Contraction between two fields in the Interaction and Heisenberg picture

Question

The m-point correlator is (including interaction): \begin{equation*} \langle{\Omega|}\mathcal{T}\left(\phi_{H}\left(t_{m} \right)\phi_{H}\left(t_{m - 1}\right) ... \phi_{H}\left(t_{1} \right) \right){|\Omega}\rangle = \frac{\langle{0|} \mathcal{T}\left(\phi_{I}\left(t_{m} \right) \phi_{I}\left(t_{m - 1} \right) ... \phi_{I}\left(t_{1} \right)S \right){|0}\rangle}{\langle{0|}S{|0}\rangle} \end{equation*} The letter $H$ denotes the operator in the Heisenberg picture, while $I$ is used for the interaction picture.

The matrix $S$ is: \begin{equation*} S = \displaystyle \lim_{T \to \infty} \mathcal{T}\left(e^{-i\int_{-T}^{T} V\left(t'\right)dt'} \right) \end{equation*}

Consider the following interaction: \begin{equation*} V\left(t\right) = \frac{\lambda}{4!}\phi^{4}\left(t\right) \end{equation*}

Let's compute $\langle{0|}S{|0}\rangle$ up to first order: \begin{equation*} \langle{0|}S{|0}\rangle = 1 + \displaystyle \lim_{T \to \infty} - \frac{i\lambda}{4!}\int_{-T}^{T}\langle{0|}\mathcal{T}\left(\phi_{I}\left(t'\right) \phi_{I}\left(t'\right) \phi_{I}\left(t'\right) \phi_{I}\left(t'\right) \right){|0}\rangle dt' \end{equation*}

$\langle{0|}\mathcal{T}\left(\phi_{I}\left(t'\right) \phi_{I}\left(t'\right) \phi_{I}\left(t'\right) \phi_{I}\left(t'\right) \right){|0}\rangle $ can be computed by using Wick's theorem. The normal-ordered terms vanish, and only the terms with two contractions contribute. There are three possible ways to contract. The equation reduces to: \begin{equation*} \langle{0|}S{|0}\rangle = 1 + \displaystyle \lim_{T \to \infty} - \frac{i\lambda}{8}\int_{-T}^{T}\langle{0|}\mathcal{T}\left(\phi_{I}\left(t'\right) \phi_{I}\left(t'\right)\right){|0}\rangle dt' \end{equation*}

My question is, why is the contraction between two scalar fields in the Interaction picture the same as for two fields in the Heisenberg picture?

$\langle{0|}\mathcal{T}\left(\phi_{I}\left(t'\right) \phi_{I}\left(t'\right)\right){|0}\rangle = \langle{0|}\mathcal{T}\left(\phi_{H}\left(t'\right) \phi_{H}\left(t'\right)\right){|0}\rangle = D_{F}\left(t' - t'\right) = D_{F}\left(0\right)$

Answer 1

The reason for this is because $\phi_I(t, \vec{x})$ has the exact same form as $\phi_H(t,\vec{x})$, since $\phi_I(t,\vec{x})$ obeys the free Klein-Gordon equation.

To avoid repetition, I will just refer you to:

Why are fields in the interaction picture the same as free fields in the Heisenberg picture?

which contains a detailed description.