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Nucleon Internal Structure: the Quark Gluon Momentum and Angular Momentum X.S.Chen, X.F.Lu Dept. of Phys., Sichuan Univ. W.M.Sun, Fan Wang NJU and PMO Joint Center for particle nuclear physics and cosmology (J-CPNPC) Outline I. Introduction II. Conflicts between Gauge invariance and Canonical Quantization III. Quantum mechanics IV. QED V. QCD VI. Nucleon internal structure VII. Summary I.Introduction • Quark gluon contribution to the nucleon observables, mass, momentum, spin, magnetic moment, etc. are unavoidable in the study of nucleon internal structure. • We never have the quark gluon momentum and angular momentum operators which satisfy both gauge invariance and canonical commutation relations. Pauli-Landau-Feynman-……. II. Conflicts between gauge invariance and canonical quantization Quantum Mechanics Even though the Schroedinger equation is gauge invariant, the matrix elements of the canonical momentum, orbital angular momentum, and Hamiltonian of a charged particle moving in eletromagnetic field are gauge dependent, orbital angular momentum especially the and Hamiltonian of the hydrogen atom are “not the measurable ones” !? It is absurd! QED The canonical momentum and orbital angular momentum of electron are gauge dependent and so their physical meaning is obscure. The canonical photon spin and orbital angular momentum operators are also gauge dependent. Their physical meaning is obscure too. Even it has been claimed that it is impossible to have photon spin and orbital angular momentum operators. Multipole radiation The multipole radiation theory is based on the decomposition of a polarized em wave into multipole radiation field with definite photon spin and orbital angular momentum coupled to a total angular momentum quantum number LM, A = ξ peik⋅r = 2π Σ∞=1i L 2L +1DMp (ϕ,θ ,0)[ALM (m) + ip ALM (e)] L L L L +1 A LM (e) = − ς L +1T LL +1M + ς L −1T LL −1M 2L + 1 2L + 1 A LM (m) = ς L T LLM Multipole radiation measurement and analysis are the basis of atomic, molecular, nuclear and hadron spectroscopy. If the orbital angular momentum of photon is gauge dependent and not measurable, then all determinations of the parity of these microscopic systems would be meaningless! QCD • Because the canonical parton (quark and gluon) momentum is “gauge dependent”, so the present analysis of parton distribution of nucleon uses the covariant derivative operator instead of the canonical momentum operator; uses the Poynting vector as the gluon momentum operator. They are not the right momentum operators! • The quark spin contribution to nucleon spin has been measured, the further study is hindered by the lack of gauge invariant quark orbital angular momentum, gluon spin and orbital angular momentum operators. The present gluon spin measurement is even under the condition that “there is not a gluon spin can be measured”. III. Quantum Mechanics Gauge is an internal degree of freedom, no matter what gauge used, the canonical coordinate and momentum of a charged particle is r and p = − i ∇ , the orbital angular momentum is ∇ , L = r× p = r× the Hamiltonian is i ( p − e A)2 H = + eϕ 2m Gauge transformation ψ → ψ ' = eieω ( x )ψ , A → A = A + ∇ω, ' ϕ → ϕ ' = ϕ − ∂ tω , The matrix elements transformed as ψ | p |ψ → ψ | p |ψ + ψ | e∇ω |ψ , ψ | L|ψ → ψ | L|ψ + ψ | er×∇ω|ψ , ψ | H |ψ → ψ | H |ψ − ψ | e∂tω |ψ , They are not gauge invariant, even though the Schroedinger equation is. New momentum operator in quantum mechanics Canonical momentum for a charged particle moving in em field: p = m r + q A = m r + q A ⊥ + q A // Its ME is not gauge invariant, but satisfies the canonical momentum commutation relation. p − q A // = m r + q A ⊥ ∇ ⋅ A⊥ = 0, ∇× A// = 0 It is both gauge invariant and satisfies canonical momentum commutation relation. We call Dphy 1 = p − qA// = ∇− qA// i i physical momentum. It is neither the canonical momentum 1 p = mr + q A = ∇ i nor the mechanical momentum 1 p − q A = mr = D i Gauge transformation ψ ' = e iq ω ( x )ψ , Aμ = Aμ + ∂ μ ω ( x), ' only affects the longitudinal part of the vector potential A// = A// + ∇ ω ( x ), ' and time component ϕ ' = ϕ − ∂ tω ( x), it does not affect the transverse part, A = A⊥, ' ⊥ so A⊥ is physical and which is used in Coulomb gauge. A // is unphysical, it is caused by gauge transformation. Hamiltonian of hydrogen atom Coulomb gauge: A = ϕ ≠ 0. c c = 0, c c A // A ≠ 0, ⊥ 0 Hamiltonian of a nonrelativistic particle c (p − qA ) ⊥ 2 Hc = + qϕ c . 2m Gauge transformed one c c A // = A + ∇ω ( x ) = ∇ω ( x ), A ⊥ = A , ϕ = ϕ c − ∂ t ω ( x ) // ⊥ c ( p − q A) ( p − q∇ω − q A ) 2 ⊥ 2 H= + qϕ = + qϕ c − q∂ tω. 2m 2m Follow the same recipe, we introduce a new Hamiltonian, c ( p − q A// − q A ) ⊥ 2 H phy = H + q∂ t ω ( x ) = + qϕ c 2m ω = ∇ −2∇ ⋅ A// which is gauge invariant, i.e., ψ | H phy | ψ = ψ c | H c | ψ c This means the hydrogen energy calculated in Coulomb gauge is gauge invariant and physical. A rigorous derivation Start from a ED Lagrangian including electron, proton and em field, under the heavy proton approximation, one can derive a Dirac equation and a Hamiltonian for electron and proved that the time evolution operator is different from the Hamiltonian exactly as we obtained phenomenologically. The nonrelativistic one is the above Schroedinger or Pauli equation. IV.QED Different approach will obtain different energy-momentum tensor and four momentum, they are not unique: Noether theorem ∇ P = ∫ d 3 x{ψ + ψ + E i ∇Ai } i Gravitational theory (Weinberg) or Belinfante tensor D P = ∫ d 3 x{ψ + ψ + E × B} i It appears to be perfect , but individual part does not satisfy the momentum algebra. Usually one supposes these two expressions are equivalent, because the integral is the same. We are experienced in quantum mechanics, so we introduce D P = ∫ d x{ψ + ψ + E i ∇A⊥ } 3 phy i i A = A // + A ⊥ D phy = ∇ − ieA// They are both gauge invariant and momentum algebra satisfied. They return to the canonical expressions in Coulomb gauge. We proved the renowned Poynting vector is not the correct momentum of em field J γ = ∫ d xr ×(E × B) = ∫ d xE × A⊥ + ∫ d xr × E ∇A ⊥ 3 3 3 i i It includes photon spin and orbital angular momentum Electric dipole radiation field i Blm = a h (kr) LYlm ,......E lm = ik Alm = ∇ × Blm lm l (1) k 1 ∗ | a11 |2 3 1+ cos2 θ sinθ Re[E11 ×B11] = ⋅ ⋅[ nr + nϕ ] 2 (kr) 16π 2 2 kr 1 i∗ | a11 |2 3 1 + cos2 θ sinθ Re[E11∇A11] = i ⋅ ⋅[ nr + nϕ ] 2 (kr) 16π 2 2 2kr dP | a11 |2 3 1 + cos 2 θ dJ z = ⋅ =k dΩ k 2 16 π 2 dΩ dJ z | a11 |2 3 = ⋅ ⋅ sin 2 θ dΩ k 3 16π J QED = S e + L e + S γ + Lγ • Each term in this decomposition satisfies the canonical angular momentum algebra, so they are qualified to be called electron spin, orbital angular momentum, photon spin and orbital angular momentum operators. • However they are not gauge invariant except the electron spin. Therefore the physical meaning is obscure. • How to reconcile these two fundamental requirements, the gauge invariance and canonical angular momentum algebra? • One choice is to keep gauge invariance and give up canonical commutation relation. J QED = S e + L ' e + J 'γ • However each term no longer satisfies the canonical angular momentum algebra except the electron spin, in this sense the second and third term is not the electron orbital and photon angular momentum operator. The physical meaning of these operators is obscure too. • One can not have gauge invariant photon spin and orbital angular momentum operator separately, the only gauge invariant one is the total angular momentum of photon. The photon spin and orbital angular momentum had been measured! Dangerous suggestion It will ruin the multipole radiation analysis used from atom to hadron spectroscopy. Where the canonical spin and orbital angular momentum of photon have been used. It is totally unphysical! J QED = S e + L e ' '+ S γ ' '+ L γ ' ' Multipole radiation Multipole radiation analysis is based on the decomposition of em vector potential in Coulomb gauge. The results are physical and gauge invariant. V. QCD ∇ P = ∫ d x{ψ ψ + E i ∇ Ai } 3 + i D P = ∫ d x{ψ ψ + E × B} 3 + i D phy P = ∫ d xψ 3 + ψ + ∫ d 3 xE i a D phy Aiphys i D phy = ∇ − ig A pure a D phy = ∇ − ig[ A pure, ] • From QCD Lagrangian, one can get the total angular momentum by Noether theorem: • One can have the gauge invariant decomposition, New decomposition '' '' '' J QCD = S q + L + S + L q g g Σ ∫ xψ ψ + S q = d 3 2 D phy L = ∫ d xψ r × '' q 3 + ψ i ∫ '' S g = d 3 x E × A phy ∫ '' L g = d 3 xEi r × a D p h y Ai phy Esential task:to define properly the pure gauge field Apure and physical one A phy Phys.Rev.Lett.100,232002(2008), arXiv:0904.0321[hep-ph] a D phy = ∇ − ig A pure A pure = T A a pure A = A pure + A phy D phy × A pure = ∇ × A pure − ig A pure × A pure = 0 a D phy ⋅ Aphy = ∇ ⋅ Aphy − ig[ Apure , Aphy ] = 0 VI. Nucleon internal structure it should be reexamined! • The present parton distribution is not the right quark and gluon momentum distribution. In the asymptotic limit, the gluon only contributes ~1/5 nucleon momentum, not 1/2 ! • The nucleon spin structure study should be based on the new decomposition and new operators. • One has to be careful when one compares experimental measured quark gluon momentum and angular momentum to the theoretical ones. The proton spin crisis is mainly due to misidentification of the measured quark axial charge to the nonrelativistic Pauli spin. Phys. Rev. D58,114032 (1998) VII. Summary • The renowned Poynting vector is not the right momentum operator of photon and gluon field. • The space time translation operators of the Fermion part are not observables. • The gauge invariant and canonical quantization rule satisfied momentum, spin and orbital angular momentum operators of the individual part do exist. • The Coulomb gauge is physical, expressions in Coulomb gauge, even with vector potential, are gauge invariant, including the hydrogen atomic Hamiltonian and multipole radiation. Thanks Nucleon Internal Structure • 1. Nucleon anomalous magnetic moment Stern’s measurement in 1933; first indication of nucleon internal structure. • 2. Nucleon rms radius Hofstader’s measurement of the charge and magnetic rms radius of p and n in 1956; Yukawa’s meson cloud picture of nucleon, p->p+ π 0 ; n+ π + ; n->n+ π 0 ; p+ π − . • 3. Gell-mann and Zweig’s quark model SU(3) symmetry: baryon qqq; meson q q . SU(6) symmetry: 1 B(qqq)= [χms(q3)ηms(q3)+χma(q3)ηma(q3)] . 2 color degree of freedom. quark spin contribution to nucleon spin, 4 1 Δu = ; Δd = − ; Δs = 0. 3 3 nucleon magnetic moments. There is no proton spin crisis but quark spin confusion The DIS measured quark spin contributions are: While the pure valence q3 S-wave quark model calculated ones are: . • It seems there are two contradictions between these two results: 1.The DIS measured total quark spin contribution to nucleon spin is about one third while the quark model one is 1; 2.The DIS measured strange quark contribution is nonzero while the quark model one is zero. • To clarify the confusion, first let me emphasize that the DIS measured one is the matrix element of the quark axial vector current operator in a nucleon state, Here a0= Δu+Δd+Δs which is not the quark spin contributions calculated in CQM. The CQM calculated one is the matrix element of the Pauli spin part only. The axial vector current operator can be expanded as • Only the first term of the axial vector current operator, which is the Pauli spin part, has been calculated in the non-relativistic quark models. • The second term, the relativistic correction, has not been included in the non-relativistic quark model calculations. The relativistic quark model does include this correction and it reduces the quark spin contribution about 25%. q • The third term, q creation and annihilation, will not contribute in a model with only valence quark configuration and so it has never been calculated in any quark model as we know. An Extended CQM with Sea Quark Components • To understand the nucleon spin structure quantitatively within CQM and to clarify the quark spin confusion further we developed a CQM with sea quark components, Where does the nucleon get its Spin • As a QCD system the nucleon spin consists of the following four terms, • In the CQM, the gluon field is assumed to be frozen in the ground state and will not contribute to the nucleon spin. • The only other contribution is the quark orbital angular momentum Lq . • One would wonder how can quark orbital angular momentum contribute for a pure S-wave configuration? • The quark orbital angular momentum operator can be expanded as, • The first term is the nonrelativistic quark orbital angular momentum operator used in CQM, which does not contribute to nucleon spin in a pure valence S-wave configuration. • The second term is again the relativistic correction, which takes back the relativistic spin reduction. • The third term is again the qq creation and annihilation contribution, which also takes back the missing spin. • It is most interesting to note that the relativistic correction and the qq creation and annihilation terms of the quark spin and the orbital angular momentum operator are exact the same but with opposite sign. Therefore if we add them together we will have where the , are the non-relativistic part of the quark spin and angular momentum operator. • The above relation tell us that the nucleon spin can be either solely attributed to the quark Pauli spin, as did in the last thirty years in CQM, and the nonrelativistic quark orbital angular momentum does not contribute to the nucleon spin; or • part of the nucleon spin is attributed to the relativistic quark spin, it is measured in DIS and better to call it axial charge to distinguish it from the Pauli spin which has been used in quantum mechanics over seventy years, part of the nucleon spin is attributed to the relativistic quark orbital angular momentum, it will provide the exact compensation missing in the relativistic “quark spin” no matter what quark model is used. • one must use the right combination otherwise will misunderstand the nucleon spin structure. VI. Summary 1.The DIS measured quark spin is better to be called quark axial charge, it is not the quark spin calculated in CQM. 2.One can either attribute the nucleon spin solely to the quark Pauli spin, or partly attribute to the quark axial charge partly to the relativistic quark orbital angular momentum. The following relation should be kept in mind, 3.We suggest to use the physical momentum, angular momentum, etc. in hadron physics as well as in atomic physics, which is both gauge invariant and canonical commutation relation satisfied, and had been measured in atomic physics with well established physical meaning. Thanks