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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.

Qmechanic
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Will
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  • The path integral integrates classical fields. – ACuriousMind Sep 02 '15 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 02 '15 at 22:26
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    Essentially a duplicate of http://physics.stackexchange.com/q/9183/2451 – Qmechanic Sep 02 '15 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 03 '15 at 05:09

2 Answers2

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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)
Prof. Legolasov
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  • 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 04 '15 at 14:27
  • @Will exactly. Proof can be found in almost any QFT textbook. – Prof. Legolasov Sep 04 '15 at 15:26
  • Are there any particular QFT books that you would recommend? – Will Sep 04 '15 at 17:13
  • @Will Peskin & Schreder would be my choice, although there might be better options – Prof. Legolasov Sep 04 '15 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. – Quantumwhisp Jun 25 '17 at 05:16
  • @Quantumwhisp to each quantum operator, we can associate a functional of a classical configuration. The details of this procedure can be found in any intro on path integrals. – Prof. Legolasov Jun 25 '17 at 08:02
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Question 1

The fields $\phi \left( x \right)$ in the generating functional $Z \left[ J \right]$ are classical fields, not operator fields. This is because the generating functional is defined as a path integral over all possible field configurations, weighted by a phase factor that depends on the action. The path integral formalism is equivalent to the operator formalism, but it does not use operators or commutation relations. Instead, it uses functional derivatives to obtain correlation functions and propagators. For example, the two-point function of a scalar field can be obtained by taking two functional derivatives of $Z \left[ J \right]$ with respect to $J \left( x \right)$ and setting $J = 0$. This gives the same result as taking the vacuum expectation value of the time-ordered product of two field operators in the operator formalism.

Question 2

The path integral does not define a mapping between field operators and classical fields, but rather a relation between correlation functions and functional integrals. The field operators are quantum objects that act on Hilbert space states and obey commutation relations, while the classical fields are functions on the configuration space that are integrated over in the path integral. The path integral formalism is a way of computing expectation values of products of field operators by using functional derivatives of a generating functional that depends on classical fields. The generating functional itself is defined as a path integral over all possible classical field configurations, weighted by a phase factor that depends on the action. The path integral formalism is equivalent to the operator formalism, but it does not use operators or commutation relations explicitly.

Hope it will help you all.

ZaellixA
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Olandelie
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