One-loop Correction to Effective Action

Question

This might be a stupid question.

In Bailin and Love's "Cosmology in gauge field theory and string theory", the authors are describing how to calculate the effective potential at a finite temperature $T$ (Section 2.3, Pg. 42). Initially, they start with the treatment in zero-temperature.

They start by saying:

In quantum field theory at zero temperature, the expectation value $\phi_c$ of a scalar field $\phi$ (also referred to as the classical field) is determined by minimizing the effective potential $V(\phi_c)$. The effective potential contains a tree-level potential term, which can be read off from the Hamiltonian density, and quantum corrections from various loop orders.

I can understand this. However, they go on to claim (with $\phi(x)=\phi_c+\tilde{\phi}(x)$),

The one-loop quantum correction is calculated by shifting the fields $\phi$ by their expectation values $\phi_c$ and isolating the terms $\mathcal{L}_\mathrm{quad}(\phi_c,\tilde{\phi})$ in the Lagrangian density which are quadratic in the shifted fields $\tilde{\phi}$.

This is the statement I am a bit confused about. Why do we only isolate terms that are quadratic in the $\tilde{\phi}$, but not higher powers? There may be self-interaction terms; do these not contribute to the one-loop correction?

Answer 1

This result can be seen diagrammatically. The effective action is computed by summing over 1PI diagrams. A term $(\phi_c)^n \tilde{\phi}^m$ in the effective action corresponds to a vertex with $n$ $\phi_c$ legs, which are the external classical field, and $m$ $\tilde{\phi}$ legs, which we are integrating over and hence appear in internal lines.

One example of a one-loop contribution is:

where I've shamelessly stolen the graphic from these lecture notes. Here the dotted lines represent $\tilde{\phi}$ legs and the solid lines represent external $\phi_c$ legs. To find the contribution to the $(\phi_c)^n$ term of the effective potential, one must sum over all such diagrams with $n$ external $\phi_c$ legs.

Note that in this diagram, all of the vertices have $m = 2$, i.e. we are only considering quadratic terms in $\tilde{\phi}$. You can convince yourself by drawing a few diagrams that any higher-order terms would require more than one loop. The presence of linear terms ($m = 1$) would mess up this counting, which is why Bailin and Love expand about the classical minimum.

Answer 2

That the one-loop quantum correction

$$ \exp\left(\frac{i}{\hbar}\Gamma_{\text{1-loop}}[\phi_{\rm cl}]\right) ~\stackrel{(13)}{=}~ {\rm Det}\left(\frac{1}{i}\frac{\delta^2 S[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k \delta \phi_{\rm cl}^{\ell}}\right)^{-1/2}$$ $$~\stackrel{\text{Gauss. int.}}{\sim}~\int\!{\cal D}\frac{\eta}{\sqrt{\hbar}} ~\exp\left(\frac{i}{2\hbar}\eta^k \frac{\delta^2 S[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k \delta \phi_{\rm cl}^{\ell}}\eta^{\ell} \right) $$ $$~=~ \int\!{\cal D}\frac{\eta}{\sqrt{\hbar}} ~\exp\left(\left.\frac{i}{\hbar} S[\phi_{\rm cl}+\eta]\right|_{\text{quadratic in }\eta} \right) $$

to the effective/proper action $\Gamma[\phi_{\rm cl}]$ is given by the determinant of the Hessian of the action $S$ is e.g. proven in eq. (13) in my Phys.SE answer here. That proof relied on the stationary phase/WKB approximation, which, in turn, explains why only quadratic fluctuations $\eta$ contribute.