# Integraldisplay d 3 x g 3 φ x t 3 33 the operators

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integraldisplay d 3 x g 3! φ ( x , t ) 3 (33) The operators φ ( x , t ) , H , H 0 and V are all in the Heisenberg picture. To do time-dependent perturbation theory we will change to the interaction picture . In the interaction picture the fields evolve only with H 0 . The interaction picture fields are just what we had been calling (and will continue to call) the free fields φ 0 ( x , t ) = e iH 0 ( t t 0 ) φ ( x ) e iH 0 ( t t 0 ) = integraldisplay d 3 p (2 π ) 3 1 2 ω p radicalbig ( a p e ipx + a p e ipx ) (34) To be precise, φ ( x ) is the Schrödinger picture field, which does not change with time. The free fields are equal to the Schrödinger picture fields and also to the Heisenberg picture fields, by def- inition, at a single reference time which we call t 0 . Using Eq. (30), we see that the Heisenberg picture fields are related to the free fields by φ ( x , t ) = S ( t, t 0 ) e iH 0 ( t t 0 ) φ 0 ( x , t ) e iH 0 ( t t 0 ) S ( t, t 0 ) (35) = U ( t, t 0 ) φ 0 ( x , t ) U ( t, t 0 ) (36) The operator U ( t, t 0 ) e iH 0 ( t t 0 ) S ( t, t 0 ) therefore relates the full Heisenberg-picture fields to the free fields at the same time t . The evolution begins from the time t 0 where the fields in the two pictures (and the Schrödinger picture) are equal. We can find a differential equation for U ( t, t 0 ) using Eq.(31): i∂ t U ( t, t 0 ) = i ( t e iH 0 ( t t 0 ) ) S ( t, t 0 ) + e iH 0 ( t t 0 ) i∂ t S ( t, t 0 ) = e iH 0 ( t t 0 ) H 0 S ( t, t 0 ) + e iH 0 ( t t 0 ) H ( t ) S ( t, t 0 ) = e iH 0 ( t t 0 ) [ H 0 + H ( t )] e iH 0 ( t t 0 ) e iH 0 ( t t 0 ) S ( t, t 0 ) = V I ( t ) U ( t, t 0 ) (37) where V I ( t ) e iH 0 ( t t 0 ) V ( t ) e iH 0 ( t t 0 ) is the original Heisenberg picture potential V ( t ) from Eq. (32), now expressed in the interaction picture. If everything commuted, the solution to Eq. (37) would be U ( t, t 0 ) = exp ( i integraltext t 0 t V I ( t ) dt ) . But V I ( t 1 ) does not necessarily commute with V I ( t 2 ) , so this is not the right answer. It turns out the right answer is very similar U ( t, t 0 ) = T braceleftbigg exp bracketleftbigg i integraldisplay t 0 t dt V I ( t ) bracketrightbiggbracerightbigg (38) where T {} is the time-ordering operator, introduced in the last lecture. This solution works because time-ordering effectively makes everything inside commute: T { A B } = T { B A } (39) 6 Section 3
Taking the derivative, you can see immediately that Eq. (38) satisfies Eq. (37). Since it has the right boundary conditions, namely U ( t, t ) = 1 , this solution is unique. Time ordering of an exponential is defined in the obvious way through its expansion U ( t, t 0 ) = 1 i integraldisplay t 0 t dt V I ( t ) 1 2 integraldisplay t 0 t dt integraldisplay t 0 t dt ′′ T { V I ( t ) V I ( t ′′ ) } + (40) This is known as a Dyson series . Dyson defined the time-ordered product and this series in his classic paper from 1949. In that paper he showed the equivalence of old-fashioned perturbation theory, or more exactly, the interaction picture method developed by Schwinger and Tomonaga based on time-dependent perturbation theory, and Feynman’s method, involving space-time dia- grams, which we are about to get to.