4
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

Perhaps you are confusing field operators in the Heisenberg picture and field operators in the interaction picture?

We usually compute Green's functions in perturbative QFT by moving to interaction picture where the field operators obey the free field equation like a free field does in Heisenberg picture, and the interaction term acts on states like in Schrodinger picture.

Now because the field operators obey the free field equation, they can be split into positive and negative frequencies in the usual fashion, leading to the usual definition of the contraction.

$\endgroup$
1
  • $\begingroup$ Oh, yes you are correct. Now I see how convenient interaction it is in dealing with interacting field. Thanks a lot. $\endgroup$
    – lsdragon
    Sep 3, 2021 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.