# What does it mean to integrate out fields from a theory?

I've done a fair bit of reading on this subject and I'm still confused about the basic principle of integrating out fields in QFT. When we have a function of 2 fields a and b, f(a,b), and we integrate out the heavy b-fields to give f(a), by what mechanism does the b-field dependence disappear? Are we basically saying that integrating out the b-fields is tantamount to solving the probability amplitude of those fields appearing and that because they are heavy their contribution is vanishingly small?

Also I've seen the contraction of fields talked about with regards to integrating out, what is the role of these contractions when integrating out? Contracting fields seems the best means of making a particular field disappear from the equations we consider!

The Feynman path integral gives you a formula for Green's functions and other amplitudes $$A = \int {\mathcal D}\phi_{\rm light} \,{\mathcal D}\phi_{\rm heavy}\,\exp(iS)\prod_{i}V_i$$ where $V_i$ are some insertions in the integral that are chosen according to the choice of the quantity (amplitude) we want to quantify. We integrate over all configurations of all fields etc. To integrate out $\phi_{\rm heavy}$ means to divide the integration process to two steps and first integrate over some degrees of freedom, namely $\phi_{\rm heavy}$ – which may be many fields – for fixed values of the remaining fields, $\phi_{\rm light}$.
The resulting i.e. remaining integral which still waits to be integrated over the remaining light fields (without insertions) is interpreted as $\exp(iS_{\rm effective})$ where $S_{\rm effective}$ only depends on the light degrees of freedom $\phi_{\rm light}$. $$A_\text{after integrating out} = \int {\mathcal D}\phi_{\rm light} \,\exp(iS_{\rm effective})\prod_{i}V_i,\\ \exp(iS_{\rm effective} [\phi_{\rm light}]) = \int {\mathcal D}\phi_{\rm heavy}\,\exp(iS)$$ In this way, we eliminate the heavy degrees of freedom and calculate the effective action that remembers all the loop effects that the now-forgotten heavy fields used to cause. However, this effective field theory with an effective action – a result of integrating out the heavy fields – is only good for asking low-energy questions, of course. The insertions $V_i$ we may insert into the simplified theory that only depends on $\phi_{\rm light}$ cannot depend on the heavy degrees of freedom anymore, of course: they have been disappeared.
• Hi @user15766, yes, at least when your rule (and the word "solved") is correctly extrapolated to higher orders. It is indeed a sort of a Born-Oppenheimer approximation in which the light (=slow) degrees of freedom are fixed and the heavy part of the theory is "solved". In practice, the calculation of the effective action leads to the evaluation of all the diagrams with at least one "heavy" propagator... In the context of the renormalization group, we also want to "integrate out" only a part of the field modes, those with energies between $E$ and $E+dE$. – Luboš Motl Apr 15 '13 at 18:05