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Electromagnetic multipole moments

Alfons Buchmann University of Tübingen

1. Introduction 2. Methods 3. Results 4. Summary

NSTAR 2017, Columbia, 19-24 August 2017 1. Introduction Elastic -nucleon scattering

e‘ N‘

Elastic form factors Q... four-momentum transfer Q Q²= -(ω²- q²) N 2 GC, M (Q ) ω...energy transfer γ N 2 q... three-momentum transfer GC (Q ) charge N 2 magnetic G M (Q ) e N

e... electron γ... photon N... nucleon (p,n) Inelastic electron-nucleon scattering

π pion e‘ N‘

Q ∆(1232) Q...four-momentum transfer of virtual photon N→Δ 2 γ G M1, E 2,C2 (Q ) transition form factors

e N

electro-pionproduction process Nucleon excitation spectrum

N*(1680) N*(1520)

N*(1440)

∆(1232) C0, M1 radial excitation CJ…charge multipoles N(939) EJ…electric multipoles MJ…magnetic multipoles

J=1/2+ J=3/2- J=3/2+ J=5/2+ …. Purpose of this talk

Explore relations ´ between nucleon ground state properties (elastic form factors)

and

transition multipole moments (inelastic form factors). Electromagnetic multipole moments

What can we learn from electromagnetic multipole moments?

Electromagnetic multipole moments are directly connected with the charge and current distributions in baryons.

Their sign and magnitude provide fundamental information on • structure, • size, • shape of baryons and their excited states. Multipole expansion of charge density ρ

–q +q +q +q –q +q –q

+q = + + + +q –q

–q –q +q –q +q

~ q0 + d1 + Q2 + Ω3 + …

𝝆𝝆 monopole octupole

J=0 J=1 J=2 J=3 Another charge configuration +q

+q +q –2q

+q –2q = + + + + …

–q –2q +q

+q ~ q + d1 + Q2 + Ω3 + … 𝜌𝜌 monopole dipole quadrupole octupole Quadrupole of baryon B

𝑸𝑸 = 3 2 2 3 𝑄𝑄 � 𝜌𝜌𝐵𝐵 𝑟𝑟⃗ 𝑧𝑧 − 𝑟𝑟 𝑑𝑑 𝑟𝑟⃗

If ρ concentrated along z-axis, 3 -term dominates → > 0 z 2 𝑧𝑧 𝑄𝑄 prolate

If ρ concentrated in x-y plane, -term dominates → < 0 z

𝑟𝑟푟 𝑄𝑄 oblate Magnetic octupole moment Ω

The magnetic octupole moment Ω measures the deviation of the distribution from spherical

Ω > 0 magnetic moment density is prolate

3 : = × 3 8 3 2 2 Ω � 𝑑𝑑 𝑟𝑟⃗ 𝑟𝑟⃗ 𝐽𝐽⃗ 𝑟𝑟⃗ 𝑧𝑧 𝑧𝑧 − 𝑟𝑟 Ω < 0 magnetic moment density is oblate What about higher multipole moments?

Presently, very little is known about magnetic octupole moments of decuplet baryons and excited N* states 2. Methods

Information needed to reveal structural details of spatial current distributions in baryons. Methods

1. SU(6) spin-flavor operator parameterization

2. 1/ expansion of operators

3. Current𝑁𝑁𝐶𝐶 operator algebra

Methods are based on the of the QCD Lagrangian.

What symmetries are relevant for bound states of u,d,s quarks ? 1. SU(6) spin-flavor symmetry analysis

SU(6) spin-flavor symmetry combines SU(3) multiplets with different spin and flavor to SU(6) spin-flavor supermultiplets.

Gürsey, Radicati, Sakita (1964) SU(6) spin-flavor supermultiplet

S

baryon supermultiplet 56 dimensional T3

56 = 8 , 2 + 10 , 4

flavor spin flavor spin Gürsey-Radicati SU(6) formula

= + + + 1 + ( + 1) 42 𝑌𝑌 𝑀𝑀 𝑀𝑀0 𝑀𝑀1 𝑌𝑌 𝑀𝑀2 𝑇𝑇 𝑇𝑇 − 𝑀𝑀3 𝐽𝐽 𝐽𝐽 hypercharge isospin SU(6) symmetry breaking term 𝑌𝑌 ⋯spin 𝑇𝑇 ⋯ 𝐽𝐽 ⋯ Relations between octet and decuplet 𝑖𝑖 𝑗𝑗 baryon 𝜎𝜎⃗ � 𝜎𝜎⃗

e.g. = ∗ ∗ 𝑀𝑀Ξ − 𝑀𝑀Σ 𝑀𝑀Ξ − 𝑀𝑀Σ General spin-flavor operator

𝑶𝑶� = A + B + C one-body two-body three-body 𝑂𝑂� 𝑂𝑂� 1 𝑂𝑂� 2 𝑂𝑂� 3

Constants A, B, C are determined from experiment.

… allowed operators in spin-flavor space = 1,2,3 � 𝑖𝑖 𝑖𝑖 𝑂𝑂 Which spin-flavor operators are allowed? Operator structures determined from SU(6) theory. SU(6) spin-flavor symmetry breaking

ei ... quark charge

Example: charge operator = σi ... quark spin

mi ... quark mass 𝑂𝑂� 𝜌𝜌�

σi σ j σi σ j

mi m j mi m j ek

ei γ γ 2-quark operator [ ] 3-quark operator [ ]

2 3 SU(6) symmetry𝜌𝜌breaking via spin and flavor𝜌𝜌 dependent two- and three-quark operators SU(6) spin-flavor symmetry and its breaking

Connection between observables of different rank = 0 = 2 e.g. charge radii ( ) and quadrupole moments ( ).𝐽𝐽 𝐽𝐽 𝐽𝐽 Example: Multipole expansion of [ ] in spin-flavor space

𝜌𝜌 2

[ ] = 3 2 3

𝜌𝜌 2 −𝐵𝐵 � 𝑒𝑒𝑖𝑖 𝜎𝜎(⃑𝑖𝑖 � 𝜎𝜎)⃑𝑗𝑗 − 𝜎𝜎𝑖𝑖𝑖𝑖 𝜎𝜎(𝑗𝑗𝑧𝑧 −) 𝜎𝜎⃑𝑖𝑖 � 𝜎𝜎⃑𝑗𝑗 𝑖𝑖≠𝑗𝑗 scalar tensor J=2 The prefactors of the Jspin=0 scalar (+2) and spin tensor ( 1) terms are determined by the SU(6) group algebra. − 2. Large expansion of QCD processes

𝑵𝑵𝑪𝑪 ... number of colors 𝑁𝑁𝐶𝐶 = = ~ strong coupling 2 2 2 𝑔𝑔 𝑄𝑄 11 12𝜋𝜋2 ln 𝟏𝟏 𝛼𝛼𝑆𝑆 𝑄𝑄 2 4𝜋𝜋 𝑄𝑄 𝑵𝑵𝒄𝒄 𝑵𝑵𝒄𝒄 − 𝑁𝑁𝑓𝑓 2 Λ two-quark operator three-quark operator

1 1 g ~ g ~ N N  C C

1 1

𝑂𝑂 1 𝑂𝑂 2 𝑁𝑁𝑐𝑐 𝑁𝑁𝑐𝑐 Example: Quadrupole moment operator

[ ] = 𝑁𝑁𝐶𝐶 3 two-body 𝐵𝐵 1 � 2 𝑖𝑖 𝑖𝑖𝑖𝑖 𝑗𝑗𝑧𝑧 𝑖𝑖 𝒋𝒋 1 𝑄𝑄 𝐶𝐶 � 𝑒𝑒 𝜎𝜎 𝜎𝜎 − 𝝈𝝈 � 𝝈𝝈 𝑁𝑁𝐶𝐶 𝑁𝑁 𝑖𝑖≠𝑗𝑗 𝑂𝑂

= 𝐶𝐶 3 [ ] 𝑁𝑁 three-body 𝐶𝐶 1 𝑄𝑄� 3 2 � 𝑒𝑒𝑘𝑘 𝜎𝜎𝑖𝑖𝑖𝑖𝜎𝜎𝑗𝑗𝑗𝑗 − 𝝈𝝈𝑖𝑖 � 𝝈𝝈𝒋𝒋 2 𝑁𝑁𝐶𝐶 𝑖𝑖≠𝑗𝑗≠𝑘𝑘 𝑂𝑂 𝑁𝑁𝐶𝐶

Note: There is no one-quark contribution. Reason: Cannot build = 2 tensor from a single = 1 matrix .

𝐽𝐽 𝑄𝑄� 𝐽𝐽 𝝈𝝈𝑖𝑖 Advantage of / analysis

𝑪𝑪 • Large NC QCD provides a perturbative𝟏𝟏 𝑵𝑵 expansion scheme • works at all energy scales • hierarchy of quark operators due to powers of 1

⁄𝑁𝑁𝐶𝐶 [ ] > [ ] > [ ]

one𝑂𝑂-�quark1 two𝑂𝑂�-quark2 three𝑂𝑂�-quark3 1 1 1 0 1 2 𝑂𝑂 ⁄𝑁𝑁𝐶𝐶 𝑂𝑂 ⁄𝑁𝑁𝐶𝐶 𝑂𝑂 ⁄𝑁𝑁𝐶𝐶 supressed

BUT: If = 0 due to selection rules dominant!

𝑂𝑂� 1 𝑂𝑂� 2 3. Current algebra method

• SU(3) generators satisfy commutation relations (Lie algebra)

𝑖𝑖 𝜆𝜆 , = 2

𝑖𝑖 𝑗𝑗 𝑖𝑖𝑖𝑖𝑖𝑖 𝑘𝑘 SU(3)𝜆𝜆 𝜆𝜆 structure𝑖𝑖constants𝑓𝑓 𝜆𝜆 i, j, k=1,…,8 • Gell-Mann Nishijima𝑓𝑓𝑖𝑖𝑖𝑖𝑖𝑖 ⋯ relation for electric charge

1 1 𝑄𝑄 = + = + 2 2 3 𝑌𝑌 𝑄𝑄 𝑇𝑇3 𝜆𝜆3 𝜆𝜆8 • Generalized electromagnetic current densities involve = , = ( + ), and = ( ) 1 1 1 𝑇𝑇3Therefore2 𝜆𝜆3 𝑇𝑇,+ they2 also𝜆𝜆1 𝑖𝑖obey𝜆𝜆2 SU(3)𝑇𝑇− Lie2algebra𝜆𝜆1−𝑖𝑖𝜆𝜆2. Electromagnetic current commutators

Electromagnetic vector currents obey SU(3) Lie algebra

, ( ) = ( ) 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝐽𝐽𝑖𝑖 𝑟𝑟⃗ 𝐽𝐽𝑗𝑗 𝑟𝑟⃗′ 𝑖𝑖 𝑓𝑓𝛼𝛼𝛼𝛼𝛼𝛼 𝛿𝛿𝑖𝑖𝑖𝑖 𝛿𝛿 𝑟𝑟⃗ − 𝑟𝑟⃗′ 𝐽𝐽0 𝑟𝑟⃗′ =0, , 3 𝛼𝛼 𝑖𝑖 𝐽𝐽• ,𝑟𝑟⃗ ⋯ four−vector current- density, 𝑖𝑖 ⋯ • , , 𝑖𝑖 𝑗𝑗 ⋯ Lorentz 4 vector indices • 𝛼𝛼 𝛽𝛽 𝛾𝛾 ⋯ SU(3) flavor indices 𝑓𝑓𝛼𝛼𝛼𝛼𝛼𝛼 ⋯ SU(3) structure1 constants = 6 2 2 𝑝𝑝 𝑝𝑝 Gell-Mann𝜇𝜇 Dashen𝑟𝑟Lee (1965) Advantage of current algebra method

Gell-Mann (1964) No matter how badly SU(3) flavor symmetry is broken, the SU(3) commutation relations between group generators are an exact law of nature. 3. Results N ransition quadrupole moment

∆ 𝐭𝐭 [ ] = 3 2 3

𝜌𝜌 2 −𝐵𝐵 � 𝑒𝑒𝑖𝑖 𝜎𝜎⃑(𝑖𝑖 � 𝜎𝜎⃑)𝑗𝑗 − 𝜎𝜎𝑖𝑖𝑖𝑖 𝜎𝜎(𝑗𝑗𝑧𝑧 −) 𝜎𝜎⃑𝑖𝑖 � 𝜎𝜎⃑𝑗𝑗 𝑖𝑖≠𝑗𝑗 scalar tensor J=0 J=2 neutron charge = 56 [ ] 56 = 4 2 𝐽𝐽=0 𝑟𝑟𝑛𝑛 𝑛𝑛 𝜌𝜌 2 𝑛𝑛 𝐵𝐵 N→∆ transition quadrupole moment = 56 [ ] 56 = 2 2 + + 𝐽𝐽=2 𝑝𝑝→∆ ∆ 𝑝𝑝 𝑄𝑄 1 𝜌𝜌 2 𝐵𝐵 = 2 + 2 𝑄𝑄𝑝𝑝→∆ 𝑟𝑟𝑛𝑛 Buchmann, Hernandez, Faessler, PRC 55 (1997) 448 spin flip

Double mechanism

C2

Simultaneously flipping the spin of two quarks via two-body exchange current

A. J. Buchmann, E. Hernandez, A. Faessler, Phys. Rev. C55 (1997) 448 N → ∆ transition quadrupole moment

Extraction of p→ ∆+(1232) transition quadrupole moment from electron-proton and photon-proton scattering data

data

∆( ) = 0.0846(33) Tiator et al., EPJ ST 198 (2011) 141

𝑄𝑄𝑁𝑁 𝑒𝑒𝑒𝑒𝑒𝑒 − fm² ∆( ) = 0.108(9) Blanpied et al., PRC 64 (2001) 025203

𝑄𝑄𝑁𝑁 exp − fm² theory 1 = = 0.0821(20) AJB, Hernandez, Faessler, PRC 55, 448 2 + 2 𝑝𝑝→∆ 𝑛𝑛 𝑄𝑄 neutron charge𝑟𝑟 radius− N quadrupole moment in

𝟏𝟏⁄𝑵𝑵𝑪𝑪 ∆ + 5 ( 1) = 2 2 + 𝐵𝐵 𝐶𝐶 𝑁𝑁𝐶𝐶 𝑁𝑁𝐶𝐶 − 𝑄𝑄𝑝𝑝→∆ − 2 Including 3-body 𝑁𝑁𝐶𝐶 𝑁𝑁𝐶𝐶 operators + 5 ( + 3) = 2 2 𝐵𝐵 𝐶𝐶 𝑁𝑁𝐶𝐶 𝑁𝑁𝐶𝐶 𝑟𝑟𝑛𝑛 − 2 𝑁𝑁𝐶𝐶 𝑁𝑁𝐶𝐶 𝑁𝑁𝐶𝐶 1 + 5 = Indicative of a 2 + 3 1 more general + 2 𝑁𝑁𝐶𝐶 𝑁𝑁𝐶𝐶 validity 𝑄𝑄𝑝𝑝→∆ 𝑟𝑟𝑛𝑛 𝑁𝑁𝐶𝐶 𝑁𝑁𝑐𝑐 − 1 AJB, J. Hester, R. Lebed,for PRD𝑁𝑁 𝐶𝐶66,=3 056002and 𝑁𝑁 (2002)𝐶𝐶=∞ N → ∆ form factor relations

p→Δ+ 2 n 2 G M1 (Q ) = − 2 G M (Q ) magnetic form factors Beg, Lee, Pais, 1964 + = − μ p→∆ 2 μ n

+ 3 2 G p→Δ (Q2 ) = − G n (Q2 ) C2 Q2 C charge form factors AJB, Phys. Rev. Lett. 93 (2004) 212301 1 2 Q →∆+ = rn p 2 Definition of C2/M1 ratio

 + q M p→Δ 2 C2 2 N GC2 (Q ) (Q ) = + p→Δ 2 M1 6 G M1 (Q )

Insert form factor relations  C 2 q M G n (Q2 ) 2 = N C (Q ) n 2 M1 Q 2Q G M (Q )

C2/M1 expressed via neutron elastic form factors

A. J. Buchmann, Phys. Rev. Lett. 93 (2004) 212301 MAID 2007 analysis

C2/M1(Q²)=S1+/M1+(Q²)

   MAID 2003  .  .  Buchmann 2004  MAID 2007

from: Drechsel, Kamalov, Tiator, EPJ A34 (2007) 69 MAID 2007 analysis

   MAID 2003  .  .  Buchmann 2004  MAID 2007

∆ JLab data analysis • MAID 2007 analysis of JLab data Interpretation in pion cloud model

Q0 > 0 Q0 < 0 prolate oblate

A. J. Buchmann and E. M. Henley, Phys. Rev. C63, 015202 (2001) Interpretation in quark model

Two-quark spin-spin operators are repulsive for quark pairs with spin 1.

In the neutron, the two down quarks In the ∆0, all quark pairs have spin 1. are in a spin 1 state, and are pushed Equal distance between down-down further apart than an up-down pair. and up-down pairs. → elongated (prolate) charge distribution → planar (oblate) charge distribution → negative neutron charge radius. → zero charge radius. d prolate oblate u 0 u ∆ n d d

2 d 0 2 Q0(n) = − rn A. J. Buchmann, Can. J. Phys. 83 (2005) Q0(∆ ) = rn Quadrupole moment commutator

, = 3 ( ) 𝛼𝛼 𝛽𝛽 3 2 2 2 𝛾𝛾 𝑄𝑄𝑧𝑧 𝑄𝑄𝑧𝑧 𝑖𝑖 𝑓𝑓𝛼𝛼𝛼𝛼𝛼𝛼 � 𝑑𝑑 𝑥𝑥 𝑧𝑧 − 𝑟𝑟 𝐽𝐽0 𝑥𝑥⃗ Evaluate between proton states

, = 3 ( ) 𝛼𝛼 𝛽𝛽 3 2 2 2 𝛾𝛾 𝑝𝑝 𝑄𝑄𝑧𝑧 𝑄𝑄𝑧𝑧 𝑝𝑝 𝑝𝑝 𝑖𝑖 𝑓𝑓𝛼𝛼𝛼𝛼𝛼𝛼 ∫ 𝑑𝑑 𝑥𝑥 𝑧𝑧 − 𝑟𝑟 𝐽𝐽0 𝑥𝑥⃗ 𝑝𝑝 • Insert allowed excited states 1232 , 1680 , on lhs ∗ ∆ 𝑁𝑁 ⋯ • Note: The rhs is nonzero even though proton has no spectroscopic quadrupole moment! ( ) transition quadrupole moment ∗ 𝑵𝑵 → 𝑵𝑵 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 4 (1680) (1680) = ( ) 5 + ( − 3 4 3 𝑧𝑧 𝑧𝑧 ( ) 0 𝑝𝑝 𝑄𝑄 𝑛𝑛 2 𝑛𝑛 𝑄𝑄 𝑝𝑝 𝑝𝑝 ∫ 𝑑𝑑 𝑥𝑥 𝑟𝑟 𝐽𝐽 𝑥𝑥⃗ 𝑝𝑝 𝐼𝐼𝐼𝐼 2 𝑄𝑄 𝑝𝑝→𝑝𝑝 1680 4 4 2 𝑟𝑟𝐼𝐼𝐼𝐼 𝑝𝑝 ( 1680 ) = ( )5 5 2 4 𝑄𝑄 𝑝𝑝 → 𝑝𝑝 𝐼𝐼𝐼𝐼 𝑟𝑟𝐼𝐼𝐼𝐼 𝑝𝑝 1 ( 1680 ) = 5 4 4 𝑄𝑄 𝑝𝑝 → 𝑝𝑝 𝐼𝐼𝐼𝐼 𝑟𝑟𝑝𝑝 − 𝑟𝑟𝑛𝑛 4th moment of the proton charge distribution 4 𝑟𝑟𝑝𝑝 ⋯4th moment of the neutron charge distribution 4 𝑟𝑟𝑛𝑛 ⋯ Comparision with data

Comparision with data (in 10 GeV ) from PDG and MAID −3 −1⁄2 = 15 ± 6, = 133 ± 12, = 44 (MAID)

𝐴𝐴1⁄2 𝑝𝑝 = −29 ± 10, 𝐴𝐴3⁄2 𝑝𝑝 = 33 ± 9, 𝑆𝑆1⁄2 𝑝𝑝 =−0 (MAID) 𝐴𝐴1⁄2 𝑛𝑛 𝐴𝐴3⁄2 𝑛𝑛 − 𝑆𝑆1⁄2 𝑛𝑛 Experiment: ( 1680 ) = 0.11 fm ( 1680 ) ( ) = 0. 60 fm2 Theory: 𝑄𝑄 𝑝𝑝 → 𝑝𝑝 𝐼𝐼𝐼𝐼 𝑒𝑒𝑒𝑒𝑒𝑒 2 Using: = 1.45 f𝑄𝑄m 𝑝𝑝and→ 𝑝𝑝 = 0.31𝐼𝐼𝐼𝐼fm𝑡𝑡푡(not𝑡𝑡𝑡𝑡 well known), and only4N*(1680) as4 intermediate4 state4on commutator side 𝑟𝑟𝑝𝑝 𝑟𝑟𝑛𝑛 −

Agreement in sign but not in magnitude. Measuring baryon magnetic octupole moments

Information on the shape of current distribution in baryons from sign and magnitude of Ω.

How can one measure Ω?

Best chance presumably by analyzing the 1680 transition form factors. ∗ 𝑁𝑁 → 𝑁𝑁 ( ) transition octupole moment ∗ 𝑵𝑵 → 𝑵𝑵 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 3 ( 1680 = 224 6 6 𝐼𝐼𝐼𝐼 𝑝𝑝 𝑛𝑛 𝛺𝛺 𝑝𝑝6 →th moment𝑝𝑝 of the proton charge𝑟𝑟distribution− 𝑟𝑟 6 𝑟𝑟𝑝𝑝 ⋯6 th moment of the neutron charge distribution 6 Use𝑟𝑟:𝑛𝑛 ⋯= 4.78 fm and = 1.18 fm (not well known) 6 and only6N*(1680)6 on commutator6 side 𝑟𝑟𝑝𝑝 𝑟𝑟𝑛𝑛 − Theory: ( 1680 ) ( ) = 0.28 fm 3 Experiment:𝛺𝛺 𝑝𝑝( → 𝑝𝑝 1680 𝐼𝐼𝐼𝐼) 𝑡𝑡푡( 𝑡𝑡𝑡𝑡 ) = 0.12 fm (PDG) 3 Agreement𝛺𝛺 𝑝𝑝 →in 𝑝𝑝sign but not𝐼𝐼𝐼𝐼 𝑒𝑒𝑒𝑒𝑒𝑒 in magnitude. Decuplet magnetic octupole moments Ω Baryon ( = ) ( ) = 𝑚𝑚𝑢𝑢

𝛀𝛀 𝒓𝒓 𝟏𝟏 𝛀𝛀 𝒓𝒓 ≠ 𝟏𝟏 𝑟𝑟 SU(3) 𝑚𝑚𝑠𝑠 − 0 0 ∆ −4𝐶𝐶 −4𝐶𝐶 breaking 0 ∆ +

∆ 4𝐶𝐶 4𝐶𝐶 parameter ++ (1 + r + r )/3 ∆ 8𝐶𝐶 8𝐶𝐶 ∗− 0 (1 + r 2r 2)/3 Σ −4𝐶𝐶 −4𝐶𝐶 ∗0 (1 + 2r r2)/3 Σ 2𝐶𝐶 − ∗+ (r + r + r2 )/3 Σ 4𝐶𝐶 4𝐶𝐶 − ∗− 0 (2r r 2 r 3)/3 Ξ −4𝐶𝐶 −4𝐶𝐶 ∗0 2r 3 Ξ 2𝐶𝐶 − − − 3 Ω A.J.B. and E.M.− Henley,4𝐶𝐶 EPJA 35, 267 (2008)−4𝐶𝐶 Decuplet magnetic octupole moments Ω

Determine constant C in pion cloud model ( = 0.003)

+ (∆ ) = = 0.012 fm³ 𝐶𝐶 −

Ω = 4𝐶𝐶/ =− 0.003 fm³ − 3 ComparisonΩ Ω −4𝐶𝐶with𝑟𝑟 other models:

+ (∆ ) = 0.0035 fm³ (Ramalho, Peña, Gross, PLB 678, 355 (2009))

Ω( ) = 0−.016 fm³ (Aliev, Aziz, Savici, PLB 681, 240 (2009)) − Ω Ω Models agree in sign but not in magnitude. 4. Summary Relation between N and ∆ form factors

+ 3 2 G p→Δ (Q2 ) = − G n (Q2 ) C2 Q2 C

N ∆ charge quadrupole form factor neutron charge form factor

AJB, Phys. Rev. Lett. 93 (2004) 212301 ( ) transition moments ∗ Relations𝑵𝑵 → between𝑵𝑵 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏electromagnetic (1680) ∗ transition moments and ground state𝑝𝑝properties→ 𝑝𝑝

1 ( 1680 ) = 5 4 4 𝑄𝑄 𝑝𝑝 → 𝑝𝑝 𝐼𝐼𝐼𝐼 𝑟𝑟𝑝𝑝 − 𝑟𝑟𝑛𝑛

3 ( 1680 = 224 6 6 𝛺𝛺 𝑝𝑝 → 𝑝𝑝 𝐼𝐼𝐼𝐼 𝑟𝑟𝑝𝑝 − 𝑟𝑟𝑛𝑛 Outlook

• Improve current algebra calculation • Calculate (1680) multipoles using 1 expansion ∗ (1680) • Calculate 𝑁𝑁quadrupole→ 𝑁𝑁 and octupole moment⁄of𝑁𝑁𝐶𝐶 ∗ 𝑁𝑁