In Feynman functional integrals why do we integrate the action over all time?

Question

Say the definition of a propagator in quantum field theory is:

$$G_F(x,y)=\int \phi(x)\phi(y) e^{i S[\phi] } D\phi$$

where $S$ is the action. Why do we integrate the Lagrangian density from $t=-\infty$ to $t=+\infty$ instead of from $x_0$ to $y_0$?

i.e.

$$S[\phi] = \int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty} {\cal{L}}(x,y,z,t)dxdydzdt$$

where $\cal{L}$ is the Lagrangian density. Surely we are interested only in the section between $t=x_0$ and $t=y_0$?

Answer 1

Short answer (assuming proper time-ordering of $x$ and $y$):

$$ \langle \Omega| \phi(x) \phi(y) | \Omega\rangle = \frac{ \langle 0 | U(-\infty, x^0) \phi_I(x) U(x^0, y^0) \phi_I(y) U(y^0, +\infty) |0 \rangle }{\langle 0 | U(-\infty, \infty) | 0 \rangle } $$

These infinities has to be in the action integral too.

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Now there is a rather deep reason for having time going to infinity in the path integral (the credit for this goes to Weinberg and his awesome book, as usual). The propagators usually has funny $i \epsilon$ in denominators, like:

$$ \frac{1}{q^2 + m^2 - i \epsilon} $$

Which makes possible integration over $q$ and helps select proper pole with Residue theorem.

Although it is usually just assumed and omitted in calculation, this $i \epsilon$ comes from the phases of field at time $\pm \infty$:

$$ \langle \Omega,out| \phi(\infty) \rangle \langle \phi(-\infty) | \Omega, in \rangle \propto e^{\epsilon \, \times \, \cdots \text{pairs of fields} } $$

for some infinitesimal $\epsilon$.

(Note that in interacting theory it is possible to calculate this product of projections only at $t = \pm \infty$, because it is assumed that the theory is free at this time.)

Given that this product is proportional to the exponential form - the product of $\epsilon$ and pairs of fields will be merged with the action! This is the reason to have $\epsilon$ in the propagator.

(You may find more details in Chapter 9 of Weinberg's QFT Volume 1)

Answer 2

This is a common theme in physics: If you have a separation of characteristic scales in the problem, physical effects associated with sub-dominant scales can be ignored.

E.g. in $S$-matrix theory, it is often assumed for simplicity, that the interactions can be neglected near the initial and final times, $t=t_i$ and $t=t_f$, and that the time interval $[t_i,t_f]$ is much bigger than all the characteristic time scales of the interactions.

In other words, if we are not interested in studying temporal boundary effects, we might as well send $t_i\to -\infty$ and $t_f\to \infty$ to simplify the problem.