# Conceptual questions on the path integral formulation of QFT

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.

• The path integral integrates classical fields. Sep 2, 2015 at 22:16
• @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?
– Will
Sep 2, 2015 at 22:26
• Essentially a duplicate of physics.stackexchange.com/q/9183/2451 Sep 2, 2015 at 23:04
• 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.
– hft
Sep 3, 2015 at 5:09

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:

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

where:

• $S[\phi]$ 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)
• 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?
– Will
Sep 4, 2015 at 14:27
• @Will exactly. Proof can be found in almost any QFT textbook. Sep 4, 2015 at 15:26
• Are there any particular QFT books that you would recommend?
– Will
Sep 4, 2015 at 17:13
• @Will Peskin & Schreder would be my choice, although there might be better options Sep 4, 2015 at 18:57
• 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. Jun 25, 2017 at 5:16