Maxwell s equations. electric field charge density. current density

Transcription

1 Maxwell s equations based on S-54 Our next task is to find a quantum field theory description of spin-1 particles, e.g. photons. Classical electrodynamics is governed by Maxwell s equations: electric field charge density current density magnetic field can be solved by writing fields in terms of a scalar potential and a vector potential 299 The potentials uniquely determine the fields, but the fields do not uniquely determine the potentials, e.g. arbitrary function of spacetime result in the same electric and magnetic fields. gauge transformation (a change of potentials that does not change the fields) More elegant relativistic notation: four-vector potential, or gauge field the field strength in components: 300

2 The first two Maxwell s equations can be written as: charge-current density 4-vector taking the four-divergence: we find that the electromagnetic current is conserved: The last two Maxwell s equations can be written as: automatically satisfied! 301 The gauge transformation in four-vector notation: The field strength transforms as: = 0 (derivatives commute) the field strength is gauge invariant! Next we want to find an action that results in Maxwell s equations as the equations of motion; it should be Lorentz invariant, gauge invariant, parity and time-reversal invariant and no more than second order in derivatives; the only candidate is: we will treat the current as an external source 302

3 obviously gauge invariant In terms of the gauge field: total divergence total divergence equations of motion: equivalent to the first two Maxwell s equations! 303 Electrodynamics in Coulomb gauge based on S-55 Next step is to construct the hamiltonian and quantize the electromagnetic field... Which should we quantize? too much freedom due to gauge invariance There is no time derivative of and so this field has no conjugate momentum (and no dynamics). To eliminate the gauge freedom we choose a gauge, e.g. an example of a manifestly relativistic gauge is Lorentz gauge: Coulomb gauge 304

4 We can impose the Coulomb gauge by acting with a projection operator: the lagrangian in terms of scalar and vector potentials: in the momentum space it corresponds to multiplying by the matrix, that projects out the longitudinal component. (also known as transverse gauge) 305 integration by parts integration by parts 0 0 equation of motion integration by parts Poisson s equation unique solution: we get the lagrangian: 306

5 the equation of motion for a free field ( ): the general solution: massless Klein-Gordon equation polarization vectors (orthogonal to k) we can choose the polarization vectors to correspond to right- and left-handed circular polarizations: in general: 307 following the procedure used for a scalar field we can express the operators in terms of fields: to find the hamiltonian we start with the conjugate momenta: the hamiltonian density is then 308

6 we impose the canonical commutation relations: with the projection operator these correspond to the canonical commutation relations for creation and annihilation operators: (the same procedure as for the scalar field) creation and annihilation operators for photons with helicity +1 (right-circular polarization) and -1 (left-circular polarization) 309 now we can write the hamiltonian in terms of creation and annihilation operators: (the same procedure as for the scalar field) 2-times the zero-point energy of a scalar field this form of the hamiltonian of electrodynamics is used in calculations of atomic transition rates,... in particle physics the hamiltonian doesn t play a special role; we start with the lagrangian with specific interactions, calculate correlation functions, plug them into LSZ to get transition amplitudes

7 LSZ reduction for photons based on S-56 Next step is to get the LSZ formula for the photon. The derivation closely follows the scalar field case; the only difference is due to the presence of polarization vectors: For a scalar field we found that in order to obtain a transition amplitude we simply replace the creation and annihilation operators in the transition amplitude by: similarly, for an incoming and outgoing photon we simply replace: 311 the LSZ formula is then valid if the field is normalized according to the free field formulae: where a single photon state is normalized according to: and the renormalization of fields results in the Z-factors in the lagrangian: we will discuss this next semester

8 Now we want to calculate correlation functions (the derivation again closely follows the scalar field case). the propagator for a free field theory: correlation functions of more fields given in terms of propagators... Next we want to calculate the path integral for the free EM field: we will treat the current as an external source 313 In the Coulomb gauge we integrate over those field configurations that satisfy ; in addition the zero s component is not dynamical we can replace it by the solution of the equation of motion and for the rest of the path integral we will guess the result based on the result we got for a scalar field: propagator 314

9 we can make it look better: where and the Coulomb term is reproduced thanks to: 315 We can simplify the propagator further... Let s define: and as a unit vector in the direction: now we can replace: and thus we get: 316

10 this looks better but we can simplify the propagator further... the momentum can be replaced by the derivative with respect to. acting on the exponential, and then integrate by parts to obtain which vanishes. and we get: = We obtained a very simple formula for the photon propagator: Feynman gauge (it would still be in the Coulomb gauge if we had kept the terms proportional to momenta) 318

11 The path integral for photons based on S-57 We will discuss the path integral for photons and the photon propagator more carefully using the Lorentz gauge: as in the case of scalar field we Fourier-transform to the momentum space: we shift integration variables so that mixed terms disappear... Problem: the matrix has zero eigenvalue and cannot be inverted. 319 To see this, note: where is a projection matrix Since and so the only allowed eigenvalues are 0 and +1 it has one 0 and three +1 eigenvalues. 320

12 We can decompose the gauge field into components aligned along a set of linearly independent four-vectors, one of which is and then this component does not contribute to the quadratic term because and it doesn t even contribute to the linear term because and so there is no reason to integrate over it; we define the path integral as integral over the remaining three basis vector; these are given by which is equivalent to Lorentz gauge 321 Within the subspace orthogonal to the projection matrix is simply the identity matrix and the inverse is straightforward; thus we get: going back to the position space propagator in the Lorentz gauge (Landau gauge) we can again neglect the term with momenta because the current is conserved and we obtain the propagator in the Feynman gauge: 322

13 Quantum electrodynamics (QED) based on S-58 Quantum electrodynamics is a theory of photons interacting with the electrons and positrons of a Dirac field: Noether current of the lagrangian for a free Dirac field we want the current to be conserved and so we need to enlarge the gauge transformation also to the Dirac field: global symmetry is promoted into local symmetry of the lagrangian and so the current is conserved no matter if equations of motion are satisfied 323 We can write the QED lagrangian as: covariant derivative (the covariant derivative of a field transforms as the field itself) and so the lagrangian is manifestly gauge invariant! Proof: 324

14 We can also define the transformation rule for D: then Now we can express the field strength in terms of D s: as required. 325 Then we simply see: no derivatives act on exponentials the field strength is gauge invariant as we already knew 326

15 To get Feynman rules we follow the usual procedure of writing the interacting lagrangian as a function of functional derivatives,... We have to make more precise statement over which field configurations we integrate because now also the Dirac fields transform under the gauge transformation (next semester). Imposing we can write it as: sum of connected Feynman diagrams with sources! (no tadpoles) 327 Feynman rules to calculate : external lines: vertex and the rest of the diagram incoming electron outgoing electron incoming positron outgoing positron incoming photon outgoing photon 328

16 vertex the arrow for the photon can point both ways one arrow in and one out draw all topologically inequivalent diagrams for internal lines assign momenta so that momentum is conserved in each vertex (the four-momentum is flowing along the arrows) propagators for each internal photon for each internal fermion 329 spinor indices are contracted by starting at the end of the fermion line that has the arrow pointing away from the vertex, write or ; follow the fermion line, write factors associated with vertices and propagators and end up with spinors or. follow arrows backwards! The vector index on each vertex is contracted with the vector index on either the photon propagator or the photon polarization vector. assign proper relative signs to different diagrams draw all fermion lines horizontally with arrows from left to right; with left end points labeled in the same way for all diagrams; if the ordering of the labels on the right endpoints is an even (odd) permutation of an arbitrarily chosen ordering then the sign of that diagram is positive (negative). sum over all the diagrams and get additional rules for counterterms and loops 330

17 Scattering in QED Let s calculate the scattering amplitude for a simple process: based on S We follow the same procedure as before: and the amplitude squared is: averaging over the initial electron and positron spins we get: 332

18 we also want to sum over the final photon polarizations: in the Coulomb gauge we found the polarization sum to be: doesn t contribute: the scattering amplitude should be invariant under a gauge transformation and so we should have: in addition, and so we find: Ward identity 333 for the spin averaged/summed amplitude we get: 334

19 we can plug the result to the formulae for differential cross section