Goal: find Lorentz-violating corrections to the spectrum of hydrogen including nonminimal effects

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2 Goal: find Lorentz-violating corrections to the spectrum of hydrogen including nonminimal effects Method: Rayleigh-Schrödinger Perturbation Theory Step 1: Find the eigenvectors ψ n and eigenvalues ε n of the unperturbed Hamiltonian h 0 ; h 0 ψ n = ε 0 ψ n Step2: Find the Lorentz-violating perturbation δh; h = h 0 + δh Step 3: Find the restriction of δh on the eigenspace of h 0 corresponding to the eigenvalue ε 0 ; ψ n h ψ m = ε 0 δ nm + ψ n δh ψ m Step 4: Find the eigenvalues δε of the restriction of δh on the eigenspace of h 0 corresponding to the eigenvalue ε 0 ; ε ε 0 + δε

3 Step 1: the unperturbed states of the hydrogen atom are unknown but they can be approximated ε(n, F, J, L) The spectrum of the hydrogen depends on the following quantum numbers Principal quantum number n The electron orbital quantum number L The electron total angular momentum quantum number J The hydrogen total angular momentum quantum number F

4 Defining the angular momentums L = r p S e S p S e J L Electron Proton Electron total angular momentum J = S e + L S p F J Hydrogen total angular momentum Electron spin Electron orbital angular momentum Proton spin F = S p + J

5 Gross structure of the spectrum of the hydrogen atom Principal quantum number n labels the different electronic shells Electron shells Proton (nucleus) n = 1 n = 2 n = 3 Spinless particle in an external classical Coulomb potential ε n = m e 2 α n 2 α is the fine-structure constant m e mass of the electron

6 Fine structure of the spectrum of the hydrogen atom Dirac fermion in an external classical Coulomb potential ε nj = 1 + m e α n J J α 2 2 Electron total angular momentum J = S e + L Electron spin Electron orbital angular momentum The energy spectrum of the hydrogen atom depends on the electron total angular momentum quantum number J

7 Corrections beyond the central field approximation Lamb shift, finite size, and recoil corrections QED radiative corrections Recoil corrections because of the finite mass of the proton Finite size corrections because of the effective radius of the proton The spectrum of hydrogen depends on the orbital angular momentum quantum number L Hyperfine structure of the spectrum of the hydrogen atom The hyperfine structure dominant effects emerge from the magnetic moment of the proton Hydrogen total angular momentum F = S p + J The spectrum of hydrogen depends on hydrogen total angular momentum quantum number F Proton spin

8 A transformation f: A A on a set A is a symmetry of A with respect to the equivalence relation if f(a) a for all a A. You need to define the equivalence relation Example 2 m 2 m Transformation A 1 m 1 m Case 1: Objects are equivalent if they have the same shape 1 m Transformation A is a symmetry Transformation B is not a symmetry 2 m Transformation B 2 m 4 m Case 2: Objects are equivalent if they have the same area Transformation A is not a symmetry Transformation B is a symmetry

9 Ahhhhhhhhhh What kind of rotation symmetry? Transformation: change the orientation of the system without changing the internal configuration of the system Equivalent: the relative behavior between the components of the system is the same Rotation symmetry Rotation violation

10 Boost symmetry: the spectrum of hydrogen is independent of the velocity of the center of mass Rotation symmetry: the spectrum of hydrogen is independent of the orientation of the atom It is independent of the quantum number m F F z Fm F = m F Fm F m F = 2 m F = 1 F F = 2 m F = 0 same spectrum F m F = 1 m F = 2

11 The approximated unperturbed states used in the calculation are the states nfjlm F The eigenspace with eigenvalue ε(nfjl) has dimension equal to 2F + 1 m F { F, F + 1,, F 1, F} nfljm F = Jm J S p m sp Fm F Lm L S e m Se Jm J m S e m Sp m Jm L nlm L S e m Se S p m sp Clebsch-Gordan coefficients j 1 m 1 j 2 m 2 j 3 m 3 r nlm L = ψ nlml (r) Schrödinger-Coulomb wave function S e m Se electron spin states S p m Sp proton spin states

12 The leading-order corrections to the spectrum of hydrogen due to Lorentz violation are expected to emerge from corrections to the propagation of the proton and the electron. Leading-order corrections to the propagation of a free fermion with flavor w due to Lorentz violation L = 1 2 ψ w i μ γ μ m w + Q ψ w + h. c. Kostelecký and Mewes, PRD 88, (2013) ip μ γ μ m W + Q ψ w = 0 Effective Dirac equation * Kept only the linear terms on the SME coefficients H = H 0 + δh LV Effective Dirac Hamiltonian * Kept only the linear terms on the SME coefficients h = h 0 + δh LV Effective one-particle Hamiltonian

13 Effective perturbation for a free spin one-half fermion in the momentum representation Kostelecký and Mewes, PRD 88, (2013) δh LV = p μ E 0 μ 1 0μ V w eff T m w eff w iμ p σ + Tw eff σ i + pi p σ E 0 + m w m w σ = σ x x + σ y y + σ z z = z x + iy x i y z p 0 = E 0 = p 2 + m w 2 w is the flavor of the fermion; w = e for the electron and w = μ for the muon In the Dirac notation the operator is represented by δh LV = d 3 p p p δh LV (p)

14 You can think of p μ μ V E w 0 eff = f(p) as a function of p. We can expand the function as f p = p μ μ V E w eff = p k 0 kjm NR Y jm p V w kjm Y jm p = Y jm (φ, θ) are the spherical harmonics p magnitude of three momentum p direction of three momentum Nonrelativistic (NR) coefficients Y jm p f p dω = k p k NR V w kjm Binomial coefficient n k = n! k! n k! V NR d 3 k w kjm = m w d q k/2 (d 3 k + 2q)/2 q (d) V w (k 2q)jm The NR coefficients include spherical coefficients of arbitrary mass dimension

15 For the spin-dependent terms the situation is more complicated Now we have two vectors σ and p. We can express the orientation of σ in terms of the orientation of p Define a helicity basis with z = p εr(p) = p ε ±(p) = 1 2 β ± iα σ = σ r εr + σ ε + + σ + ε In any frame the components of σ can be obtained in term of the direction of the momentum vector σ z = 4π 3 0Y 10 p σ r p + +1 Y 10 p σ p 1 Y 10 p σ + p

16 Spin-weighted spherical harmonics with spin weight s form a complete set of orthonormal functions on the sphere p Y jm p s e isα Y jm p s α s = 0 are the usual spherical harmonics The spin weight tells us how the phase of the function changes under a rotation around the direction of the argument. σ + (p) e iα σ + (p) σ (p) e iα σ (p) Spin weight s = ±1

17 The perturbation Kostelecký and Mewes, PRD 88, (2013) (0B) δh LV = h SI + h SD (1B) (1E) + hsd + hsd h SI = kjm p k NR V w kjm 0 Y jm (p) (0B) h SD = kjm p k NR 0B T w kjm 0 Y jm p σ r (1B) h SD = kjm p k NR 1B T w kjm +1Y jm p σ 1 Y jm p σ + (1E) h SD = kjm p k NR T 1E w kjm i +1 Y jm p σ + 1 Y jm p σ +

18 Symbol Meaning/Information NR( s P) T kjm jm s P k Spherical components Spin weight Parity type Momentum power The parity type distinguishes between the following cases Parity +1Y jm p σ 1 Y jm p σ + 1 j+1 ( +1 Y jm p σ 1 Y jm p σ + ) B-type parity i +1 Y jm p σ + 1 Y jm p σ + 1 j i +1 Y jm p σ + 1 Y jm p σ + E-type parity

19 The perturbation Kostelecký and Mewes, PRD 88, (2013) (0B) δh LV = h SI + h SD (1B) (1E) + hsd + hsd h SI = kjm p k NR V w kjm 0 Y jm (p) (0B) h SD = kjm p k NR 0B T w kjm 0 Y jm p σ r (1B) h SD = kjm p k NR 1B T w kjm +1Y jm p σ 1 Y jm p σ + (1E) h SD = kjm p k NR T 1E w kjm i +1 Y jm p σ + 1 Y jm p σ + δh H = δh LV e p e + δh LV p p p electron proton In the zero momentum frame p = p e = p p

20 Step 3: find the matrix elements of the perturbation Clebsch-Gordan coefficients nfjlm F δh H nfjlm F = A jm (nfjl) Fm F jm Fm F jm If j is an even number (spin-independent) A jm nfjl = p k nl wk If j is an odd number (spin-dependent) Λ 0E NR j V w kjm 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 Λj 1B δ we 2(L J) + δ wp 2(J F) T wkjm NR(1E) The T w kjm do not contribute to the matrix elements Kostelecký and AJV, PRD 92, (2015)

21 Matrix elements of the perturbation A jm nfjl = p k nl wk Λ j 0B 2J + 1 T wkjm Λj 1B δ we 2(L J) + δ wp 2(J F) T wkjm p 4 nl = αm r n 4 8n 2L m r = m em p m e + m p Λ j 1B F = i2(j 1) j!! (2j + 1)(F + 1/2 j + 1)! 2F j!! 2 j 1 /2 π 2F + 1 F 1/2(j + 1)! 2F + j!! Can be expressed completely in terms of Clebsch-Gordan coefficients This is one of the reason for using spherical operators

22 Useful identities that follow from the properties of the Clebsch-Gordan coefficients are A jm nfjl = 0 if j is even and j > 2J A jm nfjl = 0 if j is odd and j > 2F Which ones of the A jm (nfjl) contributes to the matrix elements in the subspace with quantum numbers F = 2, J = 5/2, and L = 2? (nd F=2 5/2 ) A jm nfjl = 0 if j is even and j > 5 A jm nfjl = 0 if j is odd and j > 4 Greatest even value of j that contribute is j = 4 Greatest odd value of j that contribute is j = 3 4 j nfjlm F δh H nfjlm F = A jm (nfjl) Fm F jm Fm F j=0 m= j Higher angular momentum states are sensitive to more coefficients

23 nfjlm F δh H nfjlm F = A jm (nfjl) Fm F jm Fm F jm Example ns 1/2 ; L = 0, J = 1/2, and F = 0 A jm nfjl = 0 if j is even and j > 2J = 1 A jm nfjl = 0 if j is odd and j > 2F = 0 n, 0,1/2,0,0 δh H n, 0,1/2,0,0 = A 00 n = A 00 n = wk p k n0 4π NR V w k00 The energy shift is given by δε = wk p k n0 4π NR V w k00

24 Example ns 1/2 ; L = 0, J = 1/2, and F = 1 A jm nfjl = 0 if j is even and j > 2J = 1 A jm nfjl = 0 if j is odd and j > 2F = 2 n, 1,1/2,0, m F δh H n, 1,1/2,0, m F = A 00 n 1m F 00 1m F + A 1m n 1m F 1m 1m F 1 m= A 00 n A 10 (n) A 11 (n) 0 A 11 (n) 2A 00 n A 11 (n) 0 A 11 (n) 2A 00 n + A 10 (n) A 00 n = wk p k n0 4π NR V w k00 A 1m n = wk p k n0 6π T w + 2T k1m w k1m

25 Step 4: find the eigenvalues of the restriction of δh H δε = A 00 (n) + ξ F 2 A(n) ξ F { 1,0,1} Effective perturbation in the eigenspace with F = 1 δh F=1 H = A 00 n + 1 F A 2 A x = 1 2 (A 11 + A 11 ) A y = i 2 (A 11 A 11 ) A z = A 10 with eigenstates nfjlξ F such that F A nfjlξ F = ξ F A nfjlξ F F=1 ns 1/2 ξ F = 1 ξ F = 0 ξ F = 1 A acts as an effective magnetic field producing a linear Zeeman shift A F ξ F = 1 ξ F = 0 F F ξ F = 1

26 Possible signal Unexpected multiple resonance peaks Three resonance peaks could be observed in the hyperfine transition of the ground state ξ F = 1 F=1 1S 1/2 F=0 1S 1/2 ξ F = 0 ξ F = 1 2πδν = ξ F 2 A Shift to the resonance frequency In the usual case there is only one resonance frequency In the Lorentz-violating case there are three resonance frequencies By modeling the line shape of the hyperfine transition of the ground state assuming the presence of multiple peaks it might be possible to constraint coefficients for Lorentz violation

27 A better approach is obtained by introducing a weak magnetic field that produces a linear Zeeman shift that dominates the Lorentz-violating energy shift Consider the following effective perturbation in the eigenspace with F = 1 μ g-factor B = e 2m e δh H F=1 μ H B = A F A + g Fμ B F B = A 00 + F 1 2 A + g Fμ B B Hydrogen magnetic dipole Applied magnetic field The eigenvalues of the perturbation are given by g F μ B B A δε + ε B = A 00 + ξ F 12 A + g Fμ B B A 00 + ξ F g F μ B B + 1 A B 2 If B = z then ξ F = m F and the effective Lorentz-violating energy shift is δε = A 00 + m F A B 2

28 Lorentz-violating energy shift of the ground state of hydrogen in a weak magnetic field δε = A 00 + m F A B 2 Hydrogen total angular momentum Applied magnetic field Lorentz-violating field The orientation of the atom is controlled by the orientation of the magnetic field B F = m F Changing the relative orientation between A and B changes the energy because it changes the orientation between the atom and the Lorentz-violating field

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31 Lab Frame ν B field Orientation Fixed frame

32 Lab Frame ν B field Orientation Fixed frame

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34 The results for hydrogen can be modified for others two-fermion atoms that have the same kind of angular momentum coupling hierarchy than hydrogen e Hydrogen H p Antihydrogen H V NR w kjm = c w kjm NR a w kjm NR NR V w kjm = cw kjm NR + a NR w kjm μ Muonium Mu e V NR pkjm = c NR pkjm a NR pkjm NR V μkjm = cμkjm NR + a NR μkjm Muonic hydrogen V NR e kjm = c NR e kjm a NR e kjm V NR μkjm = c NR μkjm NR a μkjm

35 1S F=1 1/2 Zeeman transitions in hydrogen Phillips et al., PRD 63, (2001) e H p 1S 1/2 Breit-Rabi transitions in muonium Hughes et al., PRL 87, (2001) μ Mu e NR( s P) NR( s P) NR( s P) T w kjm = gwkjm Hw kjm

36 Contain the previous minimal SME results Generalize the results presented in Bluhm, Kostelecký, and Russell, PRL 82, 2254 (1999) Bluhm, Kostelecký, and Lane, PRL 84, 1098 (2000) For the hyperfine transition the signals predicted by the minimal SME are identical than the ones predicted by the nonminimal corrections This is not true for most transitions

37 Sensitivity to k = 4 coefficients 1 p 4 The momentum of the muon in μh is 200 times larger than in muonium

38 In the presence of a weak magnetic field the energy depends on m F with B F nfjlm F = F z nfjlm F = m F nfjlm F There is no more degeneracy in the spectrum and the energy shift is given by the expectation value of δh H with respect to the unperturbed energy eigenstates δε = nfjlm F δh H nfjlm F = A jm (nfjl) jm Fm F jm Fm F = A j0 (nfjl) j Fm F j0 Fm F Example ns 1/2 ; L = 0, J = 1/2, and F = 1 δε = A 00 1m F 00 1m F + A 10 1m F 10 1m F = A 00 + m F 2 A 10 = A 00 + m F 2 A z = A 00 + m F A B 2

39 Expression for the energy shift of the spectrum of hydrogen in the presence of a weak magnetic field in the Sun-centered frame δε = j Fm F j0fm F d 0 m θ Re A Sun j m cos( m ω T ) Im A Sun j m sin( m ω T ) jm θ is the angle between the magnetic field and the rotation axis of the Earth ω = 2π is the sidereal frequency 23 h 56 m T is the sidereal time Wigner d-matrix d m m j A Sun jm is obtained by replacing the SME coefficients in A jm by the SME coefficients in the Sun-centered frame

40 Expression for the energy shift of the spectrum of hydrogen in the presence of a weak magnetic field in the Sun-centered frame δε = j Fm F j0fm F d 0 m θ Re A Sun j m cos( m ω T ) Im A Sun j m sin( m ω T ) jm The highest harmonic that contributes to the sidereal variation of the energy is given by the greatest value of j that contributes to the energy shift j max = m max Example: for nd F=2 5/2 we obtained that j max = 4 then the sidereal variation of the energy receives contributions up to the 4 th harmonic of the sidereal frequency Example: for ns 1/2 F=1 we obtained that j max = 1 then the sidereal variation of the energy receives contributions only from the first harmonic of the sidereal frequency

41 Which one of the following coefficients could be constrained by looking for sidereal variations on the second harmonic of the sidereal frequency? NR a) V w 441 NR b) V w 442 NR c) V w 443 NR V w kjm If you want to constraint the coefficient V NR w 422 at least one of the states involved in the transition must has angular momentum J equal or greater than a) J = 5/2 b) J = 3/2 A c) J = 1/2 jm = 0 for even j and j > 2J A jm = 0 for odd j and j > 2F What is the highest harmonic of the sidereal frequency that contributes to the energy shift of a state with quantum numbers F = 2 and J = 1/2? a) 1 st harmonic b) 2 nd harmonic c) 3 rd harmonic

42 H 010 H 010 g 010 NR NR g a c 000 H 450 H 450 g 450 g 450 H 011 H 011 g 011 NR NR g a c 200 H 451 H 451 g 451 g 451 H 210 H 210 g 210 NR NR g a c 400 H 452 H 452 g 452 g 452 H 211 H 211 g 211 NR NR g a c 220 H 453 H 453 g 453 g 453 H 410 H 410 g 410 NR NR g a c 221 H 454 H 454 g 454 g 454 H 411 H 411 g 411 NR NR g a c 222 H 455 H 455 g 455 g 455 H 230 H 230 g 230 NR NR g a c 420 H 231 H 231 g 231 NR NR g a c 421 H 232 H 232 g 232 NR NR g a c 422 H 233 H 233 g 232 NR NR g a c 440 H 430 H 430 g 430 NR NR g a c 441 H 431 H 431 g 431 NR NR g a c 442 H 432 H 432 g 432 NR NR g a c 443 H 433 H 433 g 433 NR NR g a c 444

43 H 010 H 010 g 010 NR NR g a c 000 H 450 H 450 g 450 g 450 H 011 H 011 g 011 NR NR g a c 200 H 451 H 451 g 451 g 451 H 210 H 210 g 210 NR NR g a c 400 H 452 H 452 g 452 g 452 H 211 H 211 g 211 NR NR g a c 220 H 453 H 453 g 453 g 453 H 410 H 410 g 410 NR NR g a c 221 H 454 H 454 g 454 g 454 H 411 H 411 g 411 NR NR g a c 222 H 455 H 455 g 455 g 455 H 230 H 230 g 230 NR NR g a c 420 Hyperfine trans. of the ground state H 231 H 231 g 231 NR NR g a c 421 H 232 H 232 g 232 NR NR g a c 422 H 233 H 233 g 232 NR NR g a c 440 H 430 H 430 g 430 NR NR g a c 441 H 431 H 431 g 431 NR NR g a c 442 H 432 H 432 g 432 NR NR g a c 443 H 433 H 433 g 433 NR NR g a c 444

44 H 010 H 010 g 010 NR NR g a c 000 H 450 H 450 g 450 g 450 H 011 H 011 g 011 NR NR g a c 200 H 451 H 451 g 451 g 451 H 210 H 210 g 210 NR NR g a c 400 H 452 H 452 g 452 g 452 H 211 H 211 g 211 NR NR g a c 220 H 453 H 453 g 453 g 453 H 410 H 410 g 410 NR NR g a c 221 H 454 H 454 g 454 g 454 H 411 H 411 g 411 NR NR g a c 222 H 455 H 455 g 455 g 455 H 230 H 230 g 230 NR NR g a c 420 Hyperfine trans. of the ground state H 231 H 231 g 231 NR NR g a c 421 1S-2S transition H 232 H 232 g 232 NR NR g a c 422 H 233 H 233 g 232 NR NR g a c 440 H 430 H 430 g 430 NR NR g a c 441 H 431 H 431 g 431 NR NR g a c 442 H 432 H 432 g 432 NR NR g a c 443 H 433 H 433 g 433 NR NR g a c 444

45 H 010 H 010 g 010 NR NR g a c 000 H 450 H 450 g 450 g 450 H 011 H 011 g 011 NR NR g a c 200 H 451 H 451 g 451 g 451 H 210 H 210 g 210 NR NR g a c 400 H 452 H 452 g 452 g 452 H 211 H 211 g 211 NR NR g a c 220 H 453 H 453 g 453 g 453 H 410 H 410 g 410 NR NR g a c 221 H 454 H 454 g 454 g 454 H 411 H 411 g 411 NR NR g a c 222 H 455 H 455 g 455 g 455 H 230 H 230 g 230 NR NR g a c 420 Hyperfine trans. of the ground state H 231 H 231 g 231 NR NR g a c 421 1S-2S transition H 232 H 232 g 232 NR NR g a c 422 2S 1/2 nd 5/2 transition H 233 H 430 H 431 H 233 H 430 H 431 g 232 g 430 g 431 g 233 NR a 440 NR c 440 g 430 NR a 441 NR c 441 NR NR g a c 442 Signals for Lorentz violation Sid. variations (1 st - 6 th harm.) H 432 H 432 g 432 NR NR g a c 443 H 433 H 433 g 433 NR NR g a c 444

46 H 010 H 010 g 010 NR NR g a c 000 H 450 H 450 g 450 g 450 H 011 H 011 g 011 NR NR g a c 200 H 451 H 451 g 451 g 451 H 210 H 210 g 210 NR NR g a c 400 H 452 H 452 g 452 g 452 H 211 H 211 g 211 NR NR g a c 220 H 453 H 453 g 453 g 453 H 410 H 410 g 410 NR NR g a c 221 H 454 H 454 g 454 g 454 H 411 H 411 g 411 NR NR g a c 222 H 455 H 455 g 455 g 455 H 230 H 231 H 232 H 230 H 231 H 232 g 230 g 231 g 232 NR NR g a c 420 NR NR g a c 421 NR NR g a c 422 Muonic hydrogen is more sensitive to some of the coefficients than muonium H 233 H 233 g 232 NR NR g a c 440 H 430 H 430 g 430 NR NR g a c 441 H 431 H 431 g 431 NR NR g a c 442 H 432 H 432 g 432 NR NR g a c 443 H 433 H 433 g 433 NR NR g a c 444

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