Density Functional Theory for two-dimensional homogeneous materials
DDENSITY FUNCTIONAL THEORY FOR TWO-DIMENSIONALHOMOGENEOUS MATERIALS
DAVID GONTIER, SALMA LAHBABI, AND ABDALLAH MAICHINE
Abstract.
We study Density Functional Theory models for systems which are transitionallyinvariant in some directions, such as a homogeneous 2-d slab in the 3-d space. We show howthe different terms of the energy are modified and we derive reduced equations in the remainingdirections. In the Thomas-Fermi model, we prove that there is perfect screening, and provideprecise decay estimates for the electronic density away from the slab. In Kohn-Sham models, weprove that the Pauli principle is replaced by a penalization term in the energy. In the reducedHartree-Fock model in particular, we prove that the resulting model is well-posed, and providesome properties for the minimizer. © Contents
1. Introduction 1Acknowledgments 32. Main results 32.1. Main results in the Thomas Fermi model 32.2. Main results for the rHF model 53. Homogeneous 2-d materials in the Thomas-Fermi model 73.1. Hartree interaction 73.2. Reduction of the Thomas Fermi model: Proof of Proposition 2.1 103.3. Existence of minimizers: Proof of Theorem 2.2 103.4. Properties of the TF density and mean-field potential 124. Homogeneous 2-d materials in the reduced Hartree-Fock model 134.1. Trace and kinetic energy per unit-surface 134.2. Reduced states: Proof of Theorem 2.6 144.3. Existence of minimizers for the reduced model: Proof of Theorem 2.8 154.4. Properties of the mean-field potential: Proof of Proposition 2.9 175. Numerical illustrations 185.1. Numerical setting and self-consistant procedure 185.2. Numerical results 185.3. Perspectives 20Appendix A. A Lieb-Thirring inequality 21References 221.
Introduction
Density Functional Theory (DFT) was first introduced [10, 11] to study the quantum energy offinite systems, such as molecules. It became an important tool also in condensed matter physics,to study infinite systems, such as crystals. Using thermodynamic limit procedure [6], it was shownthat the energy per unit cell of a crystal could be computed as the minimization of an explicit
Date : February 26, 2021. a r X i v : . [ m a t h - ph ] F e b DAVID GONTIER, SALMA LAHBABI, AND ABDALLAH MAICHINE (periodic) functional. This was first proved for the Thomas-Fermi model in [4], and for the reducedHartree-Fock model in [5].In this work, we consider the intermediate case, and study semi-infinite systems, where thesystem is infinite in s directions, but is localized in d other directions. One can think e.g. ofa nano-wire in three-dimensional space (corresponding to s = 1 and d = 2), or an infinite slab(corresponding to s = 2 and d = 1). Our interest comes from the recent developments of two-dimensional materials, such as graphene and phosphorene in the physics community [9]. Suchsystems, have been studied in [2, 3]. Our objective is to prove and study DFT models for reduceddimension systems in order to allow low computational cost.In this work, we focus on the simple case where the system is homogeneous , in the sense thatit commutes with all translations in the s variables, and we derive reduced equations in theremaining d variables. More specifically, we consider a positive charge distribution µ ≥ R s + d ,representing the semi-infinite homogeneous material, which is translation invariant in the first s dimensions: µ ( x , x , · · · , x s , x s +1 , · · · x s + d ) = µ (0 , · · · , , x s +1 , · · · , x s + d ) . (1)We are interested in the state and the (normalized) energy of the electrons in the potential gen-erated by this charge distribution.While this hypothesis may look too simple and unrealistic, the model highlights some importantfeatures for semi-infinite systems. It may not reproduce the correct physical properties of a reallife 2-dimensional material, since it does not take into account the microscopic details; however,we think that the effect of the microscopic details fade away from the slab exponentially fast, sothat our model reproduces the correct behavior of the density far from the slab. This will be theobject of future works.Let us explain briefly our main results. We focus on the Thomas-Fermi model and the reducedHartree-Fock model for simplicity, although the derivation of the reduced equations works forgeneral Kohn-Sham models. Also, in what follows, we consider a two-dimensional slab ( s = 2) inthree dimensions ( d = 1, so that s + d = 3), but the techniques apply similarly to other cases.In the Thomas-Fermi model, the energy only depends on the electronic density ρ , which weassume share the same invariance as µ in (1). Starting from the full three-dimensional periodicThomas-Fermi energy, we can define an energy per unit surface , which takes the form E TF ( ρ ) := c TF ˆ R ρ / + 12 D ( ρ − µ ) . We minimize this energy among positive densities ρ : R → R + satisfying ´ R ρ = ´ R µ . Note thatthe advantage of the new problem is that the dimension has been reduced: the problem is seton the real line R only (instead of the full space R ). Here, D is the one-dimensional Hartreeenergy, that we define below. We study this reduced model, and prove various properties such asthe existence and uniqueness of a minimizer and the perfect screening of dipolar moments. Wealso prove Sommerfeld estimates, which states that ρ ( x ) decays as | x | − away from the slab. Thisdecay implies that when studying numerically such slab systems in a box with Dirichlet boundaryconditions at distance L from the slab, one should expect a numerical error of order L − .In the reduced Hartree-Fock model, we prove similarly that one can reduce the dimension ofthe problem. Starting from the three-dimensional problem set on one-body density matrices γ satisfying the Pauli principle 0 ≤ γ ≤
1, we obtain a one-dimensional problem set on self-adjointoperators G acting on L ( R ), which are postive: G ≥
0, but which no longer need to satisfy thePauli principle G ≤
1. Instead, the Pauli principle appears as a penalization term in the energy:the three-dimensional kinetic energy Tr ( − ∆ γ ) is replaced by a one-dimensional kinetic energyof the form 12 Tr ( − ∆ G ) + π Tr ( G ) . This last term, sometime called the Tsallis entropy, prevents G from having large eigenvalues. Itis somehow a weak-form of the Pauli principle. The reduced energy then takes the form E rHF ( G ) := 12 Tr ( − ∆ G ) + π Tr ( G ) + D ( ρ G − µ ) . FT FOR 2-D HOMOGENEOUS MATERIALS 3
In this paper, we justify the two models, and study the existence, uniqueness and properties ofthe minimizers (denoted by ρ TF and G ∗ respectively). We also provide numerical simulations forthe two models. To our surprise, we found out that, with high accuracy, we have ρ TF ≈ ρ G ∗ . Inparticular, the Thomas-Fermi model is a very good approximation of the reduced Hartree-Fockmodel after this reduction of the dimensionality.The paper is structured as follows. We state our main results in Section 2. We prove the resultsconcerning the Thomas-Fermi model in Section 3, and the ones for the reduced Hartree-Fockmodel in Section 4. We explain in particular the regularization of the one-dimensional Hartreeterm in Section 3.1. Our numerical illustrations are gathered in Section 5. Finally, we provide inAppendix A a Lieb-Thirring type inequality which is related to our method for the reduction ofthe kinetic energy in the rHF model. Acknowledgments.
This work has received fundings from a CNRS international cooperationprogram (
Projet International de Collaboration Scientifique , or PICS, of D.G. and S.L.). Theresearch leading to these results has received funding from OCP grant AS70 “Towards phosphorenebased materials and devices”. 2.
Main results
Let us explain in more details our results.2.1.
Main results in the Thomas Fermi model.
In orbital-free models, such as the Thomas-Fermi (TF) model [20, 7, 21], the energy depends solely on the electronic density ρ . We referto [13] for a mathematical study of this model in the molecular case.The nuclear density µ is a positive function which is R -translation invariant, i.e. it is of theform µ ( x , x , x ) = µ (0 , , x ) = µ ( x ) ∈ L ( R ) . We denote by Z := ´ R µ > L := [ − L , L ] × R (with periodic boundary conditions), and we considerneutral systems: ρ ∈ L (Γ L ) , ˆ Γ L ρ = L Z. The supercell TF energy of such a density is E TF L ( ρ ) = c TF ˆ Γ L ρ / + 12 D ,L ( ρ − µ ) . (2)The first term is the Thomas-Fermi kinetic energy, with c TF = (3 π ) / the usual three-dimensional Thomas-Fermi constant, and the second term is the supercell three-dimensionalHartree term, describing the interaction of the electrons with the charge density µ , as well asa mean-field self-interaction of the electrons. The quadratic form D ,L (the subscript 3 refers tothe space dimension s + d = 3) is formally defined by D ,L ( f ) := ¨ (Γ L ) G L ( x − y ) f ( x ) f ( y )d x d y , where G L is the L -periodic Green’s function solution to − ∆ G L = 4 π (cid:88) ( R ,R ) ∈ L Z δ ( R ,R , , (3)with the periodic boundary conditions G L ( x + R , x + R , x ) = G L ( x , x , x ) for all ( R , R ) ∈ L Z and the symmetry condition G L ( x , x , − x ) = G L ( x , x , x ). As we prove in Proposi-tion 2.1, this model is related to the one-dimensional TF energy E TF ( ρ ) = c TF ˆ R ρ / + 12 D ( ρ − µ ) . (4) DAVID GONTIER, SALMA LAHBABI, AND ABDALLAH MAICHINE
This energy is interpreted as an energy per unit surface . It has the 3-dimensional Thomas-Fermipower 5 / D ( f ) := − π ¨ R × R | x − y | f ( x ) f ( y )d x d y. (5)This definition of D is only valid for functions that decay fast enough. We give a regularizationof this expression suitable for neutral functions in Section 3.1.2.The energy (2) is set on a three-dimensional space Γ L , while the energy (4) is set on the line R .The latter is easier to study both theoretically and numerically. Our first proposition shows thatthe minimization problems concerning the two energies are equivalent. Proposition 2.1.
Consider the minimization problems I TF L := inf (cid:26) E TF L ( ρ ) , ρ ∈ L (Γ L ) ∩ L / (Γ L ) , ρ ≥ , D ,L ( ρ − µ ) < ∞ , ˆ Γ L ρ = L Z (cid:27) and I TF := inf (cid:26) E TF ( ρ ) , ρ ∈ L ( R ) ∩ L / ( R ) , ρ ≥ , D ( ρ − µ ) < ∞ , ˆ R ρ = Z (cid:27) . Then, for all
L > , we have I TF L = L I TF . The proof can be read in Section 3.2. This justifies the reduced one-dimensional problem (4).We now focus on this reduced model (4), and prove that it is well-posed.
Theorem 2.2.
The minimization problem I TF is well-posed, and has a unique minimizer ρ TF .It is the unique solution to the Thomas-Fermi equation (cid:40) c TF ρ / = ( λ − Φ TF ) + , − Φ (cid:48)(cid:48) TF = 4 π ( ρ TF − µ ) , Φ (cid:48) TF ( ±∞ ) = 0 and Φ TF (0) = 0 , (6) where x + := max { , x } , and where λ ∈ R is chosen so that ´ R ρ TF = Z . The proof of Theorem 2.2 is presented in Section 3.3. Uniqueness comes from the strict convexityof the energy E TF . We recall that, in dimension one, there is no reference energy (the one-dimensional Green’s function does not have a limit as x → ∞ ). All potentials and all Fermi levelsare defined up to global constants. Only the difference Φ TF − λ has a physical meaning, and iscalled the mean-field Thomas-Fermi potential.Next we study the screening property of the Thomas-Fermi model. We prove that the dipolarmoment of µ is perfectly screened by the TF density ρ TF . Recall that if f is any function with ´ R f = 0 (neutral), then the potential generated by f is formally given byΦ f ( x ) := − π ˆ R f ( y ) | x − y | d y. If f is compactly supported, say in [ − R, R ] then we have ∀ x > R, Φ f ( x ) = 2 π ˆ R f ( y ) y d y, while ∀ x < − R, Φ f ( x ) = − π ˆ R f ( y ) y d y. The dipolar moment of f is defined as the differenceΦ f ( ∞ ) − Φ f ( −∞ ) = 4 π ˆ R f ( y ) y d y. The next result states that all dipolar moments are perfectly screened in the Thomas-Fermimodel (see Section 3.4.1 for the proof).
Proposition 2.3 (Screening of dipolar moments) . Assume that µ satisfies | x | µ ( x ) ∈ L ( R ) . Then,the density ρ TF satisfies | x | ρ TF ( x ) ∈ L ( R ) as well, the potential Φ TF is continuous, bounded on R , satisfies Φ TF ≤ λ and lim x → + ∞ Φ TF ( x ) = lim x →−∞ Φ TF ( x ) = λ. (7) FT FOR 2-D HOMOGENEOUS MATERIALS 5
In particular, ˆ R x ( ρ TF ( x ) − µ ( x )) d x = 0 . (8)Finally, we prove explicit decay rates of the density ρ TF in the case where µ is compactlysupported. The following result can be seen as a one-dimensional version of the Sommerfeldestimates [19, 18]. In particular, the decay rates of ρ TF and Φ TF are independent of the charge Z of the system. Proposition 2.4 (Sommerfeld estimates for a slab) . Assume that µ is a compactly supported with Supp( µ ) ⊂ [ a, b ] . Then there exist x a , x b ∈ R such that (cid:40) Φ TF ( x ) − λ = − c ( x − x a ) for x ≤ a, Φ TF ( x ) − λ = − c ( x − x b ) for x ≥ b, and (cid:40) ρ TF ( x ) = c ( x − x a ) for x ≤ a,ρ TF ( x ) = c ( x − x b ) for x ≥ b. with the constants c := 5 c π and c := 5 c π . The proof is provided in Section 3.4.2. With the usual Thomas-Fermi constant, c TF = (3 π ) / ,we find c = 5 π and c = 5 π. This result is reminiscent of the usual Sommerfeld estimate for atoms [19, 18], where similar resultshold, but with different constants c = π and c = π . We believe that these estimates stillhold when µ decays fast enough.The previous result allows to have an explicit solution for the Dirac case µ = Zδ (perfectcharged slab). Example 2.5 (The Dirac case) . In the case where µ = Zδ , the solution is explicit. By convexity,the optimal density ρ TF is even. Together with Proposition 2.4, we deduce that it is of the form ρ TF ( x ) = c ( | x | + α ) − . The value of α can be found using the fact that ´ ρ TF = Z . This gives Z = 2 ˆ ∞ c ( | x | + α ) d x = 2 c α ˆ ∞ y ) d y = 2 c α . Hence ρ TF ( x ) = c (cid:16) | x | + (cid:0) c Z (cid:1) / (cid:17) − . Main results for the rHF model.
We now turn to the reduced Hartree-Fock model. Forthe sake of simplicity, we work with spinless electrons, but our arguments can be extended tomodels with spin. In this model, the state of the electrons is described by a one-body densitymatrix γ , which is a self-adjoint operator on L ( R ) satisfying the Pauli principle0 ≤ γ ≤ . (9)We consider a setting similar to the previous section, that is, the nuclear charge distribution µ isa positive function that is invariant under translations in the first two variables, and integrable inthe third variable.For H a Hilbert space, we denote by S ( H ) the set of self-adjoint operators acting on H . Ina tube Γ L := [ − L ; L ] × R , the three-dimensional rHF energy of a one-body density matrix γ ∈ S (cid:0) L (Γ L ) (cid:1) is of the form E rHF L ( γ ) = 12 Tr ,L ( − ∆ γ ) + 12 D ,L ( ρ γ − µ ) , (10)where the last term is the Hartree energy, as in the Thomas-Fermi model, and Tr ,L ( − ∆ γ ) rep-resents the supercell kinetic energy of γ . Here and thereafter, the subscripts 3 or 1 refer to thespace dimension.The energy (10) needs to be minimized over all density matrices γ satisfying the Pauli principle0 ≤ γ ≤
1, and the neutrality condition Tr ,L ( γ ) = ZL . By convexity of the functional, and DAVID GONTIER, SALMA LAHBABI, AND ABDALLAH MAICHINE translation invariance in the first two variables, it is enough to consider density matrices whichcommute with these translations. Such operators have kernels which satisfy: γ ( x , x , x ; y , y , y ) = γ ( x − y , x − y , x ; 0 , , y ) =: γ ( x − y ; x , y ) . (11)This energy depends on γ , which is an operator acting on a three-dimensional space. In order tohave an energy depending on an operator acting on a one-dimensional space, we use the followingkey result, whose proof can be read in Section 4.2. Theorem 2.6.
For any γ ∈ S ( L (Γ L )) satisfying (11) , the Pauli principle ≤ γ ≤ , and Tr ,L ( γ ) < ∞ , there is an operator G in G := (cid:8) G ∈ S ( L ( R )) , G ≥ , Tr( G ) < ∞ (cid:9) , satisfying ρ G = ρ γ (same density), and
12 1 L Tr ,L ( − ∆ γ ) ≥
12 Tr ( − ∆ G ) + π Tr (cid:0) G (cid:1) . (12) Conversely, for any G ∈ G , there is γ ∈ P so that ρ γ = ρ G , and for which there is equality in (12) . Theorem 2.6 allows us to prove that the problem posed on 3-dimensions coincides with a problemposed on the real line. For G ∈ G , we introduce the rHF one-dimensional energy per unit surface E rHF ( G ) := 12 Tr ( − ∆ G ) + π Tr (cid:0) G (cid:1) + 12 D ( ρ G − µ ) . (13)The fact that the two models are equivalent is stated in the next Theorem, whose proof ispostponed until Section 4.2. Theorem 2.7.
Consider the minimization problems I rHF L := inf (cid:8) E rHF L ( γ ) , γ ∈ S (cid:0) L (Γ L ) (cid:1) , ≤ γ ≤ , Tr ,L ( γ ) = ZL (cid:9) and I rHF := inf (cid:8) E rHF ( G ) , G ∈ S ( L ( R )) , G ≥ , Tr ( G ) = Z (cid:9) . Then, for all
L > , we have I rHF L = L I rHF . Compared to (10), the energy (13) is 1-dimensional, which can be efficiently studied boththeoretically and numerically. Note that in the reduced problem, there is no Pauli conditionfor the operator G . It is somehow replaced by the penalty term π Tr ( G ) in the energy, whichprevents G from having large eigenvalues.Theorem 2.6 allows to obtain a Lieb-Thirring type inequality that we state in the general dimension d ∈ N \ { } in Appendix A.We now focus on the reduced problem. We first prove that it is well-posed (see Section 4.3 for theproof). Theorem 2.8.
The reduced minimization problem I rHF is well-posed, and admits a unique min-imizer G ∗ ∈ G . This minimizer satisfies the Euler-Lagrange equations G ∗ = π ( λ − H ∗ ) + H ∗ = − ∆ + Φ ∗ , − Φ (cid:48)(cid:48)∗ = 4 π ( ρ ∗ − µ ) , Φ (cid:48)∗ ( ±∞ ) = 0 and Φ ∗ (0) = 0 , where λ ∈ R is the Fermi level chosen such that Tr( G ∗ ) = Z . The uniqueness of the minimizer comes from the strict convexity of the E rHF functional, thanksto the Tr( G ) term. Compared to the usual Euler-Lagrange equations for Kohn-Sham models,we see that the optimizer if of the form G = ( λ − H ) + instead of γ = ( λ − H > G ) term in the energy. One consequence isthat, since the map λ (cid:55)→ Tr ( λ − H ) + is strictly increasing on [inf σ ( H ) , ∞ ), the Fermi level λ ∈ R is always uniquely defined.Compared to the Thomas-Fermi case, one cannot say much on the screening properties of therHF model. Still, we record the following (see Section 4.4 for the proof) FT FOR 2-D HOMOGENEOUS MATERIALS 7
Proposition 2.9.
Assume that µ satisfies | x | µ ( x ) ∈ L ( R ) . Then the density ρ ∗ of G ∗ satisfies | x | ρ ∗ ( x ) ∈ L ( R ) as well. The potential Φ ∗ is continuous, bounded on R , and satisfies lim x →±∞ Φ ∗ ( x ) = ± π ˆ R ± x ( ρ − µ ) ( x )d x. It is unclear that the limits of Φ ∗ at + ∞ and −∞ are equal, which would imply as before theperfect screening of the dipolar moment. Still, our numerical simulations seems to indicate that,even though this term may not be null, it is always very small. In the special case where µ iseven, we have ρ ∗ even as well by convexity, and Φ ∗ ( −∞ ) = Φ ∗ (+ ∞ ): there is no dipolar momentin this case. Remark 2.10 (General dimension) . In the general case, where γ is an operator acting on R s + d ,which is translation invariant with respect to its first s –variables, we obtain similar results. Letus emphasize the differences. In Theorem 2.6, the operator G now acts on the last d variables G ∈ S ( L ( R d )) , and (12) is replaced by
12 1 L s Tr s + d,L ( − ∆ s + d γ ) ≥
12 Tr d ( − ∆ d G ) + c TF ( s ) Tr d (cid:16) G s (cid:17) , with the (spinless) Thomas-Fermi constant c TF ( s ) := ss + 2 (cid:18) s | S s − | (cid:19) /s π . In Theorem 2.8, the first Euler-Lagrange equation takes the form G ∗ = s ( s + 2) 1 c TF ( s ) ( λ − H ∗ ) s/ . Homogeneous 2-d materials in the Thomas-Fermi model
We start by proving some properties of the three- and one-dimensional Hartree energy.3.1.
Hartree interaction.
Three-dimensional Hartree interaction.
Let us first give an explicit expression of the three-dimensional Green’s function G L defined in (3). Proposition 3.1.
Denoting by x = ( x , x ) ∈ R the first two variables, we have G L ( x , x ) = − πL | x | − πL (cid:88) k ∈ ( πL Z ) \{ } e −| k || x | | k | e i k · x + c, where c ∈ R . In particular, if f ( x , x ) = f ( , x ) depends only on the last variable x , then ˆ Γ L f ( y , y ) G L ( x − y , x − y )d y d y = − π ˆ R f ( y ) | x − y | d y + c ˆ R f ( x )d x . Proof.
Thanks to the periodicity of G L in the first two variables, we can make the following Ansatz G L ( x , x ) = (cid:88) k ∈ (cid:16) πL Z (cid:17) c k ( x )e i k · x . Using the two-dimensional Poisson formula in the plane ( x , x ) (cid:88) R ∈ L Z δ R ( x ) = 1 L (cid:88) k ∈ (cid:16) πL Z (cid:17) e i k · x , the equation defining G L in (3) becomes ∀ k ∈ (cid:0) πL Z (cid:1) , − c (cid:48)(cid:48) k ( x ) + | k | c k ( x ) = 4 πL δ ( x ) . DAVID GONTIER, SALMA LAHBABI, AND ABDALLAH MAICHINE
For k = , the above equation gives − c (cid:48)(cid:48) ( x ) = πL δ ( x ), whose general solution is c ( x ) = − π | x | L + λx + c, c ∈ R . By symmetry, we get λ = 0. For k (cid:54) = , we solve the equation on ( −∞ ,
0) and (0 , + ∞ ) and obtainthat ∀ x > , c k ( x ) = α + k e −| k |·| x | , and ∀ x < , c k ( x ) = α − k e −| k |·| x | . The symmetry condition implies that c k ( x ) = c k ( − x ), so α + k = α − k and c k ( x ) = α k e −| k |·| x | .We find the value of α k by looking at the singularity at x = 0. We must have − c (cid:48)(cid:48) k + | k | c k = 2 | k | α k δ = 4 πL δ ( x ) , so α k = πL | k | , and the result follows. (cid:3) Away from the slab x = 0, G L is exponentially close to − πL | x | . The Green’s function G L is defined up to a global constant c , that we take equal to 0 for simplicity. Actually, for neutralsystem, which is our main interest here, the choice of the constant is irrelevant.3.1.2. One-dimensional Hartree interaction.
We are now interested in the one-dimensional Hartreeterm. We recall that it is formally given by (we put a tilde here to emphasize that we will soonconsider another definition) (cid:102) D ( f ) := − π ¨ R × R | x − y | f ( x ) f ( y )d x d y. Remark 3.2.
In the literature, one often considers the one-dimensional Green’s function G ,solution to − ∆ G = | S | δ, with | S | = 2 , whose solution is the usual G ( x ) = −| x | . In this article, our Green’s function is rather G = 2 πG ,and satisfies − ∆ G = | S | δ with | S | = 4 π . One problem with this expression is that it is not well-defined if f has a slow decay at infinity.Another problem is that the map f (cid:55)→ (cid:102) D ( f ) is not convex in general. In our context, we considerthe Hartree interaction of f := ρ − µ , which has a null integral. We therefore adopt the followingdefinition.For f ∈ L ( R ), we set W f ( x ) = ˆ x −∞ f ( y )d y, and (cid:98) f ( k ) := 1(2 π ) / ˆ R e − i kx f ( x )d x. The function W f is a primitive of f . We define the regularized version of the one-dimensionalHartree term by D ( f ) := 4 π ˆ R | (cid:98) f ( k ) | | k | d k = 4 π ˆ R | W f ( x ) | d x. (14)This expression is well defined for f in the Coulomb space f ∈ C := (cid:8) f ∈ L ( R ) , W f ∈ L ( R ) (cid:9) = (cid:40) f ∈ L ( R ) , (cid:98) f |·| ∈ L ( R ) (cid:41) . Due to the singularity at k = 0, any f ∈ C must be neutral, in the sense W (+ ∞ ) = ˆ R f ( x )d x = (cid:98) f (0) = 0 . The two definitions (cid:102) D and D do not coincide in general: D is defined for neutral functions,while (cid:102) D ( f ) makes sense whenever f decays sufficiently fast at infinity. The next Propositionshows that they coincide for neutral functions f which decays fast enough. Proposition 3.3.
The following holds. (1)
The map
C (cid:51) f (cid:55)→ D ( f ) is strictly convex. FT FOR 2-D HOMOGENEOUS MATERIALS 9 (2) If f ∈ L ( R ) satisfies ´ R f = 0 and | x | f ( x ) ∈ L ( R ) , then f ∈ C , and (cid:102) D ( f ) = D ( f )(3) If f ∈ C , then we have D ( f ) = 4 π ¨ ( R + ) ∪ ( R − ) min {| x | , | y |} f ( x ) f ( y )d x d y = ˆ R Φ f ( x ) f ( x )d x, where Φ f ( x ) := − π ´ x W f ( y )d y is the mean-field potential, also given by Φ f ( x ) := 4 π ˆ R ± min {| x | , | y |} f ( y )d y for x ∈ R ± . (15) The function Φ f is continuous, and is the unique solution to − Φ (cid:48)(cid:48) f ( x ) = 4 πf, Φ (cid:48) f ( x ) −−−−−→ x →±∞ , Φ f (0) = 0 . Proof.
The first point comes from the Fourier representation of D , which involves a strictlypositive kernel | k | − . Let us prove the other two points. Let f ∈ L ( R ) be such that | x | f ( x ) ∈ L ( R ) and ´ R f = 0. First, we see that (cid:98) f ( k ) = (cid:98) f (0) + (cid:98) f (cid:48) (0) k + o ( k )with (cid:98) f (0) = 1(2 π ) / ˆ R f ( x )d x = 0 , and (cid:98) f (cid:48) (0) = 1(2 π ) / ˆ R xf ( x )d x ∈ R . This proves that (cid:98) f ( k ) | k | is indeed in L ( R ) (there is no singularity at k = 0), so f ∈ C . Besides, wehave for x ≥
0, using that ´ R f = 0 and Fubini,Φ f ( x ) = − π ˆ x ˆ y −∞ f ( t )d t d y = 4 π ˆ x ˆ + ∞ y f ( t )d t d y = 4 π ˆ ( R + ) y ≤ min { x,t } f ( t )d t d y = 4 π ˆ R + min { x, t } f ( t )d t = 4 π (cid:18) ˆ x tf ( t )d t + x ˆ + ∞ x f ( t )d t (cid:19) . The last equality is somehow a one-dimensional version of Newton’s theorem. A similar equalityholds for x ≤
0. A similar computation shows that D ( f ) = 4 π ˆ R | W f | = 4 π ¨ ( R + ) ∪ ( R − ) min {| x | , | y |} f ( x ) f ( y ) d x d y. Therefore D ( f ) = ˆ R Φ f ( x ) f ( x )d x. Finally, to prove that this expression is also (cid:102) D ( f ) for f ∈ C satisfying | x | f ( x ) ∈ L ( R ), we remarkthat | x | + | y | − | x − y | = (cid:40) {| x | , | y |} on ( R + ) ∪ ( R − ) . This gives D ( f ) = 2 π ˆ R ( | x | + | y | − | x − y | ) f ( x ) f ( y )d x d y = 2 π ˆ R | x | f ( x )d x ˆ R f ( y )d y + 2 π ˆ R | y | f ( y )d y ˆ R f ( x )d x + (cid:102) D ( f ) , and the first two terms vanish since ´ R f = 0 and | x | f ( x ) ∈ L ( R ). (cid:3) Reduction of the Thomas Fermi model: Proof of Proposition 2.1.
In this section,we prove Proposition 2.1: we justify that the three-dimensional Thomas-Fermi problem equals itsone-dimensional version. Recall that we defined E TF3 ,L ( ρ ) := c TF ˆ Γ L ρ / + 12 D ,L ( ρ − µ ) and E TF ( ρ ) := c TF ˆ R ρ / + 12 D ( ρ − µ ) . We also introduce (cid:103) E TF ( ρ ) := c TF ˆ R ρ / + 12 (cid:102) D ( ρ − µ ) . Let ρ : Γ L → R + be a test three-dimensional density. We define (cid:101) ρ ( x ) := 1 L ˆ [ − L , L ] ρ ( x , x ) d x . By convexity of the E TF3 ,L functional, we have E TF3 ,L ( (cid:101) ρ ) ≤ E TF3 ,L ( ρ ). In addition, using Proposition 3.1,we see that D ( (cid:101) ρ − µ ) = − π ˆ [ − L , L ] ˆ R ˆ R ( (cid:101) ρ − µ )( x )( (cid:101) ρ − µ )( y ) | x − y | d y d x = L (cid:102) D ( (cid:101) ρ − µ ) . Therefore E TF3 ,L ( ρ ) ≥ E TF3 ,L ( (cid:101) ρ ) = L (cid:103) E TF ( (cid:101) ρ ). On the other hand, if ρ : R → R + is a one-dimensionaldensity, one can extend ρ in the three dimensions setting ρ ( x , x , x ) := ρ ( x ), and we have E TF3 ,L ( ρ ) = L (cid:103) E TF ( ρ ). This proves inf E TF3 ,L = L inf (cid:103) E TF .3.3. Existence of minimizers: Proof of Theorem 2.2.
As we said before, the problem withthe (cid:102) D Hartree term turns out to be quite difficult to study. In what follows, we rather study theproblem E TF instead of (cid:103) E TF , that is with the regularized D Hartree term instead of (cid:102) D . Still, weprove in this section that if | x | µ ( x ) ∈ L ( R ), then the optimal density ρ satisfies | x | ρ ( x ) ∈ L ( R )as well. In particular, for this density, we have E TF ( ρ ) = (cid:103) E TF ( ρ ).We therefore focus on the one-dimensional Thomas-Fermi minimization probleminf (cid:8) E TF ( ρ ) , ρ ∈ R (cid:9) , with R := (cid:110) ρ ∈ L ( R ) ∩ L / ( R ) , ρ − µ ∈ C , ρ ≥ (cid:111) , and we prove Theorem 2.2.3.3.1. Existence and uniqueness of minimizer.
We first prove that the problem is well posed, andadmits a unique minimizer.We start by noting that R is not empty: for instance, we have µ ∗ e −|·| ∈ R . Since D is apositive quadratic form on C , the energy functional E TF is positive on R , thus bounded from below.Let ( ρ n ) n be a minimizing sequence in R . In particular, ( ρ n ) n is bounded in L ( R ) ∩ L / ( R )and ( W ρ n − µ ) n is bounded in L ( R ). Up to sub-sequences, there exist ρ ∈ L ( R ) ∩ L / ( R ) and W ∈ L ( R ) such that ρ n (cid:42) ρ and W ρ n − µ (cid:42) W weakly in L ( R ) ∩ L / ( R ) and L ( R ) respectively.Let us prove that W = W ρ − µ . For a test function ψ ∈ C ∞ c ( R ), we have (cid:104) W, ψ (cid:48) (cid:105) = lim n →∞ (cid:104) W ρ n − µ , ψ (cid:48) (cid:105) = − lim n →∞ ( ρ n − µ, ψ ) = −(cid:104) ρ − µ, ψ (cid:105) . We deduce that W (cid:48) = ρ − µ in the distributional sense, so W ( x ) = W ρ − µ ( x ) + c for some constant c . Since W ∈ L ( R ), we have c = lim −∞ W ( x ) = 0 hence W = W ρ − µ as wanted. This implies theneutrality condition ´ R ρ ( x )d x = ´ R µ ( x )d x = Z , so ρ ∈ R . By the lower semi-continuity of the L / ( R ) and the L ( R ) norms, we obtain E TF ( ρ ) ≤ lim inf E TF ( ρ n ) = inf (cid:8) E TF ( ρ ) , ρ ∈ R (cid:9) . Hence ρ is a minimizer. Uniqueness follows from the strict convexity of the E TF functional. FT FOR 2-D HOMOGENEOUS MATERIALS 11
The Euler-Lagrange equations.
In what follows, we denote by ρ TF the optimal density. Weprove in this section that ρ TF satisfies the Euler-Lagrange equations (6). First, we have ∀ h ∈ C ∞ ( R ) , ˆ R h = 0 , ρ + h ≥ , ∀ t ∈ [0 , , E TF ( ρ TF + th ) ≥ E TF ( ρ TF ) . Differentiating at t = 0 and using that12 D ( ρ TF + th − µ ) = 12 D ( ρ TF − µ ) + t ˆ R Φ TF h + o ( t ) , where Φ TF := Φ ρ TF − µ , we obtain ∀ h ∈ C ∞ ( R ) , ˆ R h = 0 , ρ + h ≥ , ˆ R (cid:18) c TF ρ / ( x ) + Φ TF ( x ) (cid:19) h ( x )d x ≥ , (16)As in [13], we see that on the set { x ∈ R , ρ ( x ) > } , h can locally takes positive and negativevalues, so we must have c TF ρ / ( x ) + Φ TF ( x ) = λ for some λ ∈ R , called the Fermi level. Inparticular, we have Φ TF < λ on this set. On the set { x ∈ R , ρ ( x ) = 0 } , h can take only positivevalues, and we deduce that Φ TF ≥ λ . This gives the usual Thomas-Fermi equation53 c TF ρ / ( x ) = [ λ − Φ TF ( x )] + , (17)where [ f ] + := max { , f } . The same reasoning as in [13] shows that if a density satisfies the TFequation (17), then it is the unique minimizer of the TF energy functional.Since Φ TF is continuous, the density ρ TF is also continuous. Let us prove that Φ TF ≤ λ . Werecall the following maximum principle in one-dimension. Lemma 3.4.
Let V : R → R be a continuous function such that: • for any x such that V ( x ) ≥ , we have V (cid:48)(cid:48) ( x ) ≥ , • V (cid:48) → at ±∞ .Then V ≤ or V is constant. Before proving Lemma 3.4, we show how to use it to conclude that Φ TF ≤ λ . We set V :=Φ TF − λ . For any x such that V ( x ) ≥
0, we have ρ TF ( x ) = 0 from (17), hence V (cid:48)(cid:48) ( x ) = Φ (cid:48)(cid:48) TF ( x ) = − π ( ρ ( x ) − µ ( x )) = 4 πµ ( x ) ≥
0. Besides V (cid:48) = Φ (cid:48) TF → ±∞ by Proposition 3.3. Thus V satisfies the conditions of the lemma and we conclude that either V is constant or that V =Φ TF − λ ≤
0. If V is constant, then so is ρ TF and since ρ TF is integrable, then ρ TF = 0, which isnot possible as ´ R ρ TF = Z >
0. We conclude that Φ TF ≤ λ .Now, the Euler-Lagrange equation can be written as53 c TF ρ / ( x ) + Φ TF ( x ) = λ. It remains to provide the:
Proof of Lemma 3.4.
Let us assume that there is x ∈ R such that V ( x ) >
0. Let x m ∈ R ∪{−∞} and x M ∈ R ∪ { + ∞} defined by x m = sup { x ≤ x , V ( x ) ≤ } and x M = inf { x ≥ x , V ( x ) ≤ } . By continuity of V , the open interval I = ( x m , x M ) is not empty. On this interval we have V ( x ) > V (cid:48)(cid:48) ( x ) ≥
0. There are 4 possibilities:(1) x m , x M ∈ R . In this case V ( x m ) = V ( x M ) = 0 by continuity. As V is convex on I , itfollows that V ≤ I by the maximum principle, a contradiction;(2) x m ∈ R and x M = + ∞ . In this case V ( x m ) = 0 by continuity. On I , V (cid:48)(cid:48) ≥
0, thus V (cid:48) isnon decreasing. Besides V (cid:48) → ∞ . Therefore V (cid:48) ≤ I . As V ( x m ) = 0, it followsthat V ≤ I , a contradiction;(3) x m = −∞ and x M ∈ R . This case is treated as the previous one;(4) x m = −∞ and x M = + ∞ . In this case V (cid:48)(cid:48) ≥
0, thus V (cid:48) is non decreasing on R . However, V (cid:48) → ±∞ then V (cid:48) = 0, that is V is constant.We conclude that V ≤ R or is constant. (cid:3) Properties of the TF density and mean-field potential.
In this section, we give someextra properties of ρ TF and Φ TF .3.4.1. Screening of dipolar moments: Proof of Proposition 2.3.
In what follows, we assume that | x | µ ( x ) ∈ L ( R ) (the first moment of µ is finite). Let us prove that | x | ρ TF ( x ) ∈ L ( R ) as well.We have, by equation (15), for x ≥
0, that14 π Φ TF ( x ) = ˆ ∞ ( ρ TF − µ )( y ) min { x, y } d y = (cid:18) ˆ x y ( ρ TF − µ )( y )d y + x ˆ ∞ x ( ρ TF − µ )( y )d y (cid:19) . (18)This gives ˆ x yρ TF ( y )d y ≤ π Φ TF ( x ) + ˆ x yµ ( y )d y + x ˆ ∞ x µ ( y )d y ≤ λ π + 2 ˆ ∞ | y | µ ( y )d y, where we used that xµ ( y ) ≤ yµ ( y ) for y ≥ x . This proves that ´ R + xρ ( x ) < ∞ . We can prove asimilar result for x ≤
0, which proves | x | ρ TF ( x ) ∈ L ( R ).In particular, we have, for x ≥ x ˆ ∞ x ( ρ TF − µ )( y )d y ≤ ˆ ∞ x y ( ρ TF − µ )( y )d y −−−−−→ x → + ∞ . Together with (18), we obtainlim x →∞ Φ TF ( x ) = 4 π ˆ R + y ( ρ TF − µ )( y )d y and lim x →−∞ Φ TF ( x ) = − π ˆ R − y ( ρ TF − µ )( y )d y. In particular, Φ TF have limits at ±∞ . Moreover, since Φ TF is continuous, we deduce that Φ TF isbounded. In addition, by the Euler Lagrange equation (6), we have ρ TF ( x ) → ( λ − Φ TF ( ±∞ )) / as x → ±∞ . However, since ρ TF is integrable, we must have Φ TF (+ ∞ ) = Φ TF ( −∞ ) = λ . Therefore,the total dipolar moment is null:0 = Φ TF (+ ∞ ) − Φ TF ( −∞ ) = 4 π ˆ R x ( ρ − µ ) ( x )d x. This proves Proposition 2.3.3.4.2.
Sommerfeld estimates when µ is compactly supported. Proof of Proposition 2.4. We nowconsider the special case where µ is compactly supported, say in the interval [ a, b ], and prove theSommerfeld estimates in Proposition 2.4. Outside of [ a, b ], Φ TF satisfiesΦ (cid:48)(cid:48) TF ( x ) = − π (cid:18) c TF [ λ − Φ TF ( x )] (cid:19) / . (19)We solve this ordinary differential equation explicitly. First, on the interval ( b, + ∞ ), we multi-ply (19) by Φ (cid:48) TF and integrate to obtain that12 | Φ (cid:48) TF | ( x ) = (cid:18) c TF (cid:19) / π λ − Φ TF ( x )] / + cst. As x → ∞ , we have Φ TF ( x ) → λ and Φ (cid:48) TF ( x ) →
0, so the integration constant is null. In addition,since Φ (cid:48)(cid:48) TF = − πρ ≤
0, Φ (cid:48) TF is decreasing, and goes to 0 at infinity, hence Φ (cid:48) TF ≥ b, ∞ ).Taking square roots gives Φ (cid:48) TF ( x )[ λ − Φ TF ( x )] / = 4 √ π √ (cid:18) c TF (cid:19) / . Integrating a second time shows that there is x b ∈ R so that1[ λ − Φ TF ( x )] / = √ π √ (cid:18) c TF (cid:19) / ( x − x b ) . So, for all x ≥ b , we haveΦ TF ( x ) = λ − c ( x − x b ) , with c := 5 c π . FT FOR 2-D HOMOGENEOUS MATERIALS 13
In addition, since Φ (cid:48)(cid:48) TF = − πρ , we obtain that, for all x ≥ b , we have ρ ( x ) = c ( x − x ) , with c = 5 c π . (20)We have similar results on ( −∞ , a ). In particular, we obtainlim x →∞ [ λ − Φ TF ( x )] | x | = − c , and lim x →∞ | x | ρ ( x ) = c . As we already mentioned, these limits are independent of the system under consideration. Thisconcludes the proof of Proposition 2.4.4.
Homogeneous 2-d materials in the reduced Hartree-Fock model
We now prove our results concerning the rHF model.4.1.
Trace and kinetic energy per unit-surface.
In this subsection, we define both the traceper unit surface Tr and the kinetic energy per unit surface. For R ∈ R , we denote by τ R thetranslation operator on L ( R ) given by ( τ R f )( x , x ) = f ( x − R , x ). Another way to write (11)for a density matrix γ is τ R γ = γτ R , for all R ∈ R . (21)For γ satisfying this condition, we define the trace per unit surfaceTr ( γ ) := Tr ( Γ γ Γ ) , where Γ is the tube [ − , ] × R . If, in addition, γ is locally trace-class, with density ρ γ , then ρ ( x , x , x ) = ρ ( x ) depends only on the third variable andTr ( γ ) = ˆ R ρ γ ( x )d x . In the sequel, we denote by P the following subspace P := (cid:8) γ ∈ S ( L ( R )) : 0 ≤ γ ≤ , Tr( γ ) < ∞ and τ R γ = γτ R for all R ∈ R (cid:9) . Since the elements of P commute with all R –translations, we can apply Bloch-Floquet the-ory [17, Section XIII–16], (see also [2]). Let F be the partial Fourier transform defined on C ∞ ( R )by ( F f ) ( k , z ) := 12 π ˆ R e − i k · y f ( y , z )d y , and extended by density to L ( R ). The map F is unitary on L ( R ). Since γ ∈ P commuteswith R -translations, we have F γ F − = ˆ ⊕ R γ k d k , (22)that is, for all f ∈ L ( R ), ( F γf ) ( k , · ) = γ k [( F f ) ( k , · )] . Here, ( γ k ) k ∈ R is a family of self-adjoint operators acting on L ( R ). In terms of kernels, weformally have γ ( x , x ; y , y ) = 1(2 π ) ˆ R e − i k · ( x − y ) γ k ( x , y )d k . (23)In particular, ρ γ ( x ) = 1(2 π ) ˆ R ρ γ k ( x )d k , and Tr ( γ ) = 1(2 π ) ˆ R Tr ( γ k ) d k . (24)Finally, we have F ∆ F − = | k | + ∆ , thus, the kinetic energy per unit surface of γ is given by12 Tr ( − ∆ γ ) = 12 1(2 π ) ˆ R (cid:0) | k | Tr ( γ k ) + Tr ( − ∆ γ k ) (cid:1) d k . (25) Reduced states: Proof of Theorem 2.6.
For γ ∈ P , we associate the reduced operator G γ ∈ S ( L ( R )) defined by (compare with (22): the superscript ⊕ is no longer here) ∀ f ∈ L ( R ) , ( G γ f ) := 1(2 π ) ˆ R ( γ k f )d k , (26)The kernel of G γ is given by (compare with (23)) G γ ( x , y ) = 1(2 π ) ˆ R γ k ( x , y )d k . Since γ is a positive operator, so are its fibers γ k . So that G γ ≥ ρ G γ = 1(2 π ) ˆ R ρ γ k d k = ρ γ , and Tr ( G γ ) = 1(2 π ) ˆ R Tr ( γ k )d k = Tr ( γ ) . This proves that G γ ∈ G = (cid:8) G ∈ S ( L ( R )) , G ≥ , Tr ( G ) < ∞ (cid:9) .We now prove the inequality (12). Note that for all L >
0, we have L − Tr ,L = Tr . Accordingto (25) we have 12 Tr ( − ∆ γ ) = 12 1(2 π ) ˆ R | k | Tr ( γ k )d k + 12 Tr ( − ∆ G γ ) . Unfortunately, the first term cannot be expressed directly in terms of the operator G γ . We onlyhave an inequality for this term. Since G γ is trace-class, it is compact, and it has a spectraldecomposition as follows G γ = ∞ (cid:88) j =1 g j | φ j (cid:105)(cid:104) φ j | , where ( φ j ) j is an orthonormal basis of L ( R ), composed of eigenvectors of G γ with g j ≥ (cid:80) g j < ∞ . We denote by m j ( k ) := (cid:104) φ j , γ k φ j (cid:105) . Since 0 ≤ γ k ≤
1, we have 0 ≤ m j ( k ) ≤
1. In addition, from (26), it follows that ∀ j ∈ N , g j = 1(2 π ) ˆ R m j ( k )d k . We deduce that12(2 π ) ˆ R | k | Tr ( γ k )d k = ∞ (cid:88) j =1 π ) ˆ R | k | m j ( k )d k ≥ ∞ (cid:88) j =1 inf (cid:26) π ) ˆ R | k | m ( k )d k , ≤ m ( k ) ≤ , π ) ˆ R m ( k )d k = g j (cid:27) . According to the bathtube principle (see [12, Thm 1.14]), the last minimization problem admitsa unique minimizer, of the form ( | k | ≤ k F ). The value of the radius is found with the condition ´ R m = (2 π ) g j and we get m ∗ j ( k ) := (cid:0) | k | ≤ √ πg j (cid:1) . (27)For this value, we have ∞ (cid:88) j =1 π ) ˆ R | k | m ∗ j ( k )d k = ∞ (cid:88) j =1 πg j = π Tr ( G γ ) , which proves (12). Conversely, for any G ∈ G , if we write G = (cid:80) ∞ j =1 g j | φ j (cid:105)(cid:104) φ j | and set F γ ∗ F − := ˆ ⊕ R γ k d k , with γ k := ∞ (cid:88) j =1 m ∗ j ( k ) | φ j (cid:105)(cid:104) φ j | , (28)we see that G γ ∗ = G (representability) and (12) becomes an equality. This proves Theorem 2.6. FT FOR 2-D HOMOGENEOUS MATERIALS 15
Remark 4.1.
The proof in higher dimension R d + s (see Remark 2.10) is similar. Indeed, set G = 1(2 π ) s ˆ R s γ k d k = ∞ (cid:88) j =1 g j | φ j (cid:105)(cid:104) φ j | , and, for every j ∈ N , the optimal m ∗ j is defined by m ∗ j ( k ) := (cid:16) | k | ≤ c s g /sj (cid:17) with c s := 2 π (cid:18) s | S s − | (cid:19) /s . This yields ∞ (cid:88) j =1 π ) s ˆ R s | k | m ∗ j ( k )d k = c TF ( s )Tr (cid:16) G s (cid:17) , where c TF ( s ) := ss +2 (cid:16) s | S s − | (cid:17) /s π is the s -dimensional (spinless) Thomas-Fermi constant. Thanks to the reduction of the kinetic energy, and reasoning as in the Thomas-Fermi section,we obtain that I rHF3 ,L = L I rHF , where we recall that I rHF = inf (cid:8) E rHF ( G ) , G ∈ G , Tr ( G ) = Z (cid:9) with G = (cid:8) G ∈ S ( L ( R )) , G ≥ , Tr ( G ) < ∞ (cid:9) and E rHF ( G ) = 12 Tr ( − ∆ G ) + π Tr ( G ) + 12 D ( ρ G − µ ) . This proves Theorem 2.7.4.3.
Existence of minimizers for the reduced model: Proof of Theorem 2.8.
We nowfocus on the reduced rHF problem.4.3.1.
Existence and uniqueness of the minimizer.
Let us first prove that the minimization problem I rHF is well-posed. We start by noting that G is not empty. Indeed, for ρ = µ ∗ e −| x | , we have (cid:12)(cid:12) √ ρ (cid:105)(cid:104)√ ρ (cid:12)(cid:12) ∈ G . Let ( G n ) ⊂ G be a minimizing sequence satisfying Tr ( G n ) = Z . Then E rHF ( G n )is bounded, and since it is the sum of three positive terms, there is C ≥ ( − ∆ G n ) ≤ C, Tr ( G n ) ≤ C, and D ( ρ n − µ ) ≤ C, where we set ρ n := ρ G n . In addition, (cid:107) ρ n (cid:107) L = Z , thus ( ρ n ) is bounded in L ( R ). We thendeduce that, up to a subsequence, still denoted by ( G n ) and ( ρ n ), we have the following weak-*convergences (we denote by S p := S p ( L ( R )) the p Schatten class with (cid:107) G (cid:107) p S p = Tr ( | G | p )) |∇| G n |∇| (cid:44) → T weakly-* in S G n (cid:44) → G ∗ weakly-* in S ∩ S ρ n (cid:44) → ρ ∗ weakly-* in L ( R ) ,W ρ n − µ (cid:44) → W weakly-* in L ( R )where W ρ n − µ is defined in Section 3.1.2. By standard arguments, we have that T = |∇| G ∗ |∇| ,that ρ ∗ = ρ G ∗ , and that W = W ρ ∗ − µ . In particular, the last equality shows that W ρ ∗ − µ ∈ L ( R ).In particular, we have ( ρ ∗ − µ ) ∈ C , which implies the neutrality ´ R ( ρ ∗ − µ ) = 0. Hence Tr( G ) = Z .Furthermore, we have E rHF ( G ∗ ) ≤ lim inf n E rHF ( G n ), thus G ∗ is a minimizer of E rHF .This minimizer is unique, thanks to the strict convexity of E rHF due to the Tr( G ) term. Derivation of the Euler-Lagrange equations.
We now derive the Euler-Lagrange equations.In what follows, we denote by G ∗ the unique minimizer of the E rHF functional. For all G ≥ G ) = Z , and all 0 ≤ t ≤
1, one has E rHF ((1 − t ) G ∗ + tG ) ≥ E rHF ( G ∗ ) , This gives Tr (cid:0)(cid:2) − ∆ + 2 πG ∗ + Φ ∗ (cid:3) ( G − G ∗ ) (cid:1) ≥ , (29)where Φ ∗ is the mean-field potential generated by ρ G ∗ − µ (see Proposition 3.3). Let us denote by h := H ∗ + 2 πG ∗ , where H ∗ := −
12 ∆ + Φ ∗ . Testing (29) over states of the form G = Z | φ (cid:105)(cid:104) φ | , with (cid:107) φ (cid:107) L = 1 shows that h is bounded frombelow. Let λ := inf σ ( h ) > −∞ . Since Tr( G ) = Z = Tr( G ∗ ), for all G ∈ G , the inequality (29)can also be written as Tr (( h − λ ) G ) ≥ Tr (( h − λ ) G ∗ ) , ∀ G ∈ G . In particular, the minimization probleminf { Tr (( h − λ ) G ) , G ∈ G , Tr( G ) = Z } is well-posed and admits a unique minimizer, which is G ∗ . By definition of λ , the above minimum is0. It follows that Tr (( h − λ ) G ∗ ) = 0, hence Ran G ∗ ⊂ Ker( h − λ ). In particular, Ker( h − λ ) (cid:54) = { } ,so that λ is an eigenvalue of h . Moreover, we have( h − λ ) G ∗ = G ∗ ( h − λ ) = 0 . Let us consider the spectral decomposition of G ∗ , of the form G ∗ = (cid:80) ∞ j =1 g j | φ j (cid:105)(cid:104) φ j | . Then, forall j ≥ g j >
0, we have hφ j = λφ j , that is H ∗ φ j + 2 πg j φ j = λφ j . So φ j is an eigenvector of H ∗ with corresponding eigenvalue ε j := λ − πg j . As g j >
0, then g j = 12 π ( λ − ε j ) + . This proves that G ∗ is of the form G ∗ = (cid:88) j π ( λ − ε j ) + | φ j (cid:105)(cid:104) φ j | , with H ∗ φ j = ε j φ j . (30)Conversely, if ε < λ is an eigenvalue of H ∗ with corresponding eigenvector ψ , one has0 = G ∗ ( h − λ ) ψ = G ∗ ( H ∗ − λ + 2 πG ∗ ) ψ = 2 πG ∗ ( G ∗ ψ ) + ( ε − λ ) G ∗ ψ. Denoting by φ := G ∗ ψ and g := ( λ − ε ) / π >
0, we have G ∗ φ = gφ . We claim that φ (cid:54) = 0.Otherwise, we would have hψ = H ∗ ψ = εψ , and ε would be an eigenvalue of h , smaller that λ < inf σ ( h ), a contradiction. So φ is an eigenvector of G ∗ with eigenvalue g . According to theprevious decomposition of G ∗ , φ corresponds to one of the φ j , up to a multiplicative factor. Inother words, all eigenvalues of H ∗ smaller than λ are considered in the decomposition (30). Besides,as G ∗ is a compact operator, H ∗ is a compact perturbation of h . Therefore σ ess ( H ∗ ) = σ ess ( h ),and no essential spectrum of H ∗ can lie below λ . We deduce that G ∗ = 12 π ( λ − H ∗ ) + . This ends the proof of Theorem 2.8. It is unclear whether G ∗ is finite rank: H ∗ might have aninfinity of eigenvalues below λ , in which case we would have λ = inf σ ess ( H ∗ ). FT FOR 2-D HOMOGENEOUS MATERIALS 17
Properties of the mean-field potential: Proof of Proposition 2.9.
We finally provesome properties of the mean-field potential Φ ∗ in the case where | x | µ ( x ) ∈ L ( R ). First, we notethat, for x ≥
0, Φ ∗ ( x ) = 2 π ˆ R + ( ρ − µ )( y ) min { x, y } d y ≥ − π ˆ R + µ ( y ) y d y > −∞ , and similarly for x ≤
0, so Φ ∗ is bounded from below. Let us prove that Φ ∗ is also boundedfrom above. This will imply, as in the Thomas Fermi case that | x | ρ ( x ) ∈ L ( R ) and that (seeSection 3.4.1)lim x → + ∞ Φ ∗ ( x ) = 4 π ˆ R + y ( ρ ∗ − µ )( y )d y and lim x →−∞ Φ ∗ ( x ) = − π ˆ R − y ( ρ ∗ − µ )( y )d y. Assume that lim x → + ∞ Φ( x ) = + ∞ . Then, we would also have ´ ∞ ρ ∗ ( y ) y d y = + ∞ . The contra-diction comes from Agmon’s estimates, which state that if the potential Φ( x ) goes to + ∞ , thenthe corresponding eigenvectors (hence the density ρ ) are exponentially decaying. We provide asimple proof of Agmon’s argument in the one-dimensional setting for completeness. This is notthe optimal result, and we refer to the original work [1] for details (see for instance the exampleat the end of Chapter 1 in [1]). Lemma 4.2.
Let V ∈ L ( R ) be a potential bounded from below and let E ∈ R . Assume that lim x → + ∞ V ( x ) = + ∞ and let a ∈ R be such V ( x ) > E + 1 for all x > a . Then, there is C > such that, for all u ∈ H ( R ) eigenvector of the operator H = − ∂ xx + V associated with aneigenvector E ≤ E , we have ˆ ∞ a e ( x − a ) | u | ( x )d x ≤ C. Proof.
Let ψ ∈ C ∞ ( R ) and consider the test function φ := ψ u . The equation (cid:104) φ, ( H − E ) u (cid:105) = 0becomes ˆ R u (cid:48) ( ψ u ) (cid:48) + ˆ R ( V − E ) u ψ = 0 . Together with the identity u (cid:48) ( ψ u ) (cid:48) = | ( ψu ) (cid:48) | − | ψ (cid:48) | u , this gives ˆ R | ψ (cid:48) | u = ˆ R | ( ψu ) (cid:48) | + ˆ R ( V − E ) u ψ ≥ ˆ R ( V − E ) u ψ . Assume that ψ has support in ( a, ∞ ). On this support, we have ( V − E ) ≥
1, so ˆ R ψ u ≤ ˆ R | ψ (cid:48) | u . (31)By density, this inequality remains valid for all ψ ∈ H (( a, ∞ )). Taking for instance ψ ( x ) := e ( x − a ) − x ∈ [ a, L ) η ( x ) := (cid:104)(cid:16) e ( L − a ) − (cid:17) − ε ( x − L ) (cid:105) + on [ L, + ∞ ) ψ is continuous and slowly decaying to 0 (at rate ε ) on [ L, ∞ ). Applying (31), we obtain ˆ La (e ( x − a ) − u ( x )d x + ˆ ∞ L η ( x ) u ( x )d x ≤ ˆ La e x − a u ( x )d x + ˆ ∞ L η (cid:48) ( x ) u ( x )d x. For ε small enough, we have ˆ La (cid:18)
34 e ( x − a ) − ( x − a ) + 1 (cid:19) u ( x )d x ≤ . (32)On the other hand, one can find C > ( x − a ) − ( x − a ) + 1 ≥
12 (e ( x − a ) − C ) . So that, (32) becomes ˆ La e ( x − a ) u ( x )d x ≤ C ˆ La u ( x )d x ≤ C. Letting L → ∞ proves the result. (cid:3) In our case, we consider all eigenvalues below E = λ . This gives ˆ ∞ a ρ ( x )e ( x − a ) d x = (cid:88) i ( λ − ε i ) + ˆ ∞ a e ( x − a ) | u i | ( x )d x ≤ C (cid:88) i ( λ − ε i ) + = CZ.
This contradicts the fact that ´ ∞ yρ ( y ) < ∞ .5. Numerical illustrations
In this section, we provide numerical results for our reduced one-dimensional models.5.1.
Numerical setting and self-consistant procedure.
We use a simple code with finitedifferences: all functions are evaluated on a fine grid representing some interval [ − a, a ], with N b points (we took a = 15 and N b = 5001). The operator − ∆ is computed in Fourier space (usingsparse matrices). Given a neutral function f , we evaluate the potential Φ f by performing theconvolution in (15).To solve the Thomas-Fermi problem, we use the following iterations. Recall that the optimaldensity ρ TF satisfies the Euler-Lagrange equation (6), that we write in the form ρ TF = (cid:18) c TF (Φ TF − λ ) + (cid:19) / At the point ρ n , we set Φ n = Φ ρ n − µ , and compute λ n ∈ R so that Z n ( λ n ) = Z, where Z n ( λ ) := ˆ R (cid:18) c TF (Φ n − λ ) + (cid:19) / . Since the map Z n ( · ) is decreasing, one can efficiently compute λ n using a simple dichotomy. Wethen set (cid:93) ρ n +1 := (cid:18) c TF (Φ n − λ n ) + (cid:19) / , and ρ n +1 := t (cid:93) ρ n +1 + (1 − t ) ρ n , where t is optimized to lower the Thomas-Fermi energy on the segment [0 ,
1] (linear search).To solve the rHF problem, we use a similar iterative procedure. Recall that G ∗ satisfies theEuler-Lagrange equations G ∗ = 12 π ( λ − H ∗ ) + . At point G n ∈ G , we set H n := − ∆ + Φ n and find λ n so thatTr (cid:18) π ( λ n − H n ) + (cid:19) = Z. Again, this can be solved using a dichotomy method. We then set (cid:94) G n +1 := 12 π ( λ n − H n ) + , and G n +1 := t (cid:94) G n +1 + (1 − t ) G n , For the optimization problem in t ∈ [0 , t (cid:55)→ E rHF ( t (cid:94) G n +1 + (1 − t ) G n ) isquadratic in t , so the best t ∈ [0 ,
1] is explicit.Although we believe that more sophisticated methods can be designed to study these problems,our numerical methods for both the TF and rHF problems seem to converge quite fast, which isenough for our purpose.5.2.
Numerical results.
We now provide some numerical results in three test cases. In order tocompare the (spinless) rHF and TF results, we use the spinless c TF constant c TF := / π / / forthe TF model (see Remark 4.1). FT FOR 2-D HOMOGENEOUS MATERIALS 19
Case 1: a simple slab.
For our first test case, we consider the charge µ ( x ) := ( | x | < , which models an homogeneous charged slab having some width. The results are displayed inFigure 1. In the (A) part, we display the function µ , together with the best rHF density ρ ∗ andthe generated potential Φ ∗ . The potential is scaled by a factor 10 for ease of reading. We do notdisplay the TF results here, as they are very similar to the rHF ones. In the (B) part, we displaythe difference between the TF results and the rHF ones, and we plot ρ TF − ρ ∗ and Φ TF − Φ ∗ .Although these two densities are not equal, they are very close, and they only differ by around 1%.In addition, this difference seems to be even lower far from the slab ( | x | large). The potentials Φ TF and Φ ∗ differ by a small constant as they are defined as the unique function solving the Poissonequation Φ (cid:48)(cid:48) = − π ( ρ − µ ) with the conditions Φ (cid:48) → ±∞ and Φ(0) = 0.The optimal operator G ∗ that we have numerically found has 15 positive eigenvalues, 4 of whichbeing smaller than 10 − . The remaining 11 other eigenvalues are greater than 10 − , and the largeris around 1 .
07 (which shows that G can have eigenvalues greater than 1). (a) µ (red), ρ ∗ (blue) and (Φ ∗ − λ ∗ ) /
10 (dottedblack). (b)
The difference ρ TF − ρ ∗ (blue) and Φ TF − Φ ∗ (dotted black). Figure 1.
Results for the slab µ .5.2.2. Case 2: two slabs.
For our second test case, we consider two different slabs. We take thecharge µ ( x ) := ( − < x < −
2) + 2 · (1 < x < . The main difference with the previous case is that µ has a non null dipolar moment. The resultsare displayed in Figure 2. Again, the TF results are very close to the rHF ones. This is surprising,since we expected the rHF model to exhibit some (screened) dipolar moment. However, we foundnumerically dipolar moment of order 2 · − for the rHF case, and of order 1 · − for the TFone. We believe that they come from numerical errors: recall that we are working in the finitebox [ − , TF − λ to decay as − c | x | − .We were not able to prove that the dipolar moment should vanish also in the rHF case. Still,even though it does not vanish, we believe that it can always be neglected.The optimal operator G ∗ in this case has rank 17, with two eigenvalues of order 10 − , and theremaining ones of order 10 − to 1. The highest eigenvalue is around 1 . (a) µ (red), ρ ∗ (blue) and (Φ ∗ − λ ∗ ) /
10 (dottedblack). (b)
The difference ρ TF − ρ ∗ (blue) and Φ TF − Φ ∗ (dotted black). Figure 2.
Results for the step function µ .5.2.3. Case 3: two slabs, smooth case.
Finally, we study the case where µ is smooth. We took µ ( x ) = e − ( x +2) + 2 · e − ( x − . The charge density models two slabs having non-homogeneous charge in the x -direction. Theresults are displayed in Figure 3. Again, the densities and mean-field potentials are very close.Actually, it seems that, due to smoothness, the difference ρ TF − ρ ∗ is now of order 0 . G ∗ in this case has rank 19, with 4 eigenvalues below 10 − , and theremaining ones above 10 − . The highest eigenvalue is 1 . (a) µ (red), ρ ∗ (blue) and (Φ ∗ − λ ∗ ) /
10 (dottedblack). (b)
The difference ρ TF − ρ ∗ (blue) and Φ TF − Φ ∗ (dotted black). Figure 3.
Results for the step function µ .5.3. Perspectives.
In view of these numerical results, we believe that, even for more complexKohn-Sham models, one can always approximate the reduced kinetic energy by the Thomas Fermione.More specifically, starting from a three-dimensional Kohn-Sham (KS) model of the form E KS3 ( γ ) := 12 Tr ( − ∆ γ ) + 12 D ( ρ γ − µ ) + F ( ρ γ ) , and assuming that the charge µ depends only on the third variable, one can assume that theoptimal γ will again satisfy (11) (although symmetry breaking can now happen due to the non-convexity of the models). If this is the case, then one can perform the same analysis as before,and obtain a reduced one-dimensional energy per unit surface, of the form E KS ( G ) := 12 Tr ( G ) + π Tr( G ) + 12 D ( ρ G − µ ) + F ( ρ G ) . FT FOR 2-D HOMOGENEOUS MATERIALS 21
The corresponding Thomas-Fermi model is E KS , TF ( ρ ) := c TF ˆ R ρ / + 12 D ( ρ − µ ) + F ( ρ ) . We believe that the optimal TF density ρ TF is always very close to the optimal KS one ρ KS . Theadvantage is that E KS , TF ( ρ ) is easier to optimize numerically, and to study theoretically. Appendix A. A Lieb-Thirring inequality
In this section, we explain how to use Theorem 2.6 to obtain a Lieb-Thirring type inequality [14,15]. We state our result in the general dimension d ∈ N \ { } . Proposition A.1.
Let d ∈ N \ { } and G ∈ S ( L ( R d )) be a positive operator. Then, for any s ∈ N \ { } , K (cid:18) ˆ R d ρ d + s G (cid:19) sd ≤ (cid:16) Tr d ( G s ) (cid:17) s/d Tr d ( − ∆ d G ) with the constant K := K LT ( d + s ) sd (2 c TF ( s )) sd (cid:18) sd + s (cid:19) s/d dd + s . Here, K LT ( d + s ) is the usual Lieb-Thirring constant in dimension d + s , that is the best constantin the inequality ∀ γ ∈ S (cid:0) L ( R d + s ) (cid:1) , ≤ γ ≤ , K LT ( d + s ) ˆ R d + s ρ d + s γ ≤ Tr d + s ( − ∆ γ ) . (33) Proof.
Consider G ∈ S ( L ( R d )) , G ≥ G s ) < ∞ and consider the optimal γ ∈ S ( L ( R d + s )) as in (28). Then, the Lieb-Thirring inequality (33) applied to γ gives (after theappropriate per unit surface normalization)12 Tr d ( − ∆ d G ) + c TF ( s )Tr d ( G s ) = 12 Tr d + s ( − ∆ d + s γ ) ≥ K LT ( d + s ) ˆ R d ρ d + s γ , Since ρ γ = ρ G then, for all G ∈ S s ( R d ) such that Tr( − ∆ G ) < ∞ , we get ρ G ∈ L d + s ( R d ),and Tr d ( − ∆ d G ) + 2 c TF ( s )Tr d ( G s ) ≥ K LT ( d + s ) ˆ R d ρ d + s G . Performing the scaling G λ = λG and optimizing over λ gives the result. (cid:3) Proposition A.1 corresponds to the Lieb-Thirring inequality for operators in S ( L ( R s + d )) in asemi-classical limit, when the semi-classical limit dilation is only performed in the first s variables(see also [16] for similar arguments).This type of inequalities was recently studied in [8], where it is shown that for all d ≥ ≤ p ≤ d , there is an optimal constant K p,d so that K p,d (cid:107) ρ G (cid:107) pd ( p − p ≤ (cid:107) G (cid:107) p (2 − d )+ dd ( p − S q Tr d ( − ∆ d G ) , with q := 2 p + d − dp d − dp . It is proved that this constant is somehow the dual constant of the usual Lieb-Thirring constant L γ,d . The case in Proposition A.1 corresponds to the choice p = 1 + 2 d + s with s ∈ N , so that q = 1 + 2 s . This corresponds to the constant L γ,d with γ = qq − = 1 + s . In particular, since s ≥
1, we have γ ≥ . In this regime, it is known that the best constant is the semi-classical one: L γ,d = L sc γ,d ,hence K p,d = K sc p,d . This proves for instance that12 Tr d ( − ∆ d G ) + c TF ( s )Tr d ( G s ) ≥ c TF ( d + s ) ˆ R d ρ d + s G . In other words, the energy in the rHF model is always greater than the energy in the TF model.
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CEREMADE, University of Paris-Dauphine, PSL University, 75016 Paris, France
Email address : [email protected] (Salma Lahbabi) ESSM, LRI, ENSEM, UHII, 7 Route d’El Jadida, B.P. 8118 Oasis, Casablanca; MSDA,Mohammed VI Polytechnic University, Lot 660, Hay Moulay Rachid Ben Guerir, 43150, Morocco
Email address : [email protected] (Abdallah Maichine) MSDA, Mohammed VI Polytechnic University, Lot 660, Hay Moulay Rachid BenGuerir, 43150, Morocco
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