Since we now have a continuous label for the creation and annihilation

Since we now have a continuous label for the creation

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Since we now have a continuous label for the creation and annihilation operators, we need a Lorentz-invariant way to sum over operators with different momentum. The four components of k are not independent, but satisfy k 2 k µ k µ = m 2 , and we also require positive energy, that is k 0 = ω k > 0. Taking these things into account, one is led to the integration measure integraldisplay d k integraldisplay d 4 k (2 π ) 4 2 πδ ( k 2 m 2 ) Θ ( k 0 ) = integraldisplay d 4 k (2 π ) 3 δ (( k 0 ω k )( k 0 + ω k )) Θ ( k 0 ) = integraldisplay d 4 k (2 π ) 3 1 2 ω k ( δ ( k 0 ω k ) + δ ( k 0 + ω k )) Θ ( k 0 ) = integraldisplay d 3 k (2 π ) 3 1 2 ω k . (2.19) The numerical factors are chosen such that they match those in Eq. (2.18) for the commutator of a ( k ) and a ( k ). 2.2.2 Charge and Momentum Now we have the necessary tools to construct operators which express some properties of fields and states. The first one is the operator of 4-momentum, i.e., of spatial momentum and energy. Its construction is obvious, since we interpret a ( k ) as a creation operator for a state with 4-momentum k . That means we just have to count the number of particles with each momentum and sum the contributions: P µ = integraldisplay d kk µ ( a ( k ) a ( k ) + b ( k ) b ( k ) ) . (2.20) This gives the correct commutation relations: bracketleftbig P µ ,a ( k ) bracketrightbig = k µ a ( k ) , bracketleftbig P µ ,b ( k ) bracketrightbig = k µ b ( k ) , (2.21a) bracketleftbig P µ ,a ( k ) bracketrightbig = k µ a ( k ) , bracketleftbig P µ ,b ( k ) bracketrightbig = k µ b ( k ) . (2.21b) Another important operator is the charge. Since particles and antiparticles have opposite charges, the net charge of a state is proportional to the number of particles 10
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minus the number of antiparticles: Q = integraldisplay d k ( a ( k ) a ( k ) b ( k ) b ( k ) ) , (2.22) and one easily verifies bracketleftbig Q,a ( k ) bracketrightbig = a ( k ) , bracketleftbig Q,b ( k ) bracketrightbig = b ( k ) . (2.23) We now have confirmed our intuition that a ( k ) ( b ( k ) ) creates a particle with 4- momentum k and charge +1 (-1). Both momentum and charge are conserved: The time derivative of an operator is equal to the commutator of the operator with the Hamilto- nian, which is the 0-component of P µ . This obviously commutes with the momentum operator, but also with the charge: ı d d t Q = [ Q,H ] = 0 . (2.24) So far, this construction applied to the case of a complex field. For the special case of neutral particles, one has a = b and Q = 0, i.e., the field is real. 2.2.3 Field Operator We are now ready to introduce field operators, which can be thought of as Fourier transform of creation and annihilation operators: φ ( x ) = integraldisplay d k ( e ı kx a ( k ) + e ı kx b ( k ) ) . (2.25) A spacetime translation is generated by the 4-momentum in the following way: e ı yP φ ( x ) e ı yP = φ ( x + y ) . (2.26) This transformation can be derived from the transformation of the a ’s: e ı yP a ( k ) e ı yP = a ( k ) + ı y µ bracketleftbig P µ ,a ( k ) bracketrightbig + O ( y 2 ) (2.27) = (1 + ı yk + · · · ) a ( k ) (2.28) = e ı yk a ( k ) . (2.29) The commutator with the charge operator is [ Q,φ ( x )] = φ ( x ) , bracketleftbig Q,φ bracketrightbig = φ . (2.30)
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