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?

knzhou · Accepted Answer

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.

Qmechanic · Answer

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

One-loop Correction to Effective Action

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