Gauge Freedom and Canonical Quantization Different Gauges in QED (In-)Equivalence of gauges The problem of equivalence of different gauges in external current QED Benedikt Wegener Foundations and Constructive Aspects of QFT Bergische Universit¨ at Wuppertal June 22, 2018 1 / 22
Motivation and Strategy Motivation: In classical ED: change of gauge has no influence on the experimental results. In QED this issue is more controversial.
Motivation and Strategy Motivation: In classical ED: change of gauge has no influence on the experimental results. In QED this issue is more controversial. Strategy: i) Physics Part: Gauge Freedom 1 Canonical Quantization 2 Maxwell Fields in different Gauges 3 ii) Math Part: Equivalence of Observables in different gauges 1
Gauge Freedom and Canonical Quantization Different Gauges in QED (In-)Equivalence of gauges Table of contents Gauge Freedom and Canonical Quantization 1 Gauge Freedom Canonical Quantization Different Gauges in QED 2 Gauge Freedom of Classical Electrodynamics Coulomb gauge Axial gauge (In-)Equivalence of gauges 3 Vanishing charge Non-vanishing charge 3 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Singular Systems Configuration space M , Lagrange function L : T M → C Legendre trafo ⇒ Hamiltonian: H ( q , p ) = � v i p i − L ( q , p ) i ( q i , v i ) �→ ( q i , p i := ∂ L ρ L : T M → T ∗ M , ∂ v i ) 4 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Singular Systems Configuration space M , Lagrange function L : T M → C Legendre trafo ⇒ Hamiltonian: H ( q , p ) = � v i p i − L ( q , p ) i ( q i , v i ) �→ ( q i , p i := ∂ L ρ L : T M → T ∗ M , ∂ v i ) Definition Lagrangian L is called singular if ρ L is not a local isomorphism: det( ∂ 2 L ∂ v i ∂ v j ) = 0 Problem : Hamiltonian depends linearly on some v a : H ( q i , p i , v a ) = ˜ H ( q i , p i ) − v a φ a ( q i , p i ) 4 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Singular Systems ! e.o.m. ⇒ { H , v a } = 0 ⇒ φ a = 0 With M ab := { φ a , φ b } : = 0 ⇒ d ! ! dt φ b = { φ b , ˜ H } + v a M ab φ b = 0 (1) 5 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Singular Systems ! e.o.m. ⇒ { H , v a } = 0 ⇒ φ a = 0 With M ab := { φ a , φ b } : = 0 ⇒ d ! ! dt φ b = { φ b , ˜ H } + v a M ab φ b = 0 (1) Two cases: H } � = 0, det ( M ) � = 0 ⇒ all v a are fixed by (1) { φ b , ˜ 1 H } = 0, some v a are fixed by (1) depending on rk ( M ) { φ b , ˜ 2 5 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Singular Systems ! e.o.m. ⇒ { H , v a } = 0 ⇒ φ a = 0 With M ab := { φ a , φ b } : = 0 ⇒ d ! ! dt φ b = { φ b , ˜ H } + v a M ab φ b = 0 (1) Two cases: H } � = 0, det ( M ) � = 0 ⇒ all v a are fixed by (1) { φ b , ˜ 1 H } = 0, some v a are fixed by (1) depending on rk ( M ) { φ b , ˜ 2 Definition A constraint φ α is called first class if { φ α , φ i } = 0 for every constraint function φ i , otherwise second class. 5 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Dirac Bracket Case 1: only 2 nd class constraints v a fixed: v a = − ( M − 1 ) ab { φ b , ˜ dt F = { F , ˜ H } ⇒ d H } D Definition Let F , G ∈ C ∞ ( M ). Their Dirac bracket is: { F , G } D := { F , G } − { F , φ a } ( M − 1 ) ab { φ b , G } D M phys ⊂ M with {· , ·}| M phys = {· , ·} D 6 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Gauge Freedom Case 2: 1 st class constraints φ α generate gauge transformations: δ ǫ F = ǫ α { F , φ α } { gauge orbits } ∼ = M phys Gauge fixing= intersecting each gauge orbit once ⇔ external constraints → no 1 st class cons. ⇒ Dirac bracket 1 1 Graphic from H.Itoyama, The Birth of String Theory , Progress in Experimental and Theoretical Physics, 2016 7 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Canonical Quantization Fix a Hilbert space H Any F ∈ C ∞ ( M phys ) mapped to a self adjoint operator ˆ F on H such that: { F , G } → 1 i � [ ˆ F , ˆ G ] 8 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Canonical Quantization Fix a Hilbert space H Any F ∈ C ∞ ( M phys ) mapped to a self adjoint operator ˆ F on H such that: { F , G } → 1 i � [ ˆ F , ˆ G ] Problem : {· , ·} not compatible with constraints ⇒ Solution : { F , G } D → 1 i � [ ˆ F , ˆ G ] 8 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Strategy of Canonical Quantization h : one-particle space, Bosonic Fock space: � E s ( h ⊗ n ) Γ s ( h ) = n ≥ 0 a ( f ) , a † ( f ) , f ∈ h : the usual annihilation and creation operators on Γ s 9 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom Different Gauges in QED Canonical Quantization (In-)Equivalence of gauges Strategy of Canonical Quantization h : one-particle space, Bosonic Fock space: � E s ( h ⊗ n ) Γ s ( h ) = n ≥ 0 a ( f ) , a † ( f ) , f ∈ h : the usual annihilation and creation operators on Γ s Choose H = Γ s ( h ) with h = L 2 ( R 3 ) ⊗ C 3 Find a classical representation of Dirac bracket in terms of a † a † m ( k ′ ) } D = − i δ nm δ (3) ( k − k ′ ) modes ˜ a n , ˜ n satisfying { ˜ a n ( k ) , ˜ a ( † ) → a ( † ) Quantization: ˜ n n 9 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom of Classical Electrodynamics Different Gauges in QED Coulomb gauge (In-)Equivalence of gauges Axial gauge Covariant Formulation of Maxwell Equations Vector field A ∈ Ω 1 ( Mink 4 ) and field strength tensor F µν = ∂ µ A ν − ∂ ν A µ � Current j ∈ S ( R 3 ) ⊗C 4 with charge Q = R 3 d 3 x j 0 ( x ) = � j 0 (0) Lagrange density: L = F µν F µν − j µ A µ δ L ⇒ π µ := δ∂ 0 A µ = F µ 0 = E µ ⇒ π 0 = F 00 = 0 → L is singular 10 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom of Classical Electrodynamics Different Gauges in QED Coulomb gauge (In-)Equivalence of gauges Axial gauge Gauge Freedom Two first class constraints π 0 = F 00 ≈ 0 ∇ · π + j 0 ≈ 0 (Gauss law) ⇒ two generators of gauge transformations: A 0 → A 0 + ξ A i → A i + ∂ i χ ⇒ Well knwon U (1) gauge freedom ⇒ 2 gauge conditions needed 11 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom of Classical Electrodynamics Different Gauges in QED Coulomb gauge (In-)Equivalence of gauges Axial gauge The Coulomb gauge Choose gauge conditions: (i) ∇ · A ≈ 0 , (ii) ∆ A 0 + j 0 ≈ 0 ⇒ Dirac bracket: � � δ ij − ∂ i ∂ j δ (3) ( x − y ) { A i ( x ) , π j ( y ) } D = ∆ 12 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom of Classical Electrodynamics Different Gauges in QED Coulomb gauge (In-)Equivalence of gauges Axial gauge The Coulomb gauge Choose gauge conditions: (i) ∇ · A ≈ 0 , (ii) ∆ A 0 + j 0 ≈ 0 ⇒ Dirac bracket: � � δ ij − ∂ i ∂ j δ (3) ( x − y ) { A i ( x ) , π j ( y ) } D = ∆ Let f ∈ S ( R 3 ) ⊗ R 3 , then: 1 2 P T (ˆ 1 2 P T (ˆ f )) + � k · ˆ f , � π C ( f ) = a (( ω f )) + a † (( ω 2 ) 2 ) j 0 � 2 � 2 � B C ( f ) = a ((2 ω ) − 1 1 curl ( f )) + a † ((2 ω ) curl ( f )) with P T : projection to h T := { g ∈ h ; k · ˆ g = 0 } 12 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom of Classical Electrodynamics Different Gauges in QED Coulomb gauge (In-)Equivalence of gauges Axial gauge The Axial gauge Choose gauge conditions (i) e · A ≈ 0 (ii) e · ( π − ∇ A 0 ) ≈ 0 ⇒ Dirac bracket: � � δ ij − e j ∂ i δ (3) ( x − y ) { A i ( x ) , π j ( y ) } D = e · ∇ Problem : e i ∂ j e ·∇ : S ( R 3 ) �→ L 2 ( R 3 ) 13 / 22
Gauge Freedom and Canonical Quantization Gauge Freedom of Classical Electrodynamics Different Gauges in QED Coulomb gauge (In-)Equivalence of gauges Axial gauge Smearing of the Axial gauge Extend phase space to have n copies of A → extends Gauge freedom to U (1) × · · · × U (1) Axial gauge fixing for each A i with gauge vector e i ∈ R 3 Dirac bracket: � � n � � � δ ij − 1 e i , j δ (3) ( x − y ) { A i ( x ) , π j ( y ) } D = ∂ i n e · ∇ i =1 14 / 22
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