Renormalization according to Wilson

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1 Renormalization according to Wilson Suppose we have integrated out fields with momenta > Λ. We have renormalized fields (at the scale Λ) and g(λ). Now we want to integrate out also fields with momenta between Λ and Λ (Λ Λ), and to obtain g(λ ). For example, if we consider interaction between a quark and an antiquark at a distance r, Λ 1/r.

2 Renormalization according to Wilson Suppose we have integrated out fields with momenta > Λ. We have renormalized fields (at the scale Λ) and g(λ). Now we want to integrate out also fields with momenta between Λ and Λ (Λ Λ), and to obtain g(λ ). For example, if we consider interaction between a quark and an antiquark at a distance r, Λ 1/r. Vacuum can be considered as a dielectric medium g 2 (Λ ) = g2 (Λ) ε where only fields with momenta between Λ and Λ contribute to ε. ε > 1 screening, ε < 1 antiscreening (asymptotic freedom).

3 Magnetic properties of vacuum Lorentz invariance: εµ = 1 µ > 1 (paramagnetic) asymptotic freedom

4 Magnetic properties of vacuum Lorentz invariance: εµ = 1 µ > 1 (paramagnetic) asymptotic freedom When we switch magnetic field B on E vac = ( µ 1 1 ) B 2 2 V

5 Magnetic properties of vacuum Lorentz invariance: εµ = 1 µ > 1 (paramagnetic) asymptotic freedom When we switch magnetic field B on We shall show E vac = ( µ 1 1 ) B 2 2 V E vac = β 0 g 2 (4π) µ = ε 1 = 1 + β 0 g 2 g 2 (Λ ) = [1 + β 0 g 2 Λ 2 V Λ2 B2 log 2 2 Λ2 log (4π) 2 Λ 2 Λ2 log (4π) 2 Λ 2 ] g 2 (Λ)

6 Vacuum energy First we consider electrodynamics.

7 Vacuum energy First we consider electrodynamics. Charged scalar field (particle + antiparticle) E vac = 2 i ω i 2 = i ω i

8 Vacuum energy First we consider electrodynamics. Charged scalar field (particle + antiparticle) E vac = 2 i ω i 2 = i ω i Charged fermion field energy of the Dirac sea E vac = i ω i

9 Vacuum energy First we consider electrodynamics. Charged scalar field (particle + antiparticle) E vac = 2 i ω i 2 = i ω i Charged fermion field energy of the Dirac sea E vac = i ω i Factor ( 1) 2s

10 Pauli paramagnetism Massless particle ω = k

11 Pauli paramagnetism Massless particle ω = k Switching magnetic field on g s giromagnetic ratio ω = k 2 g s s z eb

12 Pauli paramagnetism Massless particle ω = k Switching magnetic field on ω = k 2 g s s z eb g s giromagnetic ratio Massless particles only have s z = ±s V d E Pauli = ( 1) 2s 3 k [ k2 + g (2π) 3 s seb + ] k 2 g s seb 2k = ( 1) 2s V (g sseb) 2 d 3 k 1 4 (2π) 3 k = 3 ( 1)2s V (g sseb) 2 dk 8π 2 k = 2( 1) 2s (g s s) 2 e 2 Λ2 B2 log (4π) 2 Λ 2 2 V

13 Landau levels Massless charged scalar field L = (D µ ϕ) + D µ ϕ

14 Landau levels Massless charged scalar field L = (D µ ϕ) + D µ ϕ Magnetic field B along z: A y = Bx [ 2 e 2 B 2 x 2 2ieBx y + E2 ] ϕ = 0

15 Landau levels Massless charged scalar field L = (D µ ϕ) + D µ ϕ Magnetic field B along z: A y = Bx [ 2 e 2 B 2 x 2 2ieBx y + E2 ] ϕ = 0 ϕ = e i(kyy+kzz) ϕ(x)

16 Landau levels Massless charged scalar field L = (D µ ϕ) + D µ ϕ Magnetic field B along z: A y = Bx [ 2 e 2 B 2 x 2 2ieBx y + E2 ] ϕ = 0 ϕ = e i(kyy+kzz) ϕ(x) Oscillator [ 1 ] 2 2 x + ω2 2 2 x2 E n ψ n = 0 E n = ω ( ) n + 1 2

17 Landau levels Massless charged scalar field L = (D µ ϕ) + D µ ϕ Magnetic field B along z: A y = Bx [ 2 e 2 B 2 x 2 2ieBx y + E2 ] ϕ = 0 ϕ = e i(kyy+kzz) ϕ(x) Oscillator [ 1 ] 2 2 x + ω2 2 2 x2 E n ψ n = 0 E n = ω ( ) n E 2 = k 2 z + 2eB ( n ) ( ϕ = e i(kyy+kzz) ψ n x k ) y eb

18 Degeneracy Box V = L x L y L z k z = 2π L z n z dn z = L z dk z 2π

19 Degeneracy Box V = L x L y L z k y = 2π L y n y k z = 2π L z n z k y eb [0, L x] dn z = L z dk z 2π n y [ 0, eb L ] xl y 2π

20 Degeneracy Box V = L x L y L z k y = 2π L y n y Vacuum energy k z = 2π L z n z E vac = k y eb [0, L x] f ( ) n n=0 f(x) = ebv (2π) 2 + dn z = L z dk z 2π n y k 2 z + 2eBx dk z [ 0, eb L ] xl y 2π

21 Euler summation formula Smooth function (L 1) N f ( N+1 ) n f(x) dx Correction? n=0 0

22 N+1 0 f(x) dx = N 1/2 n=0 1/2 f ( n x) dx

23 N+1 0 = f(x) dx = N 1/2 n=0 1/2 N 1/2 n=0 1/2 f ( n x) dx [ f ( ) n f ( ] ) n x 2 + dx

24 N+1 0 = = f(x) dx = N 1/2 n=0 1/2 N f ( n n=0 N 1/2 n=0 1/2 f ( n x) dx [ f ( ) n f ( ] ) n x 2 + dx ) N f ( n + 2) 1 + n=0

25 N+1 0 = = = f(x) dx = N 1/2 n=0 1/2 N f ( n n=0 N 1/2 n=0 1/2 f ( n x) dx [ f ( ) n f ( ] ) n x 2 + dx ) N f ( ) n n=0 N f ( n + 2) 1 + n=0 N+1 0 f (x) dx +

26 N+1 0 = = = = f(x) dx = N 1/2 n=0 1/2 N f ( n n=0 n=0 N 1/2 n=0 1/2 f ( n x) dx [ f ( ) n f ( ] ) n x 2 + dx ) N f ( ) n N f ( n + 2) 1 + n=0 N+1 N f ( ) n f (x) n=0 0 f (x) dx + N+1 0 +

27 Euler summation formula N f ( N+1 ) n = f(x) dx 1 24 f (x) n=0 0 N+1 0 +

28 Landau diamagnetism E vac = 0 f(x) 1 24 f (x) The integral is the vacuum energy at B = 0 0

29 Landau diamagnetism E vac = 0 f(x) 1 24 f (x) The integral is the vacuum energy at B = 0 0 E Landau = 1 24 f (0) = e2 B 2 V 48π 2 dk k = 1 e 2 3 (4π) Λ 2 V Λ2 B2 log 2 2

30 Landau diamagnetism E vac = 0 f(x) 1 24 f (x) The integral is the vacuum energy at B = 0 0 E Landau = 1 24 f (0) = e2 B 2 V 48π 2 dk k = 1 e 2 3 (4π) Λ 2 V Λ2 B2 log 2 2 Full QED result e 2 E vac = β 0 β 0 = s (4π) Λ 2 V [ ( 1) 2s 2(g s s) 2 n ] s 3 Λ2 B2 log 2 2 n s the number of polarization states: n 0 = 1, n s 0 = 2

31 g 1/2 = 2 /Dψ = 0 D µ = µ iea µ /D 2 ψ = 0

32 g 1/2 = 2 /Dψ = 0 D µ = µ iea µ /D 2 ψ = 0 /D 2 = 2 e 2 A 2 2ieA µ µ ieγ µ γ ν µ A ν We suppose A = 0 /D 2 = D 2 ie 4 F µν[γ µ, γ ν ]

33 g 1/2 = 2 /Dψ = 0 D µ = µ iea µ /D 2 ψ = 0 /D 2 = 2 e 2 A 2 2ieA µ µ ieγ µ γ ν µ A ν We suppose A = 0 /D 2 = D 2 ie 4 F µν[γ µ, γ ν ] A µ = (0, 0, Bx 1, 0): F 12 = F 21 = B /D 2 = D 2 + iebγ 1 γ 2 = D 2 + 2eBs z

34 g 1/2 = 2 /Dψ = 0 D µ = µ iea µ /D 2 ψ = 0 /D 2 = 2 e 2 A 2 2ieA µ µ ieγ µ γ ν µ A ν We suppose A = 0 /D 2 = D 2 ie 4 F µν[γ µ, γ ν ] A µ = (0, 0, Bx 1, 0): F 12 = F 21 = B /D 2 = D 2 + iebγ 1 γ 2 = D 2 + 2eBs z [ 2 e 2 B 2 x 2 2ieBx y + 2eBs z + E 2 ] ψ = 0 E 2 = k 2 z + 2eB ( n + 1 2) 2eBsz g 1/2 = 2

35 Quark Chromomagnetic field along a 0 such that t a 0 is diagonal (for SU(3) a 0 = 8): the sum of squares of colour charges is Tr t a 0 t a 0 (no summation) Tr t a t b = T F δ ab

36 Quark Chromomagnetic field along a 0 such that t a 0 is diagonal (for SU(3) a 0 = 8): the sum of squares of colour charges is Tr t a 0 t a 0 (no summation) Tr t a t b = T F δ ab Contribution to β 0 (n f flavours) ( 1 1 3) 2TF n f

37 g 1 = 2 SU(2) Yang Mills equation D ν G a µν = ( ν δ ab + gε acb A cν) G b µν = 0 External field A 3 µ, linearize in A 1,2 µ.

38 g 1 = 2 SU(2) Yang Mills equation D ν G a µν = ( ν δ ab + gε acb A cν) G b µν = 0 External field A 3 µ, linearize in A 1,2 µ. For A µ = A 1 µ ia 2 µ we get D ν G µν + igg 3 µνa ν = 0 G µν = D µ A ν D ν A µ D µ = µ iga 3 µ

39 g 1 = 2 SU(2) Yang Mills equation D ν G a µν = ( ν δ ab + gε acb A cν) G b µν = 0 External field A 3 µ, linearize in A 1,2 µ. For A µ = A 1 µ ia 2 µ we get D ν G µν + igg 3 µνa ν = 0 G µν = D µ A ν D ν A µ D µ = µ iga 3 µ Gauge D µ A µ = 0 (we use [D µ, D ν ] = igg 3 µν) D 2 A µ 2igG 3 µνa ν = 0

40 g 1 = 2 SU(2) Yang Mills equation D ν G a µν = ( ν δ ab + gε acb A cν) G b µν = 0 External field A 3 µ, linearize in A 1,2 µ. For A µ = A 1 µ ia 2 µ we get D ν G µν + igg 3 µνa ν = 0 G µν = D µ A ν D ν A µ D µ = µ iga 3 µ Gauge D µ A µ = 0 (we use [D µ, D ν ] = igg 3 µν) D 2 A µ 2igG 3 µνa ν = 0 G 3 12 = G 3 21 = B, A 2 = is z A 1 (s z = ±1) [ D 2 + 2igBs z ] A 1 = 0

41 Gluon Gluons with colour a 1 such that t a 1 is diagonal don t interact with our chromomagnetic field (for SU(3) a 1 = 3). All other gluons can be arranged into pairs with positive and negative colour charges. The sum of their squares (both signs!) is C A : in the adjoint representation i.e. 2e 2 C A g 2 Tr t a t b = C A δ ab

42 Gluon Gluons with colour a 1 such that t a 1 is diagonal don t interact with our chromomagnetic field (for SU(3) a 1 = 3). All other gluons can be arranged into pairs with positive and negative colour charges. The sum of their squares (both signs!) is C A : in the adjoint representation i.e. 2e 2 C A g 2 Tr t a t b = C A δ ab β 0 = ( 4 1 3) CA ( 1 1 3) 2TF n f Pauli paramagnetism (g 1 1) 2 = 4 is stronger than Landau diamagnetism 1/3

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