Finite-temperature Field Theory

Transcription

1 Finite-temperature Field Theory Aleksi Vuorinen CERN Initial Conditions in Heavy Ion Collisions Goa, India, September 2008

2 Outline Further tools for equilibrium thermodynamics Gauge symmetry Faddeev-Popov ghosts and gauge choices Response of system to small disturbation Example: Screening of static EM fields

3 Outline Further tools for equilibrium thermodynamics Gauge symmetry Faddeev-Popov ghosts and gauge choices Response of system to small disturbation Example: Screening of static EM fields

4 The thermal propagator and self energy At leading order, observe D 0 (ω n, p) = Natural generalization: 1 ω 2 n + p 2 + m 2 = β 2 ϕ n (p)ϕ n ( p) λ=0 D(ω n, p) = β 2 ϕ n (p)ϕ n ( p) D(τ 1, x 1 ; τ 2, x 2 ) = ϕ(τ 1, x 1 )ϕ(τ 2, x 2 ) Define scalar self energy Π as correction term to inverse propagator D(ω n, p) 1 = ωn 2 + p 2 + m 2 + Π(ω n, p) = D 1 0 (1 + D 0 Π)

5 The thermal propagator and self energy At leading order, observe D 0 (ω n, p) = Natural generalization: 1 ω 2 n + p 2 + m 2 = β 2 ϕ n (p)ϕ n ( p) λ=0 D(ω n, p) = β 2 ϕ n (p)ϕ n ( p) D(τ 1, x 1 ; τ 2, x 2 ) = ϕ(τ 1, x 1 )ϕ(τ 2, x 2 ) Define scalar self energy Π as correction term to inverse propagator D(ω n, p) 1 = ωn 2 + p 2 + m 2 + Π(ω n, p) = D 1 0 (1 + D 0 Π)

6 The thermal propagator and self energy At leading order, observe D 0 (ω n, p) = Natural generalization: 1 ω 2 n + p 2 + m 2 = β 2 ϕ n (p)ϕ n ( p) λ=0 D(ω n, p) = β 2 ϕ n (p)ϕ n ( p) D(τ 1, x 1 ; τ 2, x 2 ) = ϕ(τ 1, x 1 )ϕ(τ 2, x 2 ) Define scalar self energy Π as correction term to inverse propagator D(ω n, p) 1 = ωn 2 + p 2 + m 2 + Π(ω n, p) = D 1 0 (1 + D 0 Π)

7 Self energy contains information on how the interactions modify The masses and dispersion relations of quasiparticles The interaction potential (possible screening) Self energy obtainable through computation of all connected 1PI two-point graphs Exercise: Show that Π given by ( ) δ ln ZI Π = 2 δd 0 1PI

8 Self energy contains information on how the interactions modify The masses and dispersion relations of quasiparticles The interaction potential (possible screening) Self energy obtainable through computation of all connected 1PI two-point graphs Exercise: Show that Π given by ( ) δ ln ZI Π = 2 δd 0 1PI

9 Self energy contains information on how the interactions modify The masses and dispersion relations of quasiparticles The interaction potential (possible screening) Self energy obtainable through computation of all connected 1PI two-point graphs Exercise: Show that Π given by ( ) δ ln ZI Π = 2 δd 0 1PI

10 Introducing finite chemical potentials So far in all field theory examples µ = 0 Reason: No conserved charge associated with real scalar field Consider now Dirac fermions at chemical potential µ Ĥ Ĥ µ ˆN, ˆN = d 3 xψ ψ Fermion number conserved due to global U(1) symmetry ψ e iα ψ Action changes now to β { } ] S E = dτ d d x ψ γ 0 ( τ µ) iγ i i + m ψ 0

11 Introducing finite chemical potentials So far in all field theory examples µ = 0 Reason: No conserved charge associated with real scalar field Consider now Dirac fermions at chemical potential µ Ĥ Ĥ µ ˆN, ˆN = d 3 xψ ψ Fermion number conserved due to global U(1) symmetry ψ e iα ψ Action changes now to β { } ] S E = dτ d d x ψ γ 0 ( τ µ) iγ i i + m ψ 0

12 Introducing finite chemical potentials So far in all field theory examples µ = 0 Reason: No conserved charge associated with real scalar field Consider now Dirac fermions at chemical potential µ Ĥ Ĥ µ ˆN, ˆN = d 3 xψ ψ Fermion number conserved due to global U(1) symmetry ψ e iα ψ Action changes now to β { } ] S E = dτ d d x ψ γ 0 ( τ µ) iγ i i + m ψ 0

13 Conclusion: With finite chemical potentials, Matsubara frequencies shift by iµ ω n ω n + iµ = (2n + 1)πT + iµ Exercise: Try to repeat with complex scalar theory with global U(1) symmetry Obvious instability if µ > m! Result: Bose Einstein condensation at µ = m

14 Conclusion: With finite chemical potentials, Matsubara frequencies shift by iµ ω n ω n + iµ = (2n + 1)πT + iµ Exercise: Try to repeat with complex scalar theory with global U(1) symmetry Obvious instability if µ > m! Result: Bose Einstein condensation at µ = m

15 Conclusion: With finite chemical potentials, Matsubara frequencies shift by iµ ω n ω n + iµ = (2n + 1)πT + iµ Exercise: Try to repeat with complex scalar theory with global U(1) symmetry Obvious instability if µ > m! Result: Bose Einstein condensation at µ = m

16 How to compute sum-integrals? Perturbative calculations at T 0 require performing sum-integrals S = d 3 p (2π) 3 f (ω n, p), ω n ω n = 2nπT or (2n + 1)πT + iµ Two generic tricks for evaluating the sums: Contour integrals and 3d Fourier transforms Optimal choice depends on whether fields massive or massless Real time quantities result in additional twists here, always assume imaginary time formalism

17 How to compute sum-integrals? Perturbative calculations at T 0 require performing sum-integrals S = d 3 p (2π) 3 f (ω n, p), ω n ω n = 2nπT or (2n + 1)πT + iµ Two generic tricks for evaluating the sums: Contour integrals and 3d Fourier transforms Optimal choice depends on whether fields massive or massless Real time quantities result in additional twists here, always assume imaginary time formalism

18 Contour integral trick Generic observation: May convert Matsubara sum into a contour integral via Residue theorem T f (p 0 = i 2nπT ) = 1 ( p0 ) dp 0 f (p 0 ) coth 4πi 2T n= with C circulating poles of the coth function (p 0 = i 2nπT ) in a counterclockwise direction Separating from coth a piece that vanishes at T = 0: T f (p 0 = i 2nπT ) = n= i +ε 1 2πT i +ε C ] dp 0 [f { } 1 (p 0 ) + f ( p 0 ) e βp 0 1

19 Contour integral trick Generic observation: May convert Matsubara sum into a contour integral via Residue theorem T f (p 0 = i 2nπT ) = 1 ( p0 ) dp 0 f (p 0 ) coth 4πi 2T n= with C circulating poles of the coth function (p 0 = i 2nπT ) in a counterclockwise direction Separating from coth a piece that vanishes at T = 0: T f (p 0 = i 2nπT ) = n= i +ε 1 2πT i +ε C ] dp 0 [f { } 1 (p 0 ) + f ( p 0 ) e βp 0 1

20 Advantage: Performing contour integral trick for each loop momentum, separate vacuum (T = 0) contribution from each diagram T = 0 piece easy to evaluate with standard methods Finite-T piece obviously UV safe, and can (usually) be evaluated by closing integration contour on the R.S. of complex plane Problem: Hard to do analytic calculations at high order Exercise: Repeat the above for fermions!

21 Advantage: Performing contour integral trick for each loop momentum, separate vacuum (T = 0) contribution from each diagram T = 0 piece easy to evaluate with standard methods Finite-T piece obviously UV safe, and can (usually) be evaluated by closing integration contour on the R.S. of complex plane Problem: Hard to do analytic calculations at high order Exercise: Repeat the above for fermions!

22 Advantage: Performing contour integral trick for each loop momentum, separate vacuum (T = 0) contribution from each diagram T = 0 piece easy to evaluate with standard methods Finite-T piece obviously UV safe, and can (usually) be evaluated by closing integration contour on the R.S. of complex plane Problem: Hard to do analytic calculations at high order Exercise: Repeat the above for fermions!

23 3d Fourier transforms Assume now important simplification: All T = 0 masses zero. Then... d 3 q (2π) 3 d 3 q (2π) 3 e iq r q 2 + (2nπT ) 2 = e 2 n πt r, 4πT e iq r q 2 + ((2n + 1) πt iµ) 2 = e ( 2n+1 πt iµ sign(2n+1))r 4πT Performing now the 3d momentum integrations, end up with Simple Matsubara sums: Harmonic series (Hyper)trigonometric integrals in coordinate space Result: (Almost) analytic results up to 4-loop order!

24 3d Fourier transforms Assume now important simplification: All T = 0 masses zero. Then... d 3 q (2π) 3 d 3 q (2π) 3 e iq r q 2 + (2nπT ) 2 = e 2 n πt r, 4πT e iq r q 2 + ((2n + 1) πt iµ) 2 = e ( 2n+1 πt iµ sign(2n+1))r 4πT Performing now the 3d momentum integrations, end up with Simple Matsubara sums: Harmonic series (Hyper)trigonometric integrals in coordinate space Result: (Almost) analytic results up to 4-loop order!

25 3d Fourier transforms Assume now important simplification: All T = 0 masses zero. Then... d 3 q (2π) 3 d 3 q (2π) 3 e iq r q 2 + (2nπT ) 2 = e 2 n πt r, 4πT e iq r q 2 + ((2n + 1) πt iµ) 2 = e ( 2n+1 πt iµ sign(2n+1))r 4πT Performing now the 3d momentum integrations, end up with Simple Matsubara sums: Harmonic series (Hyper)trigonometric integrals in coordinate space Result: (Almost) analytic results up to 4-loop order!

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27 of the theory As always in quantum field theories, in order to obtain finite results from perturbative calculations, we must renormalize the theory Fields and parameters appearing in Lagrangian not physical, measurable quantities Need to define parameters with renormalization corrections: ϕ Zϕ 1/2 ϕ R Simplification: Finite temperature does not generate any new divergences T = 0 renormalization sufficient Reason: Exponential suppression of finite-t contributions to integrals T does not affect UV physics

28 of the theory As always in quantum field theories, in order to obtain finite results from perturbative calculations, we must renormalize the theory Fields and parameters appearing in Lagrangian not physical, measurable quantities Need to define parameters with renormalization corrections: ϕ Zϕ 1/2 ϕ R Simplification: Finite temperature does not generate any new divergences T = 0 renormalization sufficient Reason: Exponential suppression of finite-t contributions to integrals T does not affect UV physics

29 In practice, need to introduce energy scale (Λ 2ε from yesterday) at which renormalization is performed Physical results independent of Λ however, in practice useful to choose Λ T to minimize errors in finite-order calculations

30 Gauge symmetry Faddeev-Popov ghosts and gauge choices Outline Further tools for equilibrium thermodynamics Gauge symmetry Faddeev-Popov ghosts and gauge choices Response of system to small disturbation Example: Screening of static EM fields

31 Gauge symmetry Faddeev-Popov ghosts and gauge choices Constructing the partition function Consider SU(N) YM coupled to m = 0 fundam. fermions L QCD = 1 4 F a µνf a µν + ψ /Dψ, F a µν µ A a ν ν A a µ + gf abc A b µa c ν, D µ µ iga a µt a, Tr T a T b = 1 2 δab Easy to restrict to pure Yang-Mills or QED later Theory invariant under gauge transformation A µ A a µt a Ω 1 A µ Ω + i ( µ Ω 1) Ω, g ψ Ω 1 ψ, Ω = exp [ igt a α a]

32 Gauge symmetry Faddeev-Popov ghosts and gauge choices Constructing the partition function Consider SU(N) YM coupled to m = 0 fundam. fermions L QCD = 1 4 F a µνf a µν + ψ /Dψ, F a µν µ A a ν ν A a µ + gf abc A b µa c ν, D µ µ iga a µt a, Tr T a T b = 1 2 δab Easy to restrict to pure Yang-Mills or QED later Theory invariant under gauge transformation A µ A a µt a Ω 1 A µ Ω + i ( µ Ω 1) Ω, g ψ Ω 1 ψ, Ω = exp [ igt a α a]

33 Gauge symmetry Faddeev-Popov ghosts and gauge choices Obvious issue in evaluating Z the overcounting of degrees of freedom due to gauge symmetry Famous example: Free photons in QED give twice the usual black body pressure! How to restrict to physical Hilbert space? Usual choice: Temporal A 0 = 0 gauge Coordinates and momenta now A i and Π a i δl i = Ȧa i, Resulting Hamiltonian H temp = { 1 d 3 x 2 Πa i Πa i 1 } 4 F ij a ij a ψγ i D i ψ ψ 0 ψ

34 Gauge symmetry Faddeev-Popov ghosts and gauge choices Obvious issue in evaluating Z the overcounting of degrees of freedom due to gauge symmetry Famous example: Free photons in QED give twice the usual black body pressure! How to restrict to physical Hilbert space? Usual choice: Temporal A 0 = 0 gauge Coordinates and momenta now A i and Π a i δl i = Ȧa i, Resulting Hamiltonian H temp = { 1 d 3 x 2 Πa i Πa i 1 } 4 F ij a ij a ψγ i D i ψ ψ 0 ψ

35 Gauge symmetry Faddeev-Popov ghosts and gauge choices Obvious issue in evaluating Z the overcounting of degrees of freedom due to gauge symmetry Famous example: Free photons in QED give twice the usual black body pressure! How to restrict to physical Hilbert space? Usual choice: Temporal A 0 = 0 gauge Coordinates and momenta now A i and Π a i δl i = Ȧa i, Resulting Hamiltonian H temp = { 1 d 3 x 2 Πa i Πa i 1 } 4 F ij a ij a ψγ i D i ψ ψ 0 ψ

36 Gauge symmetry Faddeev-Popov ghosts and gauge choices Gauss law not part of Hamiltonian equations of motion Must include it separately Introduce into path integral projection operator onto the space of physical states P = DΛ exp [ iβ d 3 xλ a G a], Λ( )=0 G a i F a i0 + gf abc A b i F c i0 + T a ψ ψ Result: After renaming Λ A 0, obtain expected expression Z QCD = DA µ D ψdψ exp [ β dx 0 d 3 x ( L QCD ψ µψ )] Aµ per. ψ antip. Some gauge freedom still remaining invariance under transformations periodic in τ Locality of gauge group Still infinite overcounting 0

37 Gauge symmetry Faddeev-Popov ghosts and gauge choices Gauss law not part of Hamiltonian equations of motion Must include it separately Introduce into path integral projection operator onto the space of physical states P = DΛ exp [ iβ d 3 xλ a G a], Λ( )=0 G a i F a i0 + gf abc A b i F c i0 + T a ψ ψ Result: After renaming Λ A 0, obtain expected expression Z QCD = DA µ D ψdψ exp [ β dx 0 d 3 x ( L QCD ψ µψ )] Aµ per. ψ antip. Some gauge freedom still remaining invariance under transformations periodic in τ Locality of gauge group Still infinite overcounting 0

38 Gauge symmetry Faddeev-Popov ghosts and gauge choices Gauss law not part of Hamiltonian equations of motion Must include it separately Introduce into path integral projection operator onto the space of physical states P = DΛ exp [ iβ d 3 xλ a G a], Λ( )=0 G a i F a i0 + gf abc A b i F c i0 + T a ψ ψ Result: After renaming Λ A 0, obtain expected expression Z QCD = DA µ D ψdψ exp [ β dx 0 d 3 x ( L QCD ψ µψ )] Aµ per. ψ antip. Some gauge freedom still remaining invariance under transformations periodic in τ Locality of gauge group Still infinite overcounting 0

39 Gauge symmetry Faddeev-Popov ghosts and gauge choices Removing the residual gauge freedom Standard choice for fixing residual gauge freedom: Covariant gauge condition with f a undetermined F a [A] µ A a µ f a = 0, Insert now into the path integral 1 = 1 with [A] DΩ δ[f a [A Ω ]] Ω per. SU(N) And use gauge invariance of action to obtain Z QCD = Ω per. SU(N) DΩ A Ω µ per. ψ antip. DA Ω µ D ψdψ 1 [A Ω ]δ[f a [A Ω ]] exp [ S[A Ω ] ]

40 Gauge symmetry Faddeev-Popov ghosts and gauge choices Removing the residual gauge freedom Standard choice for fixing residual gauge freedom: Covariant gauge condition with f a undetermined F a [A] µ A a µ f a = 0, Insert now into the path integral 1 = 1 with [A] DΩ δ[f a [A Ω ]] Ω per. SU(N) And use gauge invariance of action to obtain Z QCD = Ω per. SU(N) DΩ A Ω µ per. ψ antip. DA Ω µ D ψdψ 1 [A Ω ]δ[f a [A Ω ]] exp [ S[A Ω ] ]

41 Gauge symmetry Faddeev-Popov ghosts and gauge choices Removing the residual gauge freedom Standard choice for fixing residual gauge freedom: Covariant gauge condition with f a undetermined F a [A] µ A a µ f a = 0, Insert now into the path integral 1 = 1 with [A] DΩ δ[f a [A Ω ]] Ω per. SU(N) And use gauge invariance of action to obtain Z QCD = Ω per. SU(N) DΩ A Ω µ per. ψ antip. DA Ω µ D ψdψ 1 [A Ω ]δ[f a [A Ω ]] exp [ S[A Ω ] ]

42 Gauge symmetry Faddeev-Popov ghosts and gauge choices Upon change of variables A Ω A, the Ω-integral obviously factorizes Z QCD = DA µ D ψdψ 1 [A]δ[F a [A]] exp [ S[A] ] Aµ per. ψ antip. Finally, write gauge condition ( δf 1 a ) (x) F [A] = det δα b (x det M ab, ) a =0 { ) } M ab (x, y) = µ ( µ δ ab + gf abc A c µ δ(x y) in terms of anticommuting, but periodic ghost fields η, η: det M ab = [ β β D ηdη exp dx 0 d 3 x dy 0 η per. 0 0 ] d 3 y η a (x)m ab (x, y)η b (y)

43 Gauge symmetry Faddeev-Popov ghosts and gauge choices Upon change of variables A Ω A, the Ω-integral obviously factorizes Z QCD = DA µ D ψdψ 1 [A]δ[F a [A]] exp [ S[A] ] Aµ per. ψ antip. Finally, write gauge condition ( δf 1 a ) (x) F [A] = det δα b (x det M ab, ) a =0 { ) } M ab (x, y) = µ ( µ δ ab + gf abc A c µ δ(x y) in terms of anticommuting, but periodic ghost fields η, η: det M ab = [ β β D ηdη exp dx 0 d 3 x dy 0 η per. 0 0 ] d 3 y η a (x)m ab (x, y)η b (y)

44 Gauge symmetry Faddeev-Popov ghosts and gauge choices Multiplying the functional integral by [ exp 1 2ξ β 0 dx 0 d 3 x(f a (x)) 2 ] and integrating over f a, we obtain the final result [ β Z QCD = DA µ D ψdψd ηdη exp dx 0 d 3 x L eff ], Aµ per. ψ antip. L eff = L QCD + 1 2ξ ( µa a µ) 2 ψ µψ + η a ( 2 δ ab + gf abc A c µ µ ) η b 0 Feynman rules again obtained from T = 0 ones taking into account discreteness of p 0 Gluons and ghosts (despite anticommutativity!) periodic in τ (ω n = 2nπT ) Quarks antiperiodic (ωn = (2n + 1)πT )

45 Gauge symmetry Faddeev-Popov ghosts and gauge choices Multiplying the functional integral by [ exp 1 2ξ β 0 dx 0 d 3 x(f a (x)) 2 ] and integrating over f a, we obtain the final result [ β Z QCD = DA µ D ψdψd ηdη exp dx 0 d 3 x L eff ], Aµ per. ψ antip. L eff = L QCD + 1 2ξ ( µa a µ) 2 ψ µψ + η a ( 2 δ ab + gf abc A c µ µ ) η b 0 Feynman rules again obtained from T = 0 ones taking into account discreteness of p 0 Gluons and ghosts (despite anticommutativity!) periodic in τ (ω n = 2nπT ) Quarks antiperiodic (ωn = (2n + 1)πT )

46 Gauge symmetry Faddeev-Popov ghosts and gauge choices Note difference to T = 0: Even when ghosts decouple (Abelian theories), they still contribute to grand potential and other thermodynamic quantities!

47 Response of system to small disturbation Example: Screening of static EM fields Outline Further tools for equilibrium thermodynamics Gauge symmetry Faddeev-Popov ghosts and gauge choices Response of system to small disturbation Example: Screening of static EM fields

48 Response of system to small disturbation Example: Screening of static EM fields Linear response So far, only equilibrium systems considered What if system disturbed by small perturbation: Ĥ Ĥ0 + δĥ(t), with δĥ turned on at t = t 0? Goal: Want to compute shifts in expectation values Â(x, t) up to linear order in δĥ t [ ] δ Â(x, t) = dt Tr ˆρ t Â(x, t ) = i i t 0 t t 0 t t 0 [ ] dt Tr ˆρ [δĥ(t ), Â(x, t )] ] dt Tr [ˆρ [δĥ(t ), Â(x, t)] + O(δ 2 )

49 Response of system to small disturbation Example: Screening of static EM fields Linear response So far, only equilibrium systems considered What if system disturbed by small perturbation: Ĥ Ĥ0 + δĥ(t), with δĥ turned on at t = t 0? Goal: Want to compute shifts in expectation values Â(x, t) up to linear order in δĥ t [ ] δ Â(x, t) = dt Tr ˆρ t Â(x, t ) = i i t 0 t t 0 t t 0 [ ] dt Tr ˆρ [δĥ(t ), Â(x, t )] ] dt Tr [ˆρ [δĥ(t ), Â(x, t)] + O(δ 2 )

50 Response of system to small disturbation Example: Screening of static EM fields Consider scalar field coupled to external source J: δĥ(t ) = d 3 x J(x, t ) ˆϕ(x, t ) For change in ˆϕ, obtain integral of source coupled to retarded Green s function δ ˆϕ(x, t) = dt d 3 x J(x, t )D R (x, t; x, t ), [ ] D R (x, t; x, t ) Tr ˆρ [ ˆϕ(x, t), ˆϕ(x, t )] θ(t t ) or in Fourier space... δ ˆϕ(ω, p) = J(ω, p)d R (ω, p) Retarded G.F. obtained from usual thermal one through D R (ω, p) = D(iω n ω + iε, p)

51 Response of system to small disturbation Example: Screening of static EM fields Consider scalar field coupled to external source J: δĥ(t ) = d 3 x J(x, t ) ˆϕ(x, t ) For change in ˆϕ, obtain integral of source coupled to retarded Green s function δ ˆϕ(x, t) = dt d 3 x J(x, t )D R (x, t; x, t ), [ ] D R (x, t; x, t ) Tr ˆρ [ ˆϕ(x, t), ˆϕ(x, t )] θ(t t ) or in Fourier space... δ ˆϕ(ω, p) = J(ω, p)d R (ω, p) Retarded G.F. obtained from usual thermal one through D R (ω, p) = D(iω n ω + iε, p)

52 Response of system to small disturbation Example: Screening of static EM fields Consider scalar field coupled to external source J: δĥ(t ) = d 3 x J(x, t ) ˆϕ(x, t ) For change in ˆϕ, obtain integral of source coupled to retarded Green s function δ ˆϕ(x, t) = dt d 3 x J(x, t )D R (x, t; x, t ), [ ] D R (x, t; x, t ) Tr ˆρ [ ˆϕ(x, t), ˆϕ(x, t )] θ(t t ) or in Fourier space... δ ˆϕ(ω, p) = J(ω, p)d R (ω, p) Retarded G.F. obtained from usual thermal one through D R (ω, p) = D(iω n ω + iε, p)

53 Response of system to small disturbation Example: Screening of static EM fields Coupling of QED plasma to static electric field Consider coupling QED plasma to static classical background field E cl H = 1 { (E + E cl ) 2 + B 2} 2 = 1 { E 2 + B 2} + E E cl E2 cl H 0 + δh To first order in Ecl i, obtain now for the shift in E : ] δ E i (x, t) = i dt d 3 x E j cl [ˆρ (x, t) Tr [E i (x, t), E j (x, t )] θ(t t )

54 Response of system to small disturbation Example: Screening of static EM fields Coupling of QED plasma to static electric field Consider coupling QED plasma to static classical background field E cl H = 1 { (E + E cl ) 2 + B 2} 2 = 1 { E 2 + B 2} + E E cl E2 cl H 0 + δh To first order in Ecl i, obtain now for the shift in E : ] δ E i (x, t) = i dt d 3 x E j cl [ˆρ (x, t) Tr [E i (x, t), E j (x, t )] θ(t t )

55 Response of system to small disturbation Example: Screening of static EM fields Writing E i s in terms of derivatives of A µ and going to Fourier space... Enet(x) i d 3 p = eip x (2π) 3 p i p j E j cl (p)dr 00 (ω = 0, p) where we have assumed a covariant gauge. With rotational invariance (E cl p), finally obtain E net (p) = E cl(p) ɛ(p), ɛ(p) 1 + Π 00(ω = 0, p) p 2 In IR limit (p T ), obtain Π 00 (ω = 0, p) md 2 Potential between charges V (r) = e q1 q m D r 2 4πr screening!

56 Response of system to small disturbation Example: Screening of static EM fields Writing E i s in terms of derivatives of A µ and going to Fourier space... Enet(x) i d 3 p = eip x (2π) 3 p i p j E j cl (p)dr 00 (ω = 0, p) where we have assumed a covariant gauge. With rotational invariance (E cl p), finally obtain E net (p) = E cl(p) ɛ(p), ɛ(p) 1 + Π 00(ω = 0, p) p 2 In IR limit (p T ), obtain Π 00 (ω = 0, p) md 2 Potential between charges V (r) = e q1 q m D r 2 4πr screening!

57 Response of system to small disturbation Example: Screening of static EM fields Writing E i s in terms of derivatives of A µ and going to Fourier space... Enet(x) i d 3 p = eip x (2π) 3 p i p j E j cl (p)dr 00 (ω = 0, p) where we have assumed a covariant gauge. With rotational invariance (E cl p), finally obtain E net (p) = E cl(p) ɛ(p), ɛ(p) 1 + Π 00(ω = 0, p) p 2 In IR limit (p T ), obtain Π 00 (ω = 0, p) md 2 Potential between charges V (r) = e q1 q m D r 2 4πr screening!

58 Outline Further tools for equilibrium thermodynamics Gauge symmetry Faddeev-Popov ghosts and gauge choices Response of system to small disturbation Example: Screening of static EM fields

59 (so far) So far, we ve reviewed some basic formalism of FTFT s: Path integral formulation of partition function and other equilibrium thermodynamic quantities Bosons and fermions, zero and finite density Interacting theories at weak coupling: Feynman rules at finite T and evaluation of sum integrals Special issues with gauge symmetry Ghosts and gauge choices Response of system to small external perturbations Possible screening of charge due to interactions Next, we ll start applying this machinery to QCD...