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The Schwinger-Dyson equation has a delta term proportional to $h$. The origin of this term in the derivation is the commutator $[\phi, \pi]=\delta$, which is satisfied in QFTs.

In Schwartz's book, on pages 81 and 84, he says that this delta term is the reason for the loop diagrams in QFT. He also says that the absence of this term in classical field theory, because $[\phi,\pi]=0$ in CFT, is the reason for the absence of loop diagrams there. I have a few problems:

  1. Feynman diagrams in CFT appear in a completely different context. No one cares about the Schwinger Dyson equation in CFT. Instead, the Feynman diagrams appear when we're solving the interacting field equations.

  2. The perturbation expansions in QFT and CFT are not analogous at all. In QFT, the expansion is for the propagator $e^{-iH_It}$. This quantity has no counterpart in CFT. In CFT, the perturbation expansion is for the interacting field equations. These equations have a twin counterpart in QFT: the Heisenberg picture. So, if we did QFT entirely in the Heisenberg picture, then the calculation would be exactly analogous, and only tree diagrams would appear in QFT as well.

So, the appearance of loops in QFT is only because of the way we do calculations? This means that trees vs loops is not a fundamental distinction between QFT and CFT

  1. Schwartz gives (page 84) one more example of a theory having only tree diagrams: non-relativistic QM. Specifically: an electron in a classical EM field. But idk how to make sense of the Schwinger Dyson equation in non-relativistic QM, because non relativistic QM doesn't have field operators. One way out is to manually introduce non relativistic field operators in NRQM. But if we did that then $[\phi,\pi]=\delta$ will be satisfied, and we'll get the delta term in the S-D equation that is responsible for loops. ($[\phi,\pi]=\delta$ follows from $[a,a^{\dagger}]=\delta$ which electron creation and annihilation operators must obey even in NRQM)

Please address all three points.

Qmechanic
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Ryder Rude
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1 Answers1

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  1. The $\hbar$/loop-expansion shows that loops are quantum effects, cf. Ref. 1.

  2. The quantum $\delta$-term in the Schwinger-Dyson (SD) equations vanishes in the classical $\hbar\to 0$ limit.

  3. One can talk about tree-level Feynman diagrams in CFT with the very same propagators and vertices as in QFT in the path integral formalism.

    This becomes especially clear if one solves Euler-Lagrange (EL) equations perturbatively as a rooted tree diagram, cf. eqs. (5) & (6) in my Phys.SE answer here, or the mentioned eqs. (3.85) & (4.37) in Ref. 2.

  4. See also this related Phys.SE post.

References:

  1. C. Itzykson & J.B. Zuber, QFT, 1985; Section 6-2-1, p. 287-288.

  2. M.D. Schwartz, QFT & the standard model, 2014; Subsection 7.1.1, p. 81-84.

Qmechanic
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  • But this does not address my three concerns. If we solved the Heisenberg equations of motion in QFT, then only tree diagrams would appear in the perturbation expansion. This is because the Heisenberg equations are exactly identical to EL equations. Then, don't loop diagrams appear only because we work in the interaction picture as opposed to Heisenberg picture? – Ryder Rude Oct 04 '22 at 16:29