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III
Size and Structure
Mikhail Bashkanov
University of Edinburgh
UK Nuclear Physics Summer School
Outline
 Electron scattering
 Proton form factors
 Proton size puzzle
2
Electron microscopy
de Broglie wavelength of probe particle must be ~size of the object you wish to study
3
Electrons as waves
Scattering process is quantum mechanical
de Broglie wavelength: 𝜆 =
ℎ
𝑝
Electron energy: Ee = 𝑝𝑐
„resolving scale“:
𝜆 𝑓𝑚 =
2𝜋(197.3)
𝐸𝑒 [𝑀𝑒𝑉]
4
Classical Fraunhofer Diffraction
First minima: 𝑑 =
1.22𝜆
sin 𝜃
5
Example
30
′
𝑆𝑖(𝑒, 𝑒 )
1st Minimum: 1.3𝑓𝑚−1
𝜃 = 32.8°
Electron energy:
𝐸𝑒 = 454.3 𝑀𝑒𝑉
𝜆 = 2.73𝑓𝑚
1.22 ∙ 2.73
𝑑=
= 6.15𝑓𝑚
sin 32.8
𝑟 = 3.07 𝑓𝑚
6
Electron scattering. Experiment
7
Electron scattering
Fermi Golden Rule:
𝑑𝜎 2𝜋
2
=
𝑀𝑓𝑖 𝐷𝑓
𝑑Ω
ℏ
𝐷𝑓 - density of final states (phase space)
𝑀𝑓𝑖 - scattering amplitude
𝑀𝑓𝑖 =
𝜓𝑓∗ 𝑉 𝑥 𝜓𝑖 𝑑3 𝑥 =
𝑒 −𝑖𝑘𝑓 ∙𝑥 𝑉 𝑥 𝑒 𝑖𝑘𝑖∙𝑥 𝑑3 𝑥 =
𝑒 𝑖𝑞∙𝑥 𝑉 𝑥 𝑑3 𝑥
• Plane wave approximation for incoming and outgoing electrons
• Born approximation: interact only once
8
Form Factor and charge distribution
 Using Colomb potential from the charge distribution 𝝆 𝒙
𝑉 𝑥 =
𝑍𝑒 2
−
4𝜋𝜖0
𝜌 𝑥′
𝑥−𝑥 ′
𝑑3 𝑥 ′
𝑍𝑒 2
𝑀𝑓𝑖 = −
4𝜋𝜖0
𝐹 𝑞 =
𝑒 𝑖𝑞𝑥
𝑍𝑒 2
−
4𝜋𝜖0
𝑒 𝑖𝑞𝑅
𝑍𝑒 2
−
4𝜋𝜖0
𝑒 𝑖𝑞𝑅 3
𝑑 𝑅
𝑅
′
𝑖𝑞𝑥
𝑒
𝜌
𝜌 𝑥′
3 ′ 3
𝑑
𝑥 𝑑 𝑥=
′
𝑥−𝑥
𝑒
𝑖𝑞𝑥 ′
𝜌 𝑥′ 3 ′ 3
𝑑 𝑥 𝑑 𝑅=
𝑅
′
𝑖𝑞𝑥
𝑒
𝜌
𝑥 ′ 𝑑3𝑥 ′
𝑥 ′ 𝑑 3 𝑥 ′ - form factor
9
Form Factor and cross-section
 For point like particle 𝝆 𝒙 = 𝜹(𝒙) and 𝐹 𝑥 = 1 →
 Rutherford-like scattering
𝑑𝜎
𝑑Ω
𝑑𝜎
≡
𝑑Ω
𝑍𝛼
2
cos
2Θ
2
Θ
4𝐸 2 𝑠𝑖𝑛4
𝑝𝑜𝑖𝑛𝑡
𝑀𝑜𝑡𝑡
2
 Scattering from charge distribution:
𝑑𝜎
𝑑𝜎
=
∙ 𝐹 𝑞
𝑑Ω
𝑑Ω 𝑀𝑜𝑡𝑡
Θ
=
2
cos 2 - the only difference from Rutherford scattering.
2
Arises from Dirac theory for spin ½ particles
10
Experiment
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Probing the proton
 For a target with non-zero spin – form factors for
charge and magnetization:
𝑑𝜎
=
𝑑Ω
𝑍𝛼
2 cos 2 Θ
2
∙ 𝐹 𝑞 2=
Θ
4𝐸 2 𝑠𝑖𝑛4
2 𝑀𝑜𝑡𝑡
𝑑𝜎
𝜏
1
2
2
𝐺𝐸 + 𝐺𝑀
𝑑Ω 𝑀𝑜𝑡𝑡
𝜖
1+𝜏
𝑄2
1
Θ
2
𝜏=
,
= 1 + 2 1 + 𝜏 tan
2
2𝑀
𝜖
2
Θ𝑒 ′
2
′
2
𝑄 = 4𝐸𝐸 sin
+ 𝑚𝑒2
2
𝐺𝐸 - Electric form factor
𝐺𝑀 - Magnetic form factor
Sachs form factors
12
Rosenbluth separation
If we keep 𝑸𝟐 fixed and vary 𝝐 we can disentangle the magnetic and electric
form factors
Major drawback - 𝑮𝑴 is weighted with 𝑸𝟐
13
Dipole form factor
𝐺𝐷
Q2
=
1
𝑄2
1+ 2
Λ
𝜌 𝑟 = 𝜌0 𝑒 −𝑎𝑟
2,
Λ2 = 0.71 𝐺𝑒𝑉 2
with 𝑎 = 4.27 𝑓𝑚−1
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Results
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New technique: polarization transfer
• Longitudinally polarized electron+ Recoil polarimetry
• Longitudinally polarized electron+ Polarized target
𝐺𝐸
𝑃𝑡 𝐸𝑒 + 𝐸𝑒′
Θ𝑒
=−
𝑡𝑎𝑛
𝐺𝑀
𝑃𝑙 2𝑀𝑃
2
16
Data extracted using Rosenbluth separation only
17
Data extracted using polarization transfer only
B
18
All data
19
corrections to
single photon
exchange
20
21
Dipole form factor  proton radius
𝐺𝐷
Q2
=
1
𝑄2
1+ 2
Λ
𝜌 𝑟 = 𝜌0 𝑒 −𝑎𝑟
𝐺 Q2
𝑟2
2,
Λ2 = 0.71 𝐺𝑒𝑉 2
with 𝑎 = 4.27 𝑓𝑚−1
1 2 2
=1− 𝑟 𝑄 +⋯
6
𝐷𝑖𝑝𝑜𝑙𝑒
12
= 2 = 0.66 𝑓𝑚2 →
𝑎
𝑟 2 = 0.8768(69) 𝑓𝑚
𝑟2
𝑟2
𝑑𝐺 Q2
= −6
𝑑𝑄 2
𝐷𝑖𝑝𝑜𝑙𝑒
𝑄 2 =0
= 0.81 𝑓𝑚
22
Proton Charge radius
23
Bohr’s atom
𝑚𝑒 𝑐 2 𝛼
𝐸𝑛 = −
2𝑛2
𝑛 = 1,2, … ∞
Principle quantum number
𝛼≈
1
137
fine-structure constant
c – speed of light
𝑚𝑒 - mass of electron
24
Hydrogen atom
𝑎0 =
ℏ
𝑚𝑒 𝑐𝛼
- Bohr radius
25
Proton radius from
hydrogen spectroscopy
Δ𝐸𝐿𝑆 = 209.9779 49 + 5.2262𝑅𝐸2 + 0.00913𝑅3 (2) [𝑚𝑒𝑉]
3.7 𝑚𝑒𝑉
0.026 𝑚𝑒𝑉 26
Proton Charge radius
27
Proton Charge radius
28
Proton Charge radius puzzle
𝜇𝐻 data: R 𝐸 = 0.8409 ± 0.0004 fm
7𝜎 difference !!!
ep data: R 𝐸 = 0.8775 ± 0.0051 fm
29
Lamb shift:
status of known corrections
VP: vacuum polarization
SE: self energy
30
Proton radius puzzle: new physics?
 New muonic forces?
 Lepton universality violation models
 Challenges:
 New physics must also respect
𝑔 − 2 𝜇 discrepancy [1 ppm only,
compare to
10000 ppm in muonic hydrogen]
 Strong constrains from W decay
31
Proton radius puzzle: what’s next?
 H Lamb shift
 eH Lamb shift (higher accuracy)
 ep scattering analysis
 ep scattering (higher accuracy)
 MAINZ
 Jlab
 MESA
 p scattering (MUSE@PSI)
 Lepton universality (𝑒 + 𝑒 − vs 𝜇 + 𝜇− photoproduction)
32
Proton
33
Nucleon internal structure
Valence quarks
Gluons
Sea quarks
34
Valence quarks
35
Sea quarks
36
Gluons
𝜸
𝒈
37
What is inside?
𝑄2
2𝑀𝜈
𝑥=
Bjorken scaling
variable
38
Proton spin puzzle
gluons angular momentum
1 1
= ΔΣ + Δ𝐺 + 𝐿𝑞 + 𝐿𝑔
2 2
Proton spin
quarks angular momentum
quarks spin
gluon spin
ΔΣ = Δ𝑢 + Δ𝑑 + Δs
Naïve quark model:
4 1
ΔΣ = + − + 0 = 1
3 3
Reality:
ΔΣ =0.3
39
Conclusion
 Nucleon structure is complex and complicated
 A lot of puzzles and a lot of progress
 New tools, high precision, high accuracy, polarization
40