# Feynman propagator for interacting field

For scalar field, Feynman propagator is commonly defined as

$$\Delta_F(x-y) = \langle 0 | T\phi(x)\phi(y)|0 \rangle .$$

For free theory, field satisfy equation of motion is $$\phi(x) = \int\frac{dp^3}{(2\pi)^32E_k} a_ke^{-ik\cdot x}+a_k^\dagger e^{ik\cdot x} ,$$

with commutation relation $$[a_k, a_{p}^\dagger] = (2\pi)^3 2 E_k \delta(k-p)$$.

Then, Feynman propagator can be written as $$\Delta_F(x-y)= \int\frac{dp^4}{(2\pi)^4} \frac{1}{p^2-m^2+i\epsilon}.$$

For an interacting field whose Lagrangian has interaction terms such as $$\phi^4$$, Green functions such as $$G_n = \langle 0 |T \phi^1 ... \phi^n|0\rangle$$ are computed via wick expansion involving contraction which is defined as $$contraction(\phi_1,\phi_2) = \Theta(t_1-t_2)[\phi_1^\dagger,\phi_2^-]+\Theta(t_2-t_1)[\phi_2^\dagger, \phi_1^-].$$ Here, $$\phi^\dagger = \int\frac{dk^3}{(2\pi)^32E_k} a_ke^{-ik\cdot x} \\ \phi^- = \int\frac{dk^3}{(2\pi)^32E_k} a_k^\dagger e^{ik\cdot x}.$$

For free theory, it is easily to verify that $$contraction(\phi(x),\phi(y)) = \Delta_F(x-y)$$ due to commutation relations $$[a_k, a_{p}^\dagger] = (2\pi)^3 2 E_k \delta(k-p)$$. In interacting theory, fields should satisfy equation of motion in consideration of interactions. They are not necessarily the same form as that in free theory. If field operator is Fourier transformed, we can not guarantee that Fourier amplitudes always are creation and annihilation operators as in free theory. It is assumed to be true at asymptotic time in LSZ reduction form by renormalization of field.

Then why contraction is equal to Feynman propagator for interacting theory when expansion of interacting fields by creation and annihilation operators is not possible at all time from infinite past to infinite future?