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In section 4.2 of An Introduction to Quantum Field Theory by M.E.Peskin and others, it derives interaction ground state by observing the time evolution of ground state in free field theory (pg.86), and then expands the expression in eigenstate of Hamiltonian of interacting field theory, $$e^{-iHT}|0\rangle = e^{-iE_{0}T}|\Omega\rangle\langle\Omega|0\rangle + \sum_{n\ne0}{e^{-iE_{n}T}|n\rangle\langle n|0\rangle}\tag{p.86}$$ However, in the latter deriving the two-point correlator, $\langle\Omega|\text{T}\{\phi(x)\phi(y)\}|\Omega\rangle$, it is assumed that the field operator being in Heisenberg's picture, and this can be seen from eq(4.30) (here it assumes $x^{0}>y^{0}>t_0$), $$\langle\Omega|\phi(x)\phi(y)|\Omega\rangle = \lim_{T\rightarrow\infty(1-i\epsilon)}{\left(e^{iE_0(2T)}|\langle0|\Omega\rangle|^2\right)^{-1}\langle0|U(T,t_0)\,\,U^{\dagger}(x^{0}, t_{0})\phi_{\text{I}}(x)U(x^{0}, t_0)U^{\dagger}(y^{0}, t_0)\phi_{\text{I}}(y)U(y^{0}, t_0)}$$

$$U(t_0, -T)|0\rangle\tag{4.30}$$ where $U(t, t')$ is the time evolution operator in interacting picture, and $\phi_{\text{I}}(x)$ would be the field operator in interacting picture. Then, I am confused about why it first derives the interaction ground state in Schrodinger's picture and then applies it in (4.30), which is in Heisenberg's picture.

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For what it's worth, the interaction time evolution operator $\hat{U}_I(t,t^{\prime})$ defined in eqs. (4.17-26) intertwines between operators $\hat{\phi}_I(x)$ in the interaction picture and ket and bra states in the Heisenberg picture. All the kets and bras in OP's quoted equations are in the Heisenberg picture.

It should probably be mentioned that the full LHS (and hence also RHS) of OP's first equation has an interpretation as a ket state in the Schrödinger picture if we choose reference time$^1$ $t=0$ for the 3 pictures. Nevertheless, this does not invalidate P&S's further reasoning.


$^1$ P&S use reference time $t=t_0$ for the 3 pictures.

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  • $\begingroup$ I found out that the kets and bras in this section are measured at reference time $t=t_0$, where Schr$\ddot{\text{o}}$dinger's picture, Heisenberg's picture, and interaction picture coincide. So there is no problem I think. $\endgroup$ Jan 9 at 2:07

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