Dissociation of Homonuclear Relativistic Molecular Ions

Rafael Benguria

Pontificia Universidad Católica de Chile

Departamento de Física Casilla 306 Santiago 22

Chile

email: rbenguri@fis.puc.cl

Heinz Siedentop

Mathematik Universität Regensburg

93040 Regensburg

Germany

email: hkh@mathematik.uni-muenchen.de

Edgardo Stockmeyer

Pontificia Universidad Católica de Chile

Departamento de Física Casilla 306 Santiago 22

Chile

email: estockme@maxwell.fis.puc.cl

March 8, 2000

Abstract

We give lower bounds to the ‘size’ of diatomic homonuclear relativistic molecules which are modeled by the Herbst operator. We also show that – as in the non-relativistic case – the absence of sufficiently many electrons leads to the dissociation of the molecule.

To obtain these results we found new bounds for the localization error in the semi-relativistic approach.

1 Introduction

Experimentally atoms and molecules are known to have only finitely many electrons. E.g., experimentally doubly charged negative stable ions are not known. It is a mathematical challenge to show this physical fact. It has been conjectured since several years that the charge exceeding the electrically neutral molecule should be bounded by a universal constant times the number of nuclei involved. This is an unresolved question. However, many results in this direction have been obtained in the context of non-relativistic quantum mechanics, among them Lieb [5]. On the other hand it is physically obvious that – for molecules – the number of electrons cannot be too low either, since the Coulomb forces would drive the nuclei apart. A pioneering work in this direction has been done by Ruskai [10]. Later improvements are due to Solovej [11] and Alarcón et al [2]. In particular the last mentioned paper gives an upper bound on the nuclear charges of a homonuclear diatomic molecule. – The purpose of this paper is an extension of the result Alarcón et al to the case when the underlying dynamics is no longer non-relativistic but relativistic. For definiteness we now define the model to be treated, namely the (pseudo)-relativistic Herbst Hamiltonian H_{R} of N spinless electrons of mass m and charge −e in the field of two nuclei with the same nuclear charge Ze separated by a distance 2R:

\begin{aligned}H_{R} = & ∑_{i=1}^{N} (\sqrt{−ℏ^{2}c^{2}Δ_{i} + m^{2}c^{4}} − \frac{Zα}{|x_{i} − R|} − \frac{Zα}{|x_{i} + R|}) \\ & + ∑_{1≤i<j≤N} \frac{α}{|x_{i} − x_{j}|} + \frac{Z^{2}α}{2R}, \end{aligned}  (1)

which is selfadjointly realized in L^{2}(ℝ^{3}) and where R = (R, 0, 0) is the location of one nucleus and α = e^{2}/(ℏc) is the Sommerfeld fine structure constant having the physical value of about 1/137. The lowest occurring energy for a given distance 2R between the nuclei is

E(R) = inf{(ψ, H_{R}ψ) | ψ ∈ C_{0}^{∞}(ℝ^{3N}), ∥ψ∥ = 1}.  (2)

Note that we do disregard any symmetry of the underlying state space, i.e., we look at boltzonic electrons, i.e., particles for which there is no requirement on the symmetry of the states which have the same ground state energy as bosonic atoms. The ground state energy E is given by

E = inf{E(R) | R > 0}.  (3)

The binding energy of the molecule is defined as the ground state energy of H_{R} minus the energy of the split system.

E_{b}(N, R, Z) = inf{inf E(R)|R > 0} − E_{s}  (4)

where E_{s} is the lowest energy that results from separating the systems into two parts. The molecule dissociates (is instable), if E_{b} is nonnegative. Since this condition is independent of the mass (for m > 0), we use m = 1 in the following.

The structure of the paper is as follows: In Section 2 we give a minimal distance for the nuclei of a stable molecule. Technically this is expressed in Lemma 1.

In Section 3 we use Lemma 1 to find a lower bound on the nuclear charge beyond which one electron diatomic molecules dissociate.

Finally, in Section 4 we use all the results above to find a lower bound on Z/N in the general N electron case.

2 A Lower Bound on the Size of the Molecule

Lemma 1. Consider a diatomic homonuclear molecule. The nuclei of atomic number Z are located at R = (R, 0, 0) and −R. If E(R) as defined in (2) has a minimum at R_{0} > 0 and Zα < \frac{1}{4}, then

R_{0} ≥ \frac{Z^{2}α}{2N}[1 − [1 + \begin{matrix}(Zα) & 2 \\ [\frac{1}{4} + \sqrt{\frac{1}{16} − (Zα)^{2}}] & 2 \end{matrix} ]^{\frac{−1}{2}} ]^{−1}.  (5)

Note that we are working in units where m = ℏ = c = 1, i.e., α = e^{2}.

Proof. The Hamiltonian that describes the molecule is given by (1). The key ingredients in our proof are the Feynman-Hellmann formula, the virial theorem for the Herbst operator, the Rayleigh-Ritz principle and the estimate from below for the ground state of a hydrogenic atom by Martin and Roy [7]. For a shorthand notation let x ∈ ℝ^{3N} be the vector (x_{1}, x_{2}, …, x_{N}) and

V (x) = ∑_{i=1}^{N} (−\frac{Zα}{|x_{i} − Rn̂|} − \frac{Zα}{|x_{i} + Rn̂|}) + ∑_{1≤i<j≤N} \frac{α}{|x_{i} − x_{j}|}.  (6)

We denote by

x · ∇_{x}V = ∑_{i=1}^{N} x_{i} · ∇_{x_{i}}V.  (7)

It is simple to check the following identity:

\frac{1}{R}(x · ∇V + V) = ∑_{i=1}^{N}[\frac{Zα}{|x_{i} − Rn̂|^{3}} (x_{i} − Rn̂) · n̂ − \frac{Zα}{|x_{i} + Rn̂|^{3}} (x_{i} + Rn̂) · n̂]. (8)

Now, let ψ denote the ground state of H. Using the Feynman–Hellmann formula, we have

\frac{∂E(R)}{∂R} = −\frac{Z^{2}α}{2R^{2}}

+ (ψ, ∑_{i=1}^{N}[−\frac{Zα}{|x_{i} − Rn̂|^{3}} (x_{i} − Rn̂) · n̂ + \frac{Zα}{|x_{i} + Rn̂|^{3}} (x_{i} + Rn̂) · n̂]ψ) . (9)

¿From (8) and (9) we have

\frac{∂E(R)}{∂R} = −\frac{Z^{2}α}{2R^{2}} − \frac{1}{R} (ψ, (x · ∇V + V)ψ) .  (10)

According to Herbst [4], Theorem 2.4, the following Virial Theorem holds

(ψ, x · ∇V ψ) = ∑_{i=1}^{N} (ψ, \frac{p_{i} · p_{i}}{\sqrt{p_{i}^{2} + 1}}ψ) .  (11)

As quadratic forms,

\frac{p_{i} · p_{i}}{\sqrt{p_{i}^{2} + 1}} ≥ \sqrt{p_{i}^{2} + 1} − 1 ≡ K_{i}.  (12)

Combining (10), (11) and (12) we get

\frac{∂E(R)}{∂R} ≤ −\frac{Z^{2}α}{2R^{2}} − \frac{1}{R} ∑_{i=1}^{N}(ψ, K_{i}ψ) − \frac{1}{R}(ψ, V ψ).  (13)

However,

(ψ, V ψ) ≥ ∑_{i=1}^{N} (ψ, [−\frac{Zα}{|x_{i} − Rn̂|} − \frac{Zα}{|x_{i} + Rn̂|}]ψ) .  (14)

Using (13) and (14) we get

\frac{∂E(R)}{∂R} ≤ −\frac{Z^{2}α}{2R^{2}} − \frac{1}{2R} ∑_{i=1}^{N} (ψ, [K_{i} − \frac{2Zα}{|x_{i} − Rn̂|} + K_{i} − \frac{2Zα}{|x_{i} + Rn̂|}]ψ) . (15)

At this point we note that the lowest eigenvalues of the hydrogenic Herbst operator in each angular momentum channel are bounded from below by the corresponding one of the Klein-Gordon operator (Martin and Roy [7]). The latter are known explicitly.

Since K_{i} − \frac{2Zα}{|x_{i}−Rn̂|} is just the Herbst operator for a hydrogenic atom (Herbst [4]) of charge 2Z located at R (location does not matter any more), we have, using the Rayleigh-Ritz principle and following Martin and Roy that

−(ψ, K_{i} − \frac{2Zα}{|x_{i} − Rn̂|}ψ) ≤ 1 − \frac{1}{\sqrt{1 + \frac{(Zα)^{2}}{[ \frac{1}{4} +\sqrt{\frac{1}{16} −(Zα)^{2}}]^{2}}}} .  (16)

Finally, from (15) and (16)

\frac{∂E(R)}{∂R} ≤ −\frac{1}{2R^{2}} [Z^{2}α − 2RN(1 − \frac{1}{\sqrt{1 + \frac{(Zα)^{2}}{[ \frac{1}{4} +\sqrt{\frac{1}{16} −(Zα)^{2}}]^{2}}}} )].  (17)

Hence if,

R ≤ \frac{Z^{2}α}{2N}[1 − [1 + \begin{matrix}(Zα) & 2 \\ [\frac{1}{4} + \sqrt{\frac{1}{16} − (Zα)^{2}}] & 2 \end{matrix} ]^{\frac{−1}{2}} ]^{−1},  (18)

and Zα < \frac{1}{4}, we have \frac{∂E(R)}{∂R} ≤ 0.

Remark 1.

• The equation (5) is monotonically decreasing as a function of Z and has a minimum at Zα = \frac{1}{4}, thus RN ≥ 14.617 for all Zα < \frac{1}{4}.

• Observe that \frac{∂E_{b}(R)}{∂R} = \frac{∂E(R)}{∂R}, since the energy of each of the two individual atoms gotten from the splitting of the molecule does not depend on R.

• Notice also that we have treated the electrons as bosons when estimating (15) using (16).

• Because of the contraint Zα < \frac{1}{4}, the lemma is not so much an asymptotic estimate on the size of a molecule (as in [11] or in [1]) but a tool for use in subsequent estimates later.

3 The One-Electron Molecule

To demonstrate the strategy to obtain an upper bound minimal nuclear charge that prevents binding for the homonuclear diatomic molecule, we will begin with the simplest case, i.e., one electron whose kinetic energy is given by the naive quantization of the classical relativistic Hamiltonian.

Theorem 1. Consider a diatomic molecule with homonuclear nuclei with only one electron. Then instability occurs when

2.864 ≤ Z < \frac{1}{4α}.  (19)

Proof. In this situation the Hamiltonian is just

H_{R} = \sqrt{−Δ + 1} − 1 − \frac{Zα}{|x − R|} − \frac{Zα}{|x + R|} + \frac{Z^{2}α}{2R},  (20)

and the lowest occurring energy of the separated systems is given by

E_{s} = inf σ(\sqrt{−Δ + 1} − 1 − Zα/|x|).

To treat this operator we follow the general strategy of Solovej [11] (see also Alarcón et al [2]): Firstly we localize the nuclei in two half spaces and bound the electron-nuclei singularity. Secondly we use the bound on the minimal distance between the nuclei to estimate the localization error. Because of the non-local nature of the kinetic energy operator this requires some extra care.

Pick two localizing functions χ_{1}, χ_{2} : ℝ^{3} → [0, 1], χ_{1}(x) := cos(ψ(x_{1}/R)) and χ_{2}(x) := sin(ψ(x_{1}/R)) with

ψ(t) := \begin{cases}0 & t < −1 \\ \frac{π}{4} (1 + t) & t ∈ [−1, 1] \\ \frac{π}{2} & t > 1 \end{cases} .  (21)

Obviously these two functions form a Lipschitz continuous partition of unity. Note also that χ_{2}(t) = χ_{1}(−t) holds with this choice. Localizing with these functions yields

(ψ, H_{R}ψ) = (ψ, χ_{1}(\sqrt{−Δ + 1} − 1 − \frac{Zα}{| · +R|})χ_{1}ψ)
+(ψ, χ_{2}(\sqrt{−Δ + 1} − 1 − \frac{Zα}{| · −R|})χ_{2}ψ)  (22)

−(ψ, Lψ) + \frac{Z^{2}α}{2R} − Zα(ψ, (\frac{χ_{1}^{2}}{| · −R|} + \frac{χ_{2}^{2}}{| · +R|})ψ),

where L is the localization error and has the kernel

L(x, y) = \frac{K_{2}(|x − y|) sin^{2}((ψ(x_{1}) − ψ(y_{1}))/(2R))}{π^{2}|x − y|^{2}}  (23)

(Lieb and Yau [6]) where K_{2}(x) is a modified Bessel function (see [8]). Estimating the operators sandwiched between the χ_{j} in the first two lines of (22) from below by E_{s} and observing that χ_{1}^{2} + χ_{2}^{2} = 1 we get

(ψ, H_{R}ψ) − E_{s} ≥ −(ψ, Lψ) + \frac{Z^{2}α}{2R} − Zα(ψ, (\frac{χ_{1}^{2}}{| · −R|} + \frac{χ_{2}^{2}}{| · +R|})ψ).  (24)

First we use the fact that (see Appendix A for the proof)

Zα(ψ, (\frac{χ_{1}^{2}}{|x − R|} + \frac{χ_{2}^{2}}{|x + R|})ψ) ≤ \frac{Zα}{R},  (25)

yielding the following equation for instability

E_{b} ≥ −(ψ, Lψ) + \frac{Z^{2}α}{2R} − \frac{Zα}{R}.  (26)

Next we estimate the localization error. Using the Schwarz inequality we get that

(ψ, Lψ) ≤ max_{x_{1}∈ℝ} φ_{L}(x_{1})  (27)

where

φ_{L}(x_{1}) := ∫_{ℝ^{3}} L(x, y)dy.  (28)

For shorthand notation we define,

S(s, t) := 4 sin^{2}((ψ(s) − ψ(t))/2).  (29)

Now we use cylindrical coordinates, choosing the line joining the nuclei as the first coordinate axis. We have

φ_{L}(x_{1}) = \frac{1}{2π} ∫_{−∞}^{∞} dy_{1} \{∫_{0}^{∞} ρdρ\frac{K_{2}(\sqrt{(x_{1} − y_{1})^{2} + ρ^{2}})}{(x_{1} − y_{1})^{2} + ρ^{2}}\} S(\frac{x_{1}}{R} , \frac{y_{1}}{R}).  (30)

Using [3], Formula 6.596.3, we can compute the expression in { } and obtain

φ_{L}(x_{1}) = \frac{1}{2π} ∫_{−∞}^{∞} dy_{1} \frac{K_{1}(|x_{1} − y_{1}|)}{|x_{1} − y_{1}|}S(\frac{x_{1}}{R} , \frac{y_{1}}{R}).  (31)

Changing of variables, u ≡ x_{1}/R and v ≡ y_{1}/R we get,

φ_{L}(Ru) = \frac{1}{2π} ∫_{−∞}^{∞} dv \frac{K_{1}(R|u − v|)}{|u − v|}S(u, v).  (32)

Using the fact that the equation above has a maximum at u = 0 (for the proof see Appendix B) and the choice of the χ_{j} in (21) we obtain

(ψ, Lψ) ≤ \frac{2}{π} \{(1 − \frac{\sqrt{2}}{2}) ∫_{1}^{∞} \frac{K_{1}(Rv)}{v}dv + ∫_{0}^{1} \frac{K_{1}(Rv)}{v}(1 − cos(\frac{π}{4} v))dv\} . (33)

Estimating the integrals

∫_{1}^{∞} \frac{K_{1}(Rv)}{v}dv ≤ ∫_{1}^{∞} K_{1}(Rv)dv = \frac{1}{R}K_{0}(R),  (34)

and

I := ∫_{0}^{1} \frac{K_{1}(Rv)}{v}(1 − cos (\frac{π}{4} v))dv = ∫_{0}^{1} K_{1}(Rv) (\frac{π^{2}v}{32} cos(\frac{π}{4} w)) dv, (35)

where w ∈ (0, 1), and using the Taylor expansion for the cosine we get

I ≤ \frac{π^{2}}{32} ∫_{0}^{1} dvK_{1}(Rv)v ≤ \frac{π^{2}}{32} ∫_{0}^{∞} dvK_{1}(Rv)v ≤ \frac{π^{3}}{64R^{2}} .  (36)

So finally we have that,

(ψ, Lψ) ≤ \frac{1}{R} (\frac{2}{π} ) \{(1 − \frac{\sqrt{2}}{2})K_{0}(R) + \frac{π^{3}}{64R}\} := \frac{G(R)}{R}.  (37)

Note that G is a decreasing function implying that (ψ, Lψ) ≤ \frac{G(R_{0})}{R}, where R_{0} is the lower bound for R that we found in Lemma 1. To obtain a numerical value, we will suppose that Z ≤ 3. We then get that R_{0} ≥ 34.17 and we arrive to

(ψ, Lψ) ≤ \frac{0.009026}{R}.  (38)

Now using (38) to estimate (26) we hav

E_{b} ≥ −\frac{0.009026}{R} + \frac{Z^{2}α}{2R} − \frac{Zα}{R} ≥ 0,  (39)

where the last inequality holds only if Z ≥ 2.864.

Remark 2. Notice that estimate implies (24) that the molecule dissociates, i.e. E_{b} ≥ 0, when Z is big enough and R ≥ R_{0}. This is also true, if Zα < \frac{1}{4}, when R < R_{0} because \frac{∂E_{b}}{∂R} ≤ 0 by Lemma 1.

4 The N Electron Case

Theorem 2. For the N electron diatomic molecule we have dissociation (i.e. E_{b} ≥ 0) under either of the following conditions on Z and N:

a) Z ≥ N(1 + \sqrt{1 + \frac{a}{αN}}), with a = 0.7587.

b) \frac{Z}{N} ≥ min \{1 + \sqrt{1 + \frac{a}{αN}} , b\}, and Zα < \frac{1}{4} where a = 0.7587, as before, and b = 3.9.

Proof. Consider the two cluster decomposition β = (β_{1}, β_{2}) of {1, .., N}. The intercluster potential is given by,

I_{β} = ∑_{i∈β_{2}} \frac{−Zα}{|x_{i} + R|} + ∑_{i∈β_{1}} \frac{−Zα}{|x_{i} − R|} + ∑_{i∈β_{1}} ∑_{j∈β_{2}} \frac{1}{|x_{i} − x_{j}|} + \frac{Z^{2}α}{2R}.  (40)

We define the cluster Hamiltonian H_{β} = H − I_{β}. Let ψ be the ground state of H and let E_{β} = inf(σ(H_{β})) then the instability condition is given by,

E ≥ min_{β} E_{β} = E_{s}.  (41)

The partition of unity is defined as,

J_{β}(x) = ∏_{i∈β_{1}} χ_{1}(x_{i}) ∏_{j∈β_{2}} χ_{2}(x_{j}).  (42)

Now noting that ∑_{β} J_{β}(x)^{2} = 1 we insert this in the expectation value of the Hamiltonian,

(ψ, H_{R}ψ) = (ψ, (∑_{β} J_{β}H_{R}J_{β})ψ) − (ψ, (∑_{β} J_{β}[H_{R}, J_{β}])ψ),  (43)

and observe that

∑_{β} J_{β}[J_{β}, H_{R}] = ∑_{β} J_{β}[∑_{k=1}^{N} \sqrt{−Δ_{k} + 1}, J_{β}] = ∑_{k=1}^{N} ∑_{j=1}^{2} χ_{j}(x_{k})[\sqrt{−Δ_{k} + 1}, χ_{j}(x_{k})], (44)

then

(ψ, (∑_{β} J_{β}[H_{R}, J_{β}])ψ) = N(ψ, Lψ)  (45)

where the kernel of L was given in (23). Then we have

(ψ, Hψ) = (ψ, (∑_{β} J_{β}H_{β}J_{β})ψ) + ∑_{β}(ψ, (J_{β}^{2}I_{β})ψ) − N(ψ, Lψ) (46)
≥ ∑_{β} E_{β}(ψ, J_{β}^{2}ψ) + ∑_{β}(ψ, (J_{β}^{2}I_{β})ψ) − N(ψ, Lψ)  (47)
≥ E_{s} + ∑_{β}(ψ, (J_{β}^{2}I_{β})ψ) − N(ψ, Lψ).  (48)

Furthermore,

\begin{aligned}∑_{β}(ψ, (J_{β}^{2}I_{β})ψ) & ≥ −Zα ∑_{i=1}^{N}(ψ, (\frac{χ_{1}^{2}(x_{i})}{|x_{i} − R|} + \frac{χ_{2}^{2}(x_{i})}{|x_{i} + R|})ψ) + \frac{Z^{2}α}{2R} \\ & ≥ −\frac{ZαN}{R} + \frac{Z^{2}α}{2R}. \end{aligned}  (49)

Here we have dropped the inter-electronic potential and used (25). Then we obtain

E_{b} = (ψ, Hψ) − E_{s} ≥ −N(ψ, Lψ) − \frac{ZαN}{R} + \frac{Z^{2}α}{2R} ≥ 0,  (50)

where the last inequality can be written as,

\frac{Z̃^{2}α}{2} − Z̃α − \frac{R}{N} (ψ, Lψ) ≥ 0,  (51)

where Z̃ ≡ Z/N. At this point we will bound the localization error in two different ways yielding, through equation (51), the conditions a) and b) of the theorem. We start with the 1/R behavior. Observing that the Bessel function K_{1} obeys the estimate K_{1}(x) ≤ 1/x for positive x (see Appendix B) we get

∫_{1}^{∞} \frac{K_{1}(Rv)}{v}dv ≤ \frac{1}{R} ∫_{1}^{∞} \frac{1}{v^{2}} dv = \frac{1}{R},  (52)

and moreover

∫_{0}^{1} \frac{K_{1}(Rv)}{v}(1 − cos (\frac{π}{4} v))dv ≤ \frac{1}{R} ∫_{0}^{1} \frac{(1 − cos (\frac{π}{4} v))}{v^{2}}dv ≤ 0.303/R.  (53)

Thus R(ψ, Lψ) ≤ 0.3793, and using this to bound (51) we obtain that for instability we need

Z̃ ≥ 1 + \sqrt{1 + \frac{0.7587}{αN}},  (54)

which is condition a).

Finally for the 1/R^{2} behavior, we just use (37), that is

(ψ, Lψ) ≤ \frac{1}{R^{2}} (\frac{2}{π} ) \{(1 − \frac{\sqrt{2}}{2})RK_{0}(R) + \frac{π^{3}}{64} \} ≤ \frac{0.3954}{R^{2}},  (55)

since xK_{0}(x) ≤ 0.4665. Then for instability Z̃ ≥ 1 + \sqrt{1 + \frac{0.7908}{αRN}}. If we restrict ourselves to Zα < 1/4, we need only consider values of RN ≥ 14.617 (see Remarks 1 and 2 above). Therefore, we have dissociation if Z̃ ≥ 3.900. This together with (54) is condition b).

A

To prove the equation (25) we first introduce

F (t) := \frac{1}{R} \begin{cases}cos(ψ(t_{1})) & 2 \\ √ & \overline{(t_{1} − 1)^{2} + t_{2}^{2} + t_{3}^{2}} \end{cases} + \begin{matrix}sin(ψ(t_{1})) & 2 \\ √ & \overline{(t_{1} + 1)^{2} + t_{2}^{2} + t_{3}^{2}} \end{matrix} \} .  (56)

For t_{1} ≤ −1 we find that

F (t) = \frac{1}{R} \frac{1}{\sqrt{(t_{1} − 1)^{2} + t_{2}^{2} + t_{3}^{2}}} ≤ \frac{1}{|t_{1} − 1|} ≤ \frac{1}{R}.  (57)

Similarly we get F (t) ≤ 1/R for t_{1} ≥ 1. For t_{1} ∈ (−1, 1) we obtain

F (t) ≤ \frac{1}{R} \{\frac{cos(ψ(t_{1}))^{2}}{(1 − t_{1})} + \frac{sin(ψ(t_{1}))^{2}}{1 + t_{1}} \} ≤ \frac{1}{R}.  (58)

The last inequality holds because the function in the braces is a positive concave even function on (−1, 1) with maximum at t_{1} = 0 and value 1 at this point. So the equation (25) is proved by taking t := x/R.

B

To prove that the maximum of the bound for the localization error occurs at u = 0 we first observe that

\begin{matrix}φ_{L}(Ru) = & \frac{1}{2π} ∫_{−∞}^{∞} dv \frac{K_{1}(R|u − v|)}{|u − v|}S(u, v) \\ = & \frac{2}{π} ∫_{−∞}^{∞} dv \frac{K_{1}(R|u − v|)}{|u − v|}sin^{2}[(ψ(u) − ψ(v))/2], \end{matrix}  (59)

where ψ is defined in (21). We start with some simple facts.

Proposition 1. The following is true:

1. K_{1}(x) ≤ 1/x for positive x.

2. φ_{L}(u) = φ_{L}(−u) for all u ∈ ℝ.

Proof.

1. Set f(x) := xK_{1}(x). The proof is immediate from the observation that lim_{x→0} f(x) = 1 and that f′(x) < 0 for positive x.

2. Let g(u, v) := K_{1}(R|u − v|)S(u, v)/|u − v|. The function g fulfills g(u, v) = g(−u, −v). Changing the variable in the integral v := −s yields the desired result.

We will now turn to the main goal of this section, namely proving that the function φ_{L} has its maximum at zero. We will start with the massless case and extend the result to the general massive case. In the latter it is enough for us to assume R ≥ 1.

B.1 The Massless Case

Lemma 2. Υ has a unique maximum at u = 0 where

Υ(u) := ∫_{−∞}^{∞} dv \frac{sin^{2}[(ψ(u) − ψ(v))/2]}{(u − v)^{2}}

and ψ as defined in (21).

Proof. Let first suppose that u ∈ [0, 1), we start noting that

Υ(u) = ∫_{−∞}^{∞} dv \frac{sin^{2}[(ψ(u) − ψ(v + u))/2]}{v^{2}}.

Taking the derivative of Υ and separating the integral using (21) we get

Υ′(u) =\frac{π}{4} \{sin(ψ(u)) ∫_{1+u}^{∞} \frac{dv}{v^{2}} − cos(ψ(u)) ∫_{1−u}^{∞} \frac{dv}{v^{2}} \}
=\frac{π [(1 − u) sin(ψ(u)) − (1 + u) cos(ψ(u))]}{4(1 − u^{2})}.

(60)

It is clear that Υ′(0) = 0. Moreover, Υ′(u) ≤ 0 for u ∈ (0, 1) since the function between [ ] in the last equation is convex in (0, 1) and vanishes at u = 0 and u = 1. Hence the maximum of Υ on [0, 1) occurs at 0.

To extend the same result for any u we consider the extension of the second statement of Proposition 1 for the massless case and observe that for u ≥ 1

Υ(u) = \frac{1}{2} ∫_{−∞}^{−1} \frac{dv}{(u − v)^{2}} + ∫_{−1}^{1} dv \frac{sin(\frac{π}{8} (v − 1))^{2}}{(u − v)^{2}} ,

is a decreasing function of u.This together with the continuity of I proves the lemma.

B.2 The Massive Case

We will enunciate first some facts.

Proposition 2. Let

Υ_{R}(u) := ∫_{−∞}^{∞} dv \frac{K_{1}(R|u − v|)}{|u − v|} sin^{2}[(ψ(u) − ψ(v))/2],

and R ≥ 1. Then the following is true:

1. The derivative with respect to u fulfills, Υ′_{R}(u) ≤ Υ′_{R=1}(u) := Υ′_{1}(u) for all u ∈ [0, 1).

2. K_{1}(u) ≥ e^{−u}/u for all positive u.

3. K_{0}(u + 1) ≤ 5e^{−(u+1)}/4 for all non-negative u.

Proof.

1. Using the same procedure as in Lemma 2 we obtain that for u ∈ [0, 1)

Υ′_{R}(u) = \frac{π}{4} (sin(ψ(u)) ∫_{R(1+u)}^{∞} \frac{K_{1}(v)}{v}dv − cos(ψ(u)) ∫_{R(1−u)}^{∞} \frac{K_{1}(v)}{v}dv) , (61)

and then deriving with respect to R we see that the function Υ′_{R} is decreasing as R increase for R > 0 and u ∈ [0, 1).

2. Let f(u) := uK_{1}(u)e^{u}. We see that f(0) = 1 and the derivative f′(u) = ue^{u}(K_{1}(u) − K_{0}(u)) ≥ 0 for all u ∈ (0, ∞).

3. The same idea as in 2. shows that g(0) < 1 and g′(u) ≤ 0 on u ∈ [0, ∞) where g(u) := \frac{4}{5}K_{0}(1 + u)e^{(1+u)} .

Lemma 3. Assume Υ_{R} as in Proposition 2 and R ≥ 1 then Υ_{R} has a unique maximum at u = 0.

Proof. First we prove the claim for u ∈ [0, 1), we note that by (61) u = 0 is a critical point. Considering Proposition 2.1 it suffices to prove that Υ′_{1}(u) ≤ 0. Now, using Proposition 2.2 and (61) we get

Υ′_{1}(u) ≤ (sin(ψ(u)) − cos(ψ(u))) ∫_{1+u}^{∞} \frac{K_{1}(v)}{v}dv − cos(ψ(u)) ∫_{1−u}^{1+u} \frac{K_{1}(v)}{v}dv

≤ \frac{(sin(ψ(u)) − cos(ψ(u)))}{1 + u} ∫_{1+u}^{∞} K_{1}(v)dv − cos(ψ(u)) ∫_{1−u}^{1+u} \frac{e^{−v}}{v^{2}} dv

≤ \frac{(sin(ψ(u)) − cos(ψ(u)))}{1 + u}K_{0}(1 + u) − cos(ψ(u))e^{−(1+u)} (\frac{1}{1 − u} − \frac{1}{1 + u})

= \frac{K_{0}(1 + u) [sin(ψ(u)) − u sin(ψ(u)) − cos(ψ(u)) + u cos(ψ(u))]}{1 − u^{2}}

− \frac{2ue^{−(1+u)} cos(ψ(u))}{1 − u^{2}}.

Denote the expression in the numerator of the last equation by h(u). It is enough to prove that h(u) ≤ 0 for u ∈ [0, 1). Using the third statement of Proposition 2 we arrive at

h(u) ≤ e^{−(u+1)} [\frac{5}{4} sin(ψ(u)) − \frac{5u}{4} sin(ψ(u)) − \frac{5}{4} cos(ψ(u)) − \frac{3u}{4} cos(ψ(u))] ≤ 0,

the last inequality holds for all u ∈ [0, 1), and the lemma is proved for this domain. To extend our proof for all real u we use the second statement of Proposition 1 and that for u ≥ 1

Υ_{R}(u) = sin(\frac{π}{4} )^{2} ∫_{−∞}^{−1} \frac{K_{1}(R(u − v))}{u − v}dv + ∫_{−1}^{1} dv \frac{K_{1}(R(u − v))}{u − v} sin^{2} \frac{π(v − 1)}{8}.

Here clearly Υ_{R}(u) ≤ Υ_{1}(u). Noting that the derivative of K_{1}(u − v)/(u − v) with respect to u is −[2K_{1}(u − v) + (u − v)K_{0}(u − v)]/(u − v)^{2} non-positive for u ≥ 1 and that v < 1, we conclude that in this domain Υ_{R} has its maximum at 1. These facts and the continuity of Υ_{R} prove the lemma.

Acknovledgement: R. B. acknowledges partial support of the project by FONDECYT (Chile), project 199-0427; E. S. thanks CONICYT (Chile) for support through a doctoral fellowship. All authors acknowledge support by the Volkswagen Stiftung for a cooperation grant. We also are thankful for the hospitality by our partner institutions.

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