Fermion anticommutation relations

Fermion anticommutation relations {a α,a β } = a αa β + a β a α = δ α,β {a α,a β } = {a α,a β } =0 Easy to demonstrate Rewrite antsymmetric state α 1 α 2 α 3...α N = a α 1 α 2 α 3...α N = a α 1 a α 2 α 3...α N =... = a α 1 a α 2...a α N = a α i i Ensures Pauli principle α 1 α 2...α N = a α 1 a α 2...a α N = a α 2 a α 1...a α N = α 2 α 1...α N Occupation numbers n α1 =1,n α2 =0,n α3 =1, 0,..., 0,... = α 1 α 3

Examples? One-body operators in Fock space 1 particle in sp space F = α α α F β β β Operator completely determined by all α F β N N-particle space F N = F (1) + F (2) +... + F (N) = F (i) Action of F (i) on a product state F (i) α 1 α 2 α 3...α N ) = α 1 α 2... α i 1 { } β i β i F α i αi+1... α N β i = β i β i F α i α 1...α i 1 β i α i+1...α N ) i=1

One-body operators (continued) Matrix element β i F α i same for any particle (dummy variables) Then F N α 1 α 2 α 3...α N )=F(1) α 1 α 2... α N +... + α 1 α 2...F (N) α N = β 1 F α 1 β 1 α 2...α N )+... + β N F α N α 1 α 2...β N ) β 1 β N = N i=1 β i β i F α i α 1 α 2...α i 1 β i α i+1...α N ) Since F N is symmetric it commutes with the antisymmetrizer Thus N F N α 1 α 2 α 3...α N = i=1 β i β i F α i α 1 α 2...α i 1 β i α i+1...α N A

Fock-space one-body operator Consider Fock-space operator Note the ^ notation This operator accomplishes the same as F N for any N! Use ˆF = αβ α F β a αa β [ ˆF, a α i ] = α F β [a αa β,a α i ]= α F β (a αa β a α i a α i a αa β ) αβ αβ = αβ α F β a α(a β a α i + a α i a β )= αβ α F β a αδ β,αi = α α F α i a α = β i β i F α i a β i and apply ˆF α 1 α 2 α 3...α N = ˆFa α 1 a α 2...a α N = [ˆF, a α 1 ]a α 2...a α N + a α 1 ˆFa α2...a α N = [ˆF, a α 1 ]a α 2...a α N + a α 1 [ ˆF, a α 2 ]...a α N +... + a α 1 a α 2...[ ˆF, a α N ] = N β i F α i a α 1...a α i 1 a β i a α i+1...a α N i=1 β i = N β i F α i α 1...α i 1 β i α i+1...α N i=1 β i

Examples Density operator for N particles ρ N (r) = N δ(r r i ) i=1 Second-quantized form: choose { r,m s } basis In Fock space ˆρ(r) = m s,m s d 3 r 1 d 3 r 1 r 1 m s δ(r r op ) r 1m s a r 1 m s a r 1 m s = m s a rm s a rms Kinetic energy ˆT = αβ α T β a αa β = p 1 m 1 p2 op 2m p 2m 2 a p 1 m 1 a p2 m 2 p 1 m 1 p 2 m 2 = p2 1 2m a p 1 m 1 a p1 m 1 p 1 m 1

Consider ˆN = α a αa α More examples [ Determine ˆN, a αi] = α [ a α a α,a α i ] = a α i Therefore ˆN α 1...α N = N α 1...α N Change of basis a α = α = λ λ λ α = λ a λ λ α Can be done for any state in Fock space a α = λ λ α a λ Also a α = λ α λ a λ

Similar strategy N-particles Consider Two-body operators in Fock space V N = = V = αβ αβ)(αβ V γδ)(γδ γδ V (1, 2)+ V (1, 3)+ V (1, 4)+... + V (1,N)+ V (2, 3)+ V (2, 4)+... + V (2,N)+ V (3, 4)+... + V (3,N)+.... V (N 1,N) N V (i, j) = 1 N V (i, j) 2 i<j=1 Matrix elements do not depend on the selected pair i j V (i, j) α 1..α i..α j..α N )= β i β j (β i β j V α i α j ) α 1..α i 1 β i α i+1..α j 1 β j α j+1..α N ) (β i β j V α i α j ) identical for any pair as long as quantum numbers are the same, so V N α 1 α 2 α 3...α N )= N i<j=1 β i β j (β i β j V α i α j ) α 1...β i...β j...α N )

V N More on two-body operators Note: symmetric and therefore commutes with antisymmetrizer As a consequence N V N α 1 α 2 α 3...α N = (β i β j V α i α j ) α 1...β i...β j...α N i<j=1 β i β j Fock-space operator ˆV = 1 2 αβγδ (αβ V γδ)a αa β a δa γ accomplishes the same result for any particle number! Note ordering

Use [ ˆV, a α i ] = 1 2 Two-body operator (αβ V γδ)a αa β [a δa γ,a α i ] αβγδ =...... a αa β (a δa γ a α i a α i a δ a γ ) =...... a αa β (a δ(δ γ,αi a α i a γ ) a α i a δ a γ ) =...... a αa β (a δδ γ,αi δ δ,αi a γ ) = 1 (αβ V α i δ)a 2 αa β a δ 1 (αβ V γα i )a 2 αa β a γ αβδ = (αβ V α i δ)a αa β a δ = αβδ αβγ (β i β j V α i α i )a β i a β j a αi β i β j α i Note (αβ V γδ)=(βα V δγ) since V (i, j) =V (j, i)

Use to show Employ Often used with Check! ˆV = 1 4 Two-body operators ˆV α 1 α 2 α 3...α N = ˆVa α 1 a α 2...a α N = N a α 1...[ ˆV, a α i ]...a α N αβ V γδ a αa β a δa γ αβγδ = = i=1 N (β i β i V α i α i )a α 1...a β i a β a i α i a α i+1...a α N i=1 β i β i α i N N i=1 j>i β i β j (β i β j V α i α j )a α 1...a β i...a β j...a α N f(β j, α i )[a β j a αi,a α j ]= f(β j, α j )a β j β j β j α i αβ V γδ (αβ V γδ) (αβ V δγ) = αβ ˆV γδ

Most common operator Notation often used Hamiltonian Ĥ = ˆT + ˆV = α T β a αa β + 1 2 αβ ψ ms (r) a rms (αβ V γδ)a αa β a δa γ αβγδ Use and In this basis = 2 2m 2 δ(r r )δ m s,m s (r 1 m s1 r 2 m s2 V (r, r ) r 3 m s3 r 4 m s4 ) = δ(r 1 r 3 )δ(r 2 r 4 ) Ĥ = ms + 1 2 rm s T r m s = rm s p2 2m r m s msm s d 3 r ψ ms (r){ 2 d 3 r = i 2m rm s p r m s = 2 2m 2 rm s r m s 2m 2} ψ m s (r) δ m s 1,ms 3 δ m s 2,ms 4 V ( r 3 r 4 ) d 3 r ψ ms (r)ψ m s (r )V ( r r )ψ m s (r )ψ m s (r) second quantization

IPM for fermions in finite systems IPM = independent particle model Only consider Pauli principle Localized fermions Examples Hamiltonian many-body problem: with and Ĥ 0 = ˆT + Û Ĥ 1 = ˆV Û Suitably chosen auxiliary one-body potential Many-body problem can be solved for!! Also works with fixed external potential Ĥ = ˆT + ˆV = Ĥ0 + Ĥ1 Ĥ 0 U ext U Ĥ = ˆT + Ûext + ˆV = Ĥ0 + Ĥ1

Consider in the Use second quantization { λ } basis (discrete sums for simplicity) Ĥ 0 = λλ λ (T + U) λ a λ a λ = λλ ε λ δ λ,λ a λ a λ = λ ε λ a λ a λ All many-body eigenstates of with eigenvalue Ĥ 0 are of the form Φ N n = λ 1 λ 2...λ N = a λ 1 a λ 2...a λ N E N n = N i=1 ε λi

Employ [Ĥ0,a λ i ] = ε λi a λ i Explicitly and therefore Ĥ 0 λ 1 λ 2 λ 3...λ N = Ĥ0a λ 1 a λ 2...a λ N = [Ĥ0,a λ 1 ]a λ 2...a λ N + a λ 1 Ĥ 0 a λ 2...a λ N = [Ĥ0,a λ 1 ]a λ 2...a λ N + a λ 1 [Ĥ0,a λ 2 ]...a λ N +... + a λ 1 a λ 2...[Ĥ0,a λ N ] { N } = ε λi λ 1 λ 2 λ 3...λ N i=1 Corresponding many-body problem solved! Ground state Φ N 0 = λ i F a λ i Fermi sea F

208 Pb for example Empirical potential & sp energies Ĥ 0 a α 208 Pb g.s. = [ ε α + E( 208 Pb g.s. ) ] a α 208 Pb g.s. A+1: sp energies A-1: E A+1 n E A 0 directly from experiment Ĥ 0 a α 208 Pb g.s. = [ E( 208 Pb g.s. ) ε α ] aα 208 Pb g.s. also directly from E A 0 E A 1 n Shell filling for nuclei near stability follows empirical potential

Comparison with experiment Now how to explain this potential

Phenomenological! Calculated from Woods- Saxon plus spin-orbit Neutron levels as a function of A NucPhys

Closed-shells and angular momentum Atoms: consider one closed shell (argument the same for more) nlm l = lm s = 1 2,nlm l = lm s = 1 2,...nlm l = lm s = 1 2,nlm l = lm s = 1 2 Expect? Example: He (1s) 2 = 1 2 { 1s 1s ) 1s 1s )} Consider nuclear closed shell = (1s) 2 ; L =0S =0 Φ 0 = n(l 1 2)jm j = j, n(l 1 2)jm j = j 1,..., n(l 1 2)m j = j

Angular momentum and second quantization z-component of total angular momentum Ĵ z = nljm j z n l j m a nljm a n l j m nljm n l j m = nljm m a nljm a nljm Action on single closed shell Ĵ z nlj; m = j, j +1,..., j = m m a nljm a nljm nlj; m = j, j +1,..., = j Also So total angular momentum Closed shell atoms = { j m= j } m nlj; m = j, j +1,..., j = 0 nlj; m = j, j +1,..., j Ĵ ± nlj; m = j, j +1,..., j =0 J =0 L =0 S =0

Nucleon-nucleon interaction Shell structure in nuclei and lots more to be explained on the basis of how nucleons interact with each other in free space QCD Lattice calculations Effective field theory Exchange of lowest bosonic states Phenomenology Realistic NN interactions: describe NN scattering data up to pion production threshold plus deuteron properties Note: extra energy scale from confinement of nucleons

Isospin Shell closures for N and Z the same!! Also m n c 2 m p c 2 939.56 MeV vs. 938.27 MeV So strong interaction Hamiltonian (QCD) invariant for p n But weak and electromagnetic interactions are not Strong interaction dominates consequences Notation (for now) p α n α adds proton adds neutron Anticommutation relations All others 0 {p α,p β } = δ α,β {n α,n β } = δ α,β

Z proton & N neutron state Exchange all p with n Isospin α 1 α 2...α Z ; β 1 β 2...β N = p α 1 p α 2...p α Z n β 1 n β 2...n β N ˆT + = p αn α α and vice versa ˆT = α n αp α Expect [ĤS, ˆT ± ]=0 Consider ˆT 3 = 1 2 [ ˆT +, ˆT ]= 1 2 = 1 2 will also commute with αβ ( p α n α n β p β n β p βp ) αn α αβ ( p α p β δ α,β n β n αδ α,β ) = 1 2 H S ( p α p α n ) αn α α

Check Then operators obey the same algebra as Isospin so spectrum identical and simultaneously diagonal! proton neutron For this doublet and [ ˆT 3, ˆT ± ]=± ˆT ± ˆT 1 = 1 2 ˆT 2 ˆT 3 = 1 2i ( ˆT + + ˆT ) ( ˆT + ˆT ) J x,j y,j z Ĥ S, ˆT 2, ˆT 3 rm s p = rm s m t = 1 2 rm s n = rm s m t = 1 2 T 2 rm s m t = 1 2( 1 2 + 1) rm sm t T 3 rm s m t = m t rm s m t States with total isospin constructed as for angular momentum

Z=11, N=12 vs Z=12, N=11 Examples Triplet of nuclei