ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR ...
ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR OLEG SAFRONOV
The free Dirac operator on R3 is introduced by the formula (1)
D0 =
3 X
iγj
1
∂ + γ0 , ∂xj
where γj are selfadjoint 4 × 4- matrices satisfying the relations (2)
γj γl + γl γj = 2δl,j .
Obviously, if the domain of D0 is defined as H1 (R3 ; C4 ), then D0 is a selfadjoint operator in L2 (R3 ; C4 ). The spectrum of D0 is absolutely continuous, it coincides with the complement of the interval (−1, 1) as a set: σ(D0 ) = σa.c. (D0 ) = (−∞, −1] ∪ [1, ∞). The multiplicity of the spectrum is infinite. We see that R \ σ(D0 ) = (−1, 1), which means that the interval (−1, 1) is the gap in the spectrum of the operator D0 . Let us perturb the operator D0 by an electric potential. One should note that different mathematicians introduce the electric potential in different ways. One of the reasonable ways to introduce it is to add a function V to D0 . That means one sets (3)
D = D0 + V.
Such operators were studied by Birman and Laptev in [2]. Another way of defining DV is to set (4)
DV = D0 + V γ0 .
Such perturbations were, for example, studied by S. Denisov in [6]. Moreover, the free Dirac operator was also introduced in [6] differently. The matrix γ0 in the definition of D0 was omitted. A physicist would say that the Dirac operator in [6] is related to the motion of a massless particle, while the operator (4) is related to the motion of a particle whose mass is equal to 1. Our first result is the following statement. Theorem 0.1. Let V be a real valued bounded function satisfying the so called Simon’s condition Z V2 (5) dx < ∞. 2 R3 |x| Let the operators DV and D−V be defined as in (4). Suppose that the spectra of DV and D−V in (−1, 1) consist of eigenvalues Ej (V ) and Ej (−V ), satisfying X X (6) (1 − Ej2 (V ))1/2 < ∞, (1 − Ej2 (−V ))1/2 < ∞. j
j
Then the absolutely continuous spectrum of each operator DV , D−V covers the set (−∞, −1] ∪ [1, ∞). A similar statement holds for the operator (3). Below, ET (Ω) denotes the spectral measure of a self-adjoint operator T = T ∗ corresponding to a Borel set Ω ⊂ R. 1
2
OLEG SAFRONOV
Theorem 0.2. Let V be a real valued bounded function satisfying (5). Let the operator D be defined ˜j (V ), satisfying as in (3). Suppose that the spectrum of D in (−1, 1) consists of eigenvalues E X ˜ 2 (V ))1/2 < ∞. (7) (1 − E j j
Then for any subset Λ ⊂ [1, ∞) of positive Lebesgue measure, ED ({ λ : |λ| ∈ Λ}) 6= 0.
(8)
Moreover, the absolutely continuous spectrum of DV defined by (4) with the same V covers the set (−∞, −1] ∪ [1, ∞). Another result is formulated for a sign-definite V ≤ 0. Theorem 0.3. Let V ≤ 0 be a bounded function satisfying the estimate ||V ||∞ < 2. Let the operator DV be defined as in (4). Suppose that the spectrum of DV in (−1, 1) consists of eigenvalues Ej (V ), satisfying X (1 − Ej2 (V ))p < ∞, with 0 < p ≤ 1/2. j
Then Z
|V (x)|p+1/2 |x|−2 dx < ∞.
R3
Moreover, the absolutely continuous spectrum of the operator DV covers the set (−∞, −1] ∪ [1, ∞). Besides the results formulated for the operators DV and D, we state the following theorems. In one of them, we deal with the massless Dirac operator. The other result gives interesting information about the spectra of two Schr¨ odinger operators. ˜ = D − γ0 , where the Theorem 0.4. Let V be a real valued bounded function satisfying (5). Let D operator D is defined by (3). Then for any subset Λ ⊂ [0, ∞) of positive Lebesgue measure, ED˜ ({ λ : |λ| ∈ Λ}) 6= 0.
(9)
Note that a similar statement for the operator DV − γ0 was proven by S. Denisov [6]. Theorem 0.5. [cf. [19]] Let V be a real valued bounded function on Rd . Suppose that the spectra of H+ = −∆ + V and H− = −∆ − V in (−∞, 0) consist of eigenvalues λj (V ) and λj (−V ), satisfying X X (10) |λj (V )|1/2 < ∞, |λj (−V )|1/2 < ∞. j
j
Then the absolutely continuous spectrum of each operator H+ , H− covers the set [0, ∞). The presence of Theorem 0.5 in this text is not a pure coincidence. As we will see in this paper, the spectral properties of Dirac operators are closely related to the properties of Schr¨odinger operators. Therefore, we would like to draw reader’s attention to the papers [6] - [11], [14], [15], [18]-[26] containing interesting results on the absolutely continuous spectrum of multi-dimensional Schr¨odinger operators. Detailed surveys of these results are given in [12] and [27]. 1. Dirac operator with a magnetic field Besides the operator describing a massless particle, there is another Dirac operator for which one can establish the presence of the absolutely continuous spectrum under conditions similar to (5). Let
ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR
3
A : R3 7→ R3 be a vector potential with components A1 , A2 and A3 . Suppose that A ∈ L∞ (R3 ; R3 ) and consider the operator 3 ∂ X + Aj (x) + γ0 . DA = γj ∂xj 1
Our main result about the operator DA is the following statement. Theorem 1.1. Let A be a bounded vector potential satisfying the condition Z |A|2 |x|−2 dx < ∞. (11) R3
Then the absolutely continuous spectrum of the operator DA covers the set (−∞, −1] ∪ [1, ∞). The proof of this result is almost the same as the proof of Theorem 0.1. The only difference is that, additionally, one needs to use the fact that DA does not have eigenvalues in the interval (−1, 1). The 2 ≥ 1. latter property of the spectrum follows from the obvious estimate: DA 2. Unitary equivalence The operator U1 of multiplication by the matrix γ0 γ1 γ2 γ3 is unitary in L2 (R3 ; C4 ). On the other hand, U1∗ DV U1 = −DV . As a consequence, we obtain that U1∗ EDV (δ)U1 = EDV (−δ),
−δ = {λ : −λ ∈ δ},
for any Borel subset δ ⊂ R. Put differently, the spectrum is symmetric with respect to reflection about the point λ = 0. Proposition 2.1. The spectrum of DV is essentially supported by (−∞, −1] ∪ [1, ∞) if and only if the spectrum of D2V is essentially supported by [1, ∞). Let us now understand what the operator D2V is. Due to (2), D2V = −∆ + Υ + 2V + V 2 + I,
(12)
where
Υ=
3 X
iγj γ0
j=1
∂V . ∂xj
Proposition 2.2. Let Ej (±V ) be the eigenvalues of D±V inside the interval (−1, 1). Then the negative eigenvalues λj (±V ) of the operator −∆ ± 2V + V 2 satisfy the estimate Xq Xq X X (13) |λj (V )| + |λj (−V )| ≤ (1 − Ej2 (V ))1/2 + (1 − Ej2 (−V ))1/2 . j
j
j
Proof. It is obvious, that the spectrum of D±V Therefore, according to the mini-max principle, (14)
j
2
in (−∞, 1) consists of eigenvalues Ej2 (±V ).
card {j : Ej2 (±V ) < λ} = max dim F, F
where the maximum is taken over all subspaces F ⊂ Dom (D±V ) such that (15)
||D±V u||2 < λ||u||2 ,
∀u ∈ F.
4
OLEG SAFRONOV
Let P be a spectral projection of the matrix γ0 in C4 . More specifically, let γ0 P = Pγ0 = P. Then, since Pγj P = 0 for j = 1, 2, 3, we obtain that 2 P D±V P = P(−∆ ± 2V + V 2 + I)P. Consequently, if we impose the additional condition on F that PF = F, then instead of the quantity card{j : Ej2 (±V ) < λ} in the left hand side of (14) we will have card{j : λj (±V ) < λ − 1}. On the other hand, the maximum in the right hand side of (14) will become smaller. Therefore, it follows that Ej2 (±V ) − 1 < λj (±V ), ∀j. It implies (13).
3. Trace formula
In this section, V is a 4 × 4– matrix valued potential on R3 . We will assume that V ∗ = V and we will discuss the properties of the operator H = −∆ + V. Assume in this section that the support of the matrix-valued potential V is contained in the spherical layer {x ∈ R3 : 1 < |x| < R}. Let r, θ be the spherical coordinates in R3 . That means that any point x ∈ R3 can be written as x = rθ, where r > 0, θ ∈ S. Let e0 ∈ C4 be the column-vector whose first component is 1 and all the other components are equal to 0: 1 0 (16) e0 = 0 . 0 We introduce the orthogonal projection P0 acting in L2 R3 ; C4 setting Z 1 u(r, θ), e0 e0 dθ P0 u = |S| S Clearly, P0 u depends only on r and is independent of θ. Moreover, at each x ∈ R3 , the value P0 u(x) is a vector of the form α(r) 0 P0 u(x) = 0 . 0 Denote the range of the projection P0 by H0 The space H = L2 R3 ; C4 can be decomposed uniquely into the sum H = H0 ⊕ H1 We introduce the operators V0 = P0 V P0 ,
V1 = (I − P0 )V (I − P0 ),
V1,0 = (I − P0 )V P0 , ∆0 = ∆P0 , and ∆1 = ∆(I − P0 ).
ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR
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Note that the portion of V0 in the space H0 is the operator of multiplication by the spherically symmetric function Z 1 V (r, θ)e0 , e0 dθ. (17) V0 (r) = |S| S In these notations, the operator H is representable as the following matrix ∗ −∆0 + V0 V1,0 H= . V1,0 −∆1 + V1 Proposition 3.1. Let Imz 6= 0 and let f ∈ H. Then (H − z)u = P0 f
(18) implies that (19)
−∆0 + Tz − z P0 u = P0 f,
where ∗ Tz = V0 − V1,0 (−∆1 + V1 − z)−1 V1,0 .
Conversely, if (I − P0 )u = −(−∆1 + V1 − z)−1 V1,0 P0 u, then (19) implies (18). Proof. The statement follows from the fact that ∗ −∆0 + V0 − z V1,0 H −z = . V1,0 −∆1 + V1 − z Indeed, if one denotes P0 u by u0 and (I − P0 )u by u1 , then one will be able to rewrite (18) in the matrix form ∗ −∆0 + V0 − z V1,0 u0 P0 f · = . u1 0 V1,0 −∆1 + V1 − z Note that we deal with a system of two equations, which can be easily handled by the method of elimination: solve the second equation for u1 and substitute the result into the first equation. Let us introduce now the operator U from L2 R+ to H by U u(r) =
u(r) e0 . r
Note that it maps isometrically L2 R+ onto H0 . Define also operators Qz and Lz acting in L2 R+ ) by setting (20)
∗ Qz = V0 − U ∗ V1,0 (−∆1 + V1 − z)−1 V1,0 U
and (21)
Lz u = −
d2 u + Qz u, dr2
u(0) = 0.
The symbol V0 in (20) denotes the operator of multiplication by the function (17). As a consequence of the previous statement we obtain Proposition 3.2. Let Lz be the operator (21) with Qz defined by (20). Then U (Lz − z)−1 U ∗ = P0 (H − z)−1 P0 .
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OLEG SAFRONOV 2
d Proof. It is enough to observe that U ∗ −∆0 )U coincides with the operator − dr 2.
The further analysis of the operator Lz is based on the study of the so called Jost solution. The latter is introduced as the solution of the equation d2 ψk + Qz ψk = zψk , dr2 for r > R. Here k 2 = z and Imk > 0. The Jost solution also satisfies −
satisfying the property ψk = eikr the ”integral” equation
ψk (r) = eikr − k −1
Z
∞
sin k(r − s) Qz ψk (s) ds. r
By the analytic Fredholm alternative, this equation is uniquely solvable for all k except a discrete sequence of points k. Due to our assumptions about the support of V , ψk (r) = a(k)eikr + b(k)e−ikr ,
for r < 1.
The coefficients a(k) and b(k) are meromorphic functions of the variable k in the upper half-plane. They can be analytically extended as meromorphic functions into a small neighborhood of any real point k 6= 0. Proposition 3.3. The meromorphic function a(k) has the following asymptotics Z ∞ −1 V0 (r) dr + o(1/k) a(k) = 1 − (2ik) 0
as k → ∞. For the proof of this statement we refer the reader to the Appendix of the paper [15]. The function M (k) =
ψk0 (0) , ψk (0)
Im k > 0,
is called Weyl’s M -function. Observe that (22)
ImM (k) =
W (0) , 2i|ψk (0)|2
where W (r) = ψ¯k (r)ψk0 (r) − ψk (r)ψ¯k0 (r). Note that the Wronskian W does not satisfy Abel’s formula because the operator Qz is not local. Instead of that, it satisfies a certain inequality. Indeed, W 0 (r) = (¯ z − z)ψk (r)ψk (r) + ψk (r) Qz ψk (r) − ψk0 (r) Qz ψk (r), which, in combination with the property Im Qz ≤ 0, implies that Z R (23) Im(W (R) − W (0)) = ImW 0 (r) dr ≤ 0, 0
On the other hand, W = i Im W is always purely imaginary and W (R) = 2iRe ke−2Im kR . Consequently, we obtain from (22) and (23) that (24)
Im M (k) ≥
Re k e−2Im kR |ψk (0)|2
if Im k ≥ 0 and Re k > 0. The relation (23) implies also that (25)
|a(k)|2 − |b(k)|2 ≥ 1
for Imz > 0.
ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR
7
for almost all real k ∈ R. Combining (24) with (25), we obtain that Im M (k) ≥
k k ≥ , 2 |a(k) + b(k)| 4|a(k)|2
if k > 0 and Im k = +0. Now, if we view Weyl’s function as a function of the variable z, we will realize that it is analytic in the upper half-plane, simply because singularities of ψk0 and ψk occur at the same places. Moreover, according to (24), it is a function with a positive imaginary part (which also implies that it does not have poles). Therefore, there exists a positive measure µ on R, such that Z ∞ dµ(t) s.
( ¯ sin(ks), ¯ |k|−2 sin(kr) M (k) − M (−k) Gz − Gz¯ = ¯ sin(kr), ¯ |k|−2 sin(ks) M (k) − M (−k)
if if
r < s, r > s.
Consequently,
The latter implies that lim Im((H − z)−1 f, f ) = |F (λ)|2 lim ImM (k)
z→λ+i0
z→λ+i0
for all λ > 0. It remains to note that lim ImM (k) = πµ0 (λ)
z→λ+i0
due to (26) and lim Im((H − z)−1 f, f ) = π
z→λ+i0
d (EH (−∞, λ)f, f ). dλ
Proposition 3.4. Let kj and βj be zeros and poles of a(k) in the upper half-plane. Then Z Z X X π log a(k) dk = V0 (r)dr + 2π Im kj − Im βj . 2 R R j
j
8
OLEG SAFRONOV
Proof. Due to Proposition 3.3, −1
Z
log a(k) = −(2ik)
as k → ∞.
V0 (r)dr + o(1/k), R
Define B(k) =
Y k − kj Y k − β¯j . k − βj k − k¯j j
j
Then the function log[a(k)/B(k)] is analyic in the upper half-plane. Moreover, it is easy to see that Z 2i X 2i X V0 (r)dr + log a(k)/B(k) = −(2ik)−1 Im kj − Im βj + o(1/k), as k → ∞. k k R j
j
Consequently, Z
π log a(k)/B(k) dk = 2 R
Z V0 (r)dr + 2π R
X
Im kj −
j
X
Im βj .
j
Now the statement of Proposition 3.4 follows from the fact that |B(k)| = 1 for all real k ∈ R.
Corollary 3.1. Let λj be the negative eigenvalues of the operator H = −∆ + V . Then Z Z Xq π 1/2 log a(k) dk ≤ V0 (r)dr + 2π |λj | + 2π||V ||∞ . (27) 2 R R j
Proof. Unfortunately, the fact that kj is a zero of a(k) does not imply that kj2 is a negative eigenvalue ˜ of the operator H. It implies that kj2 is an eigenvalue of a slighly different self-adjoint operator H. ˜ are smaller than eigenvalues of H, they still obey the estimate Even though the eigenvalues of H |kj+1 |2 ≤ |λj |.
An estimate for the logarithmic integral of |a(k)| can be turned into an estimate of the spectral measure. Indeed, for any bounded spherically symmetric function f : R3 7→ C4 supported in the unit ball {x ∈ R3 : |x| < 1} and having the property that f = P0 f , it holds √ d k|F (λ)|2 EH (−∞, λ)f, f ≥ , λ > 0, k = λ, dλ 4π|a(k)|2 where √ Z ∞ r sin( λr) √ (f (r)e0 , e0 )dr. F (λ) = λ 0 Combining the latter inequality with the statement of Corollary 3.1, we obtain the following very important statement: Theorem 3.1. Let V be a bounded 4 × 4−matrix-valued potential on R3 . Suppose that V (x) is selfadjoint at all points x ∈ R3 , and assume that all elements of the matrix V (x) are real. Assume also that the support of V is a compact set that does not intersect the unit ball {x ∈ R3 : |x| < 1}. Let EH be the spectral measure of the operator H = −∆ + V and let λj be the negative eigenvalues of H. Then for any bounded spherically symmetric function f : R3 7→ C4 supported in the unit ball {x ∈ R3 : |x| < 1} and having the property that f = P0 f , Z b Z d Xq π (V (x)e0 , e0 ) log EH (−∞, λ)f, f λ−1/2 dλ ≥ − dx − 2π |λj | dλ 2 R3 |x|d−1 a (28) j −2π||V ||1/2 ∞ − α(f ),
ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR
9
where b > a > 0, the vector e0 is defined in (16) and Z b √λ|F (λ)|2 log α(f ) = λ−1/2 dλ. 4π a We are going to apply this theorem to the operator (12) Theorem 3.2. Let V be a bounded real potential on R3 . Assume that the support of V is a compact set that does not intersect the unit ball {x ∈ R3 : |x| < 1}. Let Ej (V ) be the negative eigenvalues of DV situated in the interval (−1, 1). Then for any bounded spherically symmetric function f : R3 7→ C4 supported in the unit ball {x ∈ R3 : |x| < 1} and having the property that f = P0 f , Z Z b d Xq 2V (x) + V 2 (x) π dx − 2π log ED2 (−∞, λ)f, f λ−1/2 dλ ≥ − 1 − Ej2 (V ) d−1 V dλ 2 |x| 3 R a (29) j −2π − α(f ), where b > a > 0 and Z α(f ) =
b
log a
√λ|F (λ)|2 4π
λ−1/2 dλ.
1/2
One should mention that ||V ||∞ appearing in (28) is now replaced by 1 , because all eigenvalues of are positive (we only needed ||V ||∞ to estimate the bottom of the spectrum of a certain operator).
D2V
4. Estmate of the potential by the eigenvalue sum The following statement also plays a very important role in the proofs of the main results. Theorem 4.1. Let W ≥ 0 be a bounded positive function on Rd having the property Z W (x) dx < ∞. d−1 |x| d R Let V : Rd 7→ R be a bounded function and let λj (±V ) be the negative eigenvalues of H± = −∆±V +W . Suppose that X |λj (V )|1/2 + |λj (−V )|1/2 < ∞. j
Then V is representable in the form V = W0 + divA(x) + |A|2 where A : Rd 7→ Rd and W0 : Rd 7→ R satisfy the condition Z (|W0 | + |A|2 )|x|1−d dx < ∞. Rd
Combining Theorem 4.1 with Proposition 2.2, we obtain the following result. Corollary 4.1. Let V be a bounded potential on Rd having the property Z V 2 (x) dx < ∞. d−1 Rd |x| Let Ej (±V ) be the eigenvalues of D±V = D0 ± V γ0 situated in the interval (−1, 1). Suppose that X (1 − Ej2 (V ))1/2 + (1 − Ej2 (−V ))1/2 < ∞. j
10
OLEG SAFRONOV
Then V is representable in the form V = W0 + divA(x) + |A|2 where A : Rd 7→ Rd and W0 : Rd 7→ R satisfy the condition Z (|W0 | + |A|2 )|x|1−d dx < ∞. Rd
Proof of Theorem 4.1 The main idea of the proof is to find the regions where the eigenfunctions of H± mostly live. The additional Dirichlet condition on the boundary of a such region will not lift the lowest eigenvalue too much. As a matter of fact, the eigenvalue will cover less than half of the distance to zero. The next statement can be proved by integration by parts. Lemma 4.1. Let φ be a real valued bounded function with bounded derivatives of first order. Suppose that ψ is a real valued solution of −∆ψ + (W ± V )ψ = λψ and the product φψ vanishes on the boundary of the domain {a < |x| < b}. Then Z Z 2 2 |∇(φψ)| + (W ± V )|φψ| dx = |∇φ|2 ψ 2 + λ|φψ|2 dx a
The free Dirac operator on R3 is introduced by the formula (1)
D0 =
3 X
iγj
1
∂ + γ0 , ∂xj
where γj are selfadjoint 4 × 4- matrices satisfying the relations (2)
γj γl + γl γj = 2δl,j .
Obviously, if the domain of D0 is defined as H1 (R3 ; C4 ), then D0 is a selfadjoint operator in L2 (R3 ; C4 ). The spectrum of D0 is absolutely continuous, it coincides with the complement of the interval (−1, 1) as a set: σ(D0 ) = σa.c. (D0 ) = (−∞, −1] ∪ [1, ∞). The multiplicity of the spectrum is infinite. We see that R \ σ(D0 ) = (−1, 1), which means that the interval (−1, 1) is the gap in the spectrum of the operator D0 . Let us perturb the operator D0 by an electric potential. One should note that different mathematicians introduce the electric potential in different ways. One of the reasonable ways to introduce it is to add a function V to D0 . That means one sets (3)
D = D0 + V.
Such operators were studied by Birman and Laptev in [2]. Another way of defining DV is to set (4)
DV = D0 + V γ0 .
Such perturbations were, for example, studied by S. Denisov in [6]. Moreover, the free Dirac operator was also introduced in [6] differently. The matrix γ0 in the definition of D0 was omitted. A physicist would say that the Dirac operator in [6] is related to the motion of a massless particle, while the operator (4) is related to the motion of a particle whose mass is equal to 1. Our first result is the following statement. Theorem 0.1. Let V be a real valued bounded function satisfying the so called Simon’s condition Z V2 (5) dx < ∞. 2 R3 |x| Let the operators DV and D−V be defined as in (4). Suppose that the spectra of DV and D−V in (−1, 1) consist of eigenvalues Ej (V ) and Ej (−V ), satisfying X X (6) (1 − Ej2 (V ))1/2 < ∞, (1 − Ej2 (−V ))1/2 < ∞. j
j
Then the absolutely continuous spectrum of each operator DV , D−V covers the set (−∞, −1] ∪ [1, ∞). A similar statement holds for the operator (3). Below, ET (Ω) denotes the spectral measure of a self-adjoint operator T = T ∗ corresponding to a Borel set Ω ⊂ R. 1
2
OLEG SAFRONOV
Theorem 0.2. Let V be a real valued bounded function satisfying (5). Let the operator D be defined ˜j (V ), satisfying as in (3). Suppose that the spectrum of D in (−1, 1) consists of eigenvalues E X ˜ 2 (V ))1/2 < ∞. (7) (1 − E j j
Then for any subset Λ ⊂ [1, ∞) of positive Lebesgue measure, ED ({ λ : |λ| ∈ Λ}) 6= 0.
(8)
Moreover, the absolutely continuous spectrum of DV defined by (4) with the same V covers the set (−∞, −1] ∪ [1, ∞). Another result is formulated for a sign-definite V ≤ 0. Theorem 0.3. Let V ≤ 0 be a bounded function satisfying the estimate ||V ||∞ < 2. Let the operator DV be defined as in (4). Suppose that the spectrum of DV in (−1, 1) consists of eigenvalues Ej (V ), satisfying X (1 − Ej2 (V ))p < ∞, with 0 < p ≤ 1/2. j
Then Z
|V (x)|p+1/2 |x|−2 dx < ∞.
R3
Moreover, the absolutely continuous spectrum of the operator DV covers the set (−∞, −1] ∪ [1, ∞). Besides the results formulated for the operators DV and D, we state the following theorems. In one of them, we deal with the massless Dirac operator. The other result gives interesting information about the spectra of two Schr¨ odinger operators. ˜ = D − γ0 , where the Theorem 0.4. Let V be a real valued bounded function satisfying (5). Let D operator D is defined by (3). Then for any subset Λ ⊂ [0, ∞) of positive Lebesgue measure, ED˜ ({ λ : |λ| ∈ Λ}) 6= 0.
(9)
Note that a similar statement for the operator DV − γ0 was proven by S. Denisov [6]. Theorem 0.5. [cf. [19]] Let V be a real valued bounded function on Rd . Suppose that the spectra of H+ = −∆ + V and H− = −∆ − V in (−∞, 0) consist of eigenvalues λj (V ) and λj (−V ), satisfying X X (10) |λj (V )|1/2 < ∞, |λj (−V )|1/2 < ∞. j
j
Then the absolutely continuous spectrum of each operator H+ , H− covers the set [0, ∞). The presence of Theorem 0.5 in this text is not a pure coincidence. As we will see in this paper, the spectral properties of Dirac operators are closely related to the properties of Schr¨odinger operators. Therefore, we would like to draw reader’s attention to the papers [6] - [11], [14], [15], [18]-[26] containing interesting results on the absolutely continuous spectrum of multi-dimensional Schr¨odinger operators. Detailed surveys of these results are given in [12] and [27]. 1. Dirac operator with a magnetic field Besides the operator describing a massless particle, there is another Dirac operator for which one can establish the presence of the absolutely continuous spectrum under conditions similar to (5). Let
ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR
3
A : R3 7→ R3 be a vector potential with components A1 , A2 and A3 . Suppose that A ∈ L∞ (R3 ; R3 ) and consider the operator 3 ∂ X + Aj (x) + γ0 . DA = γj ∂xj 1
Our main result about the operator DA is the following statement. Theorem 1.1. Let A be a bounded vector potential satisfying the condition Z |A|2 |x|−2 dx < ∞. (11) R3
Then the absolutely continuous spectrum of the operator DA covers the set (−∞, −1] ∪ [1, ∞). The proof of this result is almost the same as the proof of Theorem 0.1. The only difference is that, additionally, one needs to use the fact that DA does not have eigenvalues in the interval (−1, 1). The 2 ≥ 1. latter property of the spectrum follows from the obvious estimate: DA 2. Unitary equivalence The operator U1 of multiplication by the matrix γ0 γ1 γ2 γ3 is unitary in L2 (R3 ; C4 ). On the other hand, U1∗ DV U1 = −DV . As a consequence, we obtain that U1∗ EDV (δ)U1 = EDV (−δ),
−δ = {λ : −λ ∈ δ},
for any Borel subset δ ⊂ R. Put differently, the spectrum is symmetric with respect to reflection about the point λ = 0. Proposition 2.1. The spectrum of DV is essentially supported by (−∞, −1] ∪ [1, ∞) if and only if the spectrum of D2V is essentially supported by [1, ∞). Let us now understand what the operator D2V is. Due to (2), D2V = −∆ + Υ + 2V + V 2 + I,
(12)
where
Υ=
3 X
iγj γ0
j=1
∂V . ∂xj
Proposition 2.2. Let Ej (±V ) be the eigenvalues of D±V inside the interval (−1, 1). Then the negative eigenvalues λj (±V ) of the operator −∆ ± 2V + V 2 satisfy the estimate Xq Xq X X (13) |λj (V )| + |λj (−V )| ≤ (1 − Ej2 (V ))1/2 + (1 − Ej2 (−V ))1/2 . j
j
j
Proof. It is obvious, that the spectrum of D±V Therefore, according to the mini-max principle, (14)
j
2
in (−∞, 1) consists of eigenvalues Ej2 (±V ).
card {j : Ej2 (±V ) < λ} = max dim F, F
where the maximum is taken over all subspaces F ⊂ Dom (D±V ) such that (15)
||D±V u||2 < λ||u||2 ,
∀u ∈ F.
4
OLEG SAFRONOV
Let P be a spectral projection of the matrix γ0 in C4 . More specifically, let γ0 P = Pγ0 = P. Then, since Pγj P = 0 for j = 1, 2, 3, we obtain that 2 P D±V P = P(−∆ ± 2V + V 2 + I)P. Consequently, if we impose the additional condition on F that PF = F, then instead of the quantity card{j : Ej2 (±V ) < λ} in the left hand side of (14) we will have card{j : λj (±V ) < λ − 1}. On the other hand, the maximum in the right hand side of (14) will become smaller. Therefore, it follows that Ej2 (±V ) − 1 < λj (±V ), ∀j. It implies (13).
3. Trace formula
In this section, V is a 4 × 4– matrix valued potential on R3 . We will assume that V ∗ = V and we will discuss the properties of the operator H = −∆ + V. Assume in this section that the support of the matrix-valued potential V is contained in the spherical layer {x ∈ R3 : 1 < |x| < R}. Let r, θ be the spherical coordinates in R3 . That means that any point x ∈ R3 can be written as x = rθ, where r > 0, θ ∈ S. Let e0 ∈ C4 be the column-vector whose first component is 1 and all the other components are equal to 0: 1 0 (16) e0 = 0 . 0 We introduce the orthogonal projection P0 acting in L2 R3 ; C4 setting Z 1 u(r, θ), e0 e0 dθ P0 u = |S| S Clearly, P0 u depends only on r and is independent of θ. Moreover, at each x ∈ R3 , the value P0 u(x) is a vector of the form α(r) 0 P0 u(x) = 0 . 0 Denote the range of the projection P0 by H0 The space H = L2 R3 ; C4 can be decomposed uniquely into the sum H = H0 ⊕ H1 We introduce the operators V0 = P0 V P0 ,
V1 = (I − P0 )V (I − P0 ),
V1,0 = (I − P0 )V P0 , ∆0 = ∆P0 , and ∆1 = ∆(I − P0 ).
ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR
5
Note that the portion of V0 in the space H0 is the operator of multiplication by the spherically symmetric function Z 1 V (r, θ)e0 , e0 dθ. (17) V0 (r) = |S| S In these notations, the operator H is representable as the following matrix ∗ −∆0 + V0 V1,0 H= . V1,0 −∆1 + V1 Proposition 3.1. Let Imz 6= 0 and let f ∈ H. Then (H − z)u = P0 f
(18) implies that (19)
−∆0 + Tz − z P0 u = P0 f,
where ∗ Tz = V0 − V1,0 (−∆1 + V1 − z)−1 V1,0 .
Conversely, if (I − P0 )u = −(−∆1 + V1 − z)−1 V1,0 P0 u, then (19) implies (18). Proof. The statement follows from the fact that ∗ −∆0 + V0 − z V1,0 H −z = . V1,0 −∆1 + V1 − z Indeed, if one denotes P0 u by u0 and (I − P0 )u by u1 , then one will be able to rewrite (18) in the matrix form ∗ −∆0 + V0 − z V1,0 u0 P0 f · = . u1 0 V1,0 −∆1 + V1 − z Note that we deal with a system of two equations, which can be easily handled by the method of elimination: solve the second equation for u1 and substitute the result into the first equation. Let us introduce now the operator U from L2 R+ to H by U u(r) =
u(r) e0 . r
Note that it maps isometrically L2 R+ onto H0 . Define also operators Qz and Lz acting in L2 R+ ) by setting (20)
∗ Qz = V0 − U ∗ V1,0 (−∆1 + V1 − z)−1 V1,0 U
and (21)
Lz u = −
d2 u + Qz u, dr2
u(0) = 0.
The symbol V0 in (20) denotes the operator of multiplication by the function (17). As a consequence of the previous statement we obtain Proposition 3.2. Let Lz be the operator (21) with Qz defined by (20). Then U (Lz − z)−1 U ∗ = P0 (H − z)−1 P0 .
6
OLEG SAFRONOV 2
d Proof. It is enough to observe that U ∗ −∆0 )U coincides with the operator − dr 2.
The further analysis of the operator Lz is based on the study of the so called Jost solution. The latter is introduced as the solution of the equation d2 ψk + Qz ψk = zψk , dr2 for r > R. Here k 2 = z and Imk > 0. The Jost solution also satisfies −
satisfying the property ψk = eikr the ”integral” equation
ψk (r) = eikr − k −1
Z
∞
sin k(r − s) Qz ψk (s) ds. r
By the analytic Fredholm alternative, this equation is uniquely solvable for all k except a discrete sequence of points k. Due to our assumptions about the support of V , ψk (r) = a(k)eikr + b(k)e−ikr ,
for r < 1.
The coefficients a(k) and b(k) are meromorphic functions of the variable k in the upper half-plane. They can be analytically extended as meromorphic functions into a small neighborhood of any real point k 6= 0. Proposition 3.3. The meromorphic function a(k) has the following asymptotics Z ∞ −1 V0 (r) dr + o(1/k) a(k) = 1 − (2ik) 0
as k → ∞. For the proof of this statement we refer the reader to the Appendix of the paper [15]. The function M (k) =
ψk0 (0) , ψk (0)
Im k > 0,
is called Weyl’s M -function. Observe that (22)
ImM (k) =
W (0) , 2i|ψk (0)|2
where W (r) = ψ¯k (r)ψk0 (r) − ψk (r)ψ¯k0 (r). Note that the Wronskian W does not satisfy Abel’s formula because the operator Qz is not local. Instead of that, it satisfies a certain inequality. Indeed, W 0 (r) = (¯ z − z)ψk (r)ψk (r) + ψk (r) Qz ψk (r) − ψk0 (r) Qz ψk (r), which, in combination with the property Im Qz ≤ 0, implies that Z R (23) Im(W (R) − W (0)) = ImW 0 (r) dr ≤ 0, 0
On the other hand, W = i Im W is always purely imaginary and W (R) = 2iRe ke−2Im kR . Consequently, we obtain from (22) and (23) that (24)
Im M (k) ≥
Re k e−2Im kR |ψk (0)|2
if Im k ≥ 0 and Re k > 0. The relation (23) implies also that (25)
|a(k)|2 − |b(k)|2 ≥ 1
for Imz > 0.
ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR
7
for almost all real k ∈ R. Combining (24) with (25), we obtain that Im M (k) ≥
k k ≥ , 2 |a(k) + b(k)| 4|a(k)|2
if k > 0 and Im k = +0. Now, if we view Weyl’s function as a function of the variable z, we will realize that it is analytic in the upper half-plane, simply because singularities of ψk0 and ψk occur at the same places. Moreover, according to (24), it is a function with a positive imaginary part (which also implies that it does not have poles). Therefore, there exists a positive measure µ on R, such that Z ∞ dµ(t) s.
( ¯ sin(ks), ¯ |k|−2 sin(kr) M (k) − M (−k) Gz − Gz¯ = ¯ sin(kr), ¯ |k|−2 sin(ks) M (k) − M (−k)
if if
r < s, r > s.
Consequently,
The latter implies that lim Im((H − z)−1 f, f ) = |F (λ)|2 lim ImM (k)
z→λ+i0
z→λ+i0
for all λ > 0. It remains to note that lim ImM (k) = πµ0 (λ)
z→λ+i0
due to (26) and lim Im((H − z)−1 f, f ) = π
z→λ+i0
d (EH (−∞, λ)f, f ). dλ
Proposition 3.4. Let kj and βj be zeros and poles of a(k) in the upper half-plane. Then Z Z X X π log a(k) dk = V0 (r)dr + 2π Im kj − Im βj . 2 R R j
j
8
OLEG SAFRONOV
Proof. Due to Proposition 3.3, −1
Z
log a(k) = −(2ik)
as k → ∞.
V0 (r)dr + o(1/k), R
Define B(k) =
Y k − kj Y k − β¯j . k − βj k − k¯j j
j
Then the function log[a(k)/B(k)] is analyic in the upper half-plane. Moreover, it is easy to see that Z 2i X 2i X V0 (r)dr + log a(k)/B(k) = −(2ik)−1 Im kj − Im βj + o(1/k), as k → ∞. k k R j
j
Consequently, Z
π log a(k)/B(k) dk = 2 R
Z V0 (r)dr + 2π R
X
Im kj −
j
X
Im βj .
j
Now the statement of Proposition 3.4 follows from the fact that |B(k)| = 1 for all real k ∈ R.
Corollary 3.1. Let λj be the negative eigenvalues of the operator H = −∆ + V . Then Z Z Xq π 1/2 log a(k) dk ≤ V0 (r)dr + 2π |λj | + 2π||V ||∞ . (27) 2 R R j
Proof. Unfortunately, the fact that kj is a zero of a(k) does not imply that kj2 is a negative eigenvalue ˜ of the operator H. It implies that kj2 is an eigenvalue of a slighly different self-adjoint operator H. ˜ are smaller than eigenvalues of H, they still obey the estimate Even though the eigenvalues of H |kj+1 |2 ≤ |λj |.
An estimate for the logarithmic integral of |a(k)| can be turned into an estimate of the spectral measure. Indeed, for any bounded spherically symmetric function f : R3 7→ C4 supported in the unit ball {x ∈ R3 : |x| < 1} and having the property that f = P0 f , it holds √ d k|F (λ)|2 EH (−∞, λ)f, f ≥ , λ > 0, k = λ, dλ 4π|a(k)|2 where √ Z ∞ r sin( λr) √ (f (r)e0 , e0 )dr. F (λ) = λ 0 Combining the latter inequality with the statement of Corollary 3.1, we obtain the following very important statement: Theorem 3.1. Let V be a bounded 4 × 4−matrix-valued potential on R3 . Suppose that V (x) is selfadjoint at all points x ∈ R3 , and assume that all elements of the matrix V (x) are real. Assume also that the support of V is a compact set that does not intersect the unit ball {x ∈ R3 : |x| < 1}. Let EH be the spectral measure of the operator H = −∆ + V and let λj be the negative eigenvalues of H. Then for any bounded spherically symmetric function f : R3 7→ C4 supported in the unit ball {x ∈ R3 : |x| < 1} and having the property that f = P0 f , Z b Z d Xq π (V (x)e0 , e0 ) log EH (−∞, λ)f, f λ−1/2 dλ ≥ − dx − 2π |λj | dλ 2 R3 |x|d−1 a (28) j −2π||V ||1/2 ∞ − α(f ),
ABSOLUTELY CONTINUOUS SPECTRUM OF A DIRAC OPERATOR
9
where b > a > 0, the vector e0 is defined in (16) and Z b √λ|F (λ)|2 log α(f ) = λ−1/2 dλ. 4π a We are going to apply this theorem to the operator (12) Theorem 3.2. Let V be a bounded real potential on R3 . Assume that the support of V is a compact set that does not intersect the unit ball {x ∈ R3 : |x| < 1}. Let Ej (V ) be the negative eigenvalues of DV situated in the interval (−1, 1). Then for any bounded spherically symmetric function f : R3 7→ C4 supported in the unit ball {x ∈ R3 : |x| < 1} and having the property that f = P0 f , Z Z b d Xq 2V (x) + V 2 (x) π dx − 2π log ED2 (−∞, λ)f, f λ−1/2 dλ ≥ − 1 − Ej2 (V ) d−1 V dλ 2 |x| 3 R a (29) j −2π − α(f ), where b > a > 0 and Z α(f ) =
b
log a
√λ|F (λ)|2 4π
λ−1/2 dλ.
1/2
One should mention that ||V ||∞ appearing in (28) is now replaced by 1 , because all eigenvalues of are positive (we only needed ||V ||∞ to estimate the bottom of the spectrum of a certain operator).
D2V
4. Estmate of the potential by the eigenvalue sum The following statement also plays a very important role in the proofs of the main results. Theorem 4.1. Let W ≥ 0 be a bounded positive function on Rd having the property Z W (x) dx < ∞. d−1 |x| d R Let V : Rd 7→ R be a bounded function and let λj (±V ) be the negative eigenvalues of H± = −∆±V +W . Suppose that X |λj (V )|1/2 + |λj (−V )|1/2 < ∞. j
Then V is representable in the form V = W0 + divA(x) + |A|2 where A : Rd 7→ Rd and W0 : Rd 7→ R satisfy the condition Z (|W0 | + |A|2 )|x|1−d dx < ∞. Rd
Combining Theorem 4.1 with Proposition 2.2, we obtain the following result. Corollary 4.1. Let V be a bounded potential on Rd having the property Z V 2 (x) dx < ∞. d−1 Rd |x| Let Ej (±V ) be the eigenvalues of D±V = D0 ± V γ0 situated in the interval (−1, 1). Suppose that X (1 − Ej2 (V ))1/2 + (1 − Ej2 (−V ))1/2 < ∞. j
10
OLEG SAFRONOV
Then V is representable in the form V = W0 + divA(x) + |A|2 where A : Rd 7→ Rd and W0 : Rd 7→ R satisfy the condition Z (|W0 | + |A|2 )|x|1−d dx < ∞. Rd
Proof of Theorem 4.1 The main idea of the proof is to find the regions where the eigenfunctions of H± mostly live. The additional Dirichlet condition on the boundary of a such region will not lift the lowest eigenvalue too much. As a matter of fact, the eigenvalue will cover less than half of the distance to zero. The next statement can be proved by integration by parts. Lemma 4.1. Let φ be a real valued bounded function with bounded derivatives of first order. Suppose that ψ is a real valued solution of −∆ψ + (W ± V )ψ = λψ and the product φψ vanishes on the boundary of the domain {a < |x| < b}. Then Z Z 2 2 |∇(φψ)| + (W ± V )|φψ| dx = |∇φ|2 ψ 2 + λ|φψ|2 dx a
