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On Peskin and Schroeder's QFT book page 289, the book is trying to derive the functional formalism of $\phi^4$ theory in first three paragraphs. But the book omits many details (I thought), so I have some troubles here.

For the free Klein-Gordon theory to $\phi^4$ theory: $$ \mathcal{L}=\mathcal{L}_0-\frac{\lambda}{4 !} \phi^4. $$ Assuming $\lambda$ is small, we expand $$\exp \left[i \int d^4 x \mathcal{L}\right]=\exp \left[i \int d^4 x \mathcal{L}_0\right]\left(1-i \int d^4 x \frac{\lambda}{4 !} \phi^4+\cdots\right). $$ Here I thought the book use one approximation, since $\phi^4$ don't commute with $\mathcal{L_0}$, their have $\pi$ inside $\mathcal{L_0}$. And according to Baker-Campbell-Hausdorff (BCH) formula, the book omit order to $\lambda$. Is this right?

The book further says on p. 289:

"Making this expression in both the numerator and denominator of (9.18), we see that each is expressed entirely in terms of free-field correlation functions. Moreover, since $$ i \int d^3 x \mathcal{L}_{\mathrm{int}}=-i H_{\mathrm{int}},$$ we obtain exactly the same expression as in (4.31)."

I am really troubled for this, can anyone explain for me?

Here eq. (9.18) is $$\left\langle\Omega\left|T \phi_H\left(x_1\right) \phi_H\left(x_2\right)\right| \Omega\right\rangle=$$ $$\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\int \mathcal{D} \phi~\phi\left(x_1\right) \phi\left(x_2\right) \exp \left[i \int_{-T}^T d^4 x \mathcal{L}\right]}{\int \mathcal{D} \phi \exp \left[i \int_{-T}^T d^4 x \mathcal{L}\right]} . \tag{9.18} $$

And eq. (4.31) is $$\langle\Omega|T\{\phi(x) \phi(y)\}| \Omega\rangle=\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\left\langle 0\left|T\left\{\phi_I(x) \phi_I(y) \exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle}{\left\langle 0\left|T\left\{\exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle} . \tag{4.31} $$

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2 Answers 2

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In the path integral formalism, $\phi$ is not an operator, it is an integration variable. In other words, inside the integral $\int D\phi$, $\phi$ is just an ordinary classical field. So there’s no need to worry about Baker-Campbell-Hausdorff and such.

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  • $\begingroup$ Thank you very much, you pointed out the most essential part, later analysis is easy! $\endgroup$
    – Daren
    Sep 27, 2022 at 10:30
  • $\begingroup$ By the way, it seems that for a $\phi^3$ theory, all these expansions are zero except free two point correlation function. Since it has odd numbers of $\phi$ inside the Gaussian integral. However, I remember that Srednicki use $\phi^3$ theory as an example to derive functional integral formalism, I forgot some details about that. Do you have any comments on this? thanks. $\endgroup$
    – Daren
    Sep 27, 2022 at 10:34
  • $\begingroup$ You're making a simple mistake: the first-order term has five $\phi$'s and vanishes, but the second-order term has eight $\phi$'s and survives. Check that the first nontrivial Feynman diagram in Srednicki's analysis of the $\phi^3$ two-point function has two vertices, making it a second-order perturbation. $\endgroup$
    – Zack
    Sep 27, 2022 at 17:23
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User Zack has already answered OP's first part. Concerning the equivalence between the interaction picture (4.31) and the path integral formulation in the Heisenberg picture (9.18), note that P&S presumably assumes no derivative couplings/interactions. For the latter case, see e.g. this Phys.SE post.

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