By S. M. Bilenky (Auth.)

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9) 2 oxv dxv imply the samefieldequations. 9). 9) that the energymomentum tensor of the electromagnetic field is T μν = d~^ dA Q J)k - dA0 dxu axv μν SA Q = dxv . fr T dx = t \" SA Q SA Q , 1 ÔAQ ΘΑρ Ί ) ί-^Γ^Γ+τ^Γδ^Η (7 η ·> Let us expand the potential Αμ(χ) in plane waves. 12) where &0 = \k\. Thefirstterm is the positive-frequency part of the potential and the second term is the negative-frequency part. 12) in the form Mx) = - ^ — f —A=- [A^\k)e^+A^-\-k)c-^]dk. *(£) = (0, i), λ = 1, 2, 3. 13) 49 INTRODUCTION TO FEYNMAN DIAGRAMS [§ 7 Here e3(k) = k/ko = x is the unit vector directed alongftand ei(k) and ez(k) are unit vectors orthogonal to one another and to vector ft.

E. to require that φ+(χ) = φ(χ). 23) a(-q) = a+(q). 24) wefindfor the energy operator H = \ j [a+(q) a{q)+a{q) a+(q)] q0 dq. 25) It is obvious that this operator is Hermitian. Furthermore, it is necessary to postulate commutation relations between thefieldoperators. We assume the following commutation relations : [a(q\ a{q')} = a(q) a(q')-a{q') a{q) = 0, Mi). a+(q')] = a(q) a+{q')-a+{q') a(q) = d(q-q'). 26) From this it also follows that [a+(q), a+(q')] = 0. 25). 25) describe free particles with definite momenta.

5) then D ^ ; = 0, | ^ - = 0. 2). The potentials Αμ and Αμ give, however, the samefieldstrength tensor (gradient invariance). i l A dA A d 14 (dAA n&\ The last term in this expression is zero due to the Lorentz condition. The term 9 / , dAu Χμ \ V dXv ) does not contribute to the variation of the action. 7) and ^ = _ | ^ M . 9) 2 oxv dxv imply the samefieldequations. 9). 9) that the energymomentum tensor of the electromagnetic field is T μν = d~^ dA Q J)k - dA0 dxu axv μν SA Q = dxv . fr T dx = t \" SA Q SA Q , 1 ÔAQ ΘΑρ Ί ) ί-^Γ^Γ+τ^Γδ^Η (7 η ·> Let us expand the potential Αμ(χ) in plane waves.