Effective Field Theory and. the Nuclear Many-Body Problem

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Effective Field Theory and the Nuclear Many-Body Problem Thomas Schaefer North Carolina State University 1

Nuclear Effective Field Theory Low Energy Nucleons: Nucleons are point particles Interactions are local Long range part: pions Advantages: Systematically improvable Symmetries manifest (Chiral, gauge,...) Connection to lattice QCD 2

Effective Field Theory Effective field theory for pointlike, non-relativistic neutrons L eff = ψ ( i 0 + 2 2M ) ψ C 0 2 (ψ ψ) 2 + C [ 2 (ψψ) (ψ 2 ] ψ) + h.c. +... 16 Simplifications: neutrons only, no pions (very low energy) Scattering amplitude Effective range expansion A l = 1 p cot δ l ip p cot δ 0 = 1 a + 1 2 Λ2 n ( ) p 2 n+1 r n Λ 2 3

Low energy expansion (natural case) A = 4πa M Compare with EFT expansion [ ] 1 iap + (ar 0 /2 a 2 )p 2 + O(p 3 /Λ 3 ) Coupling constants C 0 = 4πa M C 2 = 4πa2 M r 2 4

Modified Expansion Coupling constants determined by nn interaction C 0 = 4πa M a = 18 fm C 2 = 4πa2 M r 2 r = 2.8 fm Problem: Large scattering length (ap) 1 p 10 MeV Need to sum (ap) to all orders. Small parameter Q (a 1, p,...) A 1 A 1 = 4π M 1 1 a + ip A 0 A 1 A 1 A 0 = 4π M r 0 p 2 /2 ( 1 a + ip)2 5

The Nuclear Matter Problem is Hard: Traditional View NN Potential has a very strong hard core 3-body forces, isobars, relativity,... important Saturation density too small 6

The Nuclear Matter Problem is Hard: EFT View NN Potential has a very strong hard core Short distance behavior not relevant 3-body forces, isobars, relativity,... important 3-body: Yes, Other Stuff: Hidden in Counterterms Saturation density too small Yes: NN system and nuclear matter (?) are fine tuned 7

Toy Problem (Neutron Matter) Consider limiting case ( Bertsch problem) Universal equation of state (k F a) (k F r) 0 E ( E ) A = ξ = ξ 3 ( k 2 F A 0 5 2M ) No Expansion Parameters! How to find ξ? Numerical Simulations Experiments with trapped fermions Analytic Approaches 8

Perfect Liquids sqgp (T=180 MeV) Neutron Matter (T=1 MeV) Trapped Atoms (T=5 pev) 9

Universality What do these systems have in common? dilute: rρ 1/3 1 a r k 1 F strongly correlated: aρ 1/3 1 Feshbach Resonance in 6 Li Neutron Matter 10

Low Density Expansion Finite density: L L µψ ψ Modified propagator G 0 (k) αβ = δ αβ ( θ(k k F ) k 0 k 2 /2M + iɛ + θ(k F k) ) k 0 k 2 /2M iɛ k 2 F 2M = µ Perturbative expansion E A = k2 F 2M [ 3 5 + ( 2 3π (k F a) + 4 35π 2 (11 2 log(2))(k F a) 2 ) ] +... 11

Low Density Expansion: Higher orders Effective range corrections Logarithmic terms E A = k2 F 2M 1 10π (k F a) 2 (k F r) E A = k2 F 2M (g 1)(g 2) 16 27π 3 (4π 3 3)(k F a) 4 log(k F a) related to log divergence in 3 3 scattering amplitude local counterterm D(ψ ψ) 3 exists if g 3 12

Lattice Calculation Free fermion action S free = [ ] e (m N µ)α t c i ( n)c i ( n + ˆ0) (1 6h)c i ( n)c i ( n) n,i h n,l s,i [ c i ( n)c i ( n + ˆl s ) + c i ( n)c i ( n ˆl ] s ) Contact interaction: Hubbard-Stratonovich [ ] exp Cα t a a a a ds ( = exp 1 2π 2 s2) [( exp s Cα + Cα ) ] t (a 2 a + a a ) Path Integral Tr exp [ β(h µn)] = DsDcDc exp [ S] 13

Lattice Fermions Introduce pseudo fermions: S = ψi Q ijψ j + V (s) Z = DsDφDφ exp [ S ], S = φ i Q 1 ij φ j + V (s) Hybrid Monte Carlo method C < 0 (attractive): det(q) 0 (4+1)-d Hamiltonian H(φ, s, p) = 1 2 p2 α + S (φ, s) Molecular Dynamics ṡ α = p α ṗ α = H s α Metropolis acc/rej P ([s α, p α ] [s α, p α]) = exp( H) 14

Continuum Limit Fix coupling constant at finite lattice spacing M 4πa = 1 C 0 + 1 2 Take lattice spacing b, b τ to zero p 1 E p µb τ 0 n 1/3 b 0 n 1/3 a = const Physical density fixed, lattice filling 0 Consider universal (unitary) limit n 1/3 a 15

Lattice Results 2 1.5 T = 4 MeV T = 3 MeV T = 2 MeV T = 1.5 MeV T = 1 MeV E/A*(3/5 E F ) -1 1 0.5 0.42 0.07 0 0 0.14 0.2 0.4 0.6 0.8 1 1.2 T/T F Canonical T = 0 calculation: ξ = 0.25(3) (D. Lee) Not extrapolated to zero lattice spacing 16

Green Function Monte Carlo 20 Pairing gap ( ) = 0.9 E FG E/E FG 10 odd N even N E = 0.44 N E FG 0 10 20 30 40 N Other lattice results: ξ = 0.42 (Bulgac et al.,umass) Experiment: ξ = 0.27 +0.12 0.09 [1], 0.51 ± 0.04 [2], 0.74 ± 0.07 [3] [1] Bartenstein et al., [2] Kinast et al., [3] Gehm et al. 17

Other Lattice Calculations Neutron matter with realistic interactions (pions) Sign problem returns; can be handled at T 0 Neutron matter with finite polarization Sign problem returns Nuclear Matter (neutrons and protons) No sign problem in SU(4) limit (Wigner symmetry) Need a three body force (can be handled with HS) Isospin asymmetry possible 18

Analytic Approaches: Ladder Diagrams Sum ladder diagrams at finite density (Brueckner theory) 4 3 pp ladders pp (f PP ->2) pp (f PP -> f PP ) pp (D -> 8 ) [ 3 E A = k2 F 2M 5 + 2(k F a)/(3π) 1 6 35π (11 2 log(2))(k F a) ] E/E 0 2 1 0-1 -1 0 1 2 1/k F a Independent of renormalization scale µ P DS Unitary Limit (k F a) : ξ = 0.32 19

Large N approximation: Ring Diagrams Consider N fermion species. Define x Nk F a/π [( E A = k2 F 3 2M 5 + 2x ) + 1 ( 3 3 N π H(x) 2x 3 + 4 )] (22 2 log(2))x2 35 5 4 3 ph ring (g = 2) ph ring (g-> 8 ) O((k F a) 2 ) E/(E F A) 2 1 0-1 -2-4 -2 0 2 4 C 0 ~k F a depends on PDS scale parameter µ P DS not suitable for (k F a) 20

Large d Limit In medium scattering strongly restricted by phase space P 2 + k P k 2 Find limit in which ladders are leading order λ [ ΩdC0k d 2 F M d(2π) d ] λ = const (d ) ξ = 1 2 + O(1/d) 21

Pairing in the Large d Limit BCS gap equation = C 0 2 d d p (2π) d ɛ 2 p + 2 Solution = 2e γ E F d exp ( 1 ) ( 1 + O dλ ( )) 1 d = 0.375E F Pairing Energy E A = d 4 E F ( E F ) 2 1 d. 22

Shallow Bound States For Arbitary d Upper and lower critical dimension (Nussinov & Nussinov) d = 2 : Arbitrarily weak attractive potential has a bound state ξ(d = 2) = 1 d = 4 : Bound state wave function ψ 1/r d 4. Pairs do not overlap ξ(d = 4) = 0 Conclude ξ(d = 3) 1/2? Try expansion around d = 4? 23

Epsilon Expansion Effective lagrangian for atoms Ψ = (ψ, ψ ) and dimers φ L = Ψ ( i 0 + σ 3 2 2m ) Ψ + µψ σ 3 Ψ 1 c 0 φ φ + Ψ σ + Ψφ + h.c. Nishida & Son (2006) Unitary limit c 0. Effective potential + + O(1) O(1) O(ɛ) ξ = 1 2 ɛ3/2 + 1 16 ɛ5/2 ln ɛ 0.0246 ɛ 5/2 +... = 0.475 (ɛ = 1) 24

Summary Numerical Approaches No sign problem, can compute EOS, T c,... Extensions: pions, range corrections, two flavors Analytical Approaches Large d: ξ = 1/2 + O(1/d). Higher orders? Epsilon expansion ξ = 0.47 + O(ɛ 7/2 ) Finite range corrections, pions, nuclear matter,... 25