Nonrelativistic Limit of the Dirac-Fock Equations

Nonrelativistic limit of the Dirac-Fock equations.

Maria J. Esteban and Eric Sere

CEREMADE (UMR C.N.R.S. 7534)

Universite Paris IX-Dauphine

Place du Marechal de Lattre de Tassigny

75775 Paris Cedex 16 - France

esteban or sere@ceremade.dauphine.fr

Abstract.

In this paper, the Hartree-Fock equations are proved to be the non relativistic limit of the Dirac-Fock equations as far as convergence of \stationary states" is concerned. This property is used to derive a meaningful denition of \ground state" energy and \ground state" solutions for the Dirac-Fock model.

AMS Subject Classication:

49 S 05, 35 J 60, 35 P 30, 35 Q 75, 81 Q 05, 81 V 70, 81 V 45, 81 V 55.

Key Words :

Relativistic quantum mechanics, nonrelativistic limit, quan-tum chemistry, ground state, nonlinear eigenvalue problems, Dirac-Fock equations, Hartree-Fock equations, variational methods, critical points, strongly indenite functionals, bifurcations from the essential spectrum.

1. Introduction

In this paper we prove that solutions of Dirac-Fock equations converge, in a certain sense, towards solutions of the Hartree-Fock equations when the speed of light tends to innity.

This limiting process allows us to dene a notion of ground state for the Dirac-Fock equations, valid when the speed of light is large enough.

First of all, we choose units for which

m

= 

h

= 1, where

m

is the mass of the electron, and 

h

is Planck's constant. We also impose e2

4"0 = 1, with ?

e

the charge of an electron,

"

0 the permittivity of the vacuum.

The Dirac Hamiltonian can be written as

H

c = ?

ic

r+

c

2

_;

(1)

where

c >

0 is the speed of light in the above units,



= 1I 0₀

?1I !

,



k =

_

0_k



₀k

!

(

k

= 1

;

2

;

3) and the



k are the well known Pauli

ma-trices. The operator

H

cacts on 4-spinors, i.e. functions from IR3to CI4, and

it is self-adjoint in

L

2_(IR3

_;

_CI4_{), with domain}

_H

1_(IR3

_;

_CI4_{) and form-domain}

H

1=2_(IR3

_;

_CI4_{). Its spectrum is (}

?1

;

?

c

2][[

c

2

;

+1).

Let us consider a system of

N

electrons coupled to a xed nuclear charge density

eZ

, where

e

is the charge of the proton,

Z >

0 the total number of protons and



is a probability measure dened on IR3_{. Note that in the}

particular case of

m

point-like nuclei, each one having atomic number

Z

i at

a xed location

x

i

;eZ

=Xm

i=1

eZ

i



xi and

Z

=

m

X

i=1

Z

i.

In our system of units, the Dirac-Fock equations for such a molecule are given by 8 > > > > > > > < > > > > > > > :

H

c; k:=

H

c k?

Z

(



 1 j

x

j ) k+ (



 1 j

x

j ) k ? Z IR 3

R

(

x;y

) k(

y

) j

x

?

y

j

dy

=

"

_ck k (

k

= 1

;:::N

)

;

GramL 2 = 1IN (i

:

e R IR 3  k l=



kl

;

1k

;

lN)

:

(DFc)

Here, = ( 1

;

;

N)

;

each k is a 4-spinor in

H

1=2(IR3

;

CI4) (by

bootstrap, k is also in any

W

1;p(IR3) space, 1

p <

3

=

2), and



(

x

) := N X k=1  k(

x

) k(

x

)

; R

(

x;y

) := N X k=1 k(

x

)  k(

y

)

:

(2)

We have denoted  the complex line vector whose components are the

conjugates of those of a complex (column) vector , and 

1 2 is the inner

product of two complex (column) vectors 1, 2. The

n



n

matrix Gram L

2

is dened by the usual formulas (Gram ) :=Z



k(

x

) l(

x

)

dx :

Finally,

"

_c1 

:::



"

cN are eigenvalues of

H

c;

:

Each one represents the

energy of one of the electrons, in the mean eld created by the molecule. For physical reasons, we impose 0

< "

_ck

< c

2

_:

_{Note that the scalars}

_"

_{ck can}

also be seen as Lagrange multipliers. Indeed, the Dirac-Fock equations are the Euler-Lagrange equations of the Dirac-Fock energy functional

Ec( ) = N X k=1 Z IR 3  k

H

c k?

Z





 1 j

x

j   k k +12ZZ IR3 IR 3



(

x

)



(

y

) ?tr 

R

(

x;y

)

R

(

y;x

)  j

x

?

y

j

dxdy

under the constraintsR IR

3 

k l=



kl

:

In [6] we proved that under some assumptions on

N

and

Z

, there exists an innite sequence of solutions of (DFc). More precisely:

Theorem 1 [6]

. Let

N < Z

+ 1. For any

c >

=2+2=₂ max(

Z;

3

N

?1),

there exists a sequence of solutions of (DFc),

n c;jo j1  

H

1=2_(IR3₎N , such that

(i)

0

<

Ec( c;j)

< Nc

2 ,

(ii)

_jlim !+1 Ec( c;j) =

Nc

2_,

(iii)

0

< c

2?



j

< "

c;j 1 

:::



"

c;j N

< c

2?

m

j , with



j

> m

j

>

0 independent of

c

.

The constant =2+2=₂ is related to a Hardy-type inequality obtained independently by Tix and Burenkov-Evans (see [15, 3, 16]), and which plays an important role in the proof of Theorem 1. With the physical value

c

= 137

:

037

:::

and

Z

an integer (the total number of protons in the molecule), our conditions become

N



Z; N

41

; Z

124

:

The constraint

N

41

is technical, and has no physical meaning.

Our result was recently improved by Paturel [13], who relaxed the condi-tion on

N

. Paturel obtains the same multiplicity result, assuming only that

N < Z

+ 1 and =2+2=₂ max(

Z;N

)

< c

. Taking

c

= 137

:

037

:::

, Paturel's conditions are

N



Z

124 : they cover all existing neutral atoms. This is

In [6], the critical points c;j _{are obtained by a complicated min-max}

argument (in [13], the argument is not simpler). Here, we do not give this min-max in its full detail. We just state the minimal information needed in the present paper. Let us denote

E

:=

H

1=2_(IR3

_;

_CI4_{). Since}



(

H

c) = (?1

;

?

c

2_]

[[

c

2

_;

₊

1)

;

the Hilbert space

E

can be split as

E

=

E

+c

E

? c

;

where

E

 c :=  c

E

, and  c :=



IR (

H

c). The projectors   c have a simple

expression in the Fourier domain : d

 c (



) = ^ c(



) ^(



), with b  c(



) := 12 1ICI 4  c





+ c2



p c4_{+ c}2j



j2 !

:

(4)

Proposition 2 [6, 13]

. For every

j

1, let

V

be any (

N

+

j

?1)

dimen-sional complex subspace of

E

_+c. Then, taking the notation of Theorem 1, we have

E

cj;DF :=Ec( c;j₎  sup 2(E ? c V ) N Gram L 2 1I N Ec( )

:

(5)

In the present paper, we prove three main theorems. We rst consider a sequence

c

n !+1 and a sequencef ng

n of solutions of (DFcn). For all

n

, n _{= ( n1}

_;:::;

_{nN), each nk is in}

_H

1=2_(IR3

_;

_CI4_{), with} Z

IR3  k l

dx

=



kl and

H

c n ;

n nk =

"

nk nk. Using the standard Hardy inequality, one can prove

that the functions _nk are in

H

1_(IR3

_;

_CI4_{) for}

_c

_{n large enough. We assume}

that ?1

<

lim n!+1 (

"

n1?

c

2n) lim n!+1 (

"

nN ?

c

2n)

<

0

:

(6)

A (column) vector 2CI4can be written in block form = ?'



where

'

2CI2

(respectively



2 CI

2_{) consists of the two upper (resp. lower) components}

of . This gives the splitting _nk = ?'n k

n k

with

'

_nk and



_nk in

H

1_(IR3

_;

_CI2_).

Finally, n _{splits as}? n

n

, where n _{:= (}

_'

_n1

_;:::;'

_{nN) and}



n_{:= (}



_n1

_;:::;

_nN).

Our rst result is that n₌? n

n

towards  =? 0
, where  = (

'

1

;

;

'

N) 2 

H

1_(IR3

_;

_CI2₎N is a solution of the Hartree-Fock equations:

8 > > > > > > > > < > > > > > > > > : H 

'

k = ?

'

k 2 ?

Z





 1 j

x

j 

'

k+ 



 1 j

x

j 

'

k ? Z IR3

R

(

x;y

)

'

k(

y

) j

x

?

y

j

dy

= 



k

'

k

; k

= 1

;:::N;

Z IR3

'

 k

'

l

dx

=



kl

;





k= lim_n !+1 (

"

nk?

c

2n)

:

(HF)

Here (as in the Dirac-Fock equations),



(

x

) =XN l=1

'

 l(

x

)

'

l(

x

)

; R

(

x;y

) =XN l=1

'

l(

x

)

'

 l(

y

)

:

Note that the Hartree-Fock equations are the Euler-Lagrange equations corresponding to critical points in

H

1_(IR3

_;

_CI2₎N

of the Hartree-Fock en-ergy: EHF() := N X k=1 1 2jjr

'

kjj 2 L 2 ?

Z

Z IR3 



 1 j

x

j  j

'

kj 2

_dx

(7) +12ZZ IR3 IR 3



(

x

)



(

y

)?tr(

R

(

x;y

)

R

(

y;x

)) j

x

?

y

j

dxdy ;

under the constraintZ

IR3

'



k

'

l=



kl ,

i;j

= 1

;:::N

.

Theorem 3

. Let

N < Z

+ 1. Consider a sequence

c

n ! +1 and a

sequence f ngn of solutions of (DFc

n), i.e.

n _{= ( n1}

_;

;

nN), each nk

being in

H

1=2_(IR3

_;

_CI4_{) , with} Z

IR3 

k l

dx

=



kl and

H

cn;

n nk =

"

nk nk .

Assume that the multipliers

"

_nk,

k

= 1

;:::;N;

satisfy (6). Then for

n

large enough, nk is in

H

1_(IR3

_;

_CI4_{) , and there exists a solution of (HF),}

 = (

'

1

;

;

'

N), with negative multipliers, 



1

;:::;



N, such that, after

ex-traction of a subsequence,



nk:=

"

nk?(

c

n) 2 ?! n!+1



k

; k

= 1

;:::;N ;

(8)

nk=

'



nk_nk ! ?! n!+1 

'

k 0 ! in

H

1_(IR3

_;

_CI2₎ 

H

1_(IR3

_;

_CI2₎

_;

(9)



nk+

i

2

c

n(



r)

'

nk L 2 (IR 3 ;CI 2 ) =

O

(1

=

(

c

n)3)

;

(10) and Ec n( n₎?

Nc

2n ?! n!+1 EHF()

:

(11)

As a particular case, we have

Corollary 4

. If

c

n! +1 and

N; Z; 

are xed, then for any

j

1 the

sequencef c n;j

g

n of Theorem 1 satises the assumptions of Theorem 3 (see

(iii) in Theorem 1). So it is precompact in

H

1_(IR3

_;

_CI4₎N . Up to extraction of subsequences,



c n ;j k :=

"

c n ;j k ?

c

2n?!



j k

<

0

; k

= 1

;:::;N

(12) cn;j ?! j 0 ! in 

H

1_(IR3

_;

_CI2₎N  

H

1_(IR3

_;

_CI2₎N (13) and j ₌  

'

j1

;

;

'

jn 

is a solution of the Hartree-Fock equations with multipliers 



j1

;

;



 j N. Moreover, Ec n( cn;j) ?

Nc

2n ?! n!+1 EHF(j)

:

(14)

Particular solutions of the Hartree-Fock equations are the minimizers of EHF() under the constraints Gram

L

2  = 1IN

:

They are called ground

states. Their existence was proved by Lieb and Simon [10] under the as-sumption

N < Z

+ 1 (see also [11] for the existence of excited states).

It is dicult to dene the notion of ground state for the Dirac-Fock model, sinceEc has no minimum under the constraints

R

IR3 

k l=



kl

:

Our

second main result asserts that the rst solution c;1 _{of (DFc) found in}

[6], whose energy level will be denoted

E

_c1;DF, can be considered, in some (weak) sense, as a ground state for (DFc). Indeed,

E

c1;DF ?

Nc

2 converges

to the minimum of EHF as

c

goes to innity. Moreover, for

c

large the

multipliers

"

c;1_k associated to c;1 _{are the}

_N

_{smallest positive eigenvalues of}

Theorem 5

. Let

N < Z

+ 1 and

c

suciently large. With the above notations,

E

c1;DF := Ec( c;1) = min Gram L 2=1IN EHF() +

Nc

2₊

_o

_(1)c !+1 (15)

Moreover, for any subsequence f c n;1 g n converging in 

H

1_(IR3

_;

_CI4₎N to some ? 1 0

, 1 _{is a ground state of the Hartree-Fock model, i.e.}

EHF( 1_{) = min} GramL 2=1IN EHF()

:

(16)

Furthermore, for

c

large, the eigenvalues corresponding to c;1 _{in (DFc),}

"

c;11

;:::;"

c;1N are the smallest positive eigenvalues of the linear operator

H

c; c;1

and the (

N

+1)-th positive eigenvalue of this operator is strictly larger than

"

c;1_N .

Finally, we are able to show that, for

c

large enough, the function c;1

can be viewed as an electronic ground state for the Dirac-Fock equations in the following sense: it minimizes the Dirac-Fock energy among all electronic congurations which are orthogonal to the \Dirac sea".

Theorem 6

. Fix

N;Z

with

N < Z

+1 and take

c

suciently large. Then c;1 _{is a solution of the following minimization problem:}

inffEc( ); Gram L 2 = 1IN

;

 ? = 0g (17) where ?

=



(?1;0)(

H

c; ) is the negative spectral projector of the operator

H

c; , and ? := (? 1

;

;

 ? N)

:

The constraint ?

= 0 has a physical meaning. Indeed, according

to Dirac's original ideas, the vacuum consists of innitely many electrons which completely ll up the negative space of

H

c; : these electrons form

the \Dirac sea". So, by the Pauli exclusion principle, additional electronic states should be in the positive space of the mean-eld Hamiltonian

H

c;

:

The proof of Theorem 6 will be given in Section 4. This proof uses some other interesting min-max characterizations of c;1 _{(see Lemma 9).}

2. The nonrelativistic limit

This section is devoted to the proof of Theorem 3. We rst notice that when

N < Z

+1,

N;Z

xed, and

c

is suciently large, any solution of (DFc)

is actually in (

H

1_(IR3₎₎N

. This follows from the fact that for



small, the operator

H

1?



j

x

j

is essentially self-adjoint with domain

H

1_(IR3_{) (see [14]).}

We can also obtaina prioriestimates on

H

1 norms:

Lemma 7

. Fix

N;Z

2

ZZ

+, take

c

large enough, and let c be a solution

of (DFc) . If the multipliers

"

ck associated to c satisfy 0 

"

ck 

c

2

(

k

= 1

;:::;N

), then c2(

H

1(IR

3

_;

_CI4

))N _{, and the following estimate holds}

jj c jj 2 2 + jjr c jj 2 2 

K :

The constant

K

is independent of

c

(for

c

large).

Proof.

The normalization constraint GramL 2 c_{= 1I} N implies jj c jj 2 2 =

N :

(18)

Using the (DFc) equation and the standard Hardy inequality

Z IR3

u

2 j

x

j2  4 Z IR3 jr

u

j 2

_;

(19)

one easily proves that c_{is in}

_H

1 _{, and satises:}

(

H

c c

;H

c c) =

c

4jj c jj 2 2+

c

2 jjr c jj 2 2 (20) 

c

4 jj cjj 2 2 +

`

(

Z

2₊

_N

2₎ jjr cjj 2 2+

`c

2_max(

_N;Z

₎ jjr cjj 2

;

for some

` >

0 independent of

N;Z

and

c

. The estimates (18) and (20)

prove the lemma. tu

Proof of Theorem 3.

Let us split the spinors _nk : IR3 !CI4 in blocks of

upper and lower components : _nk=

'



nk_nk

!

denote

L

:=?

i

(



r)

:

Then we can rewrite (DFc

n) in the following way: 8 > > > > > > > > > > > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > > > > > > > > > > > :

c

n

L

nk?

Z





 1 j

x

j 

'

nk+N X l=1( j

'

nlj 2 +j



nlj 2 ) 1 j

x

j 

'

nk+ (

c

n)2

'

nk ? N X l=1

'

nl(

x

) Z IR3 (

'

_nl)(

y

)

'

nk(

y

) + (



nl)(

y

)



nk(

y

) j

x

?

y

j

dy

=

"

nk

'

nk

c

n

L'

nk?

Z





 1 j

x

j 



_nk+N X l=1( j

'

nlj 2 +j



nlj 2 ) 1 j

x

j 



_nk?(

c

n) 2



_nk ? N X l=1



nl(

x

) Z IR3 (

'

_nl)(

y

)

'

ck(

y

) + (



cl)(

y

)



ck(

y

) j

x

?

y

j

dy

=

"

_nk



_nk Z IR3(

'

nk) 

'

nl+ (



nl)



nl

dx

=



kl (21) Note thatk

L

k L 2 = kr



k L 2 for all



2

H

1(IR 3

_;

_CI2

)

:

So, dividing by

c

n the

rst equation of (21), we get kr



nkk L 2 (IR 3 ;CI 2 ) =

O

(1

=c

n)

:

(22)

Dividing by 2(

c

n)2 the second equation of (21), and using the fact that

"

_nk?(

c

n)2 is a bounded sequence, we get



nk ? 1 2

c

n

L'

nk L 2 (IR 3 ;CI 2 ) = 1₍

_c

_n)2

O

XN l=1 k



nlk H 1 (IR 3 ;CI 2 ) !

:

(23)

The estimate (23) together with Lemma 7 implies

k



nkk L 2 (IR 3 ;CI 2 ) =

O

(1

=c

n)

:

(24)

Combining this with (22), we obtain

k



nkk H 1 (IR 3 ;CI 2 ) =

O

(1

=c

n)

:

(25) So XN l=1 k



nlk H 1 (IR 3 ;CI 2

) =

O

(1

=c

n), and (23) gives the estimate



nk ? 1 2

c

n

L'

nk L 2 (IR 3 ;CI 2 ) =

O

(1

=

(

c

n)3)

:

(26)

Now, the rst equation of (21), combined with (26), implies 8 > > > > > > > > > > > > > < > > > > > > > > > > > > > : ?

'

_nk 2 ?

Z

(



 1 j

x

j )

'

nk+N X l=1 j

'

nlj 2  1 j

x

j 



nk ? N X l=1

'

nl(

x

) Z IR3 (

'

_nl)(

y

)

'

nk(

y

) j

x

?

y

j

dy

=



nk

'

nk+

h

nk

;

Z IR3(

'

_nk)

'

nl =



kl+

r

nkl

;

with



_nk :=

"

_nk?(

c

n)2, and lim n!+1 jj

h

nkjj H ?1 (IR 3 ) = 0

;

nlim !+1 j

r

nklj= 0 for all

k;l

2f1

;::: ;N

g

:

Therefore n

:= (

'

n1

;:::;'

_nN) is a Palais-Smale sequence for the Hartree-Fock problem, and the multipliers



_nk satisfy limn!+1



nk

<

0

:

At this

point, we just invoke an argument used in [11] to obtain the convergence in

H

1 _{norm of some subsequence}f n

0

g towards  = (

'

1

;

;

'

N), a solution

of the Hartree-Fock equations

8 > > < > > : H  

'

k= 



k

'

k

; k

= 1

;:::N

Z IR3 

'

 k

'

l =



kl

;

where 



k= lim_n0 !+1



n 0 k .

Finally, let us prove thatEc n

0(

n0

)?

N

(

c

n0)2 converges toEHF()

:

From

Lemma 7 and the estimate (26), one easily gets

Ec n(

n₎?

Nc

2n=EHF(n) +

O

(1

=

(

c

n)2)

:

(27)

Since n0

converges in

H

1 _{norm to , the energy level} EHF(n 0

) con-verges to EHF()

:

So (27) implies the desired convergence. This ends the

3. Ground state for Dirac-Fock equations in the nonrelativistic

limit.

The aim of this section is to prove Theorem 5. The estimate given in Proposition 2 on the energyEc( c;i) and the expression of ^



c given in (4),

will be crucial.

Proof of Theorem 5

. By Corollary 4, for any sequence

c

ngoing to innity,

cn;1is precompact in

H

1 norm. If it converges, its limit is of the form (   1 0), and  Ec n( cn;1) ?

N

(

c

n)2  converges toEHF(1)

:

As a consequence, lim c!+1  Ec( c;1)?

Nc

2  inf GramL 2=1IN EHF()

:

(28)

In order to prove (15) and (16) of Theorem 5, we just have to show that lim c!+1  Ec( c;1)?

Nc

2   inf Gram L 2=1I N EHF()

:

(29) Take  = (

'

1

;

;'

N) 2 

H

1_(IR3

_;

_CI2₎N , with GramL 2 = 1IN

:

Let

V

c be the complex subspace of

E

+c dened by

V

c:= Span f+c( ' 1 0)

;:::;

+c( ' N 0 ) g

:

(30)

From formula (4) and Lebesgue's convergence theorem, one easily gets, for

k

= 1

;:::;N ;

lim c!+1 k ? c(' k 0) k H 1 = 0

:

(31)

So, for

c

suciently large, we have

dim

V

c =

N :

(32) Hence, by (5),

E

c1;DF :=Ec( c;1₎  sup 2(E ? V c) N Gram L 2 1I N Ec( )

:

(33) Let +2(

E

+c) N

;

? 2(

E

? c)N

such that GramL 2(

+₊ ?) 1I

N. By

the concavity property of Ec in the

E

?

is large enough, we have Ec( +₊ ?)  Ec( +_{) +} E 0 c( +)
? ? 1 4 N X k=1( ? k

;

p ?

c

2
+

c

4 ? k)  Ec( +) +

M

jj ? jj L 2 ?

c

2 4jj ? jj2 L 2

;

(34)

for some constant

M >

0 independent of

c

. Hence, for

c

large,

E

c1;DF  sup + 2D(V c) Ec( +_{) +} (1)c !+1

;

(35) where

D

(

V

c) := n +2(

V

c)N ; Gram L 2 +1I N o .

If

c

is large enough, it follows from Hardy's inequality (19) that the map + !Ec( +) is strictly convex on the convex set

A+:=f +2(

E

+c)N; Gram L 2 +1I N g

:

Indeed, its second derivative at any point + _of A+ is of the form E 00 c( +)[

d

+]2= 2 N X k=1(

d

k

;

p

c

4?

c

2

d

k) L 2 +

Q

(

d

+₎

with

Q

a quadratic form on (

H

1=2_(IR3

_;

_CI4₎₎N _{bounded independently of}

_c

and +2A+

:

As a consequence, sup

+ 2D(V

c)

Ec( +) is achieved by an extremal point +max

of the convex set

D

(

V

c) =A+\(

V

c)N. Being extremal in

D

(

V

c)

;

the point

_+max satises

GramL

2 +max= 1I N

:

(36)

Since _+k;max 2

V

c

;

there is a matrix

A

= (

a

kl)

1k ;lN such that, for all

l

,

+l;max = X 1kN

a

kl +c(' k 0)

:

Then

A

 Gram L 2  +c( 0) 

A

= GramL 2 +max = 1I N

:

(37)

Using the

U

(

N

) invariance of

D

(

V

c) and Ec

;

and the polar decomposition

of square matrices, one can assume, without restricting the generality, that

from (31), that GramL 2(+c ? 0
) = 1IN+o(1)

:

So (37) implies

A

2_{= 1I} N+o(1)

;

hence

A

= 1IN+ o(1)

:

Combining this with (31), we get k +k;max?( ' k 0) k H 1 =

o

(1)c !+1

:

Now, since +k;max 2

E

+c

; H

c +k;max= p

c

4?

c

2
+k;max

:

But p

c

4?

c

2


c

2?
2

:

This inequality is easily obtained in the Fourier domain: it follows from

p 1 +

x

1 + x2 (8

x

0)

:

So we get N X k=1(

H

c +k;max

;

+k;max)L 2 

Nc

2_{+ 12} XN k=1 kr +k;maxk 2 L 2

:

Combining this with (31), we nd

Ec( +max)

Nc

2₊

EHF() +(1)c !+1

:

(38)

Finally, (35) and (38) imply

E

c1;DF  Ec( +max) +(1)c !+1 

Nc

2₊ EHF() +(1)c !+1

:

(39)

Since  is arbitrary, (39) implies (29). The formulas (15), (16) of Theo-rem 5 are thus proved.

We now check the last assertion about the

"

c;1_k

;k

= 1

;:::;N;

being the smallest eigenvalues of the operator

H

c; c;1 for

c

large. By Corollary

4, we can translate this statement in the language of sequences. We take a sequence

c

n ! +1 such that f c

n;1 g n converges in 

H

1_(IR3

_;

_CI4₎N to some ? 1 0

, for

n

large enough. Let

H

n :=

H

cn;

cn;1 and H 1 := H 1

:

We have

H

n cn;1 k =

"

nk cn;1 k and H 1

'

1k= 



k

'

1k

;

with 0

< "

_n1

:::



"

nN

<

(

c

n) 2

_;

_

_

₁ 

:::





N

<

0

;



k= lim n!+1 (

"

_nk?(

c

n) 2₎

_:

Let us denote

e

n1 

:::



e

ni 


the sequence of eigenvalues of

H

n

;

in the interval (0

;c

2n)

;

counted with multiplicity. Similarly, we shall denote 



1 

:::





i 


the sequence of eigenvalues of H

(?1

;

0)

;

counted with multiplicity. Let

z

2 CIn



(H

1)

:

Then for

n

large

enough,

z

+ (

c

n)2 2CIn



(

H

n)

;

and the resolvent

R

n(

z

+ (

c

n)2) :=



(

z

+ (

c

n)2)

I

?

H

n 

?1

converges in norm towards the operator

L

(

z

) : ?'


! ?R(z)' 0

;

where 

R

(

z

) :=

z I

?H 1  ?1 is the resolvent of H

1

:

So, by the standard spectral

theory, lim_n

!+1

(

e

ni?(

c

n)

2_{) = }

_

_{i for all}

_i

1

:

We know that 1 _{is a ground state of the Hartree-Fock model. So a}

result proved in [1] tells us that 



k= 



k for all 1

k



N

, and 



N+1

>





N

:

But (

"

_nN ?(

c

n)2) converges to 



N

;

and (

e

nN+1?(

c

n)2) converges to 



N+1

;

as

n

goes to innity. So, for

n

large enough,

e

_nN+1

> "

_nN

;

hence

"

_nk=

e

_nk for all 1

k



N :

This ends the proof of Theorem 5. tu

4. Proof of Theorem 6

In this section, both  and will denote

N

-uples of 4-spinors (functions from IR3 _{into CI}4_{). As already noted above, in [6] the solution} c;1 _was

obtained by a complicated min-max argument. For

c

large, we will show that this argument can be simplied.

First of all, we introduce the notion of projector \

"

-close to +c", where +c_{= 12}j

H

cj

?1

(

H

c+j

H

cj 

is the positive free-energy projector.

Denition 8

: Let

P

+ _{be an orthogonal projector in}

_L

2_(IR3

_;

_CI4_{), whose}

restriction to

H

1

2(IR3

;

CI4) is a bounded operator on

H

1

2(IR3

;

CI4) .

Given

" >

0,

P

+ _is

_"

_{-close to +c if and only if, for all} 2

H

1 2(IR3

;

CI4),  ?

c

2
₊

_c

4 1 4 

P

+ ?+c  L2(IR 3;CI4) 

"

 ?

c

2
₊

_c

4 1 4 L2(IR 3;CI4)

:

An obvious example of projector

"

-close to _+c is _+c itself. More inter-esting examples will be given below. Let us now give a min-max principle associated to

P

+ _:

Lemma 9

. Fix

N;Z

with

N < Z

+ 1. Take

c >

0 large enough, and

P

+

a projector

"

-close to_+c, for

" >

0 small enough. Let

P

?= 1I L 2 ?P+, and dene

E

(

P

+_{) := inf} + 2(P +H 1 2) N Gram  =1I sup 2(P ?H 1 2 Span( +))N Gram =1I Ec( )

:

Then

E

(

P

+_{) does not depend on}

_P

+ _and Ec( c;1)

E

(

P

+).

Remark.

In the case

N

= 1

;

Ecis the quadratic form (

;H

) L

2 associated

to the operator

H

=

H

c?

Z

 1

jxj

:

Then

E

(+c) coincides with the min-max

level



1(

V

) dened in [4], for

V

= ?

Z



1

jxj

:

By Theorem 3.1 of [4], if

c > =

2 + 2₂

=

, then



1(

V

) is the rst positive eigenvalue of

H :

Proof of Lemma 9

.

The idea behind this lemma is inspired by [2]. Note that, under our assumptions,

E

(

P

+₎

_{< Nc}

2_{(1 +}

_K"

_{) for some}

_{K >}

_{0 independent of}

_c

and

"

. This follows from arguments similar to those used in the proof of Lemma 5.3 of [6]. In [6] the free energy projectors 

c were used. With

these projectors, it was seen that

E

(_+c)

< Nc

2 _{(thanks to a careful choice}

of +_{). When}

_P

+ _is

_"

_{-close to +c, we then get}

_E

₍

_P

+₎

_{< Nc}

2_{(1 +}

_K"

_).

To continue the proof of the lemma we perform a change of physical units. In mathematical language, this change corresponds to a dilation in space by the factor

c

, and to dividing the energies by

c

2_{. Let (}

_d

_c

_'

₎₍

_x

_{) =}

_c

3=2

_'

₍

_cx

₎

and e Ec() : = 1 c2 Ec 

d

c  = XN k=1 Z IR3 

'

k

;

(?

i

r+



)

'

k  ?

Z

c

 e



 1 j

x

j  j

'

kj 2 + 12

c

ZZ IR3 IR 3



(

x

)



(

y

)?k

R

(

x;y

)k2 j

x

?

y

j

d

3

_xd

3

_y

(40)

where ~



(

E

) =



(

c

?1

E

) for any Borel subset

E

of IR3.

The interest of this rescaled energy e

Ecis that for

c

large and Gram L 2 1I N, we have e Ec( ) = N X k=1 Z IR 3  k

;

(?

i

r+



) k  +

O

1

c

jj jj 2 (H 1=2 ) N 

:

(41) Let us denote

P

e :=

d

c?1

P

 

d

c, e  :=

d

c?1  c 

d

c=



IR   ?

i:

r+





:

Note that e

 does not depend on

c

. Now,

P

+ is

"

-close to +c if and only

if 8 > > < > > :  ?
+ 1 1 4  e

P

+? e + L2(IR 3;CI4) 

"

 ?
+ 1  1 4 L2(IR3;CI4)

;

8 2

H

1 2(IR3

;

CI4)

:

(42)

We denote 

A

the right action of an

N



N

matrix

A

= (

a

kl) 1k ;lN

on an

N

-uple  = (

'

1

;:::;'

N)2(

L

2(IR3

;

CI4))N

:

More precisely,

(

A

) := ( N X k=1

a

k1

'

k

;:::;

N X k=1

a

kN

'

k)

:

(43) Given + _{= (}

_'

₊₁

_;:::;'

_+N) 2  e

P

+

_H

1=2N

such that GramL 2 + _{= 1I} N

;

and ? 2  e

P

?

H

1=2 N

;

we dene

g

 +( ?) := (++ ?)  h GramL 2 ( +_{+ }?) i ? 1 2 = (+_{+ }?)  h 1IN + Gram L 2  ? i ? 1 2

:

(44)

We obtain a smooth map

g



+

;

from  e

P

?

H

1 2 N to  + := n 2  e

P

?

H

1 2 Span(

'

+1

;:::;'

+N) N

=

GramL 2 = 1IN o

:

In fact, the values of

g



+ lie in the following subset of   + : 0  + := n 2  +

=

Gram L 2  e

P

+ 

>

0o

:

Now, take an arbitrary 2  0

+

:

Then there is an invertible

N



N

matrix

B

such that

P

e+ = +



B

. So we may write 

B

?1= ++ e

P

? 

B

?1

:

As a consequence,

g

 +( e

P

? 

B

?1) = ( 

B

?1)  h GramL 2 ( 

B

?1) i ? 1 2

:

One easily computes GramL 2 ( 

B

?1) = (

B

)?1  GramL 2 

B

?1 = (

B B

)?1

:

Hence

g

 +( e

P

? 

B

?1 ) = ( 

B

?1 )(

B B

 )1=2 = (

B

?1 (

B B

 )1=2)

;

and nally =

g

(

P

B

1₎

_{U ;}

where

U

:= (

B B

)?1=2

B

2 U(

N

) is the unitary matrix appearing in the

polar decomposition of

B :

So we have proved that 0 + = [ ? 2( e P?H 1 2) N U2U(N)

g

 +( ?) 

U :

Now, Ec is invariant under the U(

N

) action \" , and  0

+ is dense in



+ for the norm of (

H

1=2_(IR3

_;

_CI4₎₎N _{. Hence} sup 2( e P?H 1 2 Span( +))N GramL 2 = 1IN e Ec( ) = sup ? 2( e P?H 1 2) N e Ec 

g

 +( ? )

:

(45)

We now prove Lemma 9 in three steps.

Step 1

. Let + 2 ( e

P

+

_H

1=2₎N _{be such that Gram}

L 2

+ _{= 1I}

N and such

that e

Ec(+)

N

+



, for some

 >

0 small. For

"

small and

c

large, there

is a unique ? 2  e

P

?

H

1=2 N maximizing e Ec

g



+ and lying in a small

neighborhood of 0. If we denote

k

(+_{) this maximizer, the map}

_k

_{is smooth}

from S+ = n + 2  e

P

+

_H

1=2N /GramL 2  + _{= 1I} N

;

e Ec( +₎ N +



o to e

P

?

H

1=2 N

, and equivariant for the U(

N

) action.

Proof of Step 1

. Take

r >

0. For

"; 

small and

c

large, if + 2 S+

;

? 2( e

P

?

H

1=2)N

;

and k ? k H

1=2 is not smaller than

r

, then e Ec 

g

 +( ?) 

< N

?

 ;

by (41). On the other hand, for

c

large enough, using (41) once again, one has e Ec 

g

 +(0)  = e Ec(+) 

N

?



2

:

So, if we deneVr := n ? 2  e

P

?

H

1=2 N /k ? k H 1=2 

r

o , no maximizer of e Ec

g

 + can be outside

Vr. Moreover, choosing

r

small, and then taking

c

large and

"

small, the map ? 2Vr7?! e Ec

g

 +( ?)

is strictly concave. Indeed, its second derivative at ?

2Vr is very close in

norm to the negative form ? 2( e

P

?

H

1=2 )N 7?!?2 N X i=1 k ? i k2 H 1=2 ? 2 X 1i;jN (

'

+j

;'

+i)H 1=2( ? i

;

? j )L 2

:

Step 1 immediately follows from these facts. tu

Step 2

. The min-max level

E

(

P

+_{) does not depend on}

_P

+_.

Proof of Step 2

. Take two projectors

P

₁+_,

_P

₂+_{, both}

_"

_{-close to +c. For}

i

= 1

;

2, and _+i2  e

P

_i+

_H

1=2N , with GramL 2+i= 1IN and e Ec(+i)

N

+

 ;

let

J

i_{(+i) := max} ? 2( ~P ? i H 1=2)N Gram L 2 ?=1I N e Ec 

g

i + i (?)  = e Ec

g

i + i 

k

i_(+i)

:

(46) Here,

g

i_ + and

k

i _{are the maps associated to}

_P

_i+ _{in Step 1.}

By Ekeland's variational principle [5], there is a minimizingsequence

+1;n

n0

for

J

1_{, such that (}

_J

1₎0  _+1;n ?! n?!+1 0 in 

H

?1=2 N . Let n:=

g

_1+ 1;n 

k

1_(+1;n) . Then n is a Palais-Smale sequence for e

Ec in the manifold  :=n 2 

H

1=2N /GramL 2 = 1I N o

;

with e Ec  n  

N

?



2

;

where

 >

0 is the constant of the rst step. So GramL 2  e

P

₂+ _n

>

0

:

We denote 8 > < > : _+2;n:=

P

e+ 2 n h GramL 2  e

P

₂+ _ni ? 1 2

;

? 2;n :=

P

e ? 2 n h GramL 2  e

P

₂+ _ni ? 1 2

:

(47)

One easily checks that n =

g

2

 + 2;n (? 2;n)

:

Since e Ec  n  

N

?



2

;

we have k ? 2;nk H 1=2



r

, where

r >

0 is the same as in the proof of step 1. Since

n is a Palais-Smale sequence for e

Ec, the derivative of e

?

2;n converges to 0 as

n

goes to innity. So, by the concavity properties of

e Ec

g

2  + 2;n in the domain V2;r := n ? 2  e

P

? 2

H

1=2 N /k ? k H 1=2 

r

o

(see the proof of step 1), we get

k ? 2;n?

k

2_(+2;n) k H 1=2 ?! n!+1 0 and e Ec  n  ?

J

2 _+2;n ?! n!+1 0

:

As a consequence,

E

(

P

₁+_{) = inf} + 1 2( ~P + 1 H 1=2)N Gram L 2 + 1=1IN

J

1 ₊₁  inf + 2 2( ~P + 2 H 1=2)N Gram L 2 + 2=1IN

J

2_{(+2) =}

_E

₍

_P

₂+₎

_:

Since 1

;

2 play symmetric roles in the above arguments, we conclude that

E

(

P

+_{) does not depend on}

_P

+_{, for}

_c

_{large enough and}

_"

_{small enough.} tu

Step 3

. Ec  c;1 

E

 _+c

, where c;1 _{is the rst solution of (D-F) found}

in [E-S].

Proof of Step 3

.

For

c

large enough, if ? 2 ? c

H

1=2satises GramL 2 ? 1I N, it follows

from Hardy's inequality that the map +!Ec( ++

?) is strictly convex on

W

( ?) := f + 2(+c

H

1=2)N ; Gram L 2( +₊ ?) 1I N g

:

As a consequence, for an arbitrary

N

-dimensional subspace

V

of +c

H

1=2_,

S

V( ?) := sup + 2W( ?) \V N Ec(

+ ₊ ?) is achieved by an extremal point

_+max of the convex set

W

( ?)

\

V

N. Being extremal, +max must satisfy

the contraints GramL

2( +max+ ?) = 1I N

:

So we have sup 2( ? cH 1=2 V ) N Gram L 2 1I N Ec( ) = sup ? 2( ? cH 1=2)N Gram L 2 ? 1I N

S

V( ?) = sup 2( ? cH 1=2 V ) N Gram L 2 =1IN Ec( )

:

By proposition 2, Ec  c;1  sup 2( ? cH 1=2 V ) N GramL 2 1I N Ec  

:

Finally we get, for

c

large,

Ec( c;1) inf + 2( + cH 1=2)N Gram L 2 +=1I N sup 2( ? cH 1=2 Span( +))N Gram L 2 =1IN Ec( ) =

E

(+c)

:

(The correspondence between + _and

_V

_is

_V

_{= Span(}+₎₎

_:

_{This ends the}

proof of Step 3 and of Lemma 9. tu

Thanks to Lemma 9, we are able to write the following inequalities for

c

large, and

P

+

_"

_{-close to +c,}

_"

_{small :}

E

(

P

+_{) =}

_E

_(+c)  Ec( c;1)  inf solution of (DFc) ? = 0 Ec( )  inf 8 > < > : 2(H 1=2)N Gram L 2 =1I N ? =0 Ec( ) (48)

As announced before, we now give some important examples of projectors

"

-close to _+c :

Lemma 10

. Fix

N;Z

, and take

c

large enough. Then, for any  2 

H

1=2N , with GramL 2  1I N, the projector  +  =



(0;+1) 

H

c;  is

"

-close to +c.

Proof of Lemma 10

. We adapt a method of Griesemer, Lewis, Siedentop [7] to the Hamiltonian

H

c;. Once again, it is more convenient to work in a

system of units such that

H

c; becomes

e

H

_c;~ : 7!

d

c?1

H

c;

d

c( ) =  ?

i

r+



 ?

Z

c

 e



 1 j

x

j  +1

_c





_~

_x

1  1

c

Z IR

R

~(

x;y

)

x

(

y

)

y

dy

with



e(

E

) =



(

c

?1

E

), e (

x

) =

c

?3=2(

c

?1

x

). Denoting

H

1 := ?

i

r+



, e +~ :=



(0;1)  e

H

_c;~ , e+ :=



(0;1)(

H

1),

K

_~:=

c

 e

H

_c;~?

H

1 

, we nd, as in the proof of Lemma 1 of [7],

 e _+~? e + = 1

_c

Z + 1 0

dz

h

H

21+

z

2i ?1 

H

1

K

_~

H

e c;~?

z

2

K

~ h e

H

_c;~2 +

z

2i ?1

;

and for any



2

L

2(IR3

;

CI4), following [7] (proof of Lemma 3), we get 

;

(?
+ 1)1=4( e +~? e +)  L2 

M

c

k



k L 2 k(?
+ 1)1=4 k L 2

for

c

large enough (

M

is a constant independent of

c

). As a consequence, if

c

is large enough and bigger than

M

_"

, then +_{ is}

_"

_{-close to +c. This ends}

the proof of Lemma 10. tu

Now, to end the proof of Theorem 6, we just need the following result :

Lemma 11

. Fix

N; Z

and take

c >

0 large enough. If  2 

H

1=2N , GramL 2 = 1IN,  ?  = 0 and Ec()

Nc

2, then Ec() = max n Ec( ); 2 h ? 

H

1=2Span() iN

;

GramL 2 = 1I N o

:

Proof of Lemma 11

. If ?  = 0 and GramL 2 = 1IN, then 0 is a critical

point of the map ? 2  ? 

H

1=2 N 7?!Ec 

g

( ?) 

;

with

g

( ?) =   + ?   h 1IN + Gram L 2 ? i ?1=2

:

Take

" >

0 small. By Lemma 10, + is

"

-close to +cfor

c

large enough. From the proof of Lemma

9 (Step 1), there is a unique critical point of Ec

g

 in a small neighborhood Vrof 0 in 

?

(

H

1=2) and this critical point is the unique maximizer of Ec

g



in ?

(

H

1=2). So, 0 is this maximizer. This proves Lemma 11. tu

Let us explain why Theorem 6 is now proved. We know that, for

c

large enough,

Nc

2

_{> E}

 +c  Ec( c;1)  inf 8 > < > : 2(H 1=2)N Gram L 2 =1IN ? =0 Ec( )

;

hence inf 8 > < > : 2(H 1=2)N Gram L 2 =1I N ? =0 Ec( ) = inf 8 > > > < > > > : 2(H 1=2)N GramL 2 =1IN ? =0 Ec( )Nc 2 Ec( )

:

Take

" >

0. By Lemma 10, for any 2(

H

1=2)N with Gram L

2 = 1IN,

the projector ₊ is

"

-close to _+c, if

c

is large. Hence

E

(₊ ) =

E

(_+c) (by Lemma 9), if we have chosen

"

small enough. But if also satises ?

= 0

and Ec( ) 

Nc

2, then, from Lemma 11 and from the denition of

E

(+ ),

we have

E

(_+c) =

E

(+ ₎ Ec( ). So

E

 +c  inf 8 > < > : 2(H 1=2)N Gram L 2 =1IN ? =0 Ec( )

;

and therefore,

E

 +c = Ec( c;1) = inf 8 > < > : 2(H 1=2)N GramL 2 =1IN ? =0 Ec( )

and Theorem 6 is proved. tu

Aknowledgements

. The authors are grateful to Boris Buoni for explain-ing them the work [2], and suggestexplain-ing that it might be useful in the study of the Dirac-Fock functional. The proof of Lemma 9 is inspired by this paper.

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