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(I'll work in the Euclidean for convenience)

In the path integral formulation of QFT given a field $\phi$, or a set of them if you want to, we have that the partition function is given by:

$$Z[J] = \int\mathcal{D}\phi e^{-(S[\phi] - J\phi)}$$

Where $J$ is the background/external field and $J\phi$ is called the source action.

Questions:

  1. Is the field $J$ a quantum (composite if needed) operator or is it just a classical field?

Take for example $\phi = A_\mu$, is $J$ the quantum mechanical operator $J= J^\mu = \bar{\psi}\gamma^\mu\psi$? I think in QED the operator $A_\mu J^\mu$ is already present into the action so it seems strange to me.

  1. If such an operator is just a classical field, what is his meaning as an "external" field? What does it mean to be "external"?

I think that, gravity aside, all the fields used are, in the end, quantum field operators in interactions, so i don't see why using such an instrument instead of just including the originating interaction associated to $J$ into the action.

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The status of sources $J$ in path integrals/partition functions depend on context:

  1. Often the sources $J$ are only used as a generating technique (for correlator functions and Feynman diagrams) and eventually put to zero.

  2. One can have a partition function of a subsector of a theory within a partition function of a parent theory.

    E.g. the subsector could be pure QED that sits inside a parent QED theory with matter fields. The sources $J^{\mu}$ of the pure QED partition function could then consist of matter currents originating from interaction terms ${\cal L}_{\rm int}= J^{\mu}A_{\mu}$ of the parent theory, cf. OP's example.

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  • $\begingroup$ Yes i was in fact thinking of it as a "subset" of a theory where you don't need the full dynamics. In view of trying to write down the most complete partition function of a theory, so not a subsector, the scope of $J$ is just the mere generating one, where we set it zero after derivation, right? $\endgroup$
    – LolloBoldo
    Commented Jul 2, 2023 at 14:37
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    $\begingroup$ $\uparrow$ Yes, usually. $\endgroup$
    – Qmechanic
    Commented Jul 2, 2023 at 14:47

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