I'm studying an example of the Hubbard-Stratonovich transformation in Altland and Simons' Condensed Matter Field Theory (2nd ed.), pp. 246-247.
In it they say that...
one is frequently confronted with situations where more than one Hubbard-Stratonovich field is needed to capture the full physics of the problem. To appreciate this point, consider the Coulomb interaction in momentum space. Sint[ˉψ,ψ]=12∑p1,...,p4ˉψσ,p1ˉψσ′,p3V(p1−p2)ψσ′,p4ψσ,p2δp1−p2+p3−p4.
In principle, we can decouple this interaction in any of the three channels...
discussed in the previous page. If one chooses to decouple in all three channels then the action becomes ...
Sint[ˉψ,ψ]≃12∑p,p′,q(ˉψσ,pψσ,p+qV(q)ˉψσ′,p′ψσ′,p′−q−ˉψσ,pψσ′,p+qV(p'−p)ˉψσ′,p′ψσ,p′−ˉψσ,pˉψσ′,−p+qV(p'−p)ψσ,p′ψσ′,−p′+q)
where the first term is decoupled via the
direct channel ρd,q∼∑pˉψσ,pψσ,p+q, second in the exchange channel ρx,σσ′,q∼∑pˉψσ,pψσ′,p+q, and third in the Cooper channel ρc,σσ′,q∼∑pˉψσ,pˉψσ′,−p+q.
It's generally a good strategy to decouple in all available channels when one is in doubt, then let the mean-field analysis sort out the relevant fields.
My question is, if we choose to decouple the quartic term via 3 different channels (for example) is it necessary to multiply the resulting terms by a factor of 13? This isn't discussed in the textbook and I'm confused by the liberal use of ∼ and ≃ in the examples.
Answer
No. You should not add a factor of 1/3. As you can see in page 244 of Altland and Simons, the HS transformation is done by multiplying by a unity expressed as a functional integral over an auxiliary field. In this case, they just choose to introduce 3 different fields - 1 for each term.
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