C.A. Bertulani, Texas A&M University-Commerce Lecture 3 Reactions in the continuum 1
Problem: continuum w.f. extends to infinity Matrix elements for EM multipole operators <j x l i> explode Solution: bunch states around discrete energy E i CB, Canto, NPA 539, 163 (1992) φ 0 = E 0,J 0 M 0 e ie 0 t /! φ jjm = e ie j t /! Γ( j E) E,JM de Γ i E ( ) Γ j E ( )de = δ ij Ex: histograms Γ j ( E) = 1, for (j 1)σ < E < jσ σ = 0, otherwise Or, some other sort of smooth, orthogonal functions: better 2
Application example: plane waves e iq r = 4π i l j ( l qr)y ( lm r ) * Y lm q lm ( ) r Elm = u l,e (r)y lm = 2µ " 2 E1/ 4 r ( ) π j ( l qr)y lm r ( ) ElmE'l'm' = δ ll' δ mm' δ E E' ( ) Histogram-like Γ i Smooth, orthogonal functions: better (no beats, faster convergence) Smooth function Γ i 3
Application example: dipole transitions I jl;j'l' = r 2 dr de Γ( j E) de' Γ ( * j' E' )u l,e ( r)r u ( l',e' r) I jl;j'l' =!2 µ l + l'+2 2 for histograms F jj' + δ l,l'+1 G j,j' + δ l+1,l' G j' j l l' =1 F jj' = dqγ j ( E)Γ j' E ( ) F jj' = dqγ j E ( ) d dq Γ j' ( E) E =!2 q 2 2µ E 0 =!2 η 2 2µ I 00;j1 = 2ησ π 3 / 4 E j 2! ( E 0 + E j ) 2 2µ 3 / 4 Continuum Bound state I jl;j'l' =! 2 1 µσ [ ] if j = j' 1 2 l + l'+1 ( ) j j 1 ( ) ( j+l j' l' ) 1 ( ) if j j' =1 2 j + j' 1 0 otherwise E j = ( j 1/2)σ 4
Application example: dipole transitions l = 0! l = 1 j! j + 1 j! j reorientation j! j keeping E j = 200 kev CB, Canto, NPA 539, 163 (1992) 5
From lecture 1: da k dt = i! U kj (t) = ψ r j a j (t)u kj (t) ψ k * U(t) ψ e ie n t /! ( ) = a n (t)ψ n ( r) n j d3 r e i ( E k E j )t /! n = j(e)lm (for bound and continuum states n) Amplitudes a n! occupation probabilities a n 2! wavefunctions ψ(r) (normalized if enough # n)! calculate observables 6
i! ψ t = ˆ H ψ H =!2 2µ 2 x 2 + 2 y 2 + 2 z 2 + V r ( ) + U( r,t) Bertsch, CB NPA 556, 136 (1993) PRC 49, 2839 (1994) ψ ijk (t + Δt) = e i H Δt / " ψ( x i,y j,z ) k y j ψ ijk = ψ( x i,y j,z k ) H unitary (no complex U) ψ ijk ( t + Δt) ijk 2 = ψ ijk ( t) ijk 2 x i 7
e i H Δt / " =1+ i H Δt " + i H Δt " ˆ S ψ j (t) = 2 +# Use Crank-Nicolson operator instead: Unitary and accurate to (Δt) 2 Include U: define ψ j ( t + Δt) = H Δt 1 i " + Δt S H Δt 1+ i " k U k ψ k (t) ψ j ( t) Often requires too many terms to preserve unitarity and is unstable ψ j ( t + Δt) = 1 i 1+ i H Δt " H Δt " Good to order (Δt) 2 and preserves unitarity if Δt small enough ψ j ( t) Problem: needs to invert matrix 1 H Δt 1+ i " ψ j t ( ) 8
Solution: use three-point second derivative ( ) /2(Δx) 2 Δ (2) = ψ j 1 2ψ j + ψ j+1 ( ) = Then involves only j, j-1 and j+1 ψ j t + Δt 1 iτ + Δ(2) Δt 2!τ V + Δt ˆ j S!τ 1 iτ Δ(2) + Δt ψ 2!τ V j t j 1 u j (t) = ψ H Δt j ( t) Instead of solve for ψ 1+ i j t " But u j not known! ( ) for neighboring points ( ) = 1+ i H Δt " u j (t) Solution: using 3-point formula! 3-diagonal matrix ψ 1 O 11 O 12 0 0 0 0 u 1 ψ 2 O 21 O 22 O 23 0 0 0 u 2 = 0 O 32 O 33 O 34 0 0 0 0 O 43 O 44 O 44 0 x ψ N u N 9
First operation: ψ 1 = O 11 u 1 + O 12 u 2! assume u 1 = f 1 hold fixed by a boundary condition! f 1 known, u 1, know! u 2 determined Second operation: operation involves ψ 2, u 1, u 2 and u 3! u 1 and u 2 (from previous) ψ 2 known! u 2 determined And so on:! given ψ j on lattice at t! u j obtained. Finally, ( ) = 1+ i ψ j t + Δt H Δt " u j (t) is just a matrix multiplication! One-particle in three-dimensions straightforward! Extension to many-body: e.g., Time Dependent Hartree-Fock (TDHF) 10
b V 0 CB, PRL 94, 072701 (2005) ( ) = S α ( b,z) e ik αz φ α ( r) Ψ R,r R = ( b,z), α z da k dt = i! j a j (t)u kj (t) ( )t /! a α S α with vt z e i E k E j iv z S α f α ( b,z) = U αβ ( b,z)s β ( b,z) e i ( k β k α ) z ( Q) = ik Q = K ' β ( ) δ α,0 [ ] 2π db eiq.b S α b,z = K α = jljm U αβ ( R) = ψ α ( r)u( R,r)ψ β ( r) Eikonal CDCC 11
12
13
L = L LO + L NLO + L N2 LO +! L LO = 1 v 2 µc2 c L NLO = µ4 c 2 8 L N2 LO = µc2 512 1 8 v c 4 2 Z 1Z 2 e 2 r 1 3 m 1 3 1 m 2 v c v c 6 + Z 1Z 2 e 2 16r 3 v r c 4 4 + 2 v rv c Equations of motion r(t), p(t), Θ(b), µ2 Z 1 Z 2 e 2 2m 1 m 2 r 2 + Z 1Z 2 e µc 2 r dσ dω 2 v c 2 3 v r c + v r cr 2 2 v c 2 + 4Z 1 2 Z 2 2 e 4 µ 2 c 4 r 2 14
Θ ( E,b) Θ LO (%) N 2 LO Θ LO NLO Deviations from Rutherford N 2 LO NLO 15
dσ E,θ ( ) dσ LO dσ LO % ( ) NLO N 2 LO 16
dσ dω = dσ dω elast P exc = I( πλµ ) 2 M ( fi πλ, µ ) πλµ 2 M fi ( πλµ ) = f EMOperator(λµ)i I( πλµ ) = orbital integrals I(πλμ) from Lecture 1: ( ) = dt I ω 1 Y r λ +1 t λµ ( r ˆ (t))e iω t ( ) 1 r λ +1 t ( ) Y ˆ λµ ( r (t)) complicated functions of t Possible to include retardation, Lorentz contraction exactly Aleixo, CB, NPA 505, 448 (1989) 17
Deviations from non-relativistic 40 S (100 MeV/nucleon) + Au 18
W ( γ θ ) γ =1+ B κ Q κ E γ κ=2,4 ( ) P ( κ cosθ ) γ Exact/R Deviations from from relativistic theory 38 S (100 MeV/nucleon) + Au 19
20
meson exchange, two-nucleon interaction mean field approximation, U 0 (ω exchange), U S (2π exchange) [ E V C U 0 β( mc 2 + U )] S Ψ = i!cα Ψ non-relativistic reduction!2 2m 2 + U cent +! 2mc ( ) +" U cent = m* U 0 + U S 2 1 r d dr U SO σ L ϕ = Eϕ m* =1 U 0 U S 2mc 2 +" U SO = U 0 U S +" Arnold, Clark, PLB 84, 46 (1979) Dirac phenomenology 21
Long, CB, PRC 83, 024907 (2011). Γ φ Γ ω E = d 3 r ψ a a + 1 2 ( iγ + M)ψ a d 3 rd 3 r' ψ a (r)ψ b (r') Γ φ ( r,r' )D φ ( r r' )ψ a (r)ψ b (r') φ =σ,ω,ρ,γ ( r,r' ) = g σ (r)g σ (r') ( r,r' ) = g ω γ µ ( ) = g ρ γ µ! τ Γ ρ r,r' ab ( ) r ( g ω γ ) µ r'! ( ) ( g ρ γ µ τ ) r r' σ, ω, ρ and γ exchange Γ γ ( r,r' ) = e2 [ 4 γ µ (1 τ z )] r [ γ µ (1 τ z )] r' D φ = 1 4π D γ = 1 r r' e m φ r r' r r' Lorentz transform x p = x t + b, z p = γ(z t + Rcosθ) y p = y t 22
E(A t, A p, v) = E(A t ) + E(A p,v) + E(A t, A p, v) E(A t, A p, v) = d 3 r d 3 r' ψ t, a (r)ψ p, b (r') Γ φ r,r' φ =σ,ω,ρ,γ ab ( ) D φ ( r r' ) ψ t, a (r)ψ p, b (r') Ex: σ and ω contributions E σ = 1 γ d 3 r t d 3 r p ' g σ (r t ) ρ s, t (r t ) D σ ( r r' ) ρ s, p (r ' p )g σ (r ' p ) E ω = d 3 r t d 3 ' r p g ω (r t ) ρ b, t (r t ) D ω ( r r' ) ρ b, p (r ' p )g ω (r ' p ) ρ s (r) = ψ a (r)ψ a (r), ρ b (r) = ψ a (r)γ 0 ψ a (r) a a Projectile densities boosted to the target frame 23
24
25
And the imaginary part of U opt?! No easy solution. Here put by hand 26
27
Eikonal scattering wave amplitudes ψ Eikonal = φ i (r) S i (b,z)exp(ik i R) r K 0 y i x b R r O K i z S i (K i,r) K i = 2µ R (E ε i ) /!, Energy conservation " Boundary condition S i (b,z) z δ i,0 ΔS i (b,z) 0 i! 2 K i d µ R dz S (b) i (z) = i' F (b) ii' (z) S (b) i' (z)e i(k i ' K i )z f E i,0 = E f L L L 2π 2L +1 ik i 4π im Y Lm (Ω)[S b(l;i) i,0 δ i,0 ] 28
Form factor of non-rel. E-CDCC F (b) ( c'c z) i m m' = Φ c' U nucleon T + U core T Φ c e ( )φ = F (b)λ c'c Lorentz tranform of form factor and coordinates F (b)λ c'c ( z) f λ,m' m γ F (b)λ c'c ( γz) λ z ( ) Coul f λ,m' m 1/γ, λ =1, m' m = 0 = 1, otherwise ( ) γ, λ = 2, m' m = ±1 ( ) ( ) nucl f λ,m' m =1 Assumptions # Point charges for 1, 2 and A # Neglecting far-field (r i > R) contribution # Correction to nuclear form factor Ogata, CB, PTP 121 (2009), 1399 PTP, 123 (2010) 701 29
E x = 1.5 MeV θ = 0.06 o 30
all orders σ all σ NR σ all - σ no-nuclear 31
32