Beyond Schiff: Atomic EDMs from two-photon exchange Satoru Inoue (work with Wick Haxton & Michael Ramsey-Musolf) ACFI-FRIB Workshop, October 23, 2014
Outline Very short review of Schiff theorem Multipole expansion as an expansion in powers of R N /R A 2-photon exchange gives larger atomic EDM than Schiff moment by orders of magnitude - first, naive derivation Full derivation - the final answer is essentially the same Estimation of the nuclear matrix element using toy model - EDM from 2-photon exchange is ~ hundred times larger than EDM from Schiff moment
Schiff theorem Derivation relies on 3 assumptions: 1. Constituent particles are point-like 2. Non-relativistic dynamics 3. Only electrostatic interactions
Schiff theorem Derivation relies on 3 assumptions: Loopholes 1. Constituent particles are point-like Nuclear size - Schiff moment 2. Non-relativistic dynamics Relativistic electrons - paramagnetic systems 3. Only electrostatic interactions Electromagnetic currents - topic of this talk
Natural size of interactions Schiff theorem is the cancellation between these diagrams C 1 C 1 C 1 C 1 indicating that the photon couples to the nuclear EDM
Natural size of interactions Schiff theorem is the cancellation between these diagrams C 1 C 1 C 1 C 1 indicating that the photon couples to the nuclear EDM Electron-nucleus interaction here is V C1 = 4 Z d 3 x e(~x ) 3 x 2 Y 1 (ˆx) = 4 3 R N R 2 A C A 1 C N 1 Z d 3 yy N (~y )Y 1 (ŷ) lengths can be divided out
Natural size of interactions This comes from multipole expansion of Coulomb potential ZZ V Coul = d 3 xd 3 y e(~x ) N (~y ) ~x ~y Electron is usually outside the nucleus (x > y) 1 ~x ~y = X 4 y l xl (x y)+ (y x) 2l +1 xl+1 yl+1 lm = X apple 4 y l x l 2l +1 x l+1 + y l y l+1 x l+1 (y lm Green terms survive in limit of point-like nucleus Y lm(ˆx)y lm (ŷ) x) Y lm(ˆx)y lm (ŷ) l-th multipole interaction goes as 4 R A RN R A l
Natural size of interactions This comes from multipole expansion of Coulomb potential ZZ V Coul = d 3 xd 3 y e(~x ) N (~y ) ~x ~y Electron is usually outside the nucleus (x > y) 1 ~x ~y = X 4 y l xl (x y)+ (y x) 2l +1 xl+1 yl+1 lm = X apple 4 y l x l 2l +1 x l+1 + y l y l+1 x l+1 (y lm Penetration terms are the effects of nonzero nuclear size Y lm(ˆx)y lm (ŷ) x) Y lm(ˆx)y lm (ŷ) They are suppressed by nuclear vs atomic volume RN R A 3
Natural size of interactions Schiff moment contribution comes from penetration terms Schiff Schiff Schiff moment electron-nucleus potential looks like V Sch = 4 " 3 RN X RA 4 r 3 i (~x i )+ 3 ( ~x i ) # r i S ~ R A R A ~S i - nuclear Schiff moment operator with lengths divided out Volume factor - RN R A 2 times smaller than EDM potential
Breit interaction 1-γ exchange at LO in non-relativistic expansion ZZ " # V Breit = d 3 xd 3 e N 1 ~j e ~j N +(~j e ˆn)(~j N ˆn) y ~x ~y 2 ~x ~y Current-current interaction have their own multipoles Transverse magnetic - magnetic dipole, quadrupole, etc. Z l y h i Mlm N = d 3 y Y l (ŷ) ~j N (~y ) note the powers of y lm R N Transverse electric - E N lm = i p (l + 1)(2l + 1) Z d 3 y y R N l 1 h Y l 1 (ŷ) ~j N (~y ) i lm
Multipole interactions Dividing out length scales as we did for charge multipoles, we find the following natural sizes for multipole interactions V Cl,V Ml V El V Sch 4 RN R A RN 4 R A 4 R A R A l R A l 1 RN R A 3 Can we find an interaction that gives a larger EDM than Schiff? We need a PVTV effect, and as little suppression as possible
Symmetries PCTC PVTC PCTV PVTV C even odd M odd even E odd even MQMs violate P and T, but?? we have J=0 electronic ground state, so we can t use MQM Schiff really is the leading contribution for 1-γ exchange
Breit iterated PCTC PVTC PCTV PVTV C even odd M odd even E odd even We can go to 2-γ exchange E 2 E 1 + Combining E1 and E2 multipoles can give PVTV effect They can also be recoupled to total J=1, and can give EDM
Transverse electric multipoles Transverse electric multipoles have no static nuclear moment (Siegert s theorem, to be explained later) E 2 E 1 In order to compare the effects of E1-E2 combination with Schiff, we should compare using a nuclear energy denominator V e E1 E2 V E1 1 E 0 E n V E2
Naive comparison So the E1-E2 effective interaction has size V e E1 E2 V E1 1 E 0 E n V E2 4 R A 1 4 R N E N R A R A =(4 ) 2 R N E N R 3 A Compare this with Schiff moment interaction V Sch (4 ) R3 N R 4 A VE1 e E2 4 V Sch E N R N R A R N E1-E2 is enhanced compared to Schiff moment
Problems with iterating Breit 1. Crossed diagram is ignored If electrons are relativistic, this is not small 2. Derivation of Breit interaction takes the step i k 2 = Not correct for inelastic scattering, which we have k 2 0 i ~ k 2 i ~ k 2
Time-ordered perturbation theory Old-fashioned perturbation theory, solves BOTH problems 1. Start with a normal Feynman amplitude 2. TOPT rewrites it as sum over all time-orderings of vertices i 3. Propagators turn into energy denominators k 2 Denominators show which time-orderings are important Also allows direct connection to non-relativistic calculation (derivation in Sterman s QFT textbook)
Time-ordered perturbation theory Trivial example: scalar propagator p 1 p 2 k where q ~k = ~ k2 + m 2 i k 2 m 2 + i = 1 p 1 p 2 2 ~k = 1 2 ~k i i k 0 ~k + i k 0 + ~k i apple i p 0 1 (p 0 0 1 + ~k )+i + i p 0 2 (p 0 0 2 + ~k )+i
Time-ordered perturbation theory Trivial example: scalar propagator p 1 p 2 k i k 2 m 2 + i = 1 p 1 p 2 2 ~k = 1 2 ~k Those are energy denominators i i k 0 ~k + i k 0 + ~k i apple i p 0 1 (p 0 0 1 + ~k )+i + i p 0 2 (p 0 0 2 + ~k )+i corresponding to the 2 possible time orderings of vertices
2-photon exchange in TOPT There are 4 orderings each for box & crossed diagrams Some orderings are highly suppressed by particle mass by nuclear excitation We want 1. massive particles traveling forward in time 2. only 1 intermediate state with nuclear excitations
2-photon exchange in TOPT Breit interaction comes from Iterating Breit corresponds to assuming these 4 are leading Only 1 of these is leading (1 nuclear excited state) Adjust answer by factor of 1/4
2-photon exchange in TOPT There are actually 6 leading diagrams - top middle is in Breit Left and right diagrams sum to middle ones (accident?) Crossed diagrams double the result - get back factor of 2x2 = 4
What did we learn? This shows that the naive estimate earlier is essentially correct, despite the incorrect use of the Breit interaction But we do want to make a better comparison to Schiff result 1. Siegert s theorem and giant dipole approximation 2. Toy model calculation to compare nuclear matrix elements
Siegert s theorem Transverse electric multipoles can be written using a gradient r Z " l l +1 E lm = ir N d 3 y r l ~ y Y lm# ~j N R N Partial integration makes ~r ~j N appear, which is equal to Final result is d N /dt E lm = by current conservation r l +1 R N [C N l lm,h N ] where H N is the nuclear Hamiltonian. This shows that these multipoles have no diagonal M.E. s
Siegert s theorem Excitation then de-excitation with transverse electric multipoles are simplified to an expression with charge multipoles X 1 h0 E 1 nihn E 2 0i E 0 E n n6=0 = p X 3R N 2 (E n E 0 )h0 C 1 nihn C 2 0i (schematically) n6=0 Since nuclear charge densities are much better known than current densities, this is progress.
Giant dipole approximation This expression resembles X (E n E 0 ) h0 C 1 ni 2 X (E n E 0 )h0 C 1 nihn C 2 0i n6=0 n6=0 which is known to be dominated by the giant dipole resonance. We make the ansatz that our expression is also GDR dominated X (E n E 0 )h0 C 1 nihn C 2 0i n6=0 E GDR h0 C 1 C 2 0i
Final expression With these approximations, our final expression is VE1 e E2 p 3 (4 ) 2 15 p 20 ZX RA i=1 E GDR R 3 N R 3 A We have a numerical factor, an electronic multipole, and x i 3 Y 1 (ˆx i ) C N 1,C N 2 1 a nuclear multipole, which is charge dipole and quadrupole recoupled to J=1.
Toy model calculation We compare this nuclear moment with Schiff in a toy model Shell model : nucleons of 15 N (J=1/2) in HO potential In order to get PVTV nuclear moments, we insert V CPV G p 2 2m ~ ~r core (~y) as a perturbation to the HO potential, so that h 0 Ô 0i = X n6=0 h0 Ô nihn V CPV 0i E 0 E n +c.c. is the final result.
Toy model calculation Schiff moment has overall 1/10 in the definition ~S = 1 Z d 3 y 2 ~y 5 y 2 dn ~ y 10 RN 3 3Z RN 3 N (~y ) so compare with 10 times S. Operators h0kôk0i (normalized) 10 S -10.557 C1 N,C2 N 1 11.931 Nuclear moments are the same order, as expected from the R N /R A argument.
Final comparison Comparing with V Sch = 4 10R A VE1 e E2 p 3 (4 ) 2 15 p 20 ZX RA RN R A 3 " R 4 A i=1 X i x i 3 Y 1 (ˆx i ) E GDR R 3 N R 3 A C N 1,C N 2 1 r i 3 (~x i )+ 3 ( ~x i ) r i # (10 ~ S) the ratio of the numerical factors is 4 p 15 E GDR R A 135 taking reasonable values of E GDR = 20 MeV, R A = 10 10 m
Conclusions R N /R A counting is a consistent way to estimate the sizes of EDM contributions 2-γ exchange between electrons and the nucleus allows transverse electric multipoles to generate atomic EDM, with less suppression than the Schiff moment term E1-E2 combination results in a nuclear operator similar to the Schiff moment, a charge dipole weighted by radius squared Enhancements for octupole deformed nuclei are still expected to apply because of this similarity
Backup slides
Multipoles
Multipoles Full expansion of Breit interaction V Cl = V Ml = V El = V E 0 l = 4 (2l + 1)R A 4 (2l + 1)R A 4 (2l + 1)R A 4 (2l + 1)R A RN RN R A RN R A RN R A R A l C A l l M A l l 1 E 0 A l l+1 E A l C N l M N l E N l E 0 N l There are 2 kinds of transverse electric terms, but one is larger by R N /R A counting
Multipoles List of electronic multipole operators Z l+1 Clm A d 3 RA x e (~x )Y lm(ˆx) x Z l+1 Mlm A d 3 RA x ~Y l x lm(ˆx) ~j e (~x ) Z apple 1 Elm A R l+2 A d 3 x ~r x ~ l+1 Ylm(ˆx) l ~j e (~x ) Z apple E 0 A lm RA l d 3 1 x ~r 2(2l 1)x ~ l 1 Ylm(ˆx) l E is the relevant transverse electric operator ~j e (~x )
Multipoles List of nuclear multipole operators Z l y Clm N d 3 y N (~y )Y lm(ŷ) M N lm Z E N lm 1 R l 1 N E 0 N lm 1 R l+1 N R N y d 3 y Z d 3 y Z R N d 3 y l ~Y l lm(ŷ) ~j N (~y ) h i ~r y l Y ~ l lm (ŷ) apple ~r E is the relevant transverse electric operator ~j N (~y ) y l+2 Y 2(2l + 3) ~ lm(ŷ) l ~j N (~y )
Time reversal
Time reversal Multipoles are either parity even or odd even: total charge (monopole), magnetic dipole, etc. odd: EDM, magnetic quadrupole, etc. Properties of operators under time reversal is more subtle, especially when we consider products of operators.
Time reversal The starting point is the relation ht T i = h i Notice that the bra and ket get reversed. Consider EDM in T conservation limit: hjj C N 10 jji = TC N 10 jji (T jji) = C N 10i 2j j, ji (i 2j j, ji) =hj, j C N 10 j, ji Wigner-Eckart relates these by a minus sign -> no EDM Another way to say this: C N 1 does not behave like spin under T
Time reversal For J=1 operators, TV moments come from operators that behave in the same way under T and complex conjugation Ô T ÔT 1 Ô C N 1m ( 1) m C N 1, m ( 1) m C N 1, m M N 1m ( 1) m M N 1, m ( 1) m+1 M N 1, m E N 1m ( 1) m E N 1, m ( 1) m+1 E N 1, m {E 1,E 2 } N 1m ( 1) m {E 1,E 2 } N 1, m ( 1) m {E 1,E 2 } N 1, m [E 1,E 2 ] N 1m ( 1) m [E 1,E 2 ] N 1, m ( 1) m+1 [E 1,E 2 ] N 1, m {E 1,M 1 } N 1m ( 1) m+1 {E 1,M 1 } N 1, m ( 1) m+1 {E 1,M 1 } N 1, m [E 1,M 1 ] N 1m ( 1) m+1 [E 1,M 1 ] N 1, m ( 1) m [E 1,M 1 ] N 1, m
Time reversal Multipoles are either parity even or odd even: total charge (monopole), magnetic dipole, etc. odd: EDM, magnetic quadrupole, etc. Properties of operators under time reversal is more subtle, especially when we consider products of operators.
Time-ordered perturbation theory
Time-ordered perturbation theory Steps to the derivation: 1. Write the energy-conserving delta functions at each vertex as (p 0 in p 0 out) = Z 1 1 dt e i(p0 in p0 out )t If an internal line, k, starts at vertex with time t i and ends at vertex with time t j, then there is a factor e ip0 k (t j t i )
Time-ordered perturbation theory 2. This factor, e ip0 k (t j t i ), lets us do the integration over p 0 k by closing the contour at infinity. Contour integral picks up a pole from the propagator of this line, and the pole corresponds to the on-shell energy of the particle. Result: Z d 4 p (2 ) 4 Any function of k 0 in the numerator is evaluated Z d 3 p (2 ) 3 2 ~k with on-shell energy, since that gives the residue
Time-ordered perturbation theory 3. We still have time integrals that we introduced at the start. The integrand is now e i(e in E out )t with on-shell energies. But whether the particle is in or out depends on the time ordering of the vertices. Split up the integral into regions, which correspond to orderings. Then we can perform the integrals fully. Result: an energy denominator for each intermediate state
Time-ordered perturbation theory Why is it sufficient to consider Justification comes from TOPT
Time-ordered perturbation theory These have a pure atomic excitation as the last intermediate state Compare with
Time-ordered perturbation theory for example This diagram has massive particles always traveling forward, and only 1 nuclear excitation. But atomic excited state always appears with a photon, which makes the denominator not so small