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Why does the term $\frac{i\hbar}{2}\log\det\left[\frac{\delta^2S}{\delta\phi^2}\right]$ in the expansion $$\Gamma[\phi]=S[\phi]+ \frac{i\hbar}{2}\log\det\left[\frac{\delta^2S}{\delta\phi^2}\right] +…\tag{1}$$ give only 1-loop contributions and no 2-loop? Or is this simply because we have only 1 trace?

Frobenius
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Silas
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2 Answers2

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One quick way to see that the Hessian term is a 1-loop effect is because it's 1st order in $\hbar$, cf. the $\hbar$/loop expansion.

For further arguments, see e.g. this and this related Phys.SE posts.

Qmechanic
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Loop expansion is equivalent to the expansion about $\hbar$, so the sum of the 1 loop diagrams corresponds to the linear term of $\hbar$.

Quantum effective action $\Gamma[\phi]$ is defined by $$e^{i\Gamma(\phi)}=e^{iW-i\int_x\ J\phi},$$ $$J:=-\frac{\delta\Gamma}{\delta\phi}$$ and the r.h.s. of this equation has a path integral representation as follows: \begin{align}e^{i\Gamma(\phi)}&=e^{iW-\int_x\ J\phi}\\ &=\int D\varphi e^{iS(\varphi)+i\int_x\ J(\varphi-\phi)}\\ &= \int_{1PI} D\varphi e^{iS(\varphi+\phi)}\\ &=e^{iS(\phi)}\times\int D\varphi\ e^{iS_0(\varphi)}\times\frac{\int_{1PI} D\varphi\ e^{i(S_0(\varphi)+S_{\mathrm{int}}(\phi+\varphi))}}{\int D\varphi\ e^{iS_0(\varphi)}} \end{align} where the symbol 1PI means adding up only one-particle-irreducible diagrams and $S_0$ represents the second order term of the action, i.e. free terms. Although there can be linear terms for $\varphi$, they do not constitute a 1PI diagram and can be ignored.

If we recover $\hbar$, the first term, i.e. $S(\phi)$, is the order $\hbar^0$ and the second term is the order $\hbar^{1}$, since it corresponds to the term $i\hbar\log\det(G_0)$ in $\Gamma$. Third terms are $O(\hbar^{2})$, so they are the contribution of more than two loop.

That’s why $\log\det$ term only resums the one loop diagrams.

Siam
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