# Can we obtain an exact answer when we sum Feynman diagrams to all orders?

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?

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

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.
• Hi, thanks for your answer! Which means my equation (B) is exact, right? The definition of interaction picture use free $H_0$? Commented Feb 4, 2023 at 3:50