By Peskin and Schroeder

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Note, however, that this transformation appears to be in the wrong direction compared to Eq. 2), where the transformed 60 Chapter 3 The Dirac Field eld was evaluated at 1 x. 1 we imagined that we transformed a pre-existing eld distribution that was measured by (x). Here, we are transforming the action of (x) in creating or destroying particles. These two ways of implementing the Lorentz transformation work in opposite directions. Notice, though, that the matrix acting on and the transformation of the coordinate x have the correct relative orientation, consistent with Eq.

Consider a boost along the 3-direction. First we should remind ourselves of what the boost does to the 4-momentum vector. In in nitesimal form, E m 0 1 p3 = 1 + 1 0 0 46 Chapter 3 The Dirac Field where is some in nitesimal parameter. For nite we must write E = exp m 0 1 3 1 0 p 0 m = cosh 10 01 + sinh 01 10 (3:48) 0 cosh = m m sinh : The parameter is called the rapidity. It is the quantity that is additive under successive boosts. Now apply the same boost to u(p). According to Eqs. 30), 3 pm u(p) = exp ; 12 0 ;0 3 3 = cosh( 12 ) 10 01 ; sinh( 21 ) 0 ;0 3 =2 ; 1; 3 + e =2 ; 1+ 3 e 2 2 = 0 0 hp 3; 1; 3 E +p 2 = @ hp ; 1+ 3 E + p3 2 e =2 ; 1+ 2 3 pm 0 ; 3 + e =2 1;2 p ; 3i 1 + E ; p3 1+2 p ; i A: + E ; p3 1; 3 pm (3:49) 2 The last line can be simpli ed to give p u(p) = ppp (3:50) where it is understood that in taking the square root of a matrix, we take the positive root of each eigenvalue.

1 x) = ( 1 ) (@ )( 1 x): Since the metric tensor g is Lorentz invariant, the matrices 1 obey the identity ( 1) ( 1) g = g : (3:4) Using this relation, we can compute the transformation law of the kinetic term of the Klein-Gordon Lagrangian: ; ; (@ (x))2 ! g @ 0 (x) @ 0 (x) = g ( 1 ) @ ( 1 ) @ ( 1 x) ; ; = g @ @ ( 1 x) = (@ )2 ( 1 x): Thus, the whole Lagrangian is simply transformed as a scalar: (3:5) L(x) ! L( 1 x): The action S , formed by integrating L over spacetime, is Lorentz invariant.