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

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2 Lorentz-violating energy shift for hydrogen in the presence of a weak magnetic field δε = j A j0 (nfjl) Fm F j0 Fm F Clebsch-Gordan coefficients n principal quantum number n F hydrogen total angular momentum quantum number J electron total angular momentum quantum number L electron orbital quantum number m F quantum number of the components of F in the direction of the magnetic field J = S e + L F = S p + J

3 Lorentz-violating energy shift for hydrogen in the presence of a weak magnetic field δε = A j0 (nfjl) Fm F j0 Fm F Clebsch-Gordan coefficients jm If j is an even number (spin-independent) A jm nfjl = p k nl wk Λ 0E NR j V w kjm If j is an odd number (spin-dependent) 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 A jm nfjl = p k nl wk Λ j 0B 2J + 1 T wkjm NR(0B) Λj 1B δ we 2(L J) + δ wp 2(J F) T NR(1B) wkjm

4 Energy Energy shift to the ns F 1/2 in the presence of a weak magnetic field; n, L = 0; J = 1/2, F, m F δε = A 00 n + m F 2 A 10 n = wk p k n 4π V NR w k00 m F 2 wk p k n 3π T NR(0B) NR(1B) w + 2T k10 w k10 F = 1 m F = 1 m F = ns 1/2 F = 0 m F = 1 1 2

5 Energy Energy shift to the ns F 1/2 in the presence of a weak magnetic field; n, L = 0; J = 1/2, F, m F δε = A 00 n + m F 2 A 10 n = wk p k n 4π V NR w k00 m F 2 wk p k n 3π T NR(0B) NR(1B) w + 2T k10 w k10 The spin-independent terms do not shift any of the internal transitions of the ns 1/2 state Hyperfine and Zeeman transitions are not affected F = 1 m F = 1 m F = 0 ns 1/2 F = 0 m F = 1

6 Energy Energy shift to the ns F 1/2 in the presence of a weak magnetic field; n, L = 0; J = 1/2, F, m F δε = A 00 n + m F 2 A 10 n = wk p k n 4π V NR w k00 m F 2 wk p k n 3π T NR(0B) NR(1B) w + 2T k10 w k10 The spin-independent terms do not shift any of the internal transitions of the ns 1/2 state Hyperfine and Zeeman transitions are not affected m F = 1 ns 1/2 F = 1 m F = 0 m F = 1 F = 0

7 Energy Energy shift to the ns F 1/2 in the presence of a weak magnetic field; n, L = 0; J = 1/2, F, m F δε = A 00 n + m F 2 A 10 n = wk p k n 4π V NR w k00 m F 2 wk p k n 3π T NR(0B) NR(1B) w + 2T k10 w k10 The spin-dependent terms affect the Zeeman levels The transition F = 0, m F = 0 F = 1, m F = 0 is not affected F = 1 m F = 1 m F = 0 ns 1/2 F = 0 m F = 1

8 Energy Energy shift to the ns F 1/2 in the presence of a weak magnetic field; n, L = 0; J = 1/2, F, m F δε = A 00 n + m F 2 A 10 n = wk p k n 4π V NR w k00 m F 2 wk p k n 3π T NR(0B) NR(1B) w + 2T k10 w k10 The spin-dependent terms affect the Zeeman levels The transition F = 0, m F = 0 F = 1, m F = 0 is not affected F = 1 m F = 1 m F = 0 ns 1/2 F = 0 m F = 1

9 Energy Internal transitions of the ground state F, m F F, m F δε = m F 2 3 T NR(0B) NR(1B) w T w 010 w + 5 αm 4 r T NR(0B) NR(1B) w + 2T 410 w αm 2 r T NR(0B) NR(1B) w T w 210 α = m r = m em p m e + m p F = 1 αm r 2 ~10 11 GeV 2 m F = 1 m F = 0 1S 1/2 F = 0 m F = 1

10 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 133 Cs atom to complete cycles. The meter is defined as the distance travelled by light in vacuum during a time interval of 1/ of a second. Frequencies are measured relative to other frequencies Could Lorentz symmetry affect the base units? Yes

11 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, m F = 0 F = 2, m F = 0 in 87 Rb The transition F = 3, m F = 0 F = 4, m F = 0 in 133 Cs The transition F = 0, m F = 0 F = 1, m F = 0 in H This is not surprising because the leading Lorentz-violating corrections mimic perturbations due to weak external EM fields. Lorentz-violating multi-particle operators could introduce Lorentzviolating effects to these time standards

12 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

13 αm r 2 ~10 11 GeV 2 Zeeman transitions F = 1, m F = 0 F = 1, m F = ±1 2πδv = ± 1 A B 2 Sidereal variation Standard H maser transition F = 0, m F = 0 F = 1, m F = 0 δv = 0 Reference Phillips et al., PRD 63, (2001)

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15 Transformation from the Sun-centered frame to the laboratory frame Lorentz transformation Λ μ ν = R μ αβ α ν Rotation Boost 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

16 For β 1 Orbital velocity of the Earth Velocity of the lab frame relative to the center of the Earth β L 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

17 1S-2S transition the most precisely measured transition in hydrogen ν 1S 2S = ± 10 Hz or δν ν = Parthey et al.,prl 107, (2011) 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 The Lorentz violation frequency shift to the 1S-2S is given by 1 4π w αm r 2πδν = V w200 NR 4 + αm 67 r 16 V NR w400

18 The Lorentz-violating frequency shift to the 1S-2S is given by 2πδν = 1 4π w αm 2 3 r 4 V w200 NR + αm 4 67 r 16 V NR w400 Considering the boost contribution to the transformation to the Sun-centered frame NR V,lab NR w = V,Sun (5)TTJ w + 4π 2a 200 weff + KKJ aweff β J 6 8 π c TTTJ 6 TKKJ weff + c weff β J NR + = V,Sun w + V 200 w 20 β The frequency shift in the Sun-centered frame has the form 2πδν = 1 4π w + 1 4π w αm 2 3 r 4 V w200 NR,Sun + αm 4 67 r 16 V NR,Sun w400 αm r V w20 + αm r V w40 β = S + V β

19 t The Lorentz-violating field is isotropic in this frame t t The Lorentz-violating field is anisotropic in this frame t x x x x Isotropic effects in one reference frame are not necessarily isotropic in other frames An sphere is not round in all inertial reference frames

20 frequency time δν = V β Orbital velocity of the Earth Lorentz-violating field Annual variation of the frequency

21 Matveev et al., PRL 110, (2013) 1 1 A Cs atomic fountain clock is used as the reference clock 1S-2S muonium and positronium (proposed as a gravity antimatter test) Kirch et. al., IJMPCS 30, (2014) Crivelli et. al., IJMPCS 30, (2014)

22 Optical clocks with J = 0, for example consider the transition 1S 0 3 P 0 27 Al+ ion clock In ion clock 87 Sr optical lattice clock 171 Yb optical lattice clock 199 Hg optical lattice clock Other clock-comparison experiments are more sensitive to proton and neutron coefficients 2πδv = 1 4π p2 V NR e p 4 NR p V k = p k 3 p k e P S 0 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

23 The SME allows clocks and anti-clocks to tick at different rates The isotropic Lorentz-violating frequency shift for the 1S-2S in hydrogen in the SFC 2πδν = 1 4π w αm 2 3 r 4 c w200 NR,Sun a w 200 NR,Sun + αm 4 67 r 16 c w400 NR,Sun a NR,Sun w 400 and for antihydrogen Coefficients for CPT violation Coefficients of CPT invariant operators 2πδν = 1 4π w αm 2 3 r 4 c w200 NR,Sun + a w 200 NR,Sun + αm 4 67 r 16 c w400 NR,Sun + a NR,Sun w 400 Discrepancy between hydrogen and antihydrogen 2π(δν δν) = 2 4π w 2 αm 3 r 4 a w200 NR,Sun 4 + αm 67 r 16 a NR,Sun w400

24 The SME allows clocks and anti-clocks to tick at different rates The isotropic Lorentz-violating frequency shift for the 1S-2S in hydrogen in the SFC 2πδν = 1 4π w αm 2 3 r 4 c w200 NR,Sun a w 200 NR,Sun + αm 4 67 r 16 c w400 NR,Sun a NR,Sun w 400 and for antihydrogen Coefficients for CPT violation Coefficients of CPT invariant operators 2πδν = 1 4π w αm 2 3 r 4 c w200 NR,Sun + a w 200 NR,Sun + αm 4 67 r 16 c w400 NR,Sun + a NR,Sun w 400 Discrepancy between hydrogen and antihydrogen 2π(δν δν) = 2 4π w 2 αm 3 r 4 a w200 NR,Sun 4 + αm 67 r 16 a NR,Sun w400

25 ALPHA measurement of the 1S-2S in antihydrogen 2π δν δν M. Ahmadi et al., doi.org/ /s < 5 khz First SME bound from antihydrogen spectroscopy arxiv: This bound is expected to improved in the future Using ultracold antiatoms from the GBAR antihydrogen beam in an optical lattice might improve the bound by three orders of magnitude P. Crivelli and N. Kolachevsky, arxiv:

26 40 Ca+ 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 2 Defining ε mj as the energy of the Zeeman level m J of the state D 5/2 the observable is 2πf = ε 5/2 + ε 5/2 ε 1/2 ε 1/2 The Lorentz-violating shift is given by No linear Zeeman shift 2πδf = δε 5/2 + δε 5/2 δε 1/2 δε 1/2 No contribution from spin-dependent terms Only coefficients with j = 2 and j = 4 can contribute

27 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 = π p 2 V NR e p 4 V NR e π p 4 NR V e 440 Signals for Lorentz violation From the rotation transformation Sidereal variations with the 1 st to the 4 th harmonics of the sidereal frequency Including the boost transformation Sidereal variation with the 5 th harmonic of the sidereal frequency (proportional to β L = 10 6 ) Annual variations

28 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 = π p 2 V NR e p 4 V NR e π p 4 NR V e 440 Instead of using NR coefficients all the previous signals can be observed using coefficients with mass dimensions d 8.

29 The results in the table are for 133-Cs but they can modified to 40-Ca via the map Sidereal local time T L vs. sidereal time T T L = T + φ

30 Corrections linear in the boost in the Sun-centered frame

31 Consider the transition 2S 1/2 2 D 5/2 in 88 Sr + or 40 Ca + Techniques used to eliminate systematics Averaging of the Zeeman levels m J and m J : eliminates the linear Zeeman effect and eliminates the contribution from spin-dependent terms Averaging over three Zeeman pairs (m J, m J ): eliminates the electric quadrupole shift and because 5/2 m J = 5/2 5/2m J j0 5/2m j only isotropic coefficients can contribute = 6δ j0 Using two Zeeman pairs (m J, m J ) to extrapolate the value for m J 2 = 35/12: eliminates the electric quadrupole shift and allows contributions from j = 0 and j = 4 coefficients. 2πδν = 1 4π p2 V NR e + p 4 V NR 200 e π p4 NR V e 440

32 Nucleus with even number of neutrons and even number of protons have nuclear spin I = 0 Nuclear model used 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. 7-Lithium nucleus has three protons and four neutrons Total angular momentum of the unpaired proton equal to I Protons Neutrons zero total angular momentum

33 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)

34 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 I = 7/2 implies that only nucleon coefficients with j 7 can contribute but for the case F = 3 the condition is j 6. Protons Electrons Neutrons 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 = Cesium has 55 protons and 78 neutrons In the Schmidt model the spin of the nucleus is due to the unpaired proton

35 Define the transition as F = 3, m F experiment is F = 4, m F as δν mf then the observable of the to eliminate the Zeeman shift δν c = δν 3 + δν 3 2δν 0 The shift to the observable is given by 2πδν c = π p 2 V NR p220 + p 4 V NR p π p 4 NR V p440 Signals for Lorentz violation From the rotation transformation Sidereal variations with the 1 st to the 4 th harmonics of the sidereal frequency Including the boost transformation Sidereal variation with the 5 th harmonic of the sidereal frequency (proportional to β L = 10 6 ) Annual variations

36 Estimates for sidereal variation in the first harmonic and annual variations based on the previous experiment H. Pihan-Le Bars et al., Phys. Rev. Lett. 95, (2017) Using better nuclear models contributions from the neutron coefficients are expected

37 129 Xe- 3 He comagnetometer F. Canè et al., Phys. Rev. Lett. 93, (2004) 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 δv Xe as the Zeeman transition F = 1, m 2 F = 1 F = 1, m 2 2 F = 1 2 and δv He the same transition in He. in Xe The observable in the experiment is ν = ν He γ He γ Xe ν Xe γ Xe and γ He are the gyromagnetic ratio of the corresponding ground states This observable is insensitive to the linear Zeeman effect ν He γ He B ν Xe γ Xe B

38 2πδν = 1 3π k p k He γ He γ Xe p k Xe T NR(0B) NR(1B) n + 2T k10 n k10 = A HeXe B

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