Peskin & Schroeder's QFT problem 4.1(a) I am troubled for understanding P&S's problem 4.1 (a) on p. 126. Consider a Hamiltonian for the creation of Klein-Gordon particles by a classical source:
$$H=H_0+\int d^3 x(-j(t, \mathbf{x}) \phi(x)).\tag{p.126} $$
In 4.1 (a), the book said the probability that the source create no particles is given by
$$P(0)=\left|\left\langle 0\left|T\left\{\exp \left[i \int d^4 x j(x) \phi_I(x)\right]\right\}\right| 0\right\rangle\right|^2.\tag{p.126} $$
So why they use the free ground state $| 0\rangle$, not $| \Omega\rangle$? Since according to the book's eq.(4.31):
$$\langle\Omega|T\{\phi(x) \phi(y)\}| \Omega\rangle=\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\left\langle 0\left|T\left\{\phi_I(x) \phi_I(y) \exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle}{\left\langle 0\left|T\left\{\exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle} . \tag{4.31}$$
what they show about $P(0)$ is just the denominator of eq.(4.31).
 A: I recommend you to look and understand the subsection related to the S-Matrix. There, it is explained that the overlap between an initial state of the interacting theory $|k_1 k_2\rangle_{\mathrm{in}}$ of particles of momentum $k_1,k_2$ with a final interacting state $|p_1 p_2 ...\rangle_{\mathrm{out}}$ with momentum $p_1, p_2,...$ can be computed with the corresponding momentum of the free theory, using the relation
$$ \sideset{_{\mathrm{out}}}{} \langle p_1 p_2 ... |k_1 k_2 \sideset{}{_{\mathrm{in}}}  \rangle = \langle p_1 p_2 ... | S | k_1 k_2 \rangle = \left \langle p_1 p_2 ... \left | \ \mathrm{exp} \left(T \left \{ - i \int dt H_I (t) \right \} \right ) \ \right | k_1 k_2 \right \rangle  $$
where now $|k_1 k_2\rangle$ are momentum states of the free theory, and $S$ is the $S$-matrix. The probability associated with the transition from $k_1 k_2$ to $p_1 p_2 ...$ is then the modulus square of the above expression.
This relation also holds for the vacuum in which we are interested :
$$ \sideset{_{\mathrm{out}}}{} \langle \Omega  | \Omega \sideset{}{_{\mathrm{in}}}  \rangle = \langle 0 | S | 0 \rangle  $$
So the probability that the far-past vacuum $|\Omega \rangle_{\mathrm{in}}$ goes into the far-future vacuum $|\Omega \rangle_{\mathrm{out}}$ is
$$ P(0) = |\sideset{_{\mathrm{out}}}{} \langle \Omega  | \Omega \sideset{}{_{\mathrm{in}}} \rangle |^2 = \left | \left \langle 0 \left | \ \mathrm{exp} \left(T \left \{ - i \int dt H_I (t) \right \} \right ) \ \right | 0 \right \rangle \right |^2 $$
