Introduction to Quantum Electrodynamics Peter …

1 Introduction to Quantum ElectrodynamicsPeter Pre snajderThese are lecture notes devoted to introductory chapters of QuantumElectrodynamics (QED). The notes consist of two chapters:1. The Dirac field and the relativistic invariance- The Lorentz transformations and relativistic fields- The Dirac equation and its solutions, polarization sums- Dirac field quantization, field energy and momentum, charge- Fermions, the Dirac field propagator2. Quantum Electrodynamics and Feynman rules- QED equations of motion, Gauss law, Coulomb gauge- Free transversal electromagnetic field and its quantization- The interaction picture and the perturbation theory- Self-interacting scalar field, Feynman rules- QED in Coulomb gauge, gauge invariance- The relativistic formalism and Feynman rulesThis two chapters should be followed by a part devoted to simple appli-cations of Feynman perturbation technique:3.

2 Elementary processes in QED- Scattering amplitudes and the differential cross-section- Kinematics of binary processes, decay rate of an unstable particle- The scatteringe e+ +, the square of the scattering amplitude- Unpolarized scattering and its differential cross-section- The scatteringe e , the square of the scattering amplitude- Crossing symmetry, Mandelstam variables, crossed channels- Compton scatteringe e 1- The polarization sum for photons - Ward identity- Klein-Nishina formula for the cross-section of unpolarizede scattering- The annihilatione e , crossing symmetry and cross-section1 The Dirac field and its relativistic Lorentz transformationsWe shall label the points of the Minkowski space-time as follows.

3 X= (x ) = (x0, x1, x2, x3) = (t, ~x),( )wheret=x0denotes time and~x= (x1, x2, x3) labels the space scalar product of two 4-vectorsx= (x ) ay= (y ) in Minkowskispace-time is given y =x y ,( )wherey = y , y = y .We lower or rise the indices with the help of the relativistic metric tensor( ) = diag(1, 1, 1, 1) or its inverse ( ) = diag(1, 1, 1, 1)): = = ,where is the Kronecker symbol defined by: = 1 for = and = 0for 6= . We adopt theEinstein summation convention: we sum over the2same repeated upper and lower indices, y =x0y0+x1y1+x2y2+ us consider the linear transformation which preserves the relativisticscalar y of any two 4-vectorsxandx:x 7 x , y 7 y.

4 ( )Let us rewrite the scalar product in matrix y, whereydenotes the column with 4 componentsy ,xTis a row with 4 componentsx and is 4 4 matrix with elements . Similarly the transformation law inmatrix notation can be written as follows:x7 x = xay7 y = ywhere is the 4 4 matrix with elements . The invariance of the scalarproducts induces a constraint on admissible matrices :xT y=x T y =xT T y T = .( )Here, Tis the transposed matrix of the matrix . Such matrices form a Liegroup, calledLorentz group.

5 The elements of the Lorentz group which canbe expressed in exponential form = exp ( i J) = (exp ( i J)) ( )The symbol in the exponent is a real number andJis 4 4 matrix satisfyingconditionJT + J= 0 JT= J .( )This condition is a direct consequence of ( ) and ( ). There are 6 inde-pendent 4 4 matricesJ =J satisfying ( ). Their matrix elementsare given as:(J ) =i( ).( )In this form we can freely rise and lower all indices simultaneously on any proper Lorentz transformation = exp ( iJ) the exponentJisgiven as a linear combinationJ=12 J = 0,1,2,3,( )specified by 6 real parameters =.

6 The matrixJ generate Lorentztransformations in ( )-plane in Minkowski space:3 Three generatorsJij,i, j= 1,2,3, generate rotations in 3-space (forits specification we need 3 parameters - 3 Euler angles , , ); Three generatorsJ0j,j= 1,2,3, generate boosts (the transformationto a system moving with a speed~vwith respect to the original referenceframe - this requires again 3 parameters).For infinitesimal Lorentz transformation, specified by infinitesimal pa-rameters we obtainx 7 (exp ( i2 J )) x =( i2(J ) +.)

7 X =x i2(J ) x +..x 7 x + x , = .In the last step we have used the explicit formula ( ).It can be easily shown that the matricesJ satisfy the commutationrelations for Lorentz group generators, defining relations of Lie algebraso(3,1):[J , J ] = i( J J + J J ).( )Finally, we point out that it holdsJ x =i( x x ).( )This formula is equivalent to the relationx 7 x , = exp ( i2 J )( )which tell us that 4 real numbersx= (x ), = 0,1,2,3,transform asrelativistic Relativistic scalar fields.

8 The relativistic scalar field (x) isdescribed by a (real or complex) function defined in all points of Minkowskispace-time:x7 (x). By definition, under Lorentz transformationx7 xthe field (x) is transforming in the following way: (x)7 T( ) (x) = ( 1x).( )4In ( ) 1denotes the inverse matrix of the matrix . The symbolT( ) represents the linear operator defined by the last equation. The as-signment 7 T( ) defines the Lorentz group representation because it copies the group product:T( 1)T( 2) =T( 1 2), T(1) = symbol1denotes the 4 4 unit matrix (corresponding to the unity ingroup) and the symbolIis the unit operator corresponding to the identitymap: (x)7 (x).

9 Under infinitesimal Lorentz transformationx 7 x + x the fieldtransforms as follows (x)7 ( 1x) = (x x )= (x) i2 (J )(x).Comparing the last expression with the Taylor expansion of the field on afirst line, we obtain the formula for the generator of Lorentz transformationsJ which acts on fields as a 1-st order differential operator:J = i(x x ), = ,( )where = x . It can be easily shown that the differential operatorsJ = J again satisfy the commutation relations ( ) for Lorentz relativistic fields.

10 Let us considern-component field (x) = 1(x).. n(x) with components a(x),a= 1, .. , n. We shall assume that under Lorentztransformationx xthe field components transform as follows: a(x)7 Sab( ) b( 1x) (T( ) (x))a.( )The mapping (x)7 T( ) (x) will generate the Lorentz group represen-tation:T( 1)T( 2) =T( 1 2), T(1) =I5exactly, when 7 Sba( ) will be the (n n)-matrix representation of theLorentz groupSab( 1)Sbc( 2) =Sac( 1 2), Sab(1) = : As an important example of multi-component field can servetherelativistic vector fieldV (x) which under Lorentz transformations mapsas follows:V (x)7 V ( 1x).