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Consider a $\phi^4$ theory in QFT. Following Peskin & Schroeder's QFT chapter 4, we can do some calculations of correlation functions using perturbation expansion. On their book's page 83, they consider the interaction picture field $\phi_I(t,\textbf{x})$ $$ \phi_I(t, \mathbf{x})=\left.\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\mathbf{p}}}}\left(a_{\mathbf{p}} e^{-i p \cdot x}+a_{\mathbf{p}}^{\dagger} e^{i p \cdot x}\right)\right|_{x^0=t-t_0} \tag{4.15} $$ then $$ H_I(t)=e^{i H_0\left(t-t_0\right)}\left(H_{\mathrm{int}}\right) e^{-i H_0\left(t-t_0\right)}=\int d^3 x \frac{\lambda}{4 !} \phi_I^4 \tag{4.19} $$

So (4.15) is very important. And $\text{exp}[-i\int_T^T dt H_I(t)]$ is a crucial part to calculate the correlation function. For example, (4.31) $$ \langle\Omega|T\{\phi(x) \phi(y)\}| \Omega\rangle=\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\left\langle 0\left|T\left\{\phi_I(x) \phi_I(y) \exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle}{\left\langle 0\left|T\left\{\exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle} \tag{4.31} $$

But, I think (4.15) is an $\textbf{approximation}$ in $\phi^4$ theory. Since (4.15) implies KG equation $$(\partial^2+m^2)\phi_I(t,\textbf{x})=0 \tag{A}$$ this can be derived from the Heisenberg equation $$\frac{\partial \phi_I(t,\textbf{x})}{\partial t}=i[H_0,\phi_I(t,\textbf{x})] \tag{B}$$ where $H_0$ is the free field Hamiltonian.

I think (B) is an approximation, since we need to use the full $H$, instead of $H_0$, where $H=H_0+\frac{\lambda \phi^4}{4!}$.

Does this means even though we expend (4.31) to all orders, we still cannot an exact answer, since $\phi_I(t, \mathbf{x})$ is an approximation from the first place?

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  • $\begingroup$ A stupid question that I proposed! In the interaction picture, my equation $B$ is exact. As in Sakurai QM p.321. $\endgroup$
    – Daren
    Feb 4, 2023 at 7:52

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In the interaction picture, the time evolution of operators is given by the free Hamiltonian $$ \left(\square + m^2\right) \phi_I = 0 $$ and the time evolution of states is determined by the interaction Hamiltonian $$ i \frac{\partial |\Psi\rangle_I}{\partial t} = H_I |\Psi\rangle_I $$ This is not an approximation. The interaction picture is related to the Schrodinger and Heisenberg pictures by unitary transformations. You can describe the dynamics exactly in any of these pictures.

However, summing Feynman diagrams to all orders does not give an exact answer -- but for a different reason than the one you brought up. In particular, the Feynman diagram expansion is an asymptotic expansion. The series actually diverges if you include all terms; for a given value of the coupling, there is an optimal number of terms to include that gets you close to the right answer. The ultimate issue is that the exact transition amplitudes include non-perturbative effects such as instantons that cannot be computed within perturbation theory.

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    $\begingroup$ What if we use borel transformation to sum the asymptotic series? I know there are renormalon problems, but also there are schemes that get around this problem right? Is it expected (for a theory, say QCD) that the perturbative coefficients to all orders recovers the nonperturbative behaviour? $\endgroup$ Feb 4, 2023 at 0:32
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    $\begingroup$ @QCD_IS_GOOD Yes that's a good question and an active area of research. I'm not an expert, but my understanding is that the hope is Borel resummation could be used to learn non-perturbative information about generic field theories. I am not clear on the status of these techniques applied to QCD; my impression is that there has been some success but QCD is too complicated to be completely solved this way, at least so far. $\endgroup$
    – Andrew
    Feb 4, 2023 at 0:43
  • $\begingroup$ Hi, thanks for your answer! Which means my equation (B) is exact, right? The definition of interaction picture use free $H_0$? $\endgroup$
    – Daren
    Feb 4, 2023 at 3:50
  • $\begingroup$ @Daren Correct. See Section 3.1 of damtp.cam.ac.uk/user/tong/qft.html. $\endgroup$
    – Andrew
    Feb 4, 2023 at 16:05
  • $\begingroup$ Also see the recent quanta article about Resurgence quantamagazine.org/… $\endgroup$
    – BjornW
    Apr 14, 2023 at 14:51

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