2 - The photon propagator Path integral formalism - reminder staonary phase approximaon • Classical limit ! → 0 Euler-Lagrange classical field equaon Path integral formalism • Generang Func7onal Klein-Gordon propagator ⎡ i d 4 k ⎤ Z 0 0 | 0 D exp ! k k 2 m2 ! k 1 (no interactions) 0 ( ) = J =0 = χ ⎢− 4 χ ( )( + )χ (− )⎥ = ∫ ⎢ 2 ∫ (2π ) ⎥ ⎣ ⎦ Feynman propagator Photon propagator Path integral formalism ! 4 −ik.x 1 4 ik.x ! Aµ (k) = d x e A (x), A (x) = d k e Aµ (k) ∫ µ µ 2 4 ∫ ( π ) Photon propagator Path integral formalism ! 4 −ik.x 1 4 ik.x ! Aµ (k) = d x e Aµ (x), Aµ (x) = 4 d k e Aµ (k) ∫ 2π ∫ ( ) Projec7on matrix: 2 µν k P Pµν (k)P λ (k) = Pµλ (k) ( ν ) Photon propagator 2 µν µ ν 2 µν To complete the square need to invert k g − k k ≡ k P µν ...but P kν = 0, zero eigenvalue...not invertible Photon propagator 2 µν µ ν 2 µν To complete the square need to invert k g − k k ≡ k P µν ...but P kν = 0, zero eigenvalue...not invertible In fact component of Aµ ∝ kµ doesn't appear ... sufficient to integrate DAµ over transverse components, A! µ , only ... µ ! µ ! ... equivalent to k Aµ = 0... Lorentz gauge ∂ Aµ = 0 ! iden7ty matrix in subspace Aµ µν −1 P k 2Pµν (k) = ( ) k 2 − iε k 2 Pµν Pα J k ! ! µ α ( ) χ µ (k) = Aµ (k) + 2 k − iε k 2 Pµν Pα J k ! ! µ α ( ) χ µ (k) = Aµ (k) + 2 k − iε ⎡ 4 ⎤ ! i d k ! 2 µν ! Z0 (0) = 0 | 0 J =0 = Dχ exp ⎢− 4 χ µ (k)k P χν (−k)⎥ = 1 ∫ ⎢ 2 ∫ (2π ) ⎥ ⎣ ⎦ Photon propagator In the Lorenz gauge The term iε Fields must match onto free fields at t=±∞ iS f S i = dφ e (φ) f φ(t = +∞) φ(t = −∞) i (me ordering) ∫ ψ (q) = q n n H → (1− iε)H Lim q',t ' =ψ *(q') 0 etc. t'→−∞ 0 i.e. picks out ground state in ini7al and final states Fixing the gauge Want to be able to determine the propagator is various gauges … .. instruc7ve to see how it is done in canonical formalism Klein-Gordon propagator µ 2 Want to solve : (−∂µ∂ + m )ψ = −Vψ 4 Soluon : ψ (x) = φ(x) + d x'Δ (x'− x)V (x')ψ (x') ∫ F where (∂ ∂µ − m2 )φ x = 0 µ ( ) and (∂ ∂µ − m2 )Δ x'− x = δ 4 x'− x µ F ( ) ( ) Reminder: Klein-Gordon propagator µ 2 Want to solve : (−∂µ∂ + m )ψ = −Vψ 4 Soluon : ψ (x) = φ(x) + d x'Δ (x'− x)V (x')ψ (x') ∫ F where (∂ ∂µ − m2 )φ x = 0 µ ( ) and (∂ ∂µ − m2 )Δ x'− x = δ 4 x'− x µ F ( ) ( ) Solve for propagator in momentum space by taking Fourier transform 1 −ip.(x'−x) µ 2 4 1 −ip.(x'−x) 4 4 22 e (∂ ∂ +− m )Δ (x '− x)d (x '− x) = 22 e δ (x '− x)d (x '− x) - (22π ) ∫ µ F (22π ) ∫ 2 2 ! 1 −( p + m )ΔF ( p) = 2 (2π ) −ip.x ! 1 1 1 4 1 Δ F ( p) = − 2 2 2 , Δ (x) = − 4 d p e 2 2 (2π ) p + m −iε F (2π ) ∫ p + m −iε The photon propagator • The propagators determined by terms quadrac in the fields, using the Euler Lagrange equaons. ∂ F µν = ∂ ∂µ Aν − ∂ν (∂µ A ) = jν ≡ gνλ∂2 − ∂ν ∂λ A µ µ µ ( ) λ Gauge ambiguity A → A + ∂ α µ µ 2 µ µ µ ∂ Aµ → ∂ Aµ + ∂ α The photon propagator • The propagators determined by terms quadrac in the fields, using the Euler Lagrange equaons. Choose as F µν µ Aν ν ( µ A ) jν gνλ 2 ν λ A ∂µ = ∂µ∂ − ∂ ∂ µ = ≡ ∂ − ∂ ∂ λ 1 ( ) − ∂µ A ξ µ Gauge ambiguity (gauge fixing) A → A + ∂ α µ µ 2 µ µ µ ∂ Aµ → ∂ Aµ + ∂ α i.e. with suitable “gauge” choice of α (“ξ” gauge) want to solve µ ν 1 ν µ νλ 2 1 ν λ ν ∂µ∂ A − (1− )∂ (∂µ A ) ≡ (g ∂ − (1− )∂ ∂ )Aλ = j ξ ξ Now inver7ble … in momentum space the photon propagator is −1 ⎛ 1 ⎞ i ⎛ p p ⎞ i g µν p2 (1 ) pµ pν g (1 ) µ ν − ⎜ − − ⎟ = 2 ⎜ − µν + −ξ 2 ⎟ ⎝ ξ ⎠ p ⎝ p ⎠ The photon propagator • The propagators determined by terms quadrac in the fields, using the Euler Lagrange equaons. Choose as F µν µ Aν ν ( µ A ) jν gνλ 2 ν λ A ∂µ = ∂µ∂ − ∂ ∂ µ = ≡ ∂ − ∂ ∂ λ 1 ( ) − ∂µ A ξ µ Gauge ambiguity (gauge fixing) A → A + ∂ α µ µ 2 µ µ µ ∂ Aµ → ∂ Aµ + ∂ α i.e. with suitable “gauge” choice of α (“ξ” gauge) want to solve µ ν 1 ν µ νλ 2 1 ν λ ν ∂µ∂ A − (1− )∂ (∂µ A ) ≡ (g ∂ − (1− )∂ ∂ )Aλ = j ξ ξ We have used choice of a specific gauge transformaon to modify the equaon of mo7on. The ques7on is how do you modify the Lagrangian to get this equaon of mo7on? Will need to add a ``gauge fixing” term such that it is no longer gauge invariant and gives this equaon of mo7on….what is this term? Fixing the gauge – path integral formalism path integral Suppose under , Transformaons form a group, under g, g’, Redundant integraon Want to rewrite in the form with J independent of g I = dg J ∫ Volume of the group Fixing the gauge – path integral formalism path integral Suppose under , Transformaons form a group, under g, g’, Redundant integraon Want to rewrite in the form with independent of g I = dg J J (∫ ) so that remaining path integral has no redundant variables iS(x,y) iS(x2 + y2 ) c.f. I = dx dy e ≡ dx dy e ∫ ∫ I = dθ J = (2π ) J where J = dr r eiS (r ) (∫ ) ∫ volume of group of rotaons in 2 dimensions Faddeev, Popov gauge fixing will be gauge fixing term Define Faddeev Popov determinant Dg '' = Dg i.e. Δ(A) = Δ(A ) g Faddeev, Popov gauge fixing will be gauge fixing term Define Faddeev Popov determinant Dg '' = Dg i.e. (A) (A ) iS ( A) Δ = Δ g I ≡ DA e ∫ No redundant variables changing and no7ng are invariant A → A −1 DA, S( A), Δ( A) g Fixing the electromagne7c gauge Λ ≡ g A = A − ∂ Λ g µ µ Choose f ( A) = ∂ Aµ −σ (x) µ Fixing the electromagne7c gauge Λ ≡ g A = A − ∂ Λ g µ µ Choose f ( A) = ∂ Aµ −σ (x) µ Δ( A)−1 = Dgδ ⎡ f ( A )⎤ = DΛδ (∂A− ∂2 Λ −σ ) " = " DΛδ (∂2 Λ) ∫ ⎣ g ⎦ ∫ ∫ i.e. Δ(A) is const …absorb in normalisaon Fixing the electromagne7c gauge Λ ≡ g A = A − ∂ Λ g µ µ Choose f ( A) = ∂ Aµ −σ (x) µ Δ( A)−1 = Dgδ ⎡ f ( A )⎤ = DΛδ (∂A− ∂2 Λ −σ ) " = " DΛδ (∂2 Λ) ∫ ⎣ g ⎦ ∫ ∫ i.e. Δ(A) is const …absorb in normalisaon I ≡ DA eiS ( A) ∫ Since I is independent of f, we can integrate I with an arbitrary func7onal of σ Rξ gauge 1 −1 µ ν i.e. Lgauge fixing = − ξ ∂ Aµ∂ Aν R gauge 2 ξ Propagator: −1 ⎛ 1 ⎞ i ⎛ p p ⎞ i g µν p2 (1 ) pµ pν g (1 ) µ ν − ⎜ − − ⎟ = 2 ⎜ − µν + −ξ 2 ⎟ ⎝ ξ ⎠ p ⎝ p ⎠ Propagator: −1 ⎛ 1 ⎞ i ⎛ p p ⎞ i g µν p2 (1 ) pµ pν g (1 ) µ ν − ⎜ − − ⎟ = 2 ⎜ − µν + −ξ 2 ⎟ ⎝ ξ ⎠ p ⎝ p ⎠ ξ = 1: ‘t Hoof - Feynman gauge ξ = 0 : Lorentz gauge .