The relativistic hydrogen-like atom :a theoretical laboratory for structure functions

Xavier Artru, Institut de Physique Nucléaire de Lyon, France Transversity 2005Karima Benhizia, Mentouri University, Constantine, Algeria Como, 7- 10 sept.

Theoretical environment

• pure QED• « atom » = hydrogen-like ion

(or ion with several e- , but neglecting e- - e- interactions)

• Z ~ 102 Z ~ 1 relativistic bound state• Dirac equation exact wave functions

• neglect of nuclear recoil : MN >> m

• neglect of nuclear spin (but can consider nuclear size)

• neglect of Lamb shift : (Z)4 << 1

What can we test in the atom

• positivity constraints

• sum rules- electric charge, - axial charge,- tensor charge, - magnetic moment of the atom = - e <b>

• existence of electronic and positronic sea

• T-even spin correlations : with fixed kT : CNN , CLL , Cpp , CLp , CpL

with fixed b : CNN , CLL , Cpp , C0N , CN0

• T-odd correlations : with fixed kT : C0N (Sivers) and CN0 (Boer-Mulders)

Deep-inelastic probes of the electron state

The experiment can be :

• inclusive (single-arm detector) measures k+ = k0 + kz • semi-inclusive (double-arm detector) measures also kT

• polarized or unpolarized

Compton : + e- (bound) + e- (free) at Moeller or Bhabha : eç + e- (bound) eç + e- (free) s, t, u >> m2

annihilation : e+ + e- (bound) +

Scaling limit

As scaling variable, we use :

k+ = k0 + kz = - Q2 / Q- in the atom rest frame.

( We do not use xBj = k+/P+ since it vanishes in the MN B limit )

There is no upper limit to |k+| . Typically, |k+- m| ~ (Z) m

Like with quarks, we consider :

q(k+) = unpolarized electron distribution

q(k+) = helicity distribution

q(k+) = transversity distribution,

as well as joint distributions in ( k+, kT) or ( k+, b)

Joint ( k+, kT ) distributions

Look at « infinite momentum frame », Pz >> M (replacing k+ by kz).

What is probed is the mechanical longitudinal momentum :

kzmec = kz

can – Az (x,y,z) ( kzcan = -iD = canonical )

Trouble : kzmec does not commute with kx and ky (either canonical or

mechanical) Speaking of a joint distribution in (kz ,kT) is heretical !

Nevertheless, in a gauge Az = 0 one can define joint distributions in

kTcan AND kz

can (= kzmec )

however :

• kz and kT have not « equal rights » : Az is zero but not AT …

• kTcan is not invariant under residual gauge freedom FSI included or not.

« allowed » and gauge invariant joint distributions

Quantum mechanics allows joint distributions in :

• ( z, b ) in the null-plane : ( X- , b )

• ( z, kTmec

) ( at least in our atom, where [ kxmec , ky

mec ] = i e Bz = 0 )

• ( kzmec , b ) in the null-plane : ( k+(mec) , b )

• ( kzmec , Lz )

Note : two-parton distributions ( k+1 , k+

2 , b12 ) involving a relative impact parameter are used for double parton scattering. In our case b is the relative electron – nucleus impact parameter.

Joint ( k+, b ) distribution

q( k+, |b| ), and its spin correlations, can in principle be measured in double atom + atom collisions :

b

Spin-dependent distributions in ( k+, b ) or ( k+, kT )

- without polarisation : q( k+, |b| )

- selecting an electron spin state | s > : q( k+, b , s )

- with atom polarisation <S> : q( k+, b ; <S> )

- with both polarisations : q( k+, b , s ; <S> )

Everything can be expressed in terms of q( k+, |b| ) and 7 correlation parameters :

C0N , CN0 , CNN , CLL , Cpp , CLp and CpL( k+, |b| )

* Same with ( k+, kT ). p = direction of b or kT .

Positivity constraints & dictionnary with Amsterdam

| CNN | < 1 ,

(1 ç CNN )2 > (C0N ç CN0)2 + (CLL ç Cpp)2 + (- CLp ç CpL)2

They are most easily obtained in transversity (along N) basis.After removal of kinematical factors kT / MN or kT

2 / (2MN2) :

f1 C00 = f1

f1 C0N = f1Tperp (Sivers)

f1 CN0 = - h1perp (Boer-Mulders)

f1 CLL = g1

f1 CNN = h1 - h1T

perp

f1 Cpp = h1 + h1T

perp ( Kotzinian – Mulders – Tangerman)

f1 CLp = g1T

f1 CpL = h1Lperp

Basic formula

42

31

)(r

)( )( i - i i- exp )( r b, b,

zzEzkdzk

(z/b) sinh Z - )z'y,(x, )( z

)(z

1-

)( 00

bb

b,

z

z

V'dzz

),( ),( ) ; ,( kkkq b bS b

)( )( ) ; ( s k,k,k,q b bSb

4321

0

)t( r,

gauge link :

electron density in ( k+, b ) :

spin density :

electron wave function in the atom rest frame :(depends on the atom spin direction S)

two-component spinor, after projection with 1+z :

null-plane wave function :

z0(b) arbitrary function

For ( k+, kT ) distributions, just take the Fourier transform of ( k+, b )

Results for impact parameter

CNN = 1

CLL = Cpp = (w2 - v2) / (w2 + v2)

C0N = CN0 = 2 w u / (w2 + v2)

CLp = CpL = 0

w, v : real functions of k+ and b.

After integration over b :

C0N and CN0 disappear

CNN - Cpp disappear ; ( CNN + Cpp ) /2 CTT

CTT = ( 1 + CLL ) /2 ; saturates the Soffer inequality.

Saturation of the inequalities

The spin inequalities come from the positivity of the density matrix of the (e+ atom) system in the t-channel.

When the all other commuting degrees of freedom are fixed (orbital momentum, spin of spectators, radiation field…) the (e+ atom) system is a pure state (in our mind)the density matrix is of rank one a maximal set of inequalities is saturated.

Then one predicts, without any calculation :

CNN = +1 CNN = -1(A) CLL = + Cpp OR : (B) CLL = - Cpp

C0N = + CN0 C0N = - CN0 CLp = - CpL CLp = + CpL

We have (A) at fixed b and k+ , (B) at fixed kT and k+ . If the atom is in the negative parity state P1/2, (A) and (B) are interchanged.

Charge sum rules

Integration over k+ and b yields the electric, axial and tensor charges

) 2q q (

1

31 1

q , 1

31 - 1

q q

2

2

2

2

1

,

)(Z - 1 mE , 1

Z 2

case Z = 1 : q = 1/3 (= « spin crisis »!...) , q = 2/3

with

Burkardt connection

Classically, for a particle of any spin J perpendicular to the figure:

G = centre of energy C = centre of charge

(1) d . e = normal magnetic moment 0 = (e J) / M

(2) d’ . e = anomalous magnetic moment a

(3) (d + d’) . e = total magnetic moment = 0 a

(2) or (3) OK for hydrogen-like atom. a = = -e (1+2) / (6m)

at rest after ultra-relativistic boost

.C

.G.C,G

d’d

Electron-positron seaRecall : electron density in a polarized atom =

q( k+, b ; S ) =

It is positive everywhere and non-vanishing for both signs of k+ .

On the other hand, its integral is 1.

What is the meaning of q( k+) for negative k+ ?

Why is the integral of q( k+) on positive k+ less than unity ?

Next : Interpretation in terms of deformed Dirac sea and parton-like sea

),( ),(

kk b b

Electron-positron sea (2)

Electron states are eigenstates of

H’ = H – v . p = H’0 + HI

with H’0 = . p + m – v . p , (v = atom velocity)

HI = - . A + A0 ( A(x,y,z,t) = moving field of the nucleus)

The term – v . p takes into account the recoil of the nucleus.

« m - state » = eigenstate of H’. « k - state » = eigenstate of H’0 = plane waves.

The deformed Dirac sea : all - states of negative energy’ are occupied :

| > = Pm<0 a*(m ) | 0 >

Electron-positron sea (3)

Atom in state N° 1 : |A1 > = a*(1 ) | >

DIS measures the number of electron in the plane-wave state | k > ,

N(k) = <A1 | a*(k ) a(k ) | A1 > = | ( k , 1 ) |2 + Sm<0 | ( k , m ) |2 ;

The first term is the one considered up to now. The second term exists even for a fully stripped nucleus. It represents the virtual electron cloud which may become by scattering with the probe.

DIS can also pick-up positrons in states | -k > :

Ne+(k) = <A1 | a(-k ) a*(-k ) | A1 > = Sm>1 | ( -k , m ) |2

If the nucleus is fully stripped, the sum is over all positive m.

Electron-positron sea (4)

Results :

Ne+= Sk> 0 Ne+(k) ; Ne- = Sk> 0 Ne-(k)

Sk> 0 | ( k , 1 ) |2 = Ne- (atom) - Ne- (nucleus) < 1

Sk< 0 | ( k , 1 ) |2 = Ne+ (nucleus) - Ne+ (atom) < 1

( Ne- - Ne+ )_atom – ( Ne- - Ne+ )_nucleus = 1

Second braket = renormalisation of the nucleus charge

• The hydrogen-like atom at high Z has many expected, calculable and fashionable DIS properties.• The connection between magnetic moment and average impact parameter is transparent there. • We have not yet studied the joint ( k+, kT ) distributions with final state

interaction. We only took z0(b) = 0 for the origin of the gauge link. • Positivity constaints, when saturated due to lack of spectator entropy, have a very predictive behaviour• The coulomb field generates an electron positron sea. Due to that and to charge renormalisation, neither the number of electron, nor the difference electrons – positrons is equal to one.

Conclusions

Inégalités de spin

2 | q(x) | < q(x) + q(x)

How to realise an intricate state in the t-channel (proton + antiquark X) :

p q

X

a c

(spin 0) (spin 0)

b (spin 1/2)

structure function