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πΏπ = π΄π0(ππΉπ½πΏ)
π
β¨πΉππΉπ0|πΉππΉβ© Clebsch-Gordan coefficients
Lorentz-violating energy shift for hydrogen in the presence of a weak magnetic field
β’ π principal quantum number π β’ πΉ hydrogen total angular momentum quantum number β’ π½ electron total angular momentum quantum number β’ πΏ electron orbital quantum number
β’ ππΉ quantum number of the components of πΉ in the direction of the magnetic field
π½ = Sπ + πΏ πΉ = π π + π½
πΏπ = π΄π0(ππΉπ½πΏ)
ππ
β¨πΉππΉπ0|πΉππΉβ©
If π is an even number (spin-independent)
π΄ππ ππΉπ½πΏ = β π πππΏ
π€π
Ξπ0πΈ
π±π€πππNR
If π is an odd number (spin-dependent)
π΄ππ ππΉπ½πΏ = β π πππΏ
π€π
Ξπ0π΅
2π½ + 1π―π€πππ
NR(0π΅)β Ξπ
1π΅ πΏπ€π2(πΏ β π½)
+πΏπ€π
2(π½ β πΉ)π―π€πππ
NR(1π΅)
Clebsch-Gordan coefficients
π€ = π for the electron π€ = π for the proton
Only even values of π can contribute β’ For convenience we only consider
terms with π β€ 4
Lorentz-violating energy shift for hydrogen in the presence of a weak magnetic field
ππΉ = 1
ππΉ = β1
ππΉ = 0
πS1/2
πΉ = 1
πΉ = 0
1
2ββ β ββ
1
2ββ + ββ
ββ
ββ
Energy shift to the ππ1/2πΉ in the presence of a weak magnetic field; π, πΏ = 0; π½ = 1/2, πΉ,ππΉ
En
ergy
πΏπ = π΄00 π +
ππΉ
2π΄10 π = β
π ππ
4ππ€π
π±π€π00 NR β
ππΉ
2
π ππ
3ππ€π
π―π€π10NR(0π΅) + 2π―π€π10
NR(1π΅)
πS1/2
πΉ = 1
πΉ = 0
ππΉ = 1
ππΉ = β1
ππΉ = 0
Energy shift to the ππ1/2πΉ in the presence of a weak magnetic field; π, πΏ = 0; π½ = 1/2, πΉ,ππΉ
En
ergy
πΏπ = π΄00 π +
ππΉ
2π΄10 π = β
π ππ
4ππ€π
π±π€π00 NR β
ππΉ
2
π ππ
3ππ€π
π―π€π10NR(0π΅) + 2π―π€π10
NR(1π΅)
β’ The spin-independent terms do not shift any of the internal transitions of the ππ1/2 state
β’ Hyperfine and Zeeman transitions are not affected
πS1/2 πΉ = 1
πΉ = 0
ππΉ = 1
ππΉ = β1
ππΉ = 0
Energy shift to the ππ1/2πΉ in the presence of a weak magnetic field; π, πΏ = 0; π½ = 1/2, πΉ,ππΉ
En
ergy
πΏπ = π΄00 π +
ππΉ
2π΄10 π = β
π ππ
4ππ€π
π±π€π00 NR β
ππΉ
2
π ππ
3ππ€π
π―π€π10NR(0π΅) + 2π―π€π10
NR(1π΅)
β’ The spin-independent terms do not shift any of the internal transitions of the ππ1/2 state
β’ Hyperfine and Zeeman transitions are not affected
πS1/2
πΉ = 1
πΉ = 0
ππΉ = 1
ππΉ = β1
ππΉ = 0
Energy shift to the ππ1/2πΉ in the presence of a weak magnetic field; π, πΏ = 0; π½ = 1/2, πΉ,ππΉ
En
ergy
πΏπ = π΄00 π +
ππΉ
2π΄10 π = β
π ππ
4ππ€π
π±π€π00 NR β
ππΉ
2
π ππ
3ππ€π
π―π€π10NR(0π΅) + 2π―π€π10
NR(1π΅)
β’ The spin-dependent terms affect the Zeeman levels β’ The transition πΉ = 0,ππΉ = 0 β πΉ = 1,ππΉ = 0 is not affected
Energy shift to the ππ1/2πΉ in the presence of a weak magnetic field; π, πΏ = 0; π½ = 1/2, πΉ,ππΉ
En
ergy
πΏπ = π΄00 π +
ππΉ
2π΄10 π = β
π ππ
4ππ€π
π±π€π00 NR β
ππΉ
2
π ππ
3ππ€π
π―π€π10NR(0π΅) + 2π―π€π10
NR(1π΅)
β’ The spin-dependent terms affect the Zeeman levels β’ The transition πΉ = 0,ππΉ = 0 β πΉ = 1,ππΉ = 0 is not affected
πS1/2
πΉ = 1
πΉ = 0
ππΉ = 1
ππΉ = β1
ππΉ = 0
1S1/2
πΉ = 1
πΉ = 0 Ener
gy
πΌ = 0.007
ππ =ππππ
ππ +ππ
Internal transitions of the ground state πΉ,ππΉ β· πΉβ², ππΉβ²
πΏπ = ββππΉ
2 3 π―π€010
NR(0π΅) + 2π―π€010NR(1π΅) + πΌππ
2 π―π€210NR(0π΅) + 2π―π€210
NR(1π΅)
π€
+ 5 πΌππ4 π―π€410
NR(0π΅) + 2π―π€410NR(1π΅)
ππΉ = 1
ππΉ = β1
ππΉ = 0
πΌππ2~10β11 GeV2
Frequencies are measured relative to other frequencies
Any measurement is the comparison of two physical systems
β’ 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.
β’ 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
β’ The transition πΉ = 1,ππΉ = 0 β πΉ = 2,ππΉ = 0 in 87Rb β’ The transition πΉ = 3,ππΉ = 0 β πΉ = 4,ππΉ = 0 in 133Cs β’ The transition πΉ = 0,ππΉ = 0 β πΉ = 1,ππΉ = 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)
πΌππ2~10β11 GeV2
πΉ = 1,ππΉ = 0 β· πΉ = 1,ππΉ = Β±1
2ππΏπ£ = Β±1
2π΄ β π΅
πΉ = 0,ππΉ = 0 β· πΉ = 1,ππΉ = 0
πΏπ£ = 0
Sidereal variation
Reference
Zeeman transitions
Standard H maser transition
Ξ ππ
= β πΌπΞπΌπ
Transformation from the Sun-centered frame to the laboratory frame
Boost
Rotation Lorentz transformation
For π½ βͺ 1 we can simplify the Lorentz transformation
Lab frame 0 time component π β {1,2,3} spatial components
Sun-centered frame π time component π½ β {π, π, π} spatial components
Orbital velocity of the Earth
Velocity of the lab frame relative to the center of the Earth
For π½ βͺ 1 Orbital velocity of the Earth
Velocity of the lab frame relative to the center of the Earth
π½πΏ β 10β6 sin π
Colatitude
πβ sidereal frequency Ξ©β annual frequency
πβ sidereal time
π and π are azimuthal and polar angles of π© at πβ = 0
If π΅ = π§ in the lab frame then
π1πβ2π = 2466061413187035 Β± 10 Hz or πΏπ
π= 4 Γ 10β15
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 β’ Hyperfine transitions can access frequency shifts down to 10β4 Hz β’ 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
2ππΏπ = 1
4π πΌππ
23
4π±π€200
NR + πΌππ467
16π±π€400
NR
π€
The Lorentz violation frequency shift to the 1S-2S is given by
2ππΏπ =1
4π πΌππ
23
4π±π€200
NR + πΌππ467
16π±π€400
NR
π€
The Lorentz-violating frequency shift to the 1S-2S is given by
π±π€200 NR ,lab = π±π€200
NR ,Sun + 4π 2ππ€eff
(5)πππ½+ ππ€eff
5 πΎπΎπ½π½π½
β 8 π ππ€eff6 ππππ½
+ ππ€eff6 ππΎπΎπ½
π½π½ +β― = π±π€200 NR ,Sun + π½ππ0 β π·
2ππΏπ =1
4π πΌππ
23
4π±π€200
NR,Sun + πΌππ467
16π±π€400
NR,Sun
π€
+1
4π πΌππ
23
4π½ππ0
+ πΌππ467
16π½π40
π€
β π· = π + π β π·
Considering the boost contribution to the transformation to the Sun-centered frame
The frequency shift in the Sun-centered frame has the form
π‘β²
π₯β²
π‘β²
π₯β²
π‘ π‘
π₯
π₯
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
πΏπ = π β π·
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 π½ = 0, for example consider the transition π01 β π0
3
β’ Al27 + ion clock β’ In115 + ion clock β’ Sr87 optical lattice clock β’ Yb171 optical lattice clock β’ Hg199 optical lattice clock
β’ Other clock-comparison experiments are more sensitive to proton and neutron coefficients
2ππΏπ£ = β1
4πβπ2π±π200
NR + βπ4π±π400NR
βππ = π ππ0
3 β π ππ0
1
Annual and sidereal variation due to the boost
β’ Comparing different clock types in the same laboratory frame β’ Comparing clocks in different laboratories frames
2π πΏπ = π + π β π·
The isotropic Lorentz-violating frequency shift for the 1S-2S in hydrogen in the SFC
2ππΏπ =1
4π πΌππ
23
4ππ€200
NR,Sun β ππ€200 NR,Sun + πΌππ
467
16ππ€400
NR,Sun β ππ€400 NR,Sun
π€
Coefficients for CPT violation Coefficients of CPT invariant operators
and for antihydrogen
The SME allows clocks and anti-clocks to tick at different rates
2ππΏπ =1
4π πΌππ
23
4ππ€200
NR,Sun + ππ€200 NR,Sun + πΌππ
467
16ππ€400
NR,Sun + ππ€400 NR,Sun
π€
2π(πΏπ β πΏπ) =2
4π πΌππ
23
4ππ€200
NR,Sun + πΌππ467
16ππ€400
NR,Sun
π€
Discrepancy between hydrogen and antihydrogen
The isotropic Lorentz-violating frequency shift for the 1S-2S in hydrogen in the SFC
2ππΏπ =1
4π πΌππ
23
4ππ€200
NR,Sun β ππ€200 NR,Sun + πΌππ
467
16ππ€400
NR,Sun β ππ€400 NR,Sun
π€
Coefficients for CPT violation Coefficients of CPT invariant operators
The SME allows clocks and anti-clocks to tick at different rates
2ππΏπ =1
4π πΌππ
23
4ππ€200
NR,Sun + ππ€200 NR,Sun + πΌππ
467
16ππ€400
NR,Sun + ππ€400 NR,Sun
π€
2π(πΏπ β πΏπ) =2
4π πΌππ
23
4ππ€200
NR,Sun + πΌππ467
16ππ€400
NR,Sun
π€
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 πΏπ β πΏπ < 5 kHz
This bound is expected to improved in the future
ALPHA measurement of the 1π-2π in antihydrogen
First SME bound from antihydrogen spectroscopy arXiv:1805.04499
40Ca+ experiment using entangled ions β’ Marianna discussed the experiments in details yesterday The experiment involve Zeeman levels in the state with π½ = 5/2 and that implies that electron coefficients with π < 5 could contribute
Defining πππ½ as the energy of the Zeeman level ππ½ of the state π·5/2
2 the observable is
2ππ = π5/2 + πβ5/2 β π1/2 β πβ1/2
2ππΏπ = πΏπ5/2 + πΏπβ5/2 β πΏπ1/2 β πΏπβ1/2
The Lorentz-violating shift is given by
No linear Zeeman shift
No contribution from spin-dependent terms
Only coefficients with π = 2 and π = 4 can contribute
Assumption: Electron in incomplete subshell has total angular momentum π½ = 5/2
Electrons in filled subshells form states with π½ = 0
The shift to the observable is given by
2ππΏπ =18
7 5πβ¨ π 2β©π±π220
NR + β¨ π 4β©π±π420NR +
1
7 5ππ 4 π±π440
NR
Signals for Lorentz violation β’ From the rotation transformation
β’ Sidereal variations with the 1st to the 4th harmonics of the sidereal frequency
β’ Including the boost transformation β’ Sidereal variation with the 5th harmonic of the sidereal frequency (proportional
to π½πΏ = 10β6) β’ Annual variations
Assumption: Electron in incomplete subshell has total angular momentum π½ = 5/2
Electrons in filled subshells form states with π½ = 0
The shift to the observable is given by
2ππΏπ =18
7 5πβ¨ π 2β©π±π220
NR + β¨ π 4β©π±π420NR +
1
7 5ππ 4 π±π440
NR
Instead of using NR coefficients all the previous signals can be observed using coefficients with mass dimensions π β€ 8.
Sidereal local time ππΏ vs. sidereal time πβ
The results in the table are for 133-Cs but they can modified to 40-Ca via the map
ππΏ = πβ + π
Corrections linear in the boost in the Sun-centered frame
Consider the transition S1/22 β D5/2
2 in Sr+88 or Ca+40
Techniques used to eliminate systematics β’ Averaging of the Zeeman levels ππ½ and βππ½: eliminates the linear Zeeman effect and
eliminates the contribution from spin-dependent terms
β’ Averaging over three Zeeman pairs (ππ½, βππ½): eliminates the electric quadrupole
shift and because
5/2ππ½π0|5/2ππ
5/2
ππ½=β5/2
= 6πΏπ0
only isotropic coefficients can contribute
β’ Using two Zeeman pairs (ππ½, βππ½) to extrapolate the value for ππ½2 = 35/12:
eliminates the electric quadrupole shift and allows contributions from π = 0 and π = 4 coefficients.
2ππΏπ = β1
4πβπ2π±π200
NR + π 4 π±π400NR +
1
27 πβπ4π±π440
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 πΌ = 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 πΌ
Nuclear model used
Issues with the Schmidt model β’ 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
β’ The contribution of the nucleon preferred by the model is usually dominant
Good things about the model β’ It can be easily applied to many systems and the signals for Lorentz violation
predicted by the model tend to be accurate
β’ Allow analytical expressions for the angular expectation value
β’ 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; π½ = 1/2, πΌ = 7/2, πΉ = 3 or 4
π½ = 1/2 implies that only electrons coefficients with π β€ 1 can contribute Electrons
Neutrons
Protons
πΌ = 7/2 implies that only nucleon coefficients with π β€ 7 can contribute but for the case πΉ = 3 the condition is π β€ 6.
For practical reasons we limit ourselves to π β€ 5 β’ The coefficient with the smallest mass dimension that contributes to
π = 7 has mass dimension π = 9 β’ The coefficient with the smallest mass dimension that contributes to
π = 5 has mass dimension π = 7
133-Cesium has 55 protons and 78 neutrons β’ In the Schmidt model the spin of the nucleus is due to the unpaired proton
The shift to the observable is given by
2ππΏππ = β13
14
5
πβ¨ π 2β©π±π220
NR + β¨ π 4β©π±π420NR +
45
77 ππ 4 π±π440
NR
Define the transition as πΉ = 3,ππΉ β |πΉ = 4,ππΉβ© as πΏπππΉ then the observable of the
experiment is
πΏππ = πΏπ3 + πΏπβ3 β 2πΏπ0 to eliminate the Zeeman shift
Signals for Lorentz violation β’ From the rotation transformation
β’ Sidereal variations with the 1st to the 4th harmonics of the sidereal frequency
β’ Including the boost transformation β’ Sidereal variation with the 5th harmonic of the sidereal frequency (proportional to
π½πΏ = 10β6) β’ 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 β’ The ground state of both systems have quantum numbers: π½ = 0, πΉ = πΌ = 1/2
β’ Only the nucleon coefficients with π β€ 1 contribute to the ground state β’ Only the electron coefficients with π β€ 0 contribute to the ground state
β’ In the Schmidt model the neutron carries all the nuclear spin
Define πΏπ£Xe as the Zeeman transition πΉ =1
2, ππΉ =
1
2β πΉ =
1
2, ππΉ = β
1
2 in Xe
and πΏπ£He the same transition in He.
The observable in the experiment is π = πHe βπΎHeπΎXe
πXe
πΎXe and πΎHe are the gyromagnetic ratio of the corresponding ground states
β’ This observable is insensitive to the linear Zeeman effect
πHe β πΎHπ π΅ πXe β πΎXπ π΅
F. Canè et al., Phys. Rev. Lett. 93, 230801 (2004)
2ππΏπ =β1
3π π π
HeβπΎHeπΎXe
π πXe
π
π―ππ10NR(0π΅) + 2π―ππ10
NR(1π΅) = π΄ π»πππ β π΅