I am following section 9.2 in Peskin and Schroeder in which the Feynman rules are derived for scalar fields.

They define (in eqn (9.14), page 282) the transition amplitude from $\vert\phi_a\rangle$ to $\vert\phi_b\rangle$ in time $T$ to be

$$\langle\phi_b\vert e^{-iHT}\vert\phi_a\rangle = \int \mathcal{D}\phi \exp\left[i\int_0^Td^4x\mathcal{L}\right]\tag{1}.$$

Then (on page 289) they deal with the $\phi^4$ theory and use the expansion

$$\exp\left[i\int_0^Td^4x\mathcal{L}\right] = \exp\left[i\int_0^Td^4x\mathcal{L_0}\right]\left(1-i\int d^4x \frac{\phi}{4!}\phi^4+\cdots\right)\tag{2}$$

Each term on the RHS is indeed of the form

$$\int \mathcal{D}\phi\; \phi(x_1)\cdots\phi(x_n) \exp\left[i\int_0^Td^4x\mathcal{L_0}\right],$$

which appears in an easy generalization of the formula (eqn (9.18), page 284)

$$\langle\Omega\vert T\phi_H(x_1)\phi_H(x_2)\vert\Omega\rangle = \lim_{T\to\infty(1-i\epsilon)} \left(\frac{\int \mathcal{D}\phi\; \phi(x_1)\phi(x_2) \exp\left[i\int_{-T}^Td^4x\mathcal{L_0}\right]}{\int \mathcal{D}\phi\ \exp\left[i\int_{-T}^Td^4x\mathcal{L_0}\right]}\right)\tag{3}$$

To derive the Feynman rules, we want to swap each term in (2) for a correlation function.

  1. What happens to the denominator of (3)?

  2. Also, should I worry about taking $T\to\infty(1-i\epsilon)$ instead of $T\to \infty$ (which would the natural thing to do in (1))?


2 Answers 2


I think some of your confusion stems from the fact that there are two different kinds of vacuua in QFT. First there is the vacuum of the free theory, usually denoted $|0\rangle $, second there is the full (interacting) vacuum, usually denoted $|\Omega \rangle$.

What we want to calculate are the different quantities in the full theory like: \begin{equation} \langle\Omega\vert T\phi(x_1)\phi(x_2)\vert\Omega\rangle. \end{equation}

What you have written in equation (3) is acutally the propagator in the free theory (see that you use $\mathcal{L}_0$ and not $\mathcal{L}$ as it should be). This may seem like a minor point but the difference is essential to understand. (You must have copied it wrong from eqn (9.18) in P & S, since it looks correct there.

The problem is that we do cannot calculate this directly. What we do is we basically say: What if the full theory is almost like the free theory but with a small interaction term? That is, we do perturbation theory. This is because we know the propagator of the free theory:

$$ \Delta_F(x_1-x_2) =\langle0\vert T\phi(x_1)\phi(x_2)\vert0\rangle = \frac{\int \mathcal{D}\phi\; \phi(x_1)\phi(x_2) e^{iS_0}}{\int \mathcal{D}\phi\; e^{iS_0}}.$$

The trick is then to rewrite everything in terms of the propagator of the free theory.

The actual derivation of the full 2-point function is fairly involved, but you can find it in P & S pages 82-99.


Lets start with the second question:

Also, should I worry about taking $T\to\infty(1-i\epsilon)$ instead of $T\to \infty$ (which would the natural thing to do in (1))?

And there lets start with asking ourselves "why" do we do the limit in the first place?
The answer is -- we don't want the expectation for the $|\phi_a\rangle$ and $|\phi_b\rangle$ states -- we want it for the ground state of the interacting field theory, ususlly denoted as $|\Omega\rangle$. And getting that is a tricky business, explained in detail in the Chapter 4 of the book. But the basic idea is that you take the ground state $|0\rangle$ of non-interacting QFT and evolve it for a time T.

$$e^{-iHT}|0\rangle = \sum_n e^{-iE_nT}|n\rangle\langle n|0\rangle$$

With $|n\rangle$ being the sates of interacting QFT. Then you notice that if you shift $T$ a little in a complex direction, then the exponents become decreasing. Last trick is to take a limit of large $\Re T$ so that only $E_0 = \langle\Omega|H|\Omega\rangle$ survives.

So here is your reason for all the fuss with $T\to\infty(1-i\epsilon)$

What happens to the denominator of (3)?

That is related to the previous point. While doing the procedure you'll have some extra factor, that you have to divide by (see eq. (4.27)): $$|\Omega\rangle = \lim_{T\to \infty(1-i\epsilon)}\left(e^{-iE_0T}\langle\Omega|0\rangle\right)^{-1}e^{-iHT}|0\rangle$$

So that is where the denominator comes from.

  • $\begingroup$ Thanks for your answer! I think I follow the derivation of (3), including the fuss about the imaginary part to kill off all the other energy eigenstates. My concern was that we shouldn't be doing this in (1), where the aim is to find a transition probability from $t=-\infty$ to $t=+\infty$. The other concern is that the denominator in (3) seems to disappear when we use equation (3) in (2) to find the Feynman rules for a particular theory. $\endgroup$
    – octopus
    Commented Sep 22, 2014 at 15:56
  • 1
    $\begingroup$ @octopus we simply set the vacuum-to-vacuum amplitude to 1 (in a free theory), so the denominator is equal to unity! This makes a lot of sense physically, because nothing should ever 'happen'in a free theory when you start in the ground state. $\endgroup$
    – Danu
    Commented Sep 22, 2014 at 16:26
  • $\begingroup$ Does this mean we're defining $\langle\phi_b\vert e^{-iHT}\vert\phi_a\rangle = \int \mathcal{D}\phi \exp\left[i\int_0^Td^4x\mathcal{L}\right]$differently to make the denominator go away or is there a reason why the denominator is actually one? $\endgroup$
    – octopus
    Commented Sep 22, 2014 at 17:08

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