Lagrangian formulation of the Klein Gordon equation L L d 3 x, L Klein Gordon field lagrangian density ( x) L = ( x) ( x) m2 ( x)† ( x) † } } T V L L 0 ( ) ( m2 ) 0 Manifestly Lorentz invariant Euler Lagrange equation Klein Gordon equation The Lagrangian and Feynman rules Associate with the various terms in the Lagrangian a set of propagators and vertex factors The propagators determined by terms quadratic in the fields, using the Euler Lagrange equations. The remaining terms in the Lagrangian are associated with interaction vertices. The Feynman vertex factor is just given by the coefficient of the corresponding term in iL e.g. ( ieA ) ( x) ( ieeA ) ( x) ieA ( * ( * ) ) † k, iL ie( pi p f ) A * } p p' New symmetries L = ( x) ( x) m2 ( x)† ( x) † Is invariant under ( x) ei ( x) …an Abelian (U(1)) gauge symmetry A symmetry implies a conserved current and charge. e.g. Translation Rotation Momentum conservation Angular momentum conservation What conservation law does the U(1) invariance imply? Noether current L = ( x) ( x) m2 ( x)† ( x) † Is invariant under ( x) ei ( x) i …an Abelian (U(1)) gauge symmetry i L L 0 L = ( ) ( † ) ( ) 0 (Euler lagrange eqs.) L L L i i ( † ) ( ) ( ) j 0, ie L L † j † 2 ( ) ( ) Noether current The Klein Gordon current L = ( x) ( x) m2 ( x)† ( x) † Is invariant under ( x) ei ( x) j 0, …an Abelian (U(1)) gauge symmetry ie L L † j † 2 ( ) ( ) j ie KG * * This is of the form of the electromagnetic current we used for the KG field c.f.our original interpretation as probability density … Physical interpretation of Quantum Mechanics i t 21m 2 0 Schrödinger equation (S.E.) t i ( S .E.) i (S .E.) * * = .j 0 continuity eq. j 2im ( * * ) 2 “probability density” “probability current” 2 t 2 2 m2 Klein Gordon equation i( * t t ) * Neip. x , 2 E N Negative probability? j i( * * ) j ( , j) 2 f p e ip . x 1 2 p0V Normalised free particle solutions The Klein Gordon current L = ( x) ( x) m2 ( x)† ( x) † Is invariant under ( x) ei ( x) j 0, …an Abelian (U(1)) gauge symmetry ie L L † j † 2 ( ) ( ) j ie KG * * This is of the form of the electromagnetic current we used for the KG field Q d 3 x j 0 is the associated conserved charge Suppose we have two fields with different U(1) charges : 1,2 ( x) eiQ 1,2 ( x) 1,2 L = 1 ( x) 1 ( x) m21 ( x)† 1 ( x) † 2 ( x) 2 ( x) m22 ( x)† 2 ( x) † ..no cross terms possible (corresponding to charge conservation) U(1) local gauge invariance and QED ( x) ei ( x )Q ( x) L = ( x) ( x) m2 ( x)† ( x) † not invariant due to derivatives ei ( x )Q ei ( x )Q iQei ( x )Q ( x) To obtain invariant Lagrangian look for a modified derivative transforming covariantly D ei ( x ) D Need to introduce a new vector field D iQA A A ( x) eiQ ( x ) ( x) D ei ( x ) D A A L = D ( x) D ( x) m2 ( x)† ( x) † Note : D iQA is invariant under local U(1) is equivalent to p p eA universal coupling of electromagnetism follows from local gauge invariance L = ( x) ( x) m2 ( x)† ( x) j A O(e2 ) † The electromagnetic Lagrangian 0 E1 E2 E3 F A A F F , A A LEM 14 F F j A M 2 A A E1 E2 0 B3 B3 0 B2 B1 E3 B2 B1 0 Forbidden by gauge invariance The Euler-Lagrange equations give Maxwell equations ! L L 0 A ( A ) F ` j .E , .B 0, B 0 t E B j t E The photon propagator The propagators determined by terms quadratic in the fields, using the Euler Lagrange equations. F A ( A ) j Choose as 1 A Gauge ambiguity A A (gauge fixing) A A 2 i.e. with suitable “gauge” choice of α (“ξ” gauge) want to solve 1 1 A (1 ) ( A ) ( g 2 (1 ) ) A j In momentum space 1 p p 2 1 i g p (1 ) p p g (1 ) p2 p2 (‘t Hooft Feynman gauge ξ=1)
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