Ordering ambiguity in the Feynman propagators obtained using Wick's theorem

Question

Applying Wick's theorem to a string of four field operators, $\phi_a\equiv\phi(x_a)$:

$$T(\phi_1\phi_2\phi_3\phi_4)=\{...\}, \tag{1}$$

we obtain several terms, three of which are fully contracted fields: $$\phi_1^{\bullet}\phi_2^{\bullet}\phi_3^{\bullet\bullet}\phi_4^{\bullet\bullet},\quad \phi_1^{\bullet}\phi_2^{\bullet\bullet}\phi_3^{\bullet}\phi_4^{\bullet\bullet},\quad \phi_1^{\bullet}\phi_2^{\bullet\bullet}\phi_3^{\bullet\bullet}\phi_4^{\bullet}. \tag{2}$$

Where I have given contracted fields the same number of dots. Each contracted field gives the associated Feynman propagator: $\phi_1^{\bullet}\phi_2^{\bullet}\equiv D_F(x_1-x_2)$.

My question is, when we have terms with more than one contraction, which propagator goes first? Based on what I am reading in Peskin and Schroeder we order them according to the ordering of the left-most contraction arm, however the book only demonstrates this for terms with four operators (so far), and I am unsure if this relation holds for terms with more fields.

Answer 1

For the scalar field, the order doesn't matter: it doesn't matter for the fields, why should it matter for the contractions? The only important thing is that you count every contraction just once (if you contract $(\phi_1,\phi_2)(\phi_3,\phi_4)$, you shouldn't count $(\phi_3,\phi_4)(\phi_1,\phi_2)$).

For the fermion field, it depends. For the free theory, it still doesn't matter: the only thing you need to do is untangle the contractions (and count a minus sign for every "step of untanglement"). Once you've done that, switching the place of two propagators would be like doing a quadruple switch:

$(\psi_1\bar{\psi}_2)(\psi_3\bar{\psi}_4) \rightarrow -\psi_1\psi_3\bar{\psi}_2\bar{\psi}_4 \rightarrow \psi_3\psi_1\bar{\psi}_2\bar{\psi}_4 \rightarrow -\psi_3\psi_1\bar{\psi}_4\bar{\psi}_2 \rightarrow (\psi_3\bar{\psi}_4)(\psi_1\bar{\psi}_2)$

For the interacting theory, well, it still wouldn't matter, but you have some "mandatory contractions": the field contracted with the incoming particle state must be the rightmost, while the one contracted with the outgoing particle state must be the leftmost. If you have other (internal) fermions, then their contraction order won't matter (provided that you already correctly untangled the contractions!).

Answer 2

In Wick's theorem, it is usually assumed that contractions supercommute with all pertinent operators, cf. e.g. my Phys.SE answer here. With this assumption there are no ambiguities in operator-ordering as long as one observes the sign rule for manipulating Grassmann-graded objects.

Answer 3

Mmm... For bosonic fields isn't the propagator symmetric? If so there is no ordering issue.

For fermionic fields you get a determinant or a Pfaffian instead of a hafnian or a permanent, just start with things in their original order and count the number of interchanges as you move them into adjacent pairs: a minus sign for each interchange.