Say I have a theory in four dimensions with the Lagrangian density \begin{equation} \mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu\phi - \frac{1}{2} m^2 \phi^2. \end{equation} This has the interaction Lagrangian: \begin{equation} \mathcal{L}_\text{int} = \frac{1}{3!}g\phi^3 \end{equation} where $g$ is the coupling constant. How do I show that the one-loop correction to the two-point correlator is suppressed by $\hbar$ in comparison to the free propagator? What I have tried is using the Gell-Mann-Low formula, with $\hbar\neq 1$: \begin{equation} \langle \Omega| T(\phi(x_1)\phi(x_2)\cdots )|\Omega \rangle = \langle 0|T(\phi_\text{in}(x_1) \phi_\text{in}(x_2)\cdots \exp\bigg(\frac{i}{\hbar}\int\mathcal{L}_\text{int}d^4 y\bigg))|0\rangle \end{equation} Where $|\Omega\rangle$ is the interaction vacuum and $|0\rangle$ is the free vacuum. From here I am only getting expansions in $\frac{1}{\hbar}$ and not $\hbar$. Can someone point out where I am going wrong.
2 Answers
This is a consequence of the $\hbar$/loop-expansion, i.e. that the power of $\hbar$ in a connected Feynman diagram is the number of loops. For a proof, see e.g. Ref. 1 or my Phys.SE answer here.
In particular, the generating functional $W_c[J]$ of connected diagrams is a power series (as opposed to a Laurent series) in $\hbar$.
Concerning intuition for positive powers of $\hbar$: Recall that the path integral measure is actually $${\cal D}\frac{\phi}{\sqrt{\hbar}},$$ cf. e.g. this Phys.SE post. If we substitute $\phi=\sqrt{\hbar}\phi^{\prime}$, then both positive and negative powers of $\hbar$ appear in the Boltzmann factor $$\exp\left(\frac{i}{\hbar}S_J[\sqrt{\hbar}\phi^{\prime}]\right).$$
References:
- C. Itzykson & J.B. Zuber, QFT, 1985; Section 6-2-1, p.287-288.
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Interaction terms (so, vertices in the Feynman diagrams) carry a $1/\hbar$ factor, but propagators (lines in the diagrams) carry a factor of $\hbar$ as they are the inverse of the quadratic operator in the action.
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