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Short answer to your question: yes. There's no reason for the path integral and the operator formalism to be equivalent. A simple example are all the non-Lagrangian theories. We know (some of) them through their operator algebra, but there's no corresponding Lagrangian. This might sound strange to most people, like myself not long ago, but it is not hard ...


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General comments to the question (v1): Any textbook derivation of the correspondence between $$\tag{1} \text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism}$$ is just a formal derivation, which discards contributions in the process, cf. e.g. this Phys.SE post. Rather than claiming complete understanding and existence ...


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In fact there are two different meaning for the term "global anomaly", which is a pity: global anomaly as opposed to gauge anomaly; global gauge anomaly as opposed to local gauge anomaly. An anomaly can arise from global and gauge symmetries. So here global refers to the fact that the symmetry group is not gauged: these symmetries have a physical ...


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The partition function $Z[J]$, both in QM and in CM, is underdetermined: any multiple of $Z[J]$ gives rise to the same dynamics. This means that $Z[0]$ is arbitrary, and is usually set to one: $$ Z[0]\equiv 1 \tag{1} $$ effectively getting rid of vacuum diagrams, that is, we set $H|\Omega\rangle=0$. In other words: the energy of the vacuum is not measurable ...


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In terms of Feynman diagrams, the partition is represented by the sum over so-called vacuum bubbles - diagrams with no external legs. In formulae and in terms of the interaction picture and the free vacuum $\lvert 0 \rangle$ and the interacting vacuum $\lvert \Omega \rangle$, we have that $$ \lvert \Omega\rangle = \lim_{T\to\infty(1-\mathrm{i}\epsilon)} ...



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