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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. The first two Maxwell’s equations can be written as: Classical electrodynamics is governed by Maxwell’s equations: charge-current density 4-vector electric field charge density taking the four-divergence: current density magnetic field can be solved by writing fields in terms of a scalar potential and a vector potential we find that the electromagnetic current is conserved: The last two Maxwell’s equations can be written as: automatically satisfied! 299 301 The gauge transformation in four-vector notation: The potentials uniquely determine the fields, but the fields do not uniquely determine the potentials, e.g. arbitrary function of spacetime The field strength transforms as: result in the same electric and magnetic fields. More elegant relativistic notation: gauge transformation (a change of potentials that does not change the fields) four-vector potential, or gauge field the field strength = 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: in components: we will treat the current as an external source 300 302 We can impose the Coulomb gauge by acting with a projection operator: obviously gauge invariant in the momentum space it corresponds to multiplying by the matrix , that projects out the longitudinal component. (also known as transverse gauge) total divergence In terms of the gauge field: total divergence the lagrangian in terms of scalar and vector potentials: equations of motion: equivalent to the first two Maxwell’s equations! 303 305 Electrodynamics in Coulomb gauge based on S-55 Next step is to construct the hamiltonian and quantize the electromagnetic field ... integration by parts integration by parts 0 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. 0 equation of motion integration by parts Poisson’s equation unique solution: Coulomb gauge an example of a manifestly relativistic gauge is Lorentz gauge: 304 we get the lagrangian: 306 we impose the canonical commutation relations: the equation of motion for a free field ( ): with the projection operator massless Klein-Gordon equation the general solution: polarization vectors (orthogonal to k) these correspond to the canonical commutation relations for creation and annihilation operators: (the same procedure as for the scalar field) we can choose the polarization vectors to correspond to right- and left-handed circular polarizations: in general: creation and annihilation operators for photons with helicity +1 (right-circular polarization) and -1 (left-circular polarization) 307 309 now we can write the hamiltonian in terms of creation and annihilation operators: following the procedure used for a scalar field we can express the operators in terms of fields: (the same procedure as for the scalar field) to find the hamiltonian we start with the conjugate momenta: 2-times the zero-point energy of a scalar field the hamiltonian density is then 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 ... 308 310 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: Now we want to calculate correlation functions (the derivation again closely follows the scalar field case). the propagator for a free field theory: similarly, for an incoming and outgoing photon we simply replace: 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 311 the LSZ formula is then valid if the field is normalized according to the free field formulae: 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 where a single photon state is normalized according to: and for the rest of the path integral we will guess the result based on the result we got for a scalar field: and the renormalization of fields results in the Z-factors in the lagrangian: we will discuss this next semester... 312 propagator 314 we can make it look better: this looks better but we can simplify the propagator further... where 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 the Coulomb term is reproduced thanks to: and we get: =0 315 We obtained a very simple formula for the photon propagator: We can simplify the propagator further... Let’s define: and as a unit vector in the 317 direction: Feynman gauge (it would still be in the Coulomb gauge if we had kept the terms proportional to momenta) now we can replace: and thus we get: 316 318 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: 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 as in the case of scalar field we Fourier-transform to the momentum space: 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 we shift integration variables so that mixed terms disappear... which is equivalent to Problem: the matrix has zero eigenvalue and cannot be inverted. Lorentz gauge 319 321 Within the subspace orthogonal to the projection matrix is simply the identity matrix and the inverse is straightforward; thus we get: To see this, note: where is a projection matrix and so the only allowed eigenvalues are 0 and +1 going back to the position space Since 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: it has one 0 and three +1 eigenvalues. 320 322 Quantum electrodynamics (QED) based on S-58 Quantum electrodynamics is a theory of photons interacting with the electrons and positrons of a Dirac field: We can also define the transformation rule for D: then Noether current of the lagrangian for a free Dirac field as required. we want the current to be conserved and so we need to enlarge the gauge transformation also to the Dirac field: Now we can express the field strength in terms of D’s: global symmetry is promoted into local symmetry of the lagrangian and so the current is conserved no matter if equations of motion are satisfied 323 325 We can write the QED lagrangian as: Then we simply see: covariant derivative (the covariant derivative of a field transforms as the field itself) and so the lagrangian is manifestly gauge invariant! Proof: no derivatives act on exponentials 324 the field strength is gauge invariant as we already knew 326 To get Feynman rules we follow the usual procedure of writing the interacting lagrangian as a function of functional derivatives, ... the arrow for the photon can point both ways vertex 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). 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 Imposing we can write it as: for each internal photon for each internal fermion sum of connected Feynman diagrams with sources! (no tadpoles) 327 Feynman rules to calculate 329 : vertex and the rest of the diagram external lines: 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. incoming electron outgoing electron incoming positron 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). outgoing positron incoming photon sum over all the diagrams and get outgoing photon additional rules for counterterms and loops 328 330 Scattering in QED based on S-59 we also want to sum over the final photon polarizations: Let’s calculate the scattering amplitude for a simple process: 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: 331 We follow the same procedure as before: Ward identity 333 for the spin averaged/summed amplitude we get: and the amplitude squared is: averaging over the initial electron and positron spins we get: 332 334 we can plug the result to the formulae for differential cross section... 335