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Renormalization of Dirac's Polarized Vacuum

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Renormalization of Dirac's Polarized Vacuum Renormalization of Dirac’s polarized vacuum Mathieu LEWIN [email protected] (CNRS & Universit´ede Cergy-Pontoise) joint works with C. Hainzl (Tuebingen), P. Gravejat, E.´ S´er´e (Paris) & J.-P. Solovej (Copenhagen) Ecole´ Polytechnique, Dec. 7, 2010 Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 1 / 16 Dirac operator Energy of a free electron (p i ): ←→ − ∇ non relativistic relativistic classical mechanics E = p2/(2m) E 2 = c2p2 + m2c4 quantum mechanics H = ∆/(2m) D2 = c2∆+ m2c4 − − Free Dirac operator: 3 0 2 2 D = ic αk ∂k + βmc = icα~ + βmc − − ·∇ k X=1 α~ = (α , α , α ), β = 4 4 hermitian matrices, chosen such that 1 2 3 × (D0)2 = c2∆+ m2c4. − ◮ Unbounded from below: σ(D0) = ( ; mc2] [mc2;+ ). −∞ − ∪ ∞ Units: m = c = 1, α = e2. Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 2 / 16 A nonlinear Dirac equation Stationary equation including Vacuum Polarization effects: P = χ(−∞,µ) (D) 0 −1 (1) D = D + α(ρ ν) x + XP P−1/2 − ∗ | | α = e2 (bare) fine structure constant ν charge density of nucleus (ex: ν = Zδ0) µ chemical potential chosen to fix total charge (P−1/2)(x,y) α (Hartree-Fock Theory) − |x−y| XP = 0 (Reduced Hartree-Fock Theory) ∂Fxc ∂ρ (ρP−1/2) (Relativistic Density Functional Theory) ◮ Density ρQ (x) := trC4 (Q(x, x)) [Dir34] P.A.M. Dirac. Solvay report XXV, 1934. [EngDre87] Engel and Dreizler, Phys. Rev. A, 1987. Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 3 / 16 Interpretation P = χ (D) (−∞,µ) (1) D = D0 + α(ρ ν) x −1 + 0 P−1/2 − ∗ | | ◮ ν ≡ 0 solution P = P0 := χ (D0) (free vacuum), µ ( 1, 1). − (−∞,0) ∀ ∈ − 0 1 α~ ·p+β Reason: P = = ρ 0 0. − − 2 − 2√1+|p|2 ⇒ P−−1/2 ≡ Rmk: For HF the free vacuum is different [HaiLewSol07]. Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 4 / 16 Interpretation P = χ (D) (−∞,µ) (1) D = D0 + α(ρ ν) x −1 + 0 P−1/2 − ∗ | | ◮ ν ≡ 0 solution P = P0 := χ (D0) (free vacuum), µ ( 1, 1). − (−∞,0) ∀ ∈ − 0 1 α~ ·p+β Reason: P = = ρ 0 0. − − 2 − 2√1+|p|2 ⇒ P−−1/2 ≡ Rmk: For HF the free vacuum is different [HaiLewSol07]. ◮ ν =6 0 polarized vacuum polarized vacuum + electrons pol. vacuum pol. vacuum electrons P µ = 0 σ(D) P µ σ(D) Dirac-Fock eqs including Vacuum ≃ Polarization [ChaIra89] [HaiLewSol07] Hainzl, L. and Solovej. Comm. Pure Applied Math. (2007). [ChaIra89] Chaix and Iracane. J. Phys. B (1989). Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 4 / 16 Difficulties & Outline ◮ There is no solution to the nonlinear equation (1) when ν = 0! 6 need to add an ultraviolet cut-off Λ renormalization ◮ Possible to add classical SCF magnetic field ( photons), but no ≃ mathematical result so far. Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 5 / 16 Difficulties & Outline ◮ There is no solution to the nonlinear equation (1) when ν = 0! 6 need to add an ultraviolet cut-off Λ renormalization ◮ Possible to add classical SCF magnetic field ( photons), but no ≃ mathematical result so far. Outline Existence and non-existence of solutions Renormalization Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 5 / 16 Existence, nonperturbative regime Regularization: Π = orth. projector onto H = f L2(R3; C4) : Λ Λ { ∈ supp(f ) B(0; Λ) . Define := f : f (k) 2/ k 2 < . ⊂ } C { | | | | ∞} R b b Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 6 / 16 Existence, nonperturbative regime Regularization: Π = orth. projector onto H = f L2(R3; C4) : Λ Λ { ∈ supp(f ) B(0; Λ) . Define := f : f (k) 2/ k 2 < . ⊂ } C { | | | | ∞} R Theoremb (Existence [HaiLewSer05,GraLewSer09])b For all µ ( 1, 1), α 0, Λ > 0 and ν , there exists one solution to ∈ − ≥ ∈C P = χ(−∞,µ) (D) + δ 0 1 (2) D = Π D + α(ρ 0 ν) Π ( Λ P−P− − ∗ |x| Λ where 0 δ 1 (D), such that ≤ ≤ {µ} 0 0 0 0 2 P P S , P (P P )P S , ρ 0 L . − − ∈ 2 ± − − ± ∈ 1 P−P− ∈ ∩C ◮ 0 All such solutions share the same density ρP−P− . ◮ If απ1/6211/6 ν < 1 and µ = 0, then δ = 0 and P = P2 is unique. In || ||C this case tr P0 (P P0 )P0 + P0 (P P0 )P0 = 0. − − − − + − − + [HaiLewSer05] Hainzl, L. and S´er´e. Comm. Math. Phys. + J. Phys. A: Math & Gen. (2005). [GraLewSer09] Gravejat, L. and S´er´e. Comm. Math. Phys. (2009). Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 6 / 16 Idea of proof (µ = 0): variational method 0 ◮ Subtract infinite constant, using ρ 0 0 and Q = P P P−−1/2 ≡ − − “ ν (P 1/2) ν (P0 1/2)” = EHF − −EHF − − ρ (x)ν(y) α ρ (x)ρ (y) tr D0Q α Q dx dy + Q Q dx dy := ν (Q) − x y 2 x y EBDF ZZ | − | ZZ | − | ◮ Then [BacBarHelSie99], notice Q2 = Q++ Q−− and − tr D0Q = tr D0 (Q++ Q−−)= tr D0 Q2 | | − | | ◮ Minimize the convex fn ν on the convex hull [Lie81] EBDF Q = Q∗ : Q2 Q++ Q−− S { ≤ − ∈ 1} ∗ ±± 2 ◮ The map Q = Q S (H ) : Q S (H ) ρQ L is { ∈ 2 Λ ∈ 1 Λ } 7→ ∈ ∩C continuous ◮ For µ = 0, replace D0 by D0 µ 6 − [BacBarHelSie99] Bach, Barbaroux, Helffer and Siedentop. Comm. Math. Phys. (1999). [Lie81] Lieb. Phys. Rev. Lett. (1981). Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 7 / 16 Charge ◮ We define the charge of our solution as 0 0 0 0 tr 0 tr P− (Q) := P−QP− + P+QP+ 2 0 Rmk. if P = P, then tr 0 (P P ) Z, [AvrSeiSim94,HaiLewSer05] P− − − ∈ ◮ By convexity, ν ν choosing µ ! solving E (q) := inf (Q), tr 0 (Q)= q {EBDF P− } no minimizer ∃ a minimizer E ν (q) ν ∂E (q) µ = ∂q q qm q0 qM [AvrSeiSim94] Avron, Seiler and Simon. J. Funct. Anal. (1994). Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 8 / 16 Maximum ionization Theorem (Maximum Ionization [GraLewSer-09]) Assume ν L1 with Z := ν. ∈C∩ ◮ Then Z [qm; qM ] (existence of neutral atoms). ∈ R ◮ For α ν + α(1 + log Λ) C ′ and Z 0, one has || ||C ≤ ≥ α log Λ + 1/Λ+ α ν C C || || qm 0 − 1 Cα logΛ ≤ ≤ − and 2Z + C (α log Λ + 1/Λ+ α ν C) Z qM || || . ≤ ≤ 1 Cα logΛ − Generalizes a result of Lieb in the nonrelativistic setting [Lie84]. [Lie84] Lieb. Phys. Rev. A (1984). Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 9 / 16 Renormalization ◮ (Bare) parameters: charge e (α = e2) and mass m(= 1) Regularization: UV cut-off Λ predicted observables = functions (α, Λ) ⇒ O ◮ Predicted charge / coupling constant: αph(α, Λ) = α 6 ◮ Renormalization formulae: see α as fn of (αph, Λ) αph = αph(α, Λ) = α = α(αph, Λ) ⇒ predicted observables = functions (αph, Λ) ⇒ O ◮ Renormalization: ‘remove’ Λ from observable (αph, Λ) O (possibly only perturbatively in αph) Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 10 / 16 Charge renormalization Theorem (Charge renormalization [GraLewSer-09]) Assume that ν L1(R3) with Z = ν, Λ > 0, α 0 and µ ( 1, 1). ∈ ∩C ≥ ∈ − 1 3 Then ρ 0 L (R ) and it holds P−P− ∈ R 0 tr 0 Z P− (P P−) 0 − − Z ρP−P− = (3) − R3 1+ αB Z Λ where B = 2 logΛ 5 + 2 log 2 + O(1/Λ2). Λ 3π − 9π 3π ◮ Except in neutral case, solutions are singular: P P0 / S − − ∈ 1 ◮ For ν small, µ = 0, observe αphZ and not αZ, with || ||C α αph αph = α = 1+ αB ⇐⇒ 1 αphB Λ − Λ Pb (Landau pole): αphB < 1, hence Λ αph 0. Λ →∞⇒ → Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 11 / 16 Self-consistent equation ◮ Assume µ = 0 and ker(D)= 0 . Cauchy’s formula: { } 1 ∞ 1 1 P P0 = dη − − 2π D + iη − D0 + iη −∞ We get in Fourier space Z ρ 0 (k)= αB (k) ρ 0 (k) ν(k) + F α(ρ 0 ν) (4) P−P− − Λ P−P− − Λ P−P− − b whereb b b 1 (ℓ+k/2)·(ℓ−k/2)+1−E(ℓ+k/2)E(ℓ−k/2) BΛ(k)= π2|k|2 |ℓ+k/2|≤Λ, E(ℓ+k/2)E(ℓ−k/2)(E(ℓ+k/2)+E(ℓ−k/2)) dℓ − |ℓ−k/2|≤Λ R and FΛ(f )= n≥1 F2n+1,Λ(f , ..., f ) with k P 1 ∞ 1 1 1 Fk (f , ..., fk )= ρ Π fj Π dη ,Λ 1 2π D0 + iη Λ ∗ x Λ D0 + iη −∞ j Z Y=1 | | Mathieu LEWIN (CNRS / Cergy) Renormalization of Dirac’s vacuum X, Dec. 7, 2010 12 / 16 Renormalized equation ◮ Renormalized density: ρph = ρph(αph, Λ) defined for αphBΛ < 1 by α(ν ρ 0 )= αphρph − P−P− 0 −1 such that D = D αphρph − ∗|·| ◮ We can re-express Eq. (4) in terms of physical quantities as (1 αphU (k))ρph + F (αphρph)= ν − Λ Λ b where U (k)= B (0) B (k) 0b b Λ Λ − Λ ≥ Rmk. k 2 1 z2 z4/3 2 lim UΛ(k)= | | − dz := U(k) |k|→∞ log k Λ→∞ 4π 1+ k 2(1 z2)/4 ∼ 3π | | Z0 | | − [HaiLewSer05b] C.
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