How does the Hubbard Stratonovich transformation decouple interactions?
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While reading a book i understood ⬇️⬇️⬇️⬇️, where the author suggests that
e(−βH)=e(−βe22∫drdr′ρ(r)C(r−r′)ρ(r′))=∫Dϕe(−β∫dr12ϵ(∇ϕ)2+iρeϕ)e(−βH)=e(−βe22∫drdr′ρ(r)C(r−r′)ρ(r′))=∫Dϕe(−β∫dr12ϵ(∇ϕ)2+iρeϕ)
Here, C(r−r′)C(r−r′) is named the "Coulomb" operator, but is defined as the Greens function of the Poisson equation:
∇⋅[ϵ∇C(r−r′)]=δ(r−r′)∇⋅[ϵ∇C(r−r′)]=δ(r−r′)
The other terms are what you'd expect: ρρ is a charge density, ϵ(r)ϵ(r) is a dielectric function, ee is elementary charge and ββ is 1kT1kT.
e(−βH)=e(−βe22∫drdr′ρ(r)C(r−r′)ρ(r′))=∫Dϕe(−β∫dr12ϵ(∇ϕ)2+iρeϕ)e(−βH)=e(−βe22∫drdr′ρ(r)C(r−r′)ρ(r′))=∫Dϕe(−β∫dr12ϵ(∇ϕ)2+iρeϕ)
Here, C(r−r′)C(r−r′) is named the "Coulomb" operator, but is defined as the Greens function of the Poisson equation:
∇⋅[ϵ∇C(r−r′)]=δ(r−r′)∇⋅[ϵ∇C(r−r′)]=δ(r−r′)
The other terms are what you'd expect: ρρ is a charge density, ϵ(r)ϵ(r) is a dielectric function, ee is elementary charge and ββ is 1kT1kT.
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Apparently I'm bad at math and simply can't complete the square in the right hand side to get the left hand side. How... do you do this?
My (hopefully) interesting question:
What does it mean, conceptually, to decouple an interaction by introducing a field? I interpret the left hand side of the equation above as "ρ(r)ρ(r)communicates with ρ(r′)ρ(r′) through the C(r−r′)C(r−r′)propagator" -- does this new field somehow contain all of this 'communication' information? Can I express the field ϕϕ in terms of CC
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