powerpoint presentationnr,sun− 𝑤200 nr,sun + 𝑟 4 67 16 𝑤400 nr,sun− 𝑤400 nr,sun 𝑤...

𝛿𝜖 = 𝐴𝑗0(𝑛𝐹𝐽𝐿)

𝑗

⟨𝐹𝑚𝐹𝑗0|𝐹𝑚𝐹⟩ Clebsch-Gordan coefficients

Lorentz-violating energy shift for hydrogen in the presence of a weak magnetic field

• 𝑛 principal quantum number 𝑛 • 𝐹 hydrogen total angular momentum quantum number • 𝐽 electron total angular momentum quantum number • 𝐿 electron orbital quantum number

• 𝑚𝐹 quantum number of the components of 𝐹 in the direction of the magnetic field

𝐽 = S𝑒 + 𝐿 𝐹 = 𝑆 𝑝 + 𝐽

𝛿𝜖 = 𝐴𝑗0(𝑛𝐹𝐽𝐿)

𝑗𝑚

⟨𝐹𝑚𝐹𝑗0|𝐹𝑚𝐹⟩

If 𝑗 is an even number (spin-independent)

𝐴𝑗𝑚 𝑛𝐹𝐽𝐿 = − 𝒑 𝑘𝑛𝐿

𝑤𝑘

Λ𝑗0𝐸

𝒱𝑤𝑘𝑗𝑚NR

If 𝑗 is an odd number (spin-dependent)

𝐴𝑗𝑚 𝑛𝐹𝐽𝐿 = − 𝒑 𝑘𝑛𝐿

𝑤𝑘

Λ𝑗0𝐵

2𝐽 + 1𝒯𝑤𝑘𝑗𝑚

NR(0𝐵)− Λ𝑗

1𝐵 𝛿𝑤𝑒2(𝐿 − 𝐽)

+𝛿𝑤𝑝

2(𝐽 − 𝐹)𝒯𝑤𝑘𝑗𝑚

NR(1𝐵)

Clebsch-Gordan coefficients

𝑤 = 𝑒 for the electron 𝑤 = 𝑝 for the proton

Only even values of 𝑘 can contribute • For convenience we only consider

terms with 𝑘 ≤ 4

Lorentz-violating energy shift for hydrogen in the presence of a weak magnetic field

𝑚𝐹 = 1

𝑚𝐹 = −1

𝑚𝐹 = 0

𝑛S1/2

𝐹 = 1

𝐹 = 0

1

2↑↓ − ↓↑

1

2↑↓ + ↓↑

↑↑

↓↓

Energy shift to the 𝑛𝑆1/2𝐹 in the presence of a weak magnetic field; 𝑛, 𝐿 = 0; 𝐽 = 1/2, 𝐹,𝑚𝐹

En

ergy

𝛿𝜖 = 𝐴00 𝑛 +

𝑚𝐹

2𝐴10 𝑛 = −

𝑝 𝑘𝑛

4𝜋𝑤𝑘

𝒱𝑤𝑘00 NR −

𝑚𝐹

2

𝑝 𝑘𝑛

3𝜋𝑤𝑘

𝒯𝑤𝑘10NR(0𝐵) + 2𝒯𝑤𝑘10

NR(1𝐵)

𝑛S1/2

𝐹 = 1

𝐹 = 0

𝑚𝐹 = 1

𝑚𝐹 = −1

𝑚𝐹 = 0


En

ergy

𝛿𝜖 = 𝐴00 𝑛 +

𝑚𝐹

2𝐴10 𝑛 = −

𝑝 𝑘𝑛

4𝜋𝑤𝑘


𝑚𝐹

2

𝑝 𝑘𝑛

3𝜋𝑤𝑘


NR(1𝐵)

• The spin-independent terms do not shift any of the internal transitions of the 𝑛𝑆1/2 state

• Hyperfine and Zeeman transitions are not affected

𝑛S1/2 𝐹 = 1

𝐹 = 0

𝑚𝐹 = 1

𝑚𝐹 = −1

𝑚𝐹 = 0


En

ergy

𝛿𝜖 = 𝐴00 𝑛 +

𝑚𝐹

2𝐴10 𝑛 = −

𝑝 𝑘𝑛

4𝜋𝑤𝑘


𝑚𝐹

2

𝑝 𝑘𝑛

3𝜋𝑤𝑘


NR(1𝐵)

• The spin-independent terms do not shift any of the internal transitions of the 𝑛𝑆1/2 state

• Hyperfine and Zeeman transitions are not affected

𝑛S1/2

𝐹 = 1

𝐹 = 0

𝑚𝐹 = 1

𝑚𝐹 = −1

𝑚𝐹 = 0


En

ergy

𝛿𝜖 = 𝐴00 𝑛 +

𝑚𝐹

2𝐴10 𝑛 = −

𝑝 𝑘𝑛

4𝜋𝑤𝑘


𝑚𝐹

2

𝑝 𝑘𝑛

3𝜋𝑤𝑘


NR(1𝐵)

• The spin-dependent terms affect the Zeeman levels • The transition 𝐹 = 0,𝑚𝐹 = 0 ↔ 𝐹 = 1,𝑚𝐹 = 0 is not affected


En

ergy

𝛿𝜖 = 𝐴00 𝑛 +

𝑚𝐹

2𝐴10 𝑛 = −

𝑝 𝑘𝑛

4𝜋𝑤𝑘


𝑚𝐹

2

𝑝 𝑘𝑛

3𝜋𝑤𝑘


NR(1𝐵)

• The spin-dependent terms affect the Zeeman levels • The transition 𝐹 = 0,𝑚𝐹 = 0 ↔ 𝐹 = 1,𝑚𝐹 = 0 is not affected

𝑛S1/2

𝐹 = 1

𝐹 = 0

𝑚𝐹 = 1

𝑚𝐹 = −1

𝑚𝐹 = 0

1S1/2

𝐹 = 1

𝐹 = 0 Ener

gy

𝛼 = 0.007

𝑚𝑟 =𝑚𝑒𝑚𝑝

𝑚𝑒 +𝑚𝑝

Internal transitions of the ground state 𝐹,𝑚𝐹 ⟷ 𝐹′, 𝑚𝐹′

𝛿𝜖 = −∆𝑚𝐹

2 3 𝒯𝑤010

NR(0𝐵) + 2𝒯𝑤010NR(1𝐵) + 𝛼𝑚𝑟

2 𝒯𝑤210NR(0𝐵) + 2𝒯𝑤210

NR(1𝐵)

𝑤

+ 5 𝛼𝑚𝑟4 𝒯𝑤410

NR(0𝐵) + 2𝒯𝑤410NR(1𝐵)

𝑚𝐹 = 1

𝑚𝐹 = −1

𝑚𝐹 = 0

𝛼𝑚𝑟2~10−11 GeV2

Frequencies are measured relative to other frequencies

Any measurement is the comparison of two physical systems

• The second is defined as the time it takes for the radiation emitted by the hyperfine transition of the ground state of the 133Cs atom to complete 9 192 631 770 cycles.

• The meter is defined as the distance travelled by light in vacuum during a time interval of 1/299 792 458 of a second.

Could Lorentz symmetry affect the base units? Yes

Is the second affected by Lorentz violation?

At leading order the Lorentz-violating perturbations that have been studied in the literature do not affect most of the common microwave standards such as

• The transition 𝐹 = 1,𝑚𝐹 = 0 ↔ 𝐹 = 2,𝑚𝐹 = 0 in 87Rb • The transition 𝐹 = 3,𝑚𝐹 = 0 ↔ 𝐹 = 4,𝑚𝐹 = 0 in 133Cs • The transition 𝐹 = 0,𝑚𝐹 = 0 ↔ 𝐹 = 1,𝑚𝐹 = 0 in H

Lorentz-violating multi-particle operators could introduce Lorentz-violating effects to these time standards

This is not surprising because the leading Lorentz-violating corrections mimic perturbations due to weak external EM fields.

It is important to consider the Lorentz-violating corrections to the two systems that are being compared

If the two clocks are affected by Lorentz violation in the same way then comparing the two clocks is not sensitive to Lorentz violation

Phillips et al., PRD 63, 111101 (2001)

𝛼𝑚𝑟2~10−11 GeV2

𝐹 = 1,𝑚𝐹 = 0 ⟷ 𝐹 = 1,𝑚𝐹 = ±1

2𝜋𝛿𝑣 = ±1

2𝐴 ∙ 𝐵

𝐹 = 0,𝑚𝐹 = 0 ⟷ 𝐹 = 1,𝑚𝐹 = 0

𝛿𝑣 = 0

Sidereal variation

Reference

Zeeman transitions

Standard H maser transition

Λ 𝜈𝜇

= ℛ 𝛼𝜇Β𝛼𝜈

Transformation from the Sun-centered frame to the laboratory frame

Boost

Rotation Lorentz transformation

For 𝛽 ≪ 1 we can simplify the Lorentz transformation

Lab frame 0 time component 𝑗 ∈ {1,2,3} spatial components

Sun-centered frame 𝑇 time component 𝐽 ∈ {𝑋, 𝑌, 𝑍} spatial components

Orbital velocity of the Earth

Velocity of the lab frame relative to the center of the Earth

For 𝛽 ≪ 1 Orbital velocity of the Earth

Velocity of the lab frame relative to the center of the Earth

𝛽𝐿 ≃ 10−6 sin 𝜒

Colatitude

𝜔⊕ sidereal frequency Ω⊕ annual frequency

𝑇⊕ sidereal time

𝜑 and 𝜗 are azimuthal and polar angles of 𝑩 at 𝑇⊕ = 0

If 𝐵 = 𝑧 in the lab frame then

𝜈1𝑆−2𝑆 = 2466061413187035 ± 10 Hz or 𝛿𝜈

𝜈= 4 × 10−15

Parthey et al.,PRL 107, 203001 (2011)

1S-2S transition the most precisely measured transition in hydrogen

For testing the spin-dependent terms the absolute sensitivity of the experiment is more relevant • Hyperfine transitions can access frequency shifts down to 10−4 Hz • 1S-2S transition can access frequency shifts down to 10 Hz

In the 1S-2S we can limit our attention to the spin-independent terms

2𝜋𝛿𝜈 = 1

4𝜋 𝛼𝑚𝑟

23

4𝒱𝑤200

NR + 𝛼𝑚𝑟467

16𝒱𝑤400

NR

𝑤

The Lorentz violation frequency shift to the 1S-2S is given by

2𝜋𝛿𝜈 =1

4𝜋 𝛼𝑚𝑟

23

4𝒱𝑤200

NR + 𝛼𝑚𝑟467

16𝒱𝑤400

NR

𝑤

The Lorentz-violating frequency shift to the 1S-2S is given by

𝒱𝑤200 NR ,lab = 𝒱𝑤200

NR ,Sun + 4𝜋 2𝑎𝑤eff

(5)𝑇𝑇𝐽+ 𝑎𝑤eff

5 𝐾𝐾𝐽𝛽𝐽

− 8 𝜋 𝑐𝑤eff6 𝑇𝑇𝑇𝐽

+ 𝑐𝑤eff6 𝑇𝐾𝐾𝐽

𝛽𝐽 +⋯ = 𝒱𝑤200 NR ,Sun + 𝑽𝒘𝟐0 ∙ 𝜷

2𝜋𝛿𝜈 =1

4𝜋 𝛼𝑚𝑟

23

4𝒱𝑤200

NR,Sun + 𝛼𝑚𝑟467

16𝒱𝑤400

NR,Sun

𝑤

+1

4𝜋 𝛼𝑚𝑟

23

4𝑽𝒘𝟐0

+ 𝛼𝑚𝑟467

16𝑽𝒘40

𝑤

∙ 𝜷 = 𝑆 + 𝑉 ∙ 𝜷

Considering the boost contribution to the transformation to the Sun-centered frame

The frequency shift in the Sun-centered frame has the form

𝑡′

𝑥′

𝑡′

𝑥′

𝑡 𝑡

𝑥

𝑥

The Lorentz-violating field is isotropic in this frame

The Lorentz-violating field is anisotropic in this frame

An sphere is not round in all inertial reference frames

Isotropic effects in one reference frame are not necessarily isotropic in other frames

time

freq

uen

cy

Annual variation of the frequency

Orbital velocity of the Earth Lorentz-violating field

𝛿𝜈 = 𝑉 ∙ 𝜷

Matveev et al., PRL 110, 230801 (2013) A Cs atomic fountain clock is used as the reference clock

Kirch et. al., IJMPCS 30, 1460258 (2014)

Crivelli et. al., IJMPCS 30, 1460257 (2014)

1S-2S muonium and positronium (proposed as a gravity antimatter test)

1 1

Optical clocks with 𝐽 = 0, for example consider the transition 𝑆01 − 𝑃0

3

• Al27 + ion clock • In115 + ion clock • Sr87 optical lattice clock • Yb171 optical lattice clock • Hg199 optical lattice clock

• Other clock-comparison experiments are more sensitive to proton and neutron coefficients

2𝜋𝛿𝑣 = −1

4𝜋∆𝑝2𝒱𝑒200

NR + ∆𝑝4𝒱𝑒400NR

∆𝑝𝑘 = 𝒑 𝑘𝑃0

3 − 𝒑 𝑘𝑆0

1

Annual and sidereal variation due to the boost

• Comparing different clock types in the same laboratory frame • Comparing clocks in different laboratories frames

2𝜋 𝛿𝜈 = 𝑆 + 𝑉 ∙ 𝜷

The isotropic Lorentz-violating frequency shift for the 1S-2S in hydrogen in the SFC

2𝜋𝛿𝜈 =1

4𝜋 𝛼𝑚𝑟

23

4𝑐𝑤200

NR,Sun − 𝑎𝑤200 NR,Sun + 𝛼𝑚𝑟

467

16𝑐𝑤400

NR,Sun − 𝑎𝑤400 NR,Sun

𝑤

Coefficients for CPT violation Coefficients of CPT invariant operators

and for antihydrogen

The SME allows clocks and anti-clocks to tick at different rates

2𝜋𝛿𝜈 =1

4𝜋 𝛼𝑚𝑟

23

4𝑐𝑤200

NR,Sun + 𝑎𝑤200 NR,Sun + 𝛼𝑚𝑟

467

16𝑐𝑤400

NR,Sun + 𝑎𝑤400 NR,Sun

𝑤

2𝜋(𝛿𝜈 − 𝛿𝜈) =2

4𝜋 𝛼𝑚𝑟

23

4𝑎𝑤200


16𝑎𝑤400

NR,Sun

𝑤

Discrepancy between hydrogen and antihydrogen

The isotropic Lorentz-violating frequency shift for the 1S-2S in hydrogen in the SFC

2𝜋𝛿𝜈 =1

4𝜋 𝛼𝑚𝑟

23

4𝑐𝑤200

NR,Sun − 𝑎𝑤200 NR,Sun + 𝛼𝑚𝑟

467

16𝑐𝑤400

NR,Sun − 𝑎𝑤400 NR,Sun

𝑤

Coefficients for CPT violation Coefficients of CPT invariant operators

The SME allows clocks and anti-clocks to tick at different rates

2𝜋𝛿𝜈 =1

4𝜋 𝛼𝑚𝑟

23

4𝑐𝑤200

NR,Sun + 𝑎𝑤200 NR,Sun + 𝛼𝑚𝑟

467

16𝑐𝑤400

NR,Sun + 𝑎𝑤400 NR,Sun

𝑤

2𝜋(𝛿𝜈 − 𝛿𝜈) =2

4𝜋 𝛼𝑚𝑟

23

4𝑎𝑤200


16𝑎𝑤400

NR,Sun

𝑤

Discrepancy between hydrogen and antihydrogen

and for antihydrogen

M. Ahmadi et al., doi.org/10.1038/s41586-018-0017-2

P. Crivelli and N. Kolachevsky, arXiv:1707.02214.

Using ultracold antiatoms from the GBAR antihydrogen beam in an optical lattice might improve the bound by three orders of magnitude

2𝜋 𝛿𝜈 − 𝛿𝜈 < 5 kHz

This bound is expected to improved in the future

ALPHA measurement of the 1𝑆-2𝑆 in antihydrogen

First SME bound from antihydrogen spectroscopy arXiv:1805.04499

40Ca+ experiment using entangled ions • Marianna discussed the experiments in details yesterday The experiment involve Zeeman levels in the state with 𝐽 = 5/2 and that implies that electron coefficients with 𝑗 < 5 could contribute

Defining 𝜖𝑚𝐽 as the energy of the Zeeman level 𝑚𝐽 of the state 𝐷5/2

2 the observable is

2𝜋𝑓 = 𝜖5/2 + 𝜖−5/2 − 𝜖1/2 − 𝜖−1/2

2𝜋𝛿𝑓 = 𝛿𝜖5/2 + 𝛿𝜖−5/2 − 𝛿𝜖1/2 − 𝛿𝜖−1/2

The Lorentz-violating shift is given by

No linear Zeeman shift

No contribution from spin-dependent terms

Only coefficients with 𝑗 = 2 and 𝑗 = 4 can contribute

Assumption: Electron in incomplete subshell has total angular momentum 𝐽 = 5/2

Electrons in filled subshells form states with 𝐽 = 0

The shift to the observable is given by

2𝜋𝛿𝑓 =18

7 5𝜋⟨ 𝒑 2⟩𝒱𝑒220

NR + ⟨ 𝒑 4⟩𝒱𝑒420NR +

1

7 5𝜋𝒑 4 𝒱𝑒440

NR

Signals for Lorentz violation • From the rotation transformation

• Sidereal variations with the 1st to the 4th harmonics of the sidereal frequency

• Including the boost transformation • Sidereal variation with the 5th harmonic of the sidereal frequency (proportional

to 𝛽𝐿 = 10−6) • Annual variations

Assumption: Electron in incomplete subshell has total angular momentum 𝐽 = 5/2

Electrons in filled subshells form states with 𝐽 = 0


2𝜋𝛿𝑓 =18

7 5𝜋⟨ 𝒑 2⟩𝒱𝑒220

NR + ⟨ 𝒑 4⟩𝒱𝑒420NR +

1

7 5𝜋𝒑 4 𝒱𝑒440

NR

Instead of using NR coefficients all the previous signals can be observed using coefficients with mass dimensions 𝑑 ≤ 8.

Sidereal local time 𝑇𝐿 vs. sidereal time 𝑇⊕

The results in the table are for 133-Cs but they can modified to 40-Ca via the map

𝑇𝐿 = 𝑇⊕ + 𝜑

Corrections linear in the boost in the Sun-centered frame

Consider the transition S1/22 − D5/2

2 in Sr+88 or Ca+40

Techniques used to eliminate systematics • Averaging of the Zeeman levels 𝑚𝐽 and −𝑚𝐽: eliminates the linear Zeeman effect and

eliminates the contribution from spin-dependent terms

• Averaging over three Zeeman pairs (𝑚𝐽, −𝑚𝐽): eliminates the electric quadrupole

shift and because

5/2𝑚𝐽𝑗0|5/2𝑚𝑗

5/2

𝑚𝐽=−5/2

= 6𝛿𝑗0

only isotropic coefficients can contribute

• Using two Zeeman pairs (𝑚𝐽, −𝑚𝐽) to extrapolate the value for 𝑚𝐽2 = 35/12:

eliminates the electric quadrupole shift and allows contributions from 𝑗 = 0 and 𝑗 = 4 coefficients.

2𝜋𝛿𝜈 = −1

4𝜋∆𝑝2𝒱𝑒200

NR + 𝒑 4 𝒱𝑒400NR +

1

27 𝜋∆𝑝4𝒱𝑒440

NR

The Schmidt model assumes that if a nucleus has an odd number of nucleons then all the nucleons that can be paired with other nucleons of the same kind form subshells with zero angular momentum and the spin of the nucleus is equal to the total angular momentum of the unpaired nucleon.

Nucleus with even number of neutrons and even number of protons have nuclear spin 𝐼 = 0

7-Lithium nucleus has three protons and four neutrons

Protons Neutrons

zero total angular momentum

Total angular momentum of the unpaired proton equal to 𝐼

Nuclear model used

Issues with the Schmidt model • The Schmidt model will assumes that only proton coefficients or neutron

coefficients contribute to the signals for atoms with odd number of nucleons and that is not the case

• The contribution of the nucleon preferred by the model is usually dominant

Good things about the model • It can be easily applied to many systems and the signals for Lorentz violation

predicted by the model tend to be accurate

• Allow analytical expressions for the angular expectation value

• It allows to obtain rough estimates of the sensitivity of different experiments to Lorentz violation

Better nuclear models have been used in the context of the SME

Y.V. Stadnik and V.V. Flambaum, Eur. Phys. J. C 75, 110 (2015)

Corrections to the ground state of 133 Cesium; 𝐽 = 1/2, 𝐼 = 7/2, 𝐹 = 3 or 4

𝐽 = 1/2 implies that only electrons coefficients with 𝑗 ≤ 1 can contribute Electrons

Neutrons

Protons

𝐼 = 7/2 implies that only nucleon coefficients with 𝑗 ≤ 7 can contribute but for the case 𝐹 = 3 the condition is 𝑗 ≤ 6.

For practical reasons we limit ourselves to 𝑗 ≤ 5 • The coefficient with the smallest mass dimension that contributes to

𝑗 = 7 has mass dimension 𝑑 = 9 • The coefficient with the smallest mass dimension that contributes to

𝑗 = 5 has mass dimension 𝑑 = 7

133-Cesium has 55 protons and 78 neutrons • In the Schmidt model the spin of the nucleus is due to the unpaired proton


2𝜋𝛿𝜈𝑐 = −13

14

5

𝜋⟨ 𝒑 2⟩𝒱𝑝220

NR + ⟨ 𝒑 4⟩𝒱𝑝420NR +

45

77 𝜋𝒑 4 𝒱𝑝440

NR

Define the transition as 𝐹 = 3,𝑚𝐹 → |𝐹 = 4,𝑚𝐹⟩ as 𝛿𝜈𝑚𝐹 then the observable of the

experiment is

𝛿𝜈𝑐 = 𝛿𝜈3 + 𝛿𝜈−3 − 2𝛿𝜈0 to eliminate the Zeeman shift

Signals for Lorentz violation • From the rotation transformation

• Sidereal variations with the 1st to the 4th harmonics of the sidereal frequency

• Including the boost transformation • Sidereal variation with the 5th harmonic of the sidereal frequency (proportional to

𝛽𝐿 = 10−6) • Annual variations

Estimates for sidereal variation in the first harmonic and annual variations based on the previous experiment

Using better nuclear models contributions from the neutron coefficients are expected

H. Pihan-Le Bars et al., Phys. Rev. Lett. 95, 075026 (2017)

129Xe-3He comagnetometer • The ground state of both systems have quantum numbers: 𝐽 = 0, 𝐹 = 𝐼 = 1/2

• Only the nucleon coefficients with 𝑗 ≤ 1 contribute to the ground state • Only the electron coefficients with 𝑗 ≤ 0 contribute to the ground state

• In the Schmidt model the neutron carries all the nuclear spin

Define 𝛿𝑣Xe as the Zeeman transition 𝐹 =1

2, 𝑚𝐹 =

1

2→ 𝐹 =

1

2, 𝑚𝐹 = −

1

2 in Xe

and 𝛿𝑣He the same transition in He.

The observable in the experiment is 𝜈 = 𝜈He −𝛾He𝛾Xe

𝜈Xe

𝛾Xe and 𝛾He are the gyromagnetic ratio of the corresponding ground states

• This observable is insensitive to the linear Zeeman effect

𝜈He ⊃ 𝛾H𝑒 𝐵 𝜈Xe ⊃ 𝛾X𝑒 𝐵

F. Canè et al., Phys. Rev. Lett. 93, 230801 (2004)

2𝜋𝛿𝜈 =−1

3𝜋 𝑝 𝑘

He−𝛾He𝛾Xe

𝑝 𝑘Xe

𝑘

𝒯𝑛𝑘10NR(0𝐵) + 2𝒯𝑛𝑘10

NR(1𝐵) = 𝐴 𝐻𝑒𝑋𝑒 ∙ 𝐵

powerpoint presentationnr,sun− 𝑤200 nr,sun + 𝑟 4 67 16 𝑤400 nr,sun− 𝑤400 nr,sun 𝑤...

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