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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.

Qmechanic
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QFTheorist
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2 Answers2

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  1. 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.

  2. In particular, the generating functional $W_c[J]$ of connected diagrams is a power series (as opposed to a Laurent series) in $\hbar$.

  3. 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:

  1. C. Itzykson & J.B. Zuber, QFT, 1985; Section 6-2-1, p.287-288.
Qmechanic
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  • +1 But why is $\hbar \rightarrow 0$ here called clsssical? Even the lowest order WKB wavefunction isn't classical. One gets classical mechanics only after assuming a Gaussian wavefunction which bypasses the uncertainty principle. – Ryder Rude Nov 26 '22 at 10:50
  • It's an asymptotic series. – Qmechanic Nov 26 '22 at 11:16
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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.

SolubleFish
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  • But why is the tree level called classical? I thought the $\hbar \rightarrow 0$ limit is non-trivial to take in Quantum-Classical correspondence. For example, even the lowest order WKB wavefunction isn't classical. One gets classical mechanics only after assuming a Gaussian wavefunction to bypass the uncertainty principle. – Ryder Rude Nov 26 '22 at 10:36
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    The $\hbar\to 0$ limit can be taken at the level of correlation functions as it just means keeping only tree level diagrams. Whether the wave functions converges to something meaningful in this limit is another matter – SolubleFish Nov 27 '22 at 13:54
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    Oh yeah. The limit can absolutely be taken. It's just that it gets repeated a lot that this limit is "classical" somehow. This limit is just the initial approximation of the Schrodinger equation solution, so it must still be quantum. – Ryder Rude Nov 27 '22 at 15:14