Proton-neutron electromagnetic interaction.
ISINN-22, Dubna (27-30 may 2014)
Bernard Schaeffer, Paris, France [email protected] 1 Electromagnetic interactions in a nucleus
• A proton attracts a not so neutral neutron as amber (ἤλεκτρον, elektron) attracts small neutral pieces of paper. • Nucleons repulse themselves as collinear and opposite magnets (µαγνήτης, magnetis). • Only the Coulomb repulsion between protons is taken into account in mainstream nuclear physics.
2
The nuclear shell model The binding energy of even the simplest bound nucleus, the deuteron 2H cannot be calculated with the nuclear shell model, the fundamental laws of the strong force being unknown.
The force center of the nucleus being undefined, we may assume that, in contrast with electrons, nucleons do not move, as suggested by the drawings of the atom and its nucleus.3 Estimation of 2H binding energy
Applying electric Coulomb’s law with the measured radius of the proton, rp = 0.88 fm, one obtains,
not far from 2.2 MeV, the measured value of the deuteron binding energy, indicating a possible electric origin of the nuclear interaction. 4 Proton electric charge
• It is well known that the proton contains one positive elementary electric charge, assumed here to be punctual: +e = +1.6 × 10-19 Coulomb
• The proton-proton repulsion, inexistent in the deuteron, is not the only Coulomb force in a nucleus.
5 Neutron electric charges
It is less known that the not so neutral neutron contains electric charges with no net charge, assumed for the sake of simplicity to be punctual charges +e and –e (+2e/3, -2e/3 would give a 30% error).
A proton attracts the negative charge of a nearby neutron and repulses its positive charge farther away, resulting in a net attraction, according to the 1/r Coulomb’s law.
6 Constantes
e2 =1.6 MeV 4⇡✏0rp Noyau Nuclear magnetic energy constant: •
µ0 µnµp 2 gngp | 3 | = ↵mpc | | =9.147 871 896 MeV 4⇡RP 16
3 2 eiejRP gigj RP Uem = ↵mpc 2 + | | Sij (18) 2 e rij 16 rij 3 i i=j ✓ ◆ i i=j ✓ ◆ X X6 X X6 where 4 5 S = cos (µ~ , µ~ ) 3 cos (µ~ , ~r ) cos (µ~ , ~r ) (19) ij i j i ij j ij
R g g R 3 U ↵m c2 P + | n p| P S (20) em ⇡ p r 16 r np " np ✓ np ◆ # It is more complicated for the magnetic moments: Electromagnetic contents of 2H Snp = cos (µ~ n, µ~ p) 3 cos (µ~ n, ~rnp) cos (µ~ p, ~rnp)= (21) Spin 3 electric= charges1 3 (1/r1 law):( 1) = 2 The proton repulses the +e⇥ charge⇥ of the Il su ra d’ajuster le coeneutron cient and attracts magn´etique, its -e charge, (toujoursresulting in ?)+ comprise entre 2 pour le deuton a net attraction. -e et 3). Il faut aussi tenir compte des g qui varient entre gn =+e 3.826 and gp =5.585. Formule 2 magnetic moments (1/r3 law): Proton: μ > 0 p µ > µ | p| | n| Neutron: μn < 0 Cette formule, approximativeThus, for collinear (30% and opposite d’erreur), magnetic moments, peut ´etre the g´en´eralis´ee avec des coe cients 7 ad hoc. Cette approximationresulting magnetic vient moment de ce is qu’on positive, thus a n´eglig´elarepulsive. r´epulsion entre le proton et la charge `electrique du neutron. En fait seule la partie magn´etique change. Cela veut dire que l’´energie de liaison par nucl´eon d’un noyau varie assez peu. Le principal param`etre est le nombre de liaisons neutron-proton par nucl´eon qui varie de 0,5 pour le deuton `a3 soit un rapport de 6 pour un noyau de masse infinie. La particule ↵ serait 6 fois plus ´energ´etique par nucl´eon. En fait les moments magn´etiques y sont inclin´es `a600 ce qui divise l’´energie magn´etique r´epulsive par 2, donnant 12 MeV (AVERIFIER). En fait elle est diminu´ee par la r´epulsion entre protons et neutrons n´eglig´ee provisoirement. Le probl`eme se r´eduit donc au calcul de Snp et au nombre de liaisons (neutron-proton en 1`ere approximation) par nucl´eon ou, de pr´ef´erence, par proton.
p-3 Constantes
2 e e e e U H = (1) e 4⇡✏ r + a r a 2a 0 ✓ np np ◆ Exact electric2 dipole formula 2H e 1 1 1 U2e = (2) The 2a/r approximation4⇡✏0 rnp of+ thea dipolernp isa invalid2a in the deuteron 2H where the✓ proton touches the neutron◆ (a ≈ r). 2 2 e 1 1 2a 4a Therefore,U H = the exact electric dipole formula [1] has to be (3) e 4⇡✏ r + a r a r2 ⇡ r2 used instead of0 the✓ usualnp simplifiednp formula:◆ ou e2 1 1 e2 2a (4) 4⇡✏ r + a r a ⇡ 4⇡✏ r2 0 ✓ np np ◆ 0 4He avec nn et pp • rnp : separation distance between the proton and the neutron. • 2a : separation distance between the neutron positive and negative charges 2 e e e e of the deuteron.U H = (5) e 4⇡✏ r + a r a 2a [1] B. Schaeffer, Advanced0 ✓ Electromagnetics,np np Vol. 2, No. 1,◆ September 8 2013. Potentiel ´electromagn´etique:
2 2 3 4 4R 4R 1 R 3 g g 1 g + g R U He = ↵m c2 P P + P + | n p| + p n P em p r + a r a 4 ⇥ r 4 ⇥ 16 4 ⇥ 16 r " np np pp ! ✓ np ◆ # The electromagnetic potential, for one bond (or one nucleon of the ↵ particle), is:
2 2 3 4 4R 4R 1 R 3 g g 1 g + g R U He = ↵m c2 P P + P + | n p| + p n P em p r + a r a 4 ⇥ r 4 ⇥ 16 4 ⇥ 16 r " ! ✓ ◆ # 2 Convertir ↵mpc : Fundamental constants of the nuclear energy potential • Electrostatic attraction:
2 2 938 e ↵mpc = = =6.85 MeV (6) 137 4⇡✏0RP Numerically,
2 2 3 4 3 0.210 3 0.210 3 3.83 5.59 + 5.59 +3.83 0.210 U He =6.85 ⇥ ⇥ + ⇥ em r + a r a 4 ⇥ 16 r "✓ np np ◆ ✓ np ◆ # (7) The minimum of the potential is the binding energy per nucleon, the number of neutron- proton bonds being equal to the number of nucleons.
p-1 Proton-neutron electric potential energy.
Electrostatic energy of 3 aligned punctual electric charges ep, en-, en+:
The third term is unphysically infinite for an isolated neutron where a ⋍ 0. The intuitive solution is to replace 1/2a by the exact electric dipole potential energy 1/(r+a) - 1/(r-a). The potential is thus doubbled and zero for a = 0:
The potential, being negative, is attractive. No empirical polarizability, only fundamental electric Coulomb’s law. Proton-neutron
B. Schaemagnetic↵er potential energy Magnetic dipole-dipole interaction potential energy (interactionFormule g´en´erale between du potentieltwo magnetic moments after Maxwell [2]):
µ 3(µ~ ~r )(µ~ ~r ) µ µ µ µ µ µ U = 0 µ~ µ~ n • np p • np = 0| n p| [ 1 3( 1)(1)] = 2 0| n p| (9) m 4⇡r3 n • p r2 4⇡r3 4⇡r3 np np np np The potential energy of the 2H interacting magnetic moments, eiej µ0 3(µ~i µ~j)(µ~i µ~j) µ0 3(µ~ n ~rnp)(µ~ p ~rnp) Uem = Ue+Uem = + 3 [µ~i µ~j • 2 • + 3 µ~ n µ~ p • 2 • ] collinear and opposite,4⇡✏ is0r ijpositive,4⇡r thus• repulsiver : 4⇡r • r i i=j ij ij np np ! X X6 (10)
qiqj Uem = 4⇡✏0rij µ and µ are the neutron and proton magnetic moments. i i=j n p 6 X X 10 [2] J. C. Maxwell, Aµ treatise on electricity3(µ~ andµ~ )(magnetismµ~ µ~ ) . Vol.3(µ~ II. 1995,~r )( µ~art. 384-387~r ) + 0 µ~ µ~ i • j i • j i • ij j • ij (11) 4⇡r3 i • j r2 r2 i i=j ij " ij ij # X X6
µ 3(µ~ µ~ )(µ~ µ~ ) µ µ µ U = 0 µ~ µ~ n • p n • p = 0| n p| [ 1 3 ( 1)] (12) m 4⇡r3 n • p r2 4⇡r3 ⇥ np np np
p-2 B. Schae↵er
Corrections 5-5-14 , 27-4-14 Potentiel total:
2 2 e 2 2 µ µ µ U H = +2 0| n p| em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ ✓ np ◆ 2 2 e 1 1 µ µ µ U H =2 + 0| n p| em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ ✓ np ◆ Potentiel par nucl´eon: litt´eral 2 2 e 1 1 µ µ µ U H /A = + 0| n p| em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ np B. Schae↵er 2 4 e 1 1 U He/A =4 e ⇥ 4⇡✏ r + a r a 0 ✓ np np ◆
Corrections 5-5-14 , 27-4-14 4 3 µ µ µ Proton-neutronU He/A = Coulomb-Maxwell0| n p| Potentiel total: m 2 4⇡r3 ✓ np ◆ electromagnetic2 potential energy 2 e 2 2 2 µ µ µ U 4HHe= e 1 1 +2 3 0|µ0nµnpµ|p Uemem /A =4 + |3 3 | With electric4⇡✏0⇥ attraction4⇡✏rnp0 +ranp and + ar magneticnp rnpa arepulsion2 4 ⇡4r⇡nprnp combined, the✓ total✓ binding energy ◆potential◆ ✓ ✓of 2H is: ◆◆ num´erique 2 2 e 1 1 µ µ µ U H =2 + 0| n p| em 2H 2 2 0.08493 U 4=1⇡✏0.442rnp + a rnp a +2 4⇡rnp em ✓ ◆ ✓ 3 ◆ rnp + a rnp a ⇥ rnp Potentiel par nucl´eon:or, numerically:✓ ◆ litt´eral 2H 1 1 0.0849 Uem =2 21.442 + 3 2 e r1np + a rnp1 a µrnpµ µ H ✓ ◆ 0 n p Uem /A = + | 3 | Only4 4fundamental⇡✏0 rnp + lawsa randnp constants.a 4⇡rnp He ✓ 1 1 ◆ 0.1274 Uem No/A fit,=5 empirical.76 parameter or cutoff.+ 3 11 rnp2 + a rnp a rnp 4 ✓e 1 ◆ 1 U He/A =4 Energie de liaisone d’un noyau⇥ 4 en⇡✏ g´en´eralr + a r a 0 ✓ np np ◆ Atome d’hydrog`ene: 4He 3 µ0 µnµp Um H /A = 1| 2 | 2 B = 13.62eV 4⇡↵r3mec (12) ✓ ⇡ 2 np ◆ Atome d’hydrog`ene: 2 4He e 1 1 3 µ0 µnµp U /A =4 1 2 2 + | | em ⇥ 4⇡✏ 13r .6 eV+ a ↵r mec a 2 4⇡r3 (13) 0 ✓ np ⇡ 2 np ◆ ✓ np ◆ num´erique 2 1 B H /A =1.11 MeV ↵m c2 (14) ⇡ 6 p 2 2 2 0.0849 U H =1.442 +2 em 1 2 3 r1np.11+MeVa rnp↵mac ⇥ rnp (15) ✓ ⇡ 6 p ◆ 4 1 1 0.1274 He 4He 2 Uem /AB=5/A.76=7.07 6.85 MeV = ↵m+pc (16) r +⇡ a r a r3 ✓ np np ◆ np Energie de liaison d’un noyau7.07 MeV en g´en´eral6.85 MeV = ↵m c2 (17) ⇡ p
Atome d’hydrog`ene: 2 e 2 H = ↵mpc =6.8461 9012 652MeV 4⇡✏0BRP = 13.6 eV ↵ m c (12) ⇡ 2 e Atome d’hydrog`ene: p-2 1 13.6 eV ↵2m c2 (13) ⇡ 2 e
2 1 B H /A =1.11 MeV ↵m c2 (14) ⇡ 6 p 1 1.11 MeV ↵m c2 (15) ⇡ 6 p
4 B He/A =7.07 6.85 MeV = ↵m c2 (16) ⇡ p
7.07 MeV 6.85 MeV = ↵m c2 (17) ⇡ p
2 e 2 = ↵mpc =6.846 901 65 MeV 4⇡✏0RP
e2 =1.6 MeV 4⇡✏0rp
p-2 Theoretical proton-neutron electromagnetic potentials
-2.2 MeV The equilibrium between attractive and repulsive forces is obtained at a minimum or at a horizontal saddle point.
12 Best result with the exact electric dipole formula (continuous curve). Constantes
2 2 ¯h 2 e 2 2 µ µ µ 2 + U H + +2 0| n p| = 0 (32) 2mr em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ ✓ np ◆ 2 2 2 ¯h e 2 2 µ µ µ U H = 2 + +2 0| n p| (33) em 2mr 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ ✓ np ◆ For the spherically symmetrical electrostatic potential, one has:
2 2 2 ¯h 1 2 e 2 2 µ µ µ B. Schae↵Eer H =+ + 2 0| n p| em 2m b2 rb 4⇡✏ r + a r a 4⇡r3 ✓ ◆ 0 ✓ np np ◆ ✓ np ◆
2 2 2 2µ0 µnµp 2H ¯h @ 2 @ U H =2e 2 2 Constantesµ0 µnµp E = + m+ | 3 | 2 | | em 2m @r2 r @r 4⇡✏ 4⇡rrnp + a r a 4⇡r3 ✓ ◆ ✓ 0 ✓ np ◆ np ◆ ✓ np ◆ 2H µ0 2m K Um 2 = 3 2 µnµp +4⇡2r E +⇥ | = 0| (33) 2 2 r 2 ¯h np r ¯h @ 2 @ e 2 2 µ0 µnµp 2 2 +¯h +2 2 E e 2 2 µ µ µ 2 | | =0 2 2H 2 +e U H + 2 2 +2 µ00| n p| = 03 (34) 2m @r U r @=r em 4⇡✏0 rnp + a +rnp a 3 2 µ µ 4⇡rnp ✓ em2mr ◆ 4⇡✏0 rnp✓+ a rnp a 34⇡r◆np n✓ p ◆ 4⇡✏ 0 rnp +✓a rnp a ◆ 4⇡✓ rnp ⇥ ◆ | | 2✓ 2 ◆ 2 ¯h e 2 2 µ µ µ U H = 2 + +2 0| n p| (35) Potentiel ELECTRIQUE:em 3 2 2mr 24⇡✏0 rnp + a rnp a 4⇡rnp @ 2 @ 2m e ✓ 2 ◆2 ✓ ◆µ0 µnµp For the spherically symmetrical electrostatic potential, one has: 2 + + 2 E + 2 +2 | 3 | =0 @r r @r ¯h 2H 4⇡✏e 0 rnp +2 a rnp 2 a 4⇡r 2 U = 2 np (18) 2H ¯h ✓e1 2 ✓e 2 2 ◆ µ0✓µnµp ◆◆ Eem =+ 2 4+⇡✏0 rnp + a rnp a 2 | 3 | 2m b rb 4✓⇡✏0 rnp + a rnp a ◆ 4⇡rnp ✓ ◆ ✓ ◆ ✓ ◆ 2 2m K Potentiel total: + 2 E + = 0 (34) 2 2 r 2 r 2H ¯h @ 2 @ ¯he 2 2 µ0 µnµp Eem = 2 + + ✓ ◆ 2 | 3 | 2 2m @er r @r 2 4⇡✏0 r2np + a rnp aµ µ µ4⇡r For the sphericallyU symmetricalH =✓ electrostatic◆ ✓ potential,+2 one◆0 has:| n✓ p| np ◆ em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ ✓ np ◆ ¯h2 @2 2 @ 2 e2 2 2 µ µ µ + @ + E 2 @ 2m K 2 0| n p| =0 2m @r2 r @r2 + 4⇡✏ +r + a rE +a =4⇡ 0r3 (35) ✓ 2 e ◆ 2 2 0 ✓ np 22 np ◆ µ✓ µ npµ ◆ U H = @r r @r ¯h +2r 0| n p| em 4⇡✏ r + a r ✓a ◆ 4⇡r3 2 0 np 2 np np @ 2 @ 2m ✓ e 2 2◆ ✓ µ0 µnµp ◆ 2 + + 2 E + +2 | 3 | =0 Formules approximatives@r r @r par¯h nucl´eon:4⇡✏0 rnp + a rnp a 4⇡rnp ¯h2 ✓ 1 1✓ ◆ ✓ ◆◆ Schrodinger repris HyperPhysics2 8-5-14m K (36) 2 + E + = 0 (36) 2mpb br r ¯h2 r 2 2 ✓ ◆ ✓ ◆ ¯h 2 e2 2 2 µ0 µnµp For2 the spherically+ E H symmetricale electrostatic2 potential,2 one has:µ20 µnµ|p | = 0 (19) 2mr em + 4⇡✏ r + a r +2a | 4⇡|r3 =0 @02 np2 @ 2m np K 3 np 4⇡✏0 ✓r+ + a + r Ea+ ◆ = 04✓⇡rnp ◆ (37) @r✓2 r @r ¯h2 ◆r ✓ ◆ =✓ 1◆ (20) 2 2 2H ¯h 1 1 e 2 2 µ0 µnµp E = ¯h2 1 +1 +2 | | em (38)3 2mpb 2bm b rb r 4⇡✏0 r + a r a 4⇡rnp ✓ p ✓ ◆ ◆ =✓ 0 ◆ ✓ ◆ (21) e2 2 2 µ µ µ + 2 +2 0| n p| =0 2H 1 4⇡✏e r + a 2r a 2 4⇡r3 µ0 µnµp Schrodinger repris Wiki 8-5-140 ✓ ◆ ✓ np ◆ Uem /A = +2 | 3 | 22 4⇡✏0 rnp +2 a rnp a 4⇡rnp 2H ¯h 1 1✓ e2 2 2 ◆ µ✓0 µnµp ◆ Eem = + ¯h 2 +2 | 3 | 2mEp b (rb )= 2 r 4⇡✏0 r ++ aV (rr) a (r)4⇡rnp (22) 4He ✓ e ◆ 2 ✓ 2 ◆ ✓ 3 µ0◆µnµp 2 2m r Uem /A2 =2 1 e 2 2 µ0 +µnµp | | Proton-neutronU H /A = Schrödinger+2 equation3 em ⇥ 4⇡✏0 rnp + a rnp a | 3 4| 4⇡rnp Schrodinger 2 4⇡✏0 ✓rnp + a rnp a ◆4⇡rnp ✓ ◆ With the Coulomb-Maxwell potential✓ V, the 2H Schrödinger◆ ✓ equation◆ writes: 2 2 4 e 2 2 2 3 µ µ µ ¯h 1 2 He 22 e 2 2 0 nµ0p 2 µnµp Uem /A =2H 2 + | 3 | + ¯hU ⇥ 4⇡✏0 rnpH+ a rnp a 4 4⇡rnp | | =0 2 em ✓ ◆ ✓ ◆ 3 2m b rb +4⇡✏U0em r +Va(r, ar) a= 0 4⇡ r (23) 2 2m 2 ✓ ¯h 1 ◆ 2 2H e ✓ 2 2 µ0◆ 2 µnµp ✓ ◆ With the simplest2 exponential+ Uem h well andi the Coulomb| 3 |-Maxwell =0 2m b rb 4⇡✏0 r + a r a 4⇡ r electromagnetic ✓ ◆potential,2 we obtain:✓ ◆ ✓ ◆ 2 ¯h 2 2H ¯h 1 2 2 +e U 2 V (r, a,2 b) = 0µ 2 µ µ (24) 2 H 2 em 0 n p r/b ¯h 1 2 2m 2 e 2 2 µ 2 µ µ 2 + Uem H 0 n p | r/b3 | e =0 2 + Uem | 3 | e =0 2m b 2 m rbb rb 4⇡✏4⇡✏h00 rr++a a r ar ia4⇡ 4r ⇡ r ✓ ✓ ◆ ◆ 2 ✓ ◆ ◆✓ ✓◆ ◆ B. Schae↵er ¯h +[E p-5V (r/br, a, b)] = 0 (25) Dividing the Schrödinger2m equation by e , we obtain the classical electromagnetic 2H Coulomb-Maxwell potential except for the first term. 2 2 ¯h 2 2 ¯h 1 2 2 e 2 2 µ 2 µ µ + U H +[E U] = 00 | n p| =0 (26) 2m b2 rb 2emm r4⇡✏ r +p-5 a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ 13◆ r/b 2m e 2 + [E V (r)] = 0 (27) r ¯h2 r/b = e =1 2 2m K 2 2 ¯h 1 2 2 + e 2 E2+ 2 = 0µ0 2 µnµp (28) +rU H ¯h r | | =0 2m b2 rb em 4⇡✏ r + a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ ◆ Formules exactes ENERGIE MAGNETIQUE: p-4 4 3 µ µ µ U He/A = 0| n p| m 4 4⇡r3 ✓ np ◆ Formules dipˆole a< e2 1 1 e2 2a 4⇡✏ r + a r a ⇡ 4⇡✏ r2 0 ✓ np np ◆ 0 ✓ np ◆ e2 ⇡ 4⇡✏0rnp VERIFICATION 2H 3 2 2 0.210 2 0.210 3.83 5.59 0.210 U H =6.85 ⇥ ⇥ +6.85 2 ⇥ (39) em r + a r a ⇥ ⇥ 16 r ✓ np np ◆ ✓ np ◆ 2 2 2 0.0849 U H =1.44 + (40) em r + a r a r3 ✓ np np ◆ np 4He The electromagnetic potential, for one bond of the ↵ particle, is: 4He 4He 4He Uem = Ue + Um +↵m c2 4RP 4RP p rnp+a rnp a (41) 3 2 ⇣ 3 gngp R⌘P +↵mpc | | 4 ⇥ 16 rnp Numerically, ⇣ ⌘⇣ ⌘ 4He 4 0.210 4 0.210 U =6.85 ⇥ ⇥ em rnp+a rnp a 3 (42) 3 ⇣3.83 5.59 0.210 ⌘ +6.85 ⇥ ⇥ 4 ⇥ 16 rnp ERREUR 3 corrig´epar 4 ⇣ ⌘ 4 2 2 0.0637 U He =2.88 + (43) em r + a r a r3 ✓ np np ◆ np FORMULE MAGNETIQUE 4 2 2 3 He 2 3 gngp 1 gp+gn RP U = ↵mpc | | + (44) m 4 ⇥ 16 4 ⇥ 16 rnp ⇣ ⌘⇣ ⌘ 4He 3 0.210 3 0.210 U =6.85 ⇥ ⇥ m rnp+a rnp a 3 (45) 3 ⇣3.83 5.59+5.592+3.83⌘2 0.210 +6.85 ⇥ ⇥ 4 ⇥ 16 rnp ⇣ ⌘ p-6 B. Schae↵er Proton-neutron Schrödinger equation 2 2 2 ¯h 1 2 e 2 2 µ 2 µ µ U H = + + 0 | n p| em 2m b2 rb 4⇡✏ r + a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ ◆ 2 2 ¯h 1 2 2 e 2 2 µ 2 µ µ + U H 0 | n p| =0 2m b2 rb em 4⇡✏ r + a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ ◆ r/b e Zero binding r/b energy = e =1 B. Schae2 ↵er 2 ¯h 1 2 2 e 2 2 µ 2 µ µ + U H 0 | n p| =0 2m b2 rb em 4⇡✏ r + a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ ◆ Formules exactes ENERGIE MAGNETIQUE:2H: -2.2 MeV 4 3 µ µ µ r/b U HeFor/A b= infinite,0| n p | = e =1 , the Schrödinger m 4 4⇡r3 term is ✓ zero, np simplified◆ to the electromagnetic Formules dipˆole 2 Coulomb-Maxwell2 potential (dark line), giving, at 2 a<