I'm currently trying to teach myself the path integral formulation of QFT (having studied the canonical approach previously), but I'm having some conceptual difficulties that I hope I can clear up here.

For simplicity, consider the case of a free, single real scalar field. The path integral formulation for a two point correlator in this case is given by $$\langle 0\lvert T\lbrace\hat{\phi}(x)\hat{\phi}(y)\rbrace\lvert 0\rangle =(-i^{2})\frac{1}{Z[0]}\frac{\delta^{2}Z[J]}{\delta J(x)\delta J(y)}\bigg\lvert_{J=0}$$ where $$Z[J]=\int\mathcal{D}\phi\; e^{i\int d^{4}x\left(-\frac{1}{2}\phi (\Box +m^{2})\phi+J(x)\phi (x)\right)}$$ is the generating functional for the free theory.

Here is where my issue lies. Are the fields $\phi (x)$ in the functional $Z[J]$ classical fields or are they operator fields?

If they are classical fields, then does the path integral define some sort of mapping between field operators $\hat{\phi}(x)$ and their classical (c-number) analogs?

The books I've been reading so far (Srednicki's QFT book and M. Schwartz's "QFT & the Standard Model") seem to a bit ambiguous in this area.

  • $\begingroup$ The path integral integrates classical fields. $\endgroup$
    – ACuriousMind
    Sep 2, 2015 at 22:16
  • $\begingroup$ @ACuriousMind That's what I thought, but I wanted to make sure. By "classical field" is it simply meant that the field value at a spacetime point $x^{\mu}$ is an eigenvalue of the field operator $\hat{\phi}(x)$ at that point? Also, are there any books on the path integral approach that you'd particularly recommend? $\endgroup$
    – Will
    Sep 2, 2015 at 22:26
  • 1
    $\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/9183/2451 $\endgroup$
    – Qmechanic
    Sep 2, 2015 at 23:04
  • $\begingroup$ Classical fields. The LHS is just a number (because of the expectation value <0| ... |0>). Thus the RHS also has to be just a number, so the fields can not be operator-valued on the RHS. This is an unfortunate abuse of notation. $\endgroup$
    – hft
    Sep 3, 2015 at 5:09

1 Answer 1


In the path integral, $\phi$ is not an operator, but rather a dummy integration variable which runs over all possible classical field configurations.

You can pass between the two formalisms using the following relation:

$$ \left< 0 \right| T \left\{ \hat{a} \hat{b} ... \hat{z} \right\} \left| 0 \right> = \frac{\int D\phi e^{i S[\phi] / \hbar} \cdot a(\phi) b(\phi) ... z(\phi)}{\int D\phi e^{i S[\phi] / \hbar}}, $$


  • $S[\phi]$ is the classical action
  • $a(\phi), b(\phi), ..., z(\phi)$ are some classical observables (functions on the configuration space)
  • $\hat{a}, \hat{b}, ..., \hat{z}$ are the corresponding quantum operators. There is an ordering ambiguity here, but it gets resolved by the
  • $T$ is the chronological ordering symbol (it reorders a string of operators by the time coordinate, descending)
  • $\begingroup$ On the right hand side of the equation in your answer, does this follow from the operators acting on the field eigenstates, e.g. $\hat{a}\lvert \phi\rangle = a(\phi)\lvert\phi\rangle$? Is it a kind of mapping between the operator formalism and the path integral formalism, such that they produce the same correlation functions? $\endgroup$
    – Will
    Sep 4, 2015 at 14:27
  • $\begingroup$ @Will exactly. Proof can be found in almost any QFT textbook. $\endgroup$ Sep 4, 2015 at 15:26
  • $\begingroup$ Are there any particular QFT books that you would recommend? $\endgroup$
    – Will
    Sep 4, 2015 at 17:13
  • $\begingroup$ @Will Peskin & Schreder would be my choice, although there might be better options $\endgroup$ Sep 4, 2015 at 18:57
  • $\begingroup$ In the rhs, inside the integral: Where are the $a(\phi)$ evaluated? Since they are functions of the configuration space, and not functionals, it should be something like $a(\phi(x))$ or something like that. $\endgroup$ Jun 25, 2017 at 5:16

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