The Dirac Operator with Mass m 0 ≥ 0 : Non-Existence of Zero Modes and of Threshold Eigenvalues

A simple global condition on the potential is given which excludes zero modes of the massless Dirac operator. As far as local conditions at infinity are concerned, it is shown that at energy zero the Dirac equation without mass term has no non-trivial L-solutions at infinity for potentials which are either very slowly varying or decaying at most like r−s with s ∈ (0, 1). When a mass term is present, it is proved that at the thresholds there are again no such solutions when the potential decays at most like r−s with s ∈ (0, 2). In both situations the decay rate is optimal. 2010 Mathematics Subject Classification: Primary 35P15; Secondary 81Q10.


Introduction
In their 1986 study of the stability of matter in the relativistic setting of the Pauli operator J. Fröhlich, E.H. Lieb and M. Loss recognised that there was a restriction on the nuclear charge if and only if the three-dimensional Dirac operator with mass zero has a non-trivial kernel (see [LS], Chapters 8, 9 and the references there).An example of such a zero mode was first given by M. Loss and H.T. Yau [LY]; for many more examples see [LS], p.167.Later, the Loss-Yau example was used in a completely different setting, viz. to show the necessity of certain restrictions in analogues of Hardy and Sobolev inequalities [BEU].
An observation with remarkable technological consequences is that in certain situations of non-relativistic quantum mechanics the dynamics of wave packets in crystals can be modelled by the two-dimensional massless Dirac operator (see [FW] and the references there).When the potential is spherically symmetric, a detailed spectral analysis of the Dirac operator with mass zero was given in two and three dimensions by K.M. Schmidt [S].In particular he showed that a variety of potentials with compact support can give rise to zero modes.In Theorem 2.1 of the present paper we give a simple global condition on the potential which rules out zero modes of the massless Dirac operator in any dimension.Theorem 2.7 deals with a fairly large class of massless Dirac operators under conditions on the potential solely at infinity.It is shown, for example, that for energy zero there is no non-trivial solution of integrable square at infinity if the potential is very slowly varying or decaying like r −s with s ∈ (0, 1).This decay rate is in a certain sense the best possible one (see Appendix B).To rule out non-trivial L 2 -solutions at infinity for the threshold energies ±m 0 turns out to be more complicated.Here the asymptotic analysis of Appendix B suggests 1/r 2 -behaviour as the borderline case and Theorem 2.10 indeed permits potentials which tend to zero with a rate at most like r −s with s ∈ (0, 2).Theorem 2.10 relies on a transformation of the solutions which is intimately connected with the block-structure that can be given to the Dirac matrices.In connection with this theorem we should like to draw attention to the very interesting paper [BG] where global conditions are used.In broad outline the proof of our Theorems 2.7 and 2.10 follows from the virial technique which was developed by Vogelsang [V] for the Dirac operator and later extended in [KOY], but basic differences in the assumptions on the potential are required in the situations considered here.At the beginning of §4 the general strategy of proof is outlined and comparison with [KOY] made, but the present paper is self-contained.

Acknowledgements
(2.1)This is the example of Loss and Yau [LY] mentioned in the Introduction.(Note that u is in L 2 iff n ≥ 3.) For n = 3 the potential in (2.1) is in fact the first in a hierarchy of potentials with constants larger than 3, all giving rise to zero modes; see [SU].
In contrast, we have the following result.
Theorem 2.1.Let Q : R n → C N ×N be measurable with Remark 2.2.a) In case Q is Hermitian, Theorem 2.1 can be rephrased as follows: The self-adjoint realisation H of τ with does not have the eigenvalue zero.(For the existence of such a self-adjoint realisation see [Ar] and the references therein.)b) Suppose that V is a real-valued scalar potential with Then it needs but a small additional assumption to show by means of the virial theorem that the self-adjoint realisation H of α • p + V + m 0 β (m 0 = 0) with property (2.2) does not have the eigenvalues ±m 0 (see, e.g., [L], Theorem 2.4 in conjunction with [We], p.335).c) It seems to be difficult to compare our C with the size of the compactly supported potentials in [S] which produce zero modes for n ≥ 2. Now we replace the whole space with the exterior domain where R > 0 can be arbitrarily large.On E R we consider the Dirac equation (2.3) We assume for simplicity that the vector potential b is in C 1 (E R , R n ).Further conditions will not be imposed on b but on the magnetic field Note that B can be identified with the scalar function The following result and its remark are essentially contained in [KOY], Example 6.1 and final remark on p.40.
(II) W is a measurable and bounded matrix function (not necessarily Hermitian).Suppose V , W and the magnetic field B satisfy the following conditions: a) r 1/2 V = o(1) = r∂ r V uniformly with respect to directions; b) there exist numbers K ∈ (0, 1/2) and M > 0 such that ) is a real-valued (scalar) function, condition a) can be replaced by Remark 2.5.Using a unique continuation result, e.g., the simple one [HP] or the more sophisticated one in [DO], one can conclude that u = 0 on R n .
Remark 2.6.It follows immediately from Theorem 2.3 and Remark 2.5 that the potential in (2.1) cannot create a non-zero eigenvalue.
Theorem 2.3 will now be supplemented by Theorems 2.7 and 2.10.
Remark 2.8.a) To prove Theorem 2.7, it will be important to observe that condition (i) implies b) If q is a negative bounded function with property (ii) and which satisfies [r(λ − q)] ′ ≥ δ 0 (λ − q) for some δ 0 ∈ (0, 1), then Theorem 2.7 holds for λ ≥ 0. c) In case V decays at infinity, hypothesis (H.3) demands a corresponding stronger decay of B to prevent the existence of eigenvalues.(The contrasting situation that V and B become large at infinity is considered in [MS].) Examples.For simplicity we assume V = q and W = 0. Let q 0 be a positive number.Then the functions q = q 0 [2 + sin (log log r)], (2.4) q = q 0 (log r) −s (s > 0), (2.5) q = q 0 r −s (0 < s < 1), (2.6) have the required properties (i), (ii).In addition, a magnetic field with the decay property (H.3) is allowed.As far as (2.5) and (2.6) are concerned, Remarks 2.4-2.5 already rule out any eigenvalue λ = 0.In case there is no vector potential, it follows from [S], Corollary 1 that the self-adjoint operator associated with τ = α • p + q in L 2 (R n ) N has purely absolutely continuous spectrum outside [q 0 , 3q 0 ].
Remark 2.9.More realistic potentials than (2.4) will have the property In such situations, however, it may be possible to use Theorem 2.3 or Remark 2.4 to show that which does not obey condition (i).Assuming |Bx| = o(1) uniformly w.r.t.directions, it follows from Theorem 2.3 that In particular, α • D + q has no zero mode.
For the equation our result is as follows.
we can find a sequence of functions /2 u j .We write U j rather than U j (r•) for the norm in L 2 (S n−1 ) N and similarly for the scalar product.We use the decomposition in (A.3) of Appendix A and note that the symmetric operator S with b = 0 has a purely discrete spectrum with and the assertion follows in the limit j → ∞.
4 Preliminaries to the proof of Theorem 2.7 To explain the general strategy of the proof of Theorem 2.7, let u be a solution of (2.3).We multiply U := r (n−1)/2 u by functions e ϕ , ϕ = ϕ(| • |) real-valued, and where {t k } is a sequence tending to infinity as k → ∞.Then ξ := χe ϕ U satisfies where As in proofs of unique continuation results by means of Carleman inequalities, the idea is (as it was in the earlier papers ([V], [KOY]) to prove the existence of a constant C > 0 such that for large s and in the limit holds.(For the precise inequality see (5.3) below.)If, for example ϕ = (ℓ/2)r b is permitted for some b > 0, (4.2) implies Documenta Mathematica 20 (2015) 37-64 and in the limit ℓ → ∞ the desired conclusion U = 0 on E s+1 follows.
The present paper differs from [KOY] in three important respects.Firstly, the function q in Theorem 2.7 and 2.10 is allowed to tend to zero at infinity, while it was absolutely necessary to require ±q ≥ const.> 0 in [KOY].Secondly, in contrast to [KOY], Proposition 3.1, the virial relation (A.8) from which we set out here does not contain a term involving q ′ /q.Such a term arose in [KOY] as it was necessary to divide ξ by (±q) 1/2 in order to cope with the case that the potential (and possibly a variable mass) became large at infinity.Thirdly, we use a more refined cutoff function.Given t k := 2 k and s < t k , there exists a function χ ∈ C ∞ (0, ∞) with ] and all k ∈ N .Moreover, for ℓ = 2 and ℓ = 3.Estimates of the five terms in (4.4) will lead us to inequality (5.1) below, from which an inequality of type (4.3) will eventually emerge with the help of a bootstrap argument.We start with the left-hand side of (4.4) and write by means of (i) and (H.1) in Theorem 2.7.Since 0 < q/(q − λ) ≤ 1 and (q − λ) −1 ≤ O(r 1−δ0 ) at infinity in view of Remark 2.8, we have The four terms I 1 , I 2 , I 3 and T 1 on the right-hand side of (4.4) are estimated as follows.
With hypothesis (H.2) we see that it is The same is true of the third term in (4.8), since (4.9) and (4.10) hold and j = o(1) by Remark 2.8 a) and the first part of assumption (ii).This leaves us with which, by assumptions (ii) and Remark 2.8 a), is again o(1).Collecting terms, we finally obtain (4.6) from which (4.7) follows, employing the properties of our cutoff function and Remark 2.8 a).
Remark 4.4.Before proving Theorem 2.7 we prepare the following Proposition 5.1.Suppose f > 0, g ≥ 0 are functions on (0, ∞) with and f is continuous and non-decreasing.Let Proof.Assume to the contrary that there are numbers ε 0 > 0 and N ∈ N such that gives the desired contradiction.
Proof of Theorem 2.7 From (4.4) and (4.5) and Lemma 4.1-4.3we see where for all s > R and ℓ > 0. Let j ∈ N .We choose ϕ = (j/2) log log r in (5.1) and note , the first integral on the right-hand side of (5.1) can be majorised by The last integral on the right-hand side of (5.1) can be estimated by Thus there is a sub-sequence {t k ℓ } ∞ ℓ=1 on which (5.2) tends to zero in view of Proposition 5.1.This proves Moreover, for some c 0 > 0 the inequality (5.1) implies Since we can let L → ∞, this establishes the claim.

Documenta Mathematica 20 (2015) 37-64
We insert ϕ = (jb/2) log r into (5.1) and observe In view of Part a) there is a sub-sequence {t k ℓ } with Using Remark 2.8 a) again, we see that (5.1) implies For b ∈ (0, δ 0 ) we can again move the first term of the right-hand side to the left and let L → ∞.
c) In order to show that U vanishes a.e. on E R1 for some R 1 > R, we choose ϕ = (ℓ/2)r b where ℓ > 0 and b ∈ (0, δ 0 ).From we observe (ϕ ′ /r) + ϕ ′′ > 0, so that the last part of the first integral on the right-hand side of (5.1) can be discarded.On account of Part b) there is a sequence {t k ℓ } on which the last integral vanishes.Finally we note Now there is an R 1 > R with the property that the left-hand side of (5.3) can be estimated from below by for s > R 1 .The assertion therefore follows in the limit ℓ → ∞.
Lemma 6.2.Let ϕ ′ ≥ 0. Then with a constant which is independent of ϕ and t k .
Proof.Abbreviating φ := e ϕ ζ, we have On the interval [s − 1, s] the integral can be estimated by using (6.3).On [t k , t k+1 ] we majorise the integral by and note that it is permitted to use (A.9), (A.11) with χ = 1 (i.e., g = 0) and The integral (6.5) can therefore be estimated by const.
which concludes the proof.
Summing up, we have (We refer to the references in [AKS] for this theorem which goes back to Perron, Lettenmeyer and Hartman-Wintner.)Hence (B.2) has a solution which is in L 2 at infinity if rq(r) → 0 as r → ∞ and 2|k| > 1.Note that (B.4) has a fundamental system of solutions The constant matrix in (B.6) has the eigenvalues So, if k = 0, 1, and if r 2 q → 0 r → ∞, (B.6) has a fundamental system of solutions w ± with |w ± (r)| = r µ±+o(1) as r → ∞.