_1 \, 2^{2/3}\pi ^{4/3} \, \Vert |V|\Vert _{L^3(\mathbb {R}^3)} + c_2 \, |\mathfrak {R}z| \, \frac{2^{10/3}}{3 \pi ^{1/3}} \, \Vert |V|\Vert _{L^{3/2}(\mathbb {R}^{3})} \right) \Vert \phi \Vert _\mathcal {H}\Vert \psi \Vert _\mathcal {H}\,. \end{aligned} \end{aligned}$$

Here, the first inequality follows by Lemma 2 (where the exponential is estimated by 1) and the second estimate is due to the Hardy–Littlewood–Sobolev inequality with optimal constants [13, Thm. 4.3] together with the Hölder inequality. $\square$

Theorem 1 (respectively, Theorem 2) follows as a consequence of Corollary 1 and Lemma 4 (respectively, Lemma 5).