From Special Relativity to Feynman Diagrams.pdf

# P e 2 i e p t d 3 p 2 π 3 v e p a p a p b p b p 1138

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p ) e 2 i E p t = d 3 p ( 2 π ) 3 V E p ( a ( p ) a ( p ) + b ( p ) b ( p )). (11.38) where we have used the definition of E p in ( 11.26 ). The Hamiltonian operator has therefore the following form, ˆ H = d 3 p ( 2 π ) 3 V E p ( a ( p ) a ( p ) + b ( p ) b ( p )) = d 3 p ( 2 π ) 3 V E p a ( p ) a ( p ) + b ( p ) b ( p ) + ( 2 π ) 3 V δ 3 ( p p ) . (11.39) The Dirac delta function appearing in the last term of the right hand side is an infinite constant devoid of physical significance since it associates with the vacuum an infinite

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11.2 Quantization of the Klein–Gordon Field 369 energy. This is apparent if we consider the particle in a finite-size box, with volume V . The momentum becomes discretized and ( 11.39 ) will have then the form: ˆ H = p E p a ( p ) a ( p ) + b ( p ) b ( p ) + 1 . (11.40) The vacuum energy part would read p E p = ∞ . This inessential infinite constant can be formally eliminated in the same way as we did for the electromagnetic field in Chap.6 , that is by introducing the normal ordering prescription when computing physical quantities. Let us recall the definition of “normal ordering”: An operator product is normal ordered if all the creation operators stand to the left of all destruc- tion operators. For instance: : a ( p ) a ( p ) : = : a ( p ) a ( p ) := a ( p ) a ( p ), : b ( p ) b ( p ) : = : b ( p ) b ( p ) := b ( p ) b ( p ). (11.41) With the normal order prescription the Hamiltonian ( 11.9 ) is replaced by ˆ H = d 3 x : ˆ π ˆ π + c 2 ˆ φ · ˆ φ m 2 c 4 2 ˆ φ ˆ φ : (11.42) As a consequence ( 11.39 ) takes the following form: ˆ H = d 3 p ( 2 π ) 3 V E p : a ( p ) a ( p ) + b ( p ) b ( p ) : = d 3 p ( 2 π ) 3 V E p a ( p ) a ( p ) + b ( p ) b ( p ) . (11.43) where no infinite constant appears. For finite volume V the normal ordered Hamiltonian reads ˆ H = p E p : a ( p ) a ( p ) + b ( p ) b ( p ) : = p E p a ( p ) a ( p ) + b ( p ) b ( p ) . (11.44) It is instructive to compare the above expression with the corresponding one ( 6.61 ) found in Chap.6 for the electromagnetic field. Identifying ω k with the energy E p of a photon of momentum p = k , we recognize that the two expressions for the energy are quite similar. The only differences between ( 11.44 ) and ( 6.61 ) consist, on the one hand, in the absence in the former of the polarization index, as it must be the case for a spinless field, (recall that the electromagnetic field has spin 1 and therefore has a polarization index related to the helicity of the photon); on the other hand we have the presence, on the right hand side of ( 11.44 ), of additional operators
370 11 Quantization of Boson and Fermion Fields b , b , which, as will be shown in the following, are always present for a charge field . They are not present in the electromagnetic field due to the hermiticity of ˆ A μ ( x ). We can proceed in the same way to evaluate the total quantum momentum of the field ˆ P i , given in ( 11.14 ), in terms of the operators a ( p ), b ( p ) and their hermitian conjugates. Using ( 11.35 ), ( 11.21 ) and ( 11.22 ) we find: ˆ P = d 3 p ( 2 π ) 3 V 2 p a ( p ) a ( p ) + b ( p ) b ( p ) + a ( p ) b ( p ) e 2 i E p t b ( p ) a ( p ) e 2 i E p t + h . c ., where h . c . denotes the hermitian conjugate terms. In the last term on the right hand

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• Fall '17
• Chris Odonovan

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