5
$\begingroup$

My question is about the loopwise expansion of the effective action $\Gamma(\varphi)$ up to 1-loop contributions. I've understood well the results for both $Z[J]$ and $W[J]$ functionals loopwise expansions. But then something is missing when I follow the path towards the expansion of the effective action. Following this excerpt from Zinn-Justin's "Quantum field theory and critical phenomena":

Zinn-Justin "Quantum field theory and critical phenomena"

I don't understand the statement: "Therefore a correction of order $\hbar$ to the relation between $J(x)$ and $\varphi(x)$ will produce a change of order $\hbar^2$ to the r.h.s. of equation (6.47)"

Could someone please write down some more details?

My undestanding at this stage is that the relation between $J(x)$ and $\varphi$ is fixed by the stationarity of the (6.47) (as for any Legendre transformation). But then, how to proceed?

$\endgroup$
4
$\begingroup$

The problem is to evaluate $\Gamma[\phi]$ at fixed $\phi$ given an expansion of $W[J]$ in powers of $\hbar$, where $\Gamma[\phi]$ is the Legendre transform of $W[J]$. By definition, $$ \Gamma[\phi]=\sup_{J}\Big[\phi\cdot J-W[J]\Big]. $$ Suppose that the expression in the RHS attains its maximum at some $\hat{J}$. Then formally, we can expand $\phi\cdot J-W[J]$ about $\hat{J}$ to obtain $$ \phi\cdot (\hat{J}+\delta J)-W[\hat{J}+\delta J]=\Gamma[\phi]+\phi\cdot\delta J-\frac{\delta W}{\delta J}\Big|_{\hat{J}}\cdot\delta J+\mathcal{O}(\delta J^2). $$ Since a maximum is attained at $\hat{J}$, the linear term $(\phi-\delta_J W)\cdot\delta J$ vanishes for arbitrary $\delta J$, and we see that corrections start at order $\delta J^2$.
Now we return to the expansion in $\hbar$. If $W[J]$ has an expansion, then it is reasonable that $\hat J$ has an expansion too after solving $\delta_J W=\phi$. In fact, if $\hat J$ has an expansion in $\hbar$ then the zeroth order term must be the solution to $\delta_J W_0 = \phi$, where $W=W_0+\hbar W_1+\dots$. Hence, we can write $\hat J_0=\hat J-\hbar \delta\hat J$, where $\delta \hat J$ includes all higher order corrections. From the above argument, evaluating $\phi\cdot J-W[J]$ at $\hat J_0$ will differ from the evaluation at $\hat J$ by terms of order $\hbar^2$. Hence, to order $\hbar$ we can use $\hat J_0$ in evaluating $\Gamma[\phi]$.

$\endgroup$
  • $\begingroup$ Thanks. I was missing the second part of the explanation, connecting clearly the expansion of $W$ to the relation between $J$ and $\phi$. $\endgroup$ – Riccardo Buscicchio Jun 13 '15 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.