Relation of field creation operators to path integral?
Answers
Applying two field creation operators to a vacuum I get: ψ^†(x)ψ^†(y)|0⟩=(ϕ^(x)ϕ^(y)−s−1(x−y))|0⟩ where the quantum field creation operators are defined by (see https://en.wikipedia.org/wiki/Second_quantization#Creation_and_annihilation_operators) ψ^†(x)=∫a†(k)eix.kdk3 and s(x−y)=∫eik.(x−y)k2+m2−−−−−−−√dk3 But the field operators are a sum of creation and annihilation operators: ϕ^(x)=∫(a†(k)eix.k+a(k)e−ix.k)dk3 π^(x)=δδϕ(x)=∫k2+m2−−−−−−−√(a†(k)eix.k−a(k)e−ix.k)dk3 and so on. Where ⟨ϕ|ϕ^(x)|0⟩=ϕ(x)⟨ϕ|0⟩. Presumably ϕ^(x)=ψ(x)+ψ†(x) ?? and ψ†(x)=ϕ(x)+∫s−1(x−y)δδϕ(y)dy3 But in the path integral for an input state starting with two particles it is just ϕ(x)ϕ(y) and not ϕ(x)ϕ(y)−s(x−y). e.g. ∫ϕ(x,t)ϕ(y,t)ϕ(z,t′)ϕ(w,t′)eS[ϕ]Dϕ Or in other words we calculate ⟨0|ϕ^(x)ϕ^(y)..|0⟩ not ⟨0|ψ^(x)ψ^(y)..|0⟩. Why is this? What happened to the missing s(x−y). Why do we use field operators ϕ and not field creation operators ψ? Question: Is it equivalent and if so how can this be proven? Am I doing something wrong here? Why this is bothering me is that a complete set of orthogonal states would be given by the field creation operators ψ^(x)ψ^(y)..|0⟩ but not I presume the field operators ϕ. To put it another way. If ϕ represents a photon of light and we are calculating the scattering of two photons. Should we use: ⟨0|ϕ(x)ϕ(y)ϕ(z)ϕ(w)|0⟩ or ⟨0|(ϕ(x)ϕ(y)−s(x−y))(ϕ(z)ϕ(w)−s(w−z))|0⟩ or, for example, are the effects of s so small not to matter when the photons are far away from the scattering point? But which one is "theoretically" correct? I know in scattering experiments where x and y are presumed to start infinitely far apart this doesn't matter but is it correct?
◆⟨0|ϕ(x)ϕ(y)ϕ(z)ϕ(w)|0⟩⟨0|ϕ(x)ϕ(y)ϕ(z)ϕ(w)|0⟩
◆⟨0|(ϕ(x)ϕ(y)−s(x−y))(ϕ(z)ϕ(w)−s(w−z))|0⟩