Asymptotic Neutrality of Large IonsCharles L. Fe�erman*andLuis A. Seco**Department of Mathematics, Princeton University

Consider the Hamiltonian for a nucleus of charge Z and N quantized electrons,HZ;N = NXi=1�(��xi)� Zjxij�+ 12Xi6=j 1jxi � xj j = ��+ VCoulomb:The ground state energy is thenE(Z) = infN E(Z;N) = infN inf 2Hjj jj2=1 hHZ;N ; iwhere H = ^Ni=1(L2(R3)Cq) is the space of antisymmetric wave functions with q spins.Throughout this paper we will simply refer to them as \antisymmetric" wave functions.For each Z, call N(Z) the smallest number for which E(Z) = E(Z;N).It is an interesting problem to obtain sharp estimates for N(Z). The sharpest knownresult appears in [8], where the reader will �nd a discussion of the history of the problem.In particular, N(Z)=Z ! 1 as Z ! 1, although there were no estimates for the rate ofconvergence. Our main result is the following:* Partially supported by a NSF grant at Princeton University** Supported by a Sloan Foundation Dissertation Fellowship at Princeton University1

Theorem : N(Z) = Z +O(Z�) for � = 4756We announced this result in [1]. We are grateful to V. Bach for pointing out a minor errorin our announcement. The rest of the paper is devoted to the proof of this theorem.For the proof we will be interested only in the case Z � N � 2Z. Lieb ([6] and [7]) hasgiven a simple argument to handle the case N > 2Z.Thomas-Fermi theory will play a central role in the proof. We recall a few fundamentalfacts. For a nice detailed discussion see [5].1. E(Z) � CTF Z7=3 as Z !1.2. Let �TF be Thomas-Fermi density. Then:a. RR3 �TF(x) dx = Zb. For jxj > Z�1=3+", �TF(x) � Cjxj�6.c. Rjxj>R �TF(x) dx � C�TFR�3 for R > Z�1=3 , for some constant C�TF .1. Key EstimateFix N points in R3, x1; : : : ; xN . Take a radially symmetric function �, supported inB(0; 1100Z�2=3), R � = 1, sup j�j � CZ2, and set �x1;:::;xN (x) =Pi �(x� xi).Observe that the subharmonicity and positive-de�niteness of the Coulomb potential impliesthatXi<j 1jxi � xjj � 12 ZZ �x1;:::;xN (x)�x1;:::;xN (y)jx� yj dx dy � CZ2=3 �N= 12 ZZ ��x1;:::;xN � �TF�(x)��x1;:::;xN � �TF�(y)jx� yj dx dy+ ZZ �x1;:::;xN (x)�TF(y)jx� yj dx dy � 12 ZZ �TF(x)�TF(y)jx� yj dx dy � CZ2=3 �N= c Z j�j�2j�̂x1;:::;xN (�)� �̂TF(�)j2 d�2

+Xj W (xj)� 12 ZZ �TF(x)�TF(y)jx� yj dx dy � CZ2=3 �Nwith W (x) = ZZ �(x� z)�TF(y)jz � yj dz dyLet's set K(x1; : : : ; xN ) = 12 ZZ (�x1;:::;xN � �TF)(x)(�x1;:::;xN � �TF)(y)jx� yj dx dyand note that K � 0 pointwise. This provides the operator inequalityHZ;N � K + NXi=1�(��xi)� Zjxij +W (xi)�� 12 ZZ �TF(x)�TF(y)jx� yj dx dy � CZ5=3 :for N � 2Z.We point out that similar inequalities were used in [2] and [4].Let �1; �2; � � � be the negative eigenvalues of(��x)� Zjxj +W (x)and let ~E(Z) = q 1Xi=1 �i � 12 ZZ �TF(x)�TF(y)jx� yj dx dyIt follows then that HZ;N � ~E(Z) + O(Z5=3)for all N between Z and 2Z. From [3] we know that for some b between 1=3 and 2=3 we have~E(Z) � CTF Z7=3 + q8Z2 + O(Z7=3�b)A careful exposition of Hughes' proof appears in [10].Similarly, it follows from [9] thatE(Z;N) � CTF Z7=3 + q8Z2 + O(Z7=3�b)for all N � Z. 3

Putting all this together we see that~E(Z) � E(Z;N) +O(Z7=3�b)for any N between Z and 2Z. In particular, we haveHZ;N � E0(Z) +K(x1; : : : ; xN)� CHSW(Z7=3�b) (1)for E0(Z) = infN�(1+�])ZE(Z;N)where �] will be picked later, and some constant CHSW.The constant b plays a crucial role in the analysis of the best possible power of Z for theexcess charge. From [3] and [9] it follows that we can take b = 3=8: this implies that wecan take � = 47=56 in the statement of the theorem. Notice however that this value of bhas been obtained using a much stronger result, namely the correct asymptotics for theenergy. It is clear that one can do better and maybe one can take b = 2=3, which wouldallow us to take � = 5=7.2. Estimates for a BallWhat we are planning to do now is conclude that if the number of electrons a particularstate puts inside a ball is too di�erent from what Thomas-Fermi theory predicts, then thisstate will have too much energy. This will be then generalized to random variables otherthan the number of electrons inside a ball. We need a few de�nitions.Consider a ball B(0; R). De�neNR(x1; : : : ; xN ) = number of xi 2 B(0; R)Also, take a smooth function � = � 1 for jxj < 4=3R0 for jxj > 5=3Rthat we will call �R whenever we want to make it explicit which R is being used, and letN�(x1; : : : ; xN ) =Xi � � �(xi) = Z �x1;:::;xN (x) � �(x) dx4

Observe that NR=2(x1; : : : ; xN ) � N�(x1; : : : ; xN) � N2R(x1; : : : ; xN ), for all x1; : : : ; xN ,provided R > 3Z�2=3 , which will certainly hold in our case, since we will be working withR � Z�1=3 .De�ne NTF� = Z �TF(x)�(x) dx:Now, note thatN�(x1; : : : ; xN )�NTF� = Z ��x1;:::;xN � �TF�(x)�(x) dx = Z ��̂x1;:::;xN � �̂TF�(�)�̂(�) d�:HencejN� �NTF� j2 � Z j�̂(�)j2 � j�j2 d� Z j(�̂x1;:::;xN � �̂TF)j2 � j�j�2 d� � CR �K(x1; : : : ; xN)therefore, (1) impliesDHZ;N ; E � E0(Z) + CK jN� �NTF� j2 ; �R � CHSW(Z7=3�b):for some constant CK. In particular, Cauchy-Schwarz impliesDHZ;N ; E � E0(Z) + CK jN� ; ��NTF� j2R � CHSW(Z7=3�b) (A)for any antisymmetric , jj jj2 = 1.The previous argument can be generalized to yield the following result:Lemma 2.1: Given any function '(x) 2 L2(R3), we haveDHZ;N ; E � E0(Z) + CK ��(� � �� �TF) ; '���2jjr'jj22 � CHSW(Z7=3�b)for 2 H, jj jj2 = 1.Here, � = N Z j (x; x2; : : : ; xN )j2 dx2 : : : dxNin the case of a fully antisymmetric (i.e. q = 1) and in general� (x) = NXi=1 Z j (x1; : : : ; xi�1; x; xi+1; : : : ; xN )j2 Yi6=j dxj5

Now, we de�ne a working variant of estimate (A).De�nition: We say that Estimate(��; �; R) holds if for a nucleus of charge Z at the originand N quantized electrons con�ned to the ball B(0; R) we havehHZ;N ; i � E0(Z) + ��ZR (N � (1 + �)Z)for normalized , where E0(Z) = inf0�N�(1+�])ZE(Z;N)for �] to be picked later. By N quantized electrons con�ned to the ball B(0; R) we meanthat = (x1; : : : ; xN ) is supported in the set xi 2 B(0; R) for all i = 1; � � � ; N .Now it will be necessary to introduce two parameters 1 and 2, in the proof. They arerelated to b by the relation1. 1 = 37 b.2. 2 = b7 .The signi�cance of 1 is that it represents the excess charge. PreciselyN(Z) = Z +O(Z1� 1):On the other hand, 2 is related to the radius of the largest ball for which we can obtainfavorable estimates for its excess charge. This is clearly seen in the following lemma.Lemma 2.2: There exist constants C0 and c0 independent of Z, such thatEstimate(��; �; R) holds for1. �] � � � C0Z� 12. �� � c0Z� 13. R � C0Z�1=3+ 2Proof: Pick R, � and �� within this range. Say con�nes N electrons to B(0; R).6

If N � (1 + �)Z, thenhHZ;N ; i � E(Z;N) � E0(Z)� E0(Z) + ��ZR (N � (1 + �)Z)If N > (1 + �)Z, then hN� ; i > (1 + �)Zfor � = �2R. Since NTF� < Z,CK ��hN� ; i �NTF� ��22R > CK jN � Zj22R > CK �2Z22R> CKC202C0 Z7=3�(2 1+ 2) � 2CHSWZ7=3�bfor C0 large enough; so, DHZ;N ; E � E0(Z) + CK jN � Zj24ROn the other hand,CK jN � Zj24R > ��ZR (N � (1 + �)Z) for N � (1 + �)Z (2)To see this, observe that it holds trivially at N = (1 + �)Z, because the right hand side iszero; if now we di�erentiate both sides with respect to N , we obtain for the left hand sideCK(N � Z)2R > CK2 � �ZRand ��Z=R for the right hand side. So Estimate(��; �; R) holds taking c0 small enough.Throughout the proof we will need a couple more conditions on C0 and c0 that will forceus to take them to be larger and smaller respectively than what we needed for this lemma.Lemma 2.3: Let R = Z�1=3+ 2 , and � = �R. Say that for a number Y ,hN� ; i = Z + Y Z1� 1Then, for some universal constant c1hHZ;N ; i � E0(Z) + CK (jY j � c1)2+ Z2�2 14R � CHSWZ7=3�b7

Proof:Observe that we haveNTF� � Z � Zjxj�R=2 �TF(x) dx � Z � 8C�TFZ1�3 2 = Z � 8C�TFZ1� 1and NTF� � ZThus, ��hN� ; i �NTF� ��2R � �hN� ; i �NTF� �2+R � (Y )2+ Z2�2 1Rand ��hN� ; i �NTF� ��2R � �NTF� � hN� ; i�2+R � (�Y � 8C�TF)2+ Z2�2 1R :Applying (A) to both cases and averaging the resulting inequalities, we gethHZ;N ; i � E0(Z) + CKZ2�2 1 (�Y � 8C�TF)2+ + (Y )2+2R � CHSWZ7=3�bElementary calculus says that(�x� c)2+ + (x)2+ � (jxj � c0)2+2and this implies the conclusion of the lemma.The previous estimates could have been done for R of the form Z�1=3+ for � 2, with theonly e�ect of decreasing 1 and thus worsening a little the estimates for the excess charge.However, they cannot be carried out with these techniques for > 2, since whatever termwe want to estimate will give a contribution so small to the energy that it will simply belost in the O(Z7=3�b). For radii this big we need a di�erent approach in which we use ina more direct way the properties of the Coulomb potential. The goal of next section is toanalyze how estimates for a given ball imply corresponding estimates for its double.3. Estimates for Spherical ShellsIn this section we consider a system of N quantized electrons con�ned to B(0; R), andN 0 electrons con�ned to B(0; 2R)� B(0;R=2). That is, we will be considering wave func-tions (x1; : : : ; xN ; x01; : : : ; x0N 0) supported on the set x1; : : : ; xN 2 B(0; R), x01; : : : ; x0N 0 2B(0; 2R)�B(0;R=2). We have to impose the extra condition that for �xed x01; : : : ; x0N 0 , is8

antisymmetric in the x1; : : : ; xN and viceversa: that is, we will be considering vector-valued , with domain x01; : : : ; x0N 0 2 B(0; 2R)� B(0;R=2), and values in H and vector-valued ,with domain x1; : : : ; xN 2 B(0; R), and values in H.To stress the di�erent role of the two sets of electrons, we rewrite the hamiltonian asHZ;N+N 0 = HZ;N +Hextra = (��x1;:::;xN + V ) + (��extra + Vextra)with �extra = �x01;:::;x0N0V = � Xi=1;:::;N Zjxij + 12Xi6=j 1jxi � xj jVextra = � Xi=1;:::;N 0 Zjx0ij + Xi=1;:::;N0j=1;:::;N 1jx0i � xj j + 12 Xi=1;:::;N0j=1;:::;N0i6=j 1jx0i � x0j jAlso, we restrict our attention to the case R > C0R� = C0Z�1=3+ 2 .The content of the following lemma is as follows:We know from previous estimates that approximately Z of the electrons will organizethemselves to be close to the nucleus; this will have the important e�ect of \screening" thenucleus. That is, all the other electrons will hardly feel any negative electrostatic potential;this has as a consequence that a lot more than Z electrons will only make the energy ofthe system grow above the ground state energy.Lemma 3.1: Assume that N +N 0 � (1 + �)Z forC0Z� 1 � � < �1 + 10�6�C0Z� 1 :Then, for some constant c, hVextra ; i � c�ZR N 0or else hHZ;N+N 0 ; i � E0(Z) + cZ7=3�7 2Proof: We will assume that N 0 6= 0 or else there is nothing to prove.Recall that R� = Z�1=3+ 2 . Let � = �R� . 9

Note that N� = Xi=1;:::;N � � �(xi) + Xj=1;:::;N 0 � � �(x0j) = Xi=1;:::;N � � �(xi)as operators on our space of functions, since � � � and have disjoint support in the x0jvariables.Let Vextra =Pj Vj(x1; : : : ; xN ; x01; : : : ; x0N 0) forVj = � Zjx0j j + Xi=1;:::;N 1jx0j � xij + 12 Xi=1;:::;N0i6=j 1jx0i � x0j jNote that NXi=1 1jx0j � xij = NXi=1 � � �(xi)jx0j � xij + NXi=1 1� � � �(xi)jx0j � xij� Zjx0j j = �Z �N�jx0j j � NXi=1 � � �(xi)jx0j jand writeVj(x1; : : : ; xN ; x01; : : : ; x0N 0) = �Z �N�jx0jj+ NXi=1 1� � � �(xi)jx0j � xij+ 12 Xi=1;:::;N0i6=j 1jx0j � x0ij+ NXi=1 1jx0j � xij � 1jx0jj! � (� � �) (xi)= T1 + T2 + T3 + T4Let's analyze this term by term:�) hT1 ; i � �jZ � hN� ; ijR=2�) T2 � N �N�8R pointwise. 10

�) hT3 ; i � N 0 � 18R .Therefore �T1 + T2 + T3� ; � � N +N 0 � hN� ; i � 16 jZ � hN� ; ij � 18R� �Z � 20 jZ � hN� ; ij+ 9Rfor = N +N 0 � �Z � Z. Note that by hypothesis, � 0.Summing over all j we obtainN 0Xj=1 �T1 + T2 + T3� ; � � �Z � 20 jZ � hN� ; ij+ 9R N 0 (3)If we could prove that jZ � hN� ; ij < cZ1� 1 for some constant independent of C0, thenwe would have �T1 + T2 + T3� ; � � �Z � 20 jZ � hN� ; ij+ 9R� �Z10Rby simply taking C0 large enough.So, the result will follow if we can prove that eitherhHZ;N+N 0 ; i � E0(Z) + cZ7=3�7 2or else: ������Xj hT4 � ; i������ < �Z16RN 0and jZ � hN� ; ij � cZ1� 1with c independent of C0.In order to analyze this, let's de�ne x01;:::;x0N0 (x1; : : : ; xN) = (x1; : : : ; xN ; x01; : : : ; x0N 0)together with its normalized version� x01;:::;x0N0 (x1; : : : ; xN ) = ������ x01;:::;x0N0 �������1L2(dx1 ��� dxN ) x01;:::;x0N0 (x1; : : : ; xN )11

de�ned on the set E = n(x01; : : : ; x0N 0) �� ������ x01;:::;x0N0 ������2 6= 0o :Also, as in Lemma 2.1, de�ne� x01;:::;x0N0 (y) = NXi=1 Z j (x1; : : : ; xi�1; y; xi+1; : : : ; xN ; x01; : : : ; x0N 0)j2 Yi6=j dxj� � x01;:::;x0N0 (y) = NXi=1 Z j � (x1; : : : ; xi�1; y; xi+1; : : : ; xN ; x01; : : : ; x0N 0)j2 Yi6=j dxjNote also that since �TF and � are radially symmetricZ 1jx0j � yj � 1jx0jj! �TF(y)�(y) dy = 0ThenZ T4 �� � (x1; : : : ; xN ; x01; : : : ; x0N 0)��2 dx1 : : : dxN == Z 1jx0j � yj � 1jx0jj! (� � �) (y) � � x01;:::;x0N0 (y) dy= ZZ 1jx0j � yj � 1jx0j j! �(z)�(y � z) � � x01;:::;x0N0 (y) dy dz= ZZ 1jx0j � zj � 1jx0jj! �(z)�(y � z) � � x01;:::;x0N0 (y) dy dz� ZZ 1jx0j � zj � 1jx0j � yj! �(z)�(y � z) � � x01;:::;x0N0 (y) dy dz= Z 1jx0j � yj � 1jx0jj! �(y) �� � � � x01;:::;x0N0� (y) dy� ZZ 1jx0j � zj � 1jx0j � yj! �(z)�(y � z) � � x01;:::;x0N0 (y) dy dzNow observe that 12

�����ZZ 1jx0j � zj � 1jx0j � yj! �(z)�(y � z) � � x01;:::;x0N0 (y) dy dz������ 10 ZZ jz � yjR2 �(z)�(y � z) � � x01;:::;x0N0 (y) dy dz� 10N ��������Z jz � yjR2 �(z)�(y � z) dz��������L1(dy)� N Z�2=3R2 Z �(z) dz � 2 Z1=3R2Summing over all j we obtain that������Xj ZZ 1jx0j � zj � 1jx0j � yj! �(z)�(y � z) � � x01;:::;x0N0 (y) dy dz������ � 2Z1=3R2 N 0Note that Z1=3R2 N 0 � Z2=3� 2R N 0 << �ZR N 0 (4)as long as 2=3 � 2 < 1� 3 2, that is, 2 < 1=6, or b < 7=6, which certainly holds in this case.So,������Xj Z T4 �� � (x1; : : : ; xN ; x01; : : : ; x0N 0)��2 dx1 : : : dxN ������ �� ������Xj Z 1jx0j � yj � 1jx0j j! �(y) �� � x01;:::;x0N0 � �� (y) dy������+ 2Z1=3R2 N 0= ������Xj Z 1jx0j � yj � 1jx0j j! �(y) �� � x01;:::;x0N0 � �(y)� �TF(y)� dy������+ 2Z1=3R2 N 0= ������*Xj 1jx0j � yj � 1jx0j j! � �;�� � x01;:::;x0N0 � �� �TF�+������+ 2Z1=3R2 N 0So, if we de�ne '(y) =Xj 1jx0j � yj � 1jx0j j! � �(y)13

we have������Xj Z T4 �� � (x1; : : : ; xN ; x01; : : : ; x0N 0)��2 dx1 : : : dxN ������ �� �����';�� � x01;:::;x0N0 � �� �TF������+ 2Z1=3R2 N 0In particular0@ ������Xj Z T4 �� � (x1; : : : ; xN ; x01; : : : ; x0N 0)��2 dx1 : : : dxN ������� 2Z1=3R2 N 01A2+ �� �����';�� � x01;:::;x0N0 � �� �TF������2Next, observe thatjry'j � 8>>>>>>>>><>>>>>>>>>:

CN 0jx0j � yj2 for jyj � R� Cjx0j � yj2 + CR�R!N 0 for R� � jyj � 2 �R�0 for jyj > 2 �R�In any case, jry'j � CN 0RR�therefore, jjry'jj22 � CN 02R�R2By Lemma 2.1 applied to � , we have for (x01; : : : ; x0N 0) 2 E,DHZ;N � x01;:::;x0N0 ; � x01;:::;x0N0E �� E0(Z) + C 0 R2N 02R� ������ � x01;:::;x0N0 � �� �TF; '�����2 � CHSWZ7=3�b� E0(Z) + C 0 R2N 02R� 0@������*Xj T4 � x01;:::;x0N0 ; � x01;:::;x0N0+������� 2Z1=3R2 N 01A2+ � CHSWZ7=3�b14

in other words,DHZ;N x01;:::;x0N0 ; x01;:::;x0N0E �� ������ x01;:::;x0N0 ������22 � E0(Z)� CHSWZ7=3�b + C 0 R2N 02R� �� ����DPj T4 x01;:::;x0N0 ; x01;:::;x0N0E���� 2Z1=3R2 N 0 � ������ x01;:::;x0N0 ������22�2+������ x01;:::;x0N0 ������42 != �E0(Z)� CHSWZ7=3�b� ������ x01;:::;x0N0 ������22+ C 0 R2N 02R� ����DPj T4 x01;:::;x0N0 ; x01;:::;x0N0E���� 2Z1=3R2 N 0 � ������ x01;:::;x0N0 ������22�2+������ x01;:::;x0N0 ������22Now, integrate with respect to (x01; : : : ; x0N 0) 2 E to obtainDHZ;N ; E � E0(Z)� CHSWZ7=3�b + C 0 R2N 02R� �� ZE ����DPj T4 x01;:::;x0N0 ; x01;:::;x0N0E���� 2Z1=3R2 N 0 � ������ x01;:::;x0N0 ������22�2+������ x01;:::;x0N0 ������22 dx01 � � � dx0N 0Now use Cauchy-Schwarz and the fact thatZ �f(x)�+ dx � �Z f(x) dx�+ � 0to realize that0@������*Xj T4 ; +������� 2Z1=3R2 N 01A2+ � 15

� 0@ZE0@������*Xj T4 x01;:::;x0N0 ; x01;:::;x0N0+������� 2Z1=3R2 N 0 � ������ x01;:::;x0N0 ������221A dx01 � � � dx0N 01A2+� 0@ZE0@������*Xj T4 x01;:::;x0N0 ; x01;:::;x0N0+������� 2Z1=3R2 N 0 � ������ x01;:::;x0N0 ������221A+ dx01 � � � dx0N 01A2� �Z ������ x01;:::;x0N0 ������22 dx01 � � � dx0N 0�� ZE ����DPj T4 x01;:::;x0N0 ; x01;:::;x0N0E���� 2Z1=3R2 N 0 � ������ x01;:::;x0N0 ������22�2+������ x01;:::;x0N0 ������22 dx01 � � � dx0N 0= ZE ����DPj T4 x01;:::;x0N0 ; x01;:::;x0N0E���� 2Z1=3R2 N 0 � ������ x01;:::;x0N0 ������22�2+������ x01;:::;x0N0 ������22 dx01 � � � dx0N 0thereforeDHZ;N ; E � E0(Z) + C 0 R2N 02R� 0@������*Xj T4 ; +������� 2Z1=3R2 N 01A2+ � CHSWZ7=3�bRecall that HZ;N +Hextra = HZ;N+N 0to obtainDHZ;N+N 0 ; E � E0(Z) + hHextra ; i+ C 0 R2N 02R� 0@������*Xj T4 ; +������� 2Z1=3R2 N 01A2+ � CHSWZ7=3�bUsing (3) we getDHZ;N+N 0 ; E � E0(Z) + �Z � 20 jZ � hN� ; ij+ 9R N 0 +*Xj T4 ; ++ C 0 R2N 02R� 0@������*Xj T4 ; +������� 2Z1=3R2 N 01A2+ � CHSWZ7=3�b (5)16

Similarly, since hN� ; i = Z DN� x01;:::;x0N0 ; x01;:::;x0N0E dx01 � � � dx0N 0if we let, for (x01; : : : ; x0N 0) 2 E,DN� � x01;:::;x0N0 ; � x01;:::;x0N0E = Z + Z1� 1Y (x01; : : : ; x0N 0)and hN� ; i = Z + Y Z1� 1then Y = ZE Y (x01; : : : ; x0N 0) ������ x01;:::;x0N0 ������22 dx01 � � � dx0N 0Now, Lemma 2.3 implies thatDHZ;N � x01;:::;x0N0 ; � x01;:::;x0N0E � E0(Z)+ CK4 Z7=3�b (jY (x01; : : : ; x0N 0)j � c1)2+ � CHSWZ7=3�bfor (x01; : : : ; x0N 0) 2 E.Arguing just as before we see that(jY j � c1)2+ � �ZE (jY (x01; : : : ; x0N 0)j � c1) ������ x01;:::;x0N0 ������22 dx01 � � � dx0N 0�2+� �ZE (jY (x01; : : : ; x0N 0)j � c1)+ ������ x01;:::;x0N0 ������22 dx01 � � � dx0N 0�2� ZE (jY (x01; : : : ; x0N 0)j � c1)2+ ������ x01;:::;x0N0 ������22 dx01 � � � dx0N 0therefore, hHZ;N ; i � E0(Z) + �CK4 (jY j � c1)2+ � CHSW�Z7=3�band using (3) we see thatDHZ;N+N 0 ; E � E0(Z) + �Z � 20 jZ � hN� ; ij+ 9R N 0 +*Xj T4 ; ++ �CK4 (jY j � c1)2+ � CHSW�Z7=3�b17

Therefore, averaging this expression with (5), we get thatDHZ;N+N 0 ; E � E0(Z) + �Z � 20 jZ � hN� ; ij+ 9R N 0 +*Xj T4 ; ++ �C 0 (jY j � c1)2+ � CHSW�Z7=3�b+ C 0 R2R� 0@������ 1N 0 *Xj T4 ; +������� 2Z1=3R2 1A2+ (6)possibly with a di�erent constant C 0.Let's say now that, for some numbers S and V*Xj T4 � ; + = �S �ZN 0RN 0 = V �Z RR� > C0V �ZThus V � 0.Note that if jSj < 1=16 and jY j is bounded above independently of C0 we are done. Thisfollows from the remarks following (3).Observe that = N 0 +N � Z � �Z� (N 0 + hN� ; i � Z � �Z)+� �C0V � 2jY jC0 � 1�+ �ZUsing (4) we can rewrite (6) to obtainDHZ;N+N 0 ; E � E0(Z) + F (S; Y; V ) � Z7=3�b:forF (S; Y; V ) = C20V9 � 209 �1 + 10�6�C0V jY j+ �C30V 2 � 2C0V jY j � C20V9 �+� �1 + 10�6�2 C20 jSjV + C 0 �(jY j � c1)2+ + �C0jSj � 10�6�2+�� CHSW� 19�C20V � 30C0V jY j+ �C30V 2 � 2C0V jY j � C20V �+ � 10C20 jSjV+ C 00 �(jY j � c1)2+ + �C0jSj � 10�6�2+�� 9CHSW�18

Observe that we can assume that C 00 < 1 and c1 is so big thatc21C 00 � 18CHSW > 2: (7)The rest of the lemma is devoted to proving that either jSj < 1=16 and jY j < some constantor else F > some other constant. In order to understand why this is so, we deal withdi�erent cases:case 1: jSj > 1=16 jY j < 2c1 V < C0024 jSjSince jSj > 1=16, for C0 > 64 we have (C0jSj � 10�6)2+ > 1=2C20S2. Thus,9F � �52C 00c1C0jSj � 512C 00C20S2 + 12C 00C20S2 � 9CHSW� �40C 00c1C0S2 + 112C 00C20S2 � 9CHSW� C 00C0S2��40c1 + 112C0�� 9CHSW� C 00162C0 ��40c1 + 112C0�� 9CHSWand pick C0 large enough so that this is at least 1.case 2: jSj > 1=16 jY j < 2c1 V � C0024 jSj9F � �60c1V C0 + (C30V 2 � 4c1C0V � C20V )+ � 10V C20 jSj � 9CHSWPick now C0 large enough so that(C30V 2 � 4c1C0V � C20V )+ > 1=2C30V 2 for V > C 0024 � 16then 9F � �60c1V C0 + 1=2C30V 2 � 240C 00 V 2C20 � 9CHSW� V 2 ��60 � 24 � 16c1C 00 C0 + 1=2C30 � 240C 00 C20�� 9CHSWand again pick C0 large enough so that this is at least 1 for V > C0024�16 .19

case 3: All jSj jY j > 2c1 V < 10�6C 00If jSj � 1=16, observe that �10C20V jSj + C00=2C20 jSj2 is increasing in jSj as long as V <C 00=160, which is true in our case. So it is enough to consider the case jSj < 1=16. Observethat C20V � 10C20V jSj � 3=8C20V if jSj < 1=16(jY j � c1)2+ � Y 24 if jY j � 2c1 (8)therefore 9F � 3=8C20V � 30C0V jY j+ C 004 Y 2 � 9CHSW= 3=8C20V + rC 004 Y � 15C0pC 00=4V!2 � 900C 00 C20V 2 � 9CHSW� C20 �3=8V � 900C 00 V 2�� 9CHSW� C20 �3=8V � 900C 00 10�6C 00V �� 9CHSW� C20V �3=8 � 900 � 10�6�� 9CHSWNow, if V > 100CHSWC�20 we are done. Otherwise,9F � �3000CHSWC�10 jY j+ C 004 Y 2 � 9CHSW� Y 2��3000CHSW2c1C0 + C 004 �� 9CHSW� C 008 Y 2 � 9CHSW � c21C 002 � 9CHSW � 1by (7) and for C0 large enough.case 4: jSj < 1=16 jY j > 2c1 10�13C 002C1=30 � V � 10�6C 00If jY j > C5=30 , by (8) 9F � �30C0V jY j+ C 004 Y 2 � 9CHSW� �3 � 10�12C 002C4=30 jY j+ C 004 Y 2 � 9CHSW:20

Di�erentiate with respect to jY j to realize that for C0 > 216 � 10�36C 003 the right handside is increasing for jY j � C5=30 . So,9F � �3 � 10�12C 002C9=30 + C 004 C10=30 � 9CHSW (9)and pick C0 big enough so that this is at least 1.If, on the contrary, jY j � C5=30 , we have2C0V jY j+ C20V � 10�13C 002 �2C30 + C7=30 � � 3 � 10�13C 002C30 � 3=10C30V 2so, using (8) again9F � �30C0V jY j+ 1=2C30V 2 � 9CHSW � �30C8=30 V + 1=2C30V 2 � 9CHSW (10)and pick C0 so that this is bigger than 1 for V � 10�6C 00.case 5: jSj � 1=16 jY j > 2c1 10�13C 002C1=30 � V � 10�6C 00In this case, argue as in case 4, with the only di�erence that C20 (V � 10jSjV + 1=2C 00S2)need no longer be positive, and we have to include it in (9) and (10), which will be replacedrespectively by9F � �3 � 10�12C 002C9=30 + C 004 C10=30 + C20 (V � 10jSjV + 1=2C 00S2)� 9CHSWand 9F � �30C8=30 V + 1=2C30V 2 + C20 (V � 10jSjV + 1=2C 00S2)� 9CHSWNote that C20 (V � 10jSjV + 1=2C 00S2) � �100C 00 C20V 2 (11)since, for given V , minjSj �V � 10jSjV + 1=2C 00jSj2�is attained when 10V = C 00jSj: so,minjSj �V � 10jSjV + 1=2C 00jSj2� � �100C 00 V 2Therefore, in our range of V , (11) is at least �10�24C 003C8=30 so, for C0 big enough it doesn'ta�ect the result since V is bounded below by a constant independent of C0.21

case 6: All jSj jY j > 2c1 10�13C 002C1=30 � VIf jY j > 200C00C0V , and jSj < 1=16, by (8)9F � �30C0V jY j+ C 004 Y 2 � 9CHSWAgain the derivative of this with respect to jY j is positive for jY j > 200C00C0V . So, pluggingin for jY j the value jY j = 200C00C0V , we obtain9F � C20V 2 ��6000C 00 + 10;000C 00 �� 9CHSW � 4000C 00 C20V 2 � 9CHSWIf, jSj � 1=16, by (11) we have to subtract 100C00C20V 2, that does not alter the result.Now, if jY j � 200C00C0V , observe thatC30V 2 � 2C0V jY j � C20V � C30V 2 � 400C 00 C20V 2 � C20V � 1=2C30V 2for C0 big enough, since V � 1. So, for jSj < 1=16 we have9F � �30C0V jY j+ 1=2C30V 2 � 9CHSW� �6000C 00 C20V 2 + 1=2C30V 2 � 9CHSWand as usual we pick C0 so that this is at least 1.If jSj � 1=16, we have to subtract 100C00C20V 2, which again is harmless for C0 big enough.This proves that either both jSj � 1=16 and jY j � 2c1 orDHZ;N+N 0 ; E � E0(Z) + cZ7=3�b:which concludes the lemma.4. The Bootstrap 22

In section two we obtained estimates for wave functions supported on balls, where the ballwas to be of a certain size. The estimates from the previous section will help us obtainessentially the same kind of estimates for ball of twice their size, and by induction, to allballs in R3.Lemma 4.1: Let R � C0Z�1=3+ 2 , �� = c0Z� 1andC0Z� 1 � � � �1 + 10�12�C0Z� 1If Estimate(��; �; R) holds, then Estimate(��; �0; 2R) also holds, with�0 = �+ 2C��RZprovided that �0 � �] < 1.Proof: We consider a partition of unity given by two smooth functions, �0 and �1, satis-fying �0(x) = � 1 if jxj < R=20 if jxj > R�20(x) + �21(x) = 1Given a wave function (x1; : : : ; xN) supported on B(0; 2R), and given any sequencei1; : : : ; iN of 0's and 1's, we de�ne i1;:::;iN = �i1(x1) � � ��iN (xN ) (x1; : : : ; xN )Assume for simplicity that ij = 0 for j = 1; : : : ; N1 and ij = 1 for j = N1 + 1; : : : ; N ; letN2 = N �N1.We de�ne xN1+1;:::;xN to be where the variables xN1+1; : : : ; xN are �xed. It is thus anantisymmetric wave function supported on �B(0; R)�N1 . Since Estimate(��; �; R) holds,we haveHZ;N1 xN1+1;:::;xN ; xN1+1;:::;xN � � �E0(Z) + ��ZR (N1 � (1 + �)Z)� ���� xN1+1;:::;xN ����22Integrate this against dxN1+1 : : : dxN to obtainhHZ;N1 i1;:::;iN ; i1;:::;iN i � �E0(Z) + ��ZR (N1 � (1 + �)Z)� jj i1;:::;iN jj22 (12)23

Our goal now is to prove thathHZ;N1+N2 i1;:::;iN ; i1;:::;iN i � �E0(Z) + ��ZR (N1 +N2 � (1 + �)Z)� jj i1;:::;iN jj22 (13)This is trivial if N = N1 +N2 � (1 + �)Z, since � � �].If N � (1 + �)Z, we can apply Lemma 3:1, with � = �.If hHZ;N1+N2 i1;:::;iN ; i1;:::;iN i � �E0(Z) + cZ7=3�b� jj i1;:::;iN jj22then, either cZ7=3�b > ��ZR (N � (1 + �)Z)in which case (13) is proved, orcZ7=3�b � ��ZR (N � (1 + �)Z) (14)If this is the case, just like in (2) note that provided c0 < cCK=(8CHSW) where c is theconstant in Lemma 3.1 (we can assume c < 1 and CHSW > 1),cCK jN � Zj28CHSWR > ��ZR (N � (1 + �)Z) : (15)(14) then implies that CK jN � Zj24R > 2CHSWZ7=3�b:Estimate (A) then with � = �2R implies thathHZ;N1+N2 i1;:::;iN ; i1;:::;iN i � E0(Z) + CK jN � Zj28R ! jj i1;:::;iN jj22� E0(Z) + cCK jN � Zj28CHSWR ! jj i1;:::;iN jj22and (15) again implies (13).The alternative left from Lemma 3.1 is thathHextra i1;:::;iN ; i1;:::;iN i � c�ZR N2 jj i1;:::;iN jj22 (16)24

Since HZ;N1+N2 = HZ;N1 +Hextraand c� > �� (for c0 < c � C0), (12) and (16) implyhHZ;N1+N2 i1;:::;iN ; i1;:::;iN i � �E0(Z) + ��ZR (N1 +N2 � (1 + �)Z)� � jj i1;:::;iN jj22and (13) is proved.Putting all these estimates together, we see thatXi1;:::;iN hHZ;N1+N2 i1;:::;iN ; i1;:::;iN i � E0(Z) + ��ZR (N1 +N2 � (1 + �)Z)Now, Xi1;:::;iN hVCoulomb i1;:::;iN ; i1;:::;iN i = hVCoulomb ; iFor ��, we haveXi1;:::;iN h��x1;:::;xN i1;:::;iN ; i1;:::;iN i == Xi1;:::;iNXk h��xk i1;:::;iN ; i1;:::;iN i= Xi1;:::;iNXk *Yj 6=k �2ij (xj) � (��xk�ik(xk) ) ; (�xk(xk) )+=Xk Xik h��xk (�ik(xk) ) ; (�ik(xk) )i=Xi;k �2i (xk) (��xk ) ; �+Xi;k h (��xk�i(xk)) ; �i(xk) i� 2Xi;k hrxk � rxk�i(xk); �i(xk)i= h�� ; i+*Xi;k �i(xk) (��xk�i(xk)) ; +�Xi;k rxk�2i (xk) � rxk ; �The last term on the right is zero, sinceXi rxk�2i (xk) = rxkXi �2i (xk) = r1 = 025

Hence, if we de�ne W (x) =Xi �i(x)��i(x)we get Xi1;:::;iN h��x1;:::;xN i1;:::;iN ; i1;:::;iN i = h�� ; i �*Xk W (xk) ; +As a result, we gethHZ;N ; i �*Xk W (xk) ; + � E0(Z) + ��ZR (N � (1 + �)Z):Observe now that jW (x)j � CR2 :Thus it follows that �����*Xk W (xk) ; +����� � CNR2 :Thus we have hHZ;N ; i � E0(Z) + ��ZR (N � (1 + �)Z)� CNR2 :So we are only left to investigate when��ZR (N � (1 + �)Z)� CNR2 � ��Z2R (N � (1 + �0)Z) (17)Observe that the derivative with respect to N of the left hand side is bigger than thederivative of the right hand side. This amounts to checking that��ZR � CR2 � ��Z2Rwhich is equivalent to CR � ��Z2which will hold as long as R � 2C��Z . This certainly holds in our case, sinceR > Z�1=3+ 2 >> Z�1+ 1So it is enough to prove that (17) holds for the smallest value of N in which we areinterested. For N = (1 + �0)Z, (17) is equivalent toC 1 + �0R � ��Z (�0 � �)26

i.e. �0 � �+ C(1 + �0)��ZRSo, Estimate(��; �0; 2R) holds with �0 = �+ 2C��ZRCorollary: There exist ��], �] such that Estimate(��], �]; R) holds for all R �C0Z�1=3+ 2Proof: De�ne �0 = C0Z� 1��0 = c0Z� 1R0 = C0Z�1=3+ 2�n = �n�1 + 2C��0ZRnRn = 2nR0Note that �n � �0 + 1Xk=0 4C��0Z2kR0 < C]�0def= �]for Z large enough, provided b < 14=15 (which is true in our discussion), with C] ��1 + 10�12�.By Lemma 2.2 we see that Estimate(��0, �0; R0) holds.Therefore, by the previous Lemma, if we make ��] = ��0, Estimate(��0; �n; Rn) holds forall n, and the corollary follows.This implies that hHZ;N ; i � E0(Z)and therefore N(Z) � Z + �]Z = Z + O(Z1� 1):From [11] if follows that N(Z) � Z27

and therefore, N(Z) = Z + O(Z1� 1)which proves the theorem.

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References[1] Fe�erman, C., Seco, L. (1989) \An Upper Bound for the Number of Electrons in aLarge Ion" To appear in Proceedings of the Nat. Acad. Sci., USA[2] Hertel, P., Lieb, E., Thirring, W. (1975) \Lower Bound to the Energy of ComplexAtoms" Journal of Chemical Physics Vol 62 no. 8 3355.[3] Hughes, W. To appear in Advances in Mathematics.[4] Lieb, E. H. (1979) \A Lower Bound for Coulomb Energies" Physics Letters Vol 70Ano. 5,6 444.[5] Lieb, E. H. (1981) \Thomas{Fermi and Related Theories of Atoms and Molecules"Reviews of Modern Physics Vol. 53, no. 4.[6] Lieb, E. H. (1984) \Atomic and Molecular Negative Ions" Phys. Rev. Lett. 52, 315.[7] Lieb, E. H. (1984) \Bound on the Maximum Negative Ionization of Atoms and Mole-cules" Phys. Rev. A29, 3018{3028[8] Lieb, E. H, Sigal, I., Simon, B. & Thirring, W. (1988) \Approximate Neutrality ofLarge-Z Ions" Communications in Mathematical Physics 116(4) 635-644.[9] Siedentop, H., Weikard, R. (1987) \On the Leading Energy Correction for the StatisticalModel of the Atom: Interacting Case" Communications in Mathematical Physics 112471-490[10] Siedentop, H., Weikard, R. To appear in Inven. Math.[11] Zhislin, G. (1960) \Discussion of the Spectrum of Schr�odinger Operator for System ofMany Particles". Tr. Mosk. Mat. Obs. 9, 81-128.

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