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After reading from Yeh's notes on Advanced Condensed Matter Theory, Section II.9, I've decided that Kleinert was right about the numerator, but the denominator actually comes from normalization.
So, you start from the definition of the Green's function.
$$G^{(n)}(x_1,\cdots, x_n)=\bra{\Omega}\mathcal{T}\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\ket{\Omega}$$
Tangent...regarding my question on the use of $\mathcal{T}$ to change pictures:
Note that $\mathcal{T}$ in the expression above is implicitly defined to time-order all of the visible symbols inside of its grasp, that is, it operates symbolically on the $\phi_{iH}(t)$. You can't use it to "cheat" like this sketchy misproof:
$$\phi_I(t)=\mathcal{T}\left[\phi_I(t)\right]\\=\mathcal{T}\left[e^{i\int_0^t dt'H_0(t')}\phi_Se^{-i\int_0^t dt'H_0(t')}\right]\\\neq \mathcal{T}\left[e^{i\int_0^t dt'H_0(t')}e^{-i\int_0^t dt'H_0(t')}\phi_S\right]\\=\mathcal{T}[\phi_S]=\phi_S$$
The middle "equality" actually fails because the mistaken proofwriter accidentally changed definitions of $\mathcal{T}$ without noticing. The first line works assuming $\mathcal{T}$ is defined to rearrange the set of explicitly visible operators (in this case, just trivially $\phi_I(t)$). But the middle equality would only be possible if the $\mathcal{T}$ had been intended to order all the embedded $H_0(t)$ as well. Since the $H_0(t)$ in line 2 do not happen to be all ordered already, one cannot make this switch of definitions for $\mathcal{T}$ without changing the value of the expression. One should thus be careful to be consistent on how $\mathcal{T}$ is defined to avoid contradictions. It doesn't operate on the value inside the brackets, it operates in a specific manner on the symbols.
Back to the flow of the proof
We want to substitute in our Gell-Mann and Low expression into the definition of the Green's function, but notice that $\ket{\Omega}$ is not just $\ket{\Psi^{\pm}}$... because $\ket{\Psi^{\pm}}$ is not normalized. So really
$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{\Psi^+}\mathcal{T}\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\ket{\Psi^-}}{\braket{\Psi^+|\Psi^-}}$$
(where I'm using the fact that actually $\ket{\Psi^+}=\ket{\Psi^-}$ and are just different expressions for the same ket.) Expanding, we find
$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}U_I(\infty,0)\mathcal{T}\left[\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\right]U_I(0,-\infty)\ket{0}}{\braket{0|U_I(\infty,0)|0}\braket{0|U_I(0,-\infty)|0}}\\\quad \times 1/\frac{\braket{0|U_I(\infty,0)U_I(0,-\infty)|0}}{\braket{0|U_I(\infty,0)|0}\braket{0|U_I(-0,\infty)|0}}$$
The unpaired Moller operators in the denominator cancel:
$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}U_I(\infty,0)\mathcal{T}\left[\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\right]U_I(0,-\infty)\ket{0}}{\braket{0|U_I(\infty,-\infty)|0}}$$
What we CAN do with $\mathcal{T}$
Now we just have to clean up the numerator. Assume WLOG that, for a given $\{x_i\}$, those Heisenberg operators are time ordered when written from 1 to $n$ as above. Then write them out in terms of the interaction operators using
$$O_H(t)=U^\dagger(t,0)U_0(t,0)O_I(t)U_0^\dagger(t,0)U(t,0)$$
and
$$U_I(t, t')=U_0^{\dagger}(t, 0)U(t, t')U_0(t', 0)$$
(where subscripts indicate full, free, and interaction evolution operators) to reach
$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}\mathcal{T}\left[U(\infty,t_1)\phi_{1H}(x_1)U_I(t_1,t_2)\phi_{2H}(x_2)\cdots U_I(t_{n-1},t_n)\phi_{nH}(x_n)U(t_n,-\infty)\right]\ket{0}}{\braket{0|U(\infty,-\infty)|0}}$$
I played some tricks here; let me explain. Formally, the $\mathcal{T}$ was only assumed at the start to order the $\phi_{iH}(t)$ around as unbreakable chunks (as discussed above). But since each of the individual evolution operators contains only the already time-ordered $H_I(t)$ from $t_i$ to $t_{i+1}$, then, if the $\phi_{iH}(t)$ are in order, that means all of the various implicit $H_I(t)$ in the entire expression are actually in time-order for us. So now we can actually treat the $\mathcal{T}$ as ordering the $\phi_{iH}(t)$ and the $H_I(t)$ without changing the value of our expression. Then we can also bring the outermost evolution operators inside the $\mathcal{T}$ since they are already at earlier/later times than all other operators.
Now, having played all those games with the definition of the time-ordering operator to show that we can have it apply to all of the various symbols inside the brakets for this expression...we are free to rearrange all those symbols ourselves. Let's put all the evolution operators together to the right, and voila
$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}\mathcal{T}\left[\phi_{1H}(x_1)\cdots \phi_{nH}(x_n)S\right]\ket{0}}{\braket{0|S|0}}$$
where the $\mathcal{T}$ is understood to order together all of the $\phi_I(t)$ and the $H_I(t)$ inside the Dyson series for the $S$-matrix (even though the $H_I(t)$ inside the $S$-matrix are, of course, already time-ordered within themselves by the Dyson series!) Phew!
The first question we have to ask is: what is a one particle state in an interacting theory? It is reasonable to require that they are states that are both momentum eigenstates and energy eigenstates. (In fact, as the Hamiltonian and the momentum operator commute, these are not two different conditions.) Weinberg, in his famous textbook, says that particle states are those which transform under an irreducible representation of the Poincare group, but we need not fuss around with the Poincare group here.
All we will say is that, in the interacting theory, there are some single particle states, labelled by
$$|\lambda k \rangle$$
where $k$ is the four-momentum, and $\lambda$ is whatever other labels we need for our particles. (In this answer I will be working with just a real scalar field, but even in the spin-0 case there can still be extra data that distinguishes our particles in an interacting theory.)
Now, we know know we have a set of momentum and energy eigenstates $|\lambda k\rangle$ that represent the stable particles of our theory. We can now "smudge out" these definite momentum states into wave packets, using a Gaussian window function $f_W$ that has some momentum uncertainty $\kappa$. We will denote these smudged out, approximate energy and momentum eigenstates with a subscript $W$ for "window."
$$|\lambda k\rangle_W \equiv \int d^3{\mathbf{k}'} f_W(\mathbf{k} - \mathbf{k}') |\lambda k\rangle$$
We will come back to these.
Now, the free vacuum $|0\rangle$ of $\hat H_0$ and the true vacuum $|\Omega\rangle$ of $\hat H = \hat H_0 + \hat H_{\rm int}$ are very different states. Particles in the interacting theory must indeed be defined to be formed from the action of the "creation operator" on the true vacuum, as long as we properly define what we mean by the "creation operator" in the interacting theory.
To create an annihilate particles, we will use the Klein Gordon inner product. (We suppress $\hbar$ and $c$.)
$$(\psi_1, \psi_2)_{KG} \equiv i \int d^3 x (\psi^*_1 \partial_t \psi_2 - \partial_t \psi^*_1 \psi_2)$$
The motivation for defining this is that in the FREE theory, the Klein Gordon inner product gives us an inner product between single particle states. If we have two single particle states (in the free theory) $|\Psi_1\rangle$ and $|\Psi_2 \rangle$, we have
$$\langle \Psi_1 | \Psi_2 \rangle = ( \psi_1, \psi_2 )_{KG}$$
where we used the "single particle wave functions" of the states defined by
$$\psi_i(x) \equiv \langle 0| \hat \phi(x) |\Psi_i\rangle$$
The niceties of free field theory come from the simple algebra of the creation and annihilation operators, combined with the fact that the annihilation operator annihilates the vacuum. We will try to recreate those relationships using the Klein Gordon inner product. However, to do this, we will need to use widely separated wave packets.
From here on out, everything will be in the interacting theory.
For a given function $\psi$, we define the creation and annihilation operators that "create" the state corresponding to that wave function as follows.
$$ \hat a^\dagger_i (t) \equiv -\big( \psi^*_i(t, \cdot), \hat \phi(t, \cdot) \big)_{KG} $$
$$\hat a_i(t) = \big( \psi_i(t, \cdot), \hat \phi(t, \cdot) \big)_{KG}$$
(In the free theory, this creation operator literally would create the single particle state with the single particle wave function $\psi_1$.)
(Something I must mention about these operators is their time evolution. It is a point of notational confusion that $\hat a^\dagger_{1}(t)$ depends explicitly on a time $t$, given that we usually have defined time dependence such that $e^{i \hat H t} \hat{O}(t') e^{-i \hat H t} = \hat {\mathcal{O}}(t'+t)$. This is not the case here.)
Now, sadly, in the interacting theory, the annihilation operator defined above will not annihilate the vacuum. However, we can recover something close:
$$\langle \Omega| \hat a_1(t) |\Omega\rangle = i \int d^3{x}\langle \Omega| \big( \psi_1^*(t, \vec x) \partial_t \hat \phi (t, \vec x) - \partial_t \psi_1^*(t, \vec x) \hat \phi(t, \vec x) \big) |\Omega\rangle$$
$$= i \int d^3{x} \big( \psi_1^*(t, \vec x) \partial_t \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle - \partial_t \psi^*_1(t, \vec x) \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle \big)$$
$$= i \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle \int d^3{x} (- \partial_t \psi_1^*(t, \vec x))$$
The fact that $\partial_t \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle = 0$ follows directly from the fact that the vacuum state has zero energy, so $e^{- i \hat H t} |\Omega\rangle = |\Omega\rangle$. Now as we want $\langle \Omega| \hat a_1(t) |\Omega\rangle = 0$ for any $\psi_1$, we can see that this is achieved if and only if $\langle \Omega| \hat \phi(x) |\Omega\rangle = \langle \Omega| \hat \phi(0) |\Omega\rangle = 0$. We will assume this is the case.
In the free theory, $\langle 0| \hat a_1(t) \hat a_2^\dagger(t) |0\rangle = \langle \Psi_1 | \Psi_2\rangle = (\psi_1, \psi_2)_{KG}$. In an interacting theory, for any $\hat a_1$ and state $|\Psi_2\rangle$ (not just a single particle state) we have
$$\langle \Omega| \hat a_1(t) |\Psi_2\rangle = \langle \Omega| \big( \psi_1(t, \cdot) , \hat \phi(t, \cdot) \big)_{KG}|\Psi_2\rangle$$
$$= \big( \psi_1(t, \cdot), \langle \Omega| \hat \phi(t, \cdot) |\Psi_2\rangle\big)_{KG} $$
$$= \big( \psi_1(t, \cdot), \psi_2(t, \cdot) \big)_{KG}$$
$$\langle\Psi_2| \hat a_1(t) |\Omega\rangle = \big( \psi_1(t, \cdot) , \psi_2^*(t, \cdot) \big)_{KG}$$
Remember our single particle states? We're now going consider the "single particle wave function" of those states. Namely, they have to be plane waves.
$$\langle \Omega| \hat \phi(x) |\lambda k\rangle = C_\lambda e^{-ikx}$$
where $C_\lambda$ is a constant that depends on $\lambda$.
We now want to see what our states $\hat a^\dagger_1|\Omega\rangle$ have to do with these true particle wave packets $| \lambda k \rangle_W$. To do this, we will see what the inner product of these two states are. Just from our simple algebra above, for an annihilation operator $\hat a_{\lambda_1 k_1} = (\psi_{k_1}, \hat \phi)_{KG}$ where $k_1^2 = m_{\lambda_1}^2$, we have
\begin{equation*}
\begin{split}
\langle \Omega | \hat a_{\lambda_1 k_1} (t) |\lambda_2 k_2\rangle_W = \big( \psi_{ k_1}(t, \cdot), \langle \Omega |\hat \phi(t, \cdot)|\lambda_2 k_2\rangle_W \big)_{KG} = C_{\lambda_2}\big(\psi_{ k_1}(t, \cdot), \psi_{ k_2 }(t, \cdot) \big)_{KG}\\
{}_W \langle \lambda_2 k_2 | \hat a_{\lambda_1 k_1}(t) |\Omega\rangle = \big( \psi_{ k_1}(t, \cdot), {}_W \langle \lambda_2 k_2 |\hat \phi(t, \cdot) |\Omega\rangle \big)_{KG} = C_{\lambda_2}\big(\psi_{ k_1}(t, \cdot), \psi^*_{ k_2 }(t, \cdot) \big)_{KG}.
\end{split}
\end{equation*}
We desire for the top expression to be $\propto \delta_{\lambda_1 \lambda_2} \delta^3(\mathbf{k}_1 - \mathbf{k}_2)$ and for the bottom expression to be 0. If this were the case, then the only single particle state $\hat a^\dagger_{ k_1}(t) |\Omega\rangle$ would overlap with would be $|\lambda_1 k_1\rangle$, and $\hat a_{k_1}(t)$ could still functionally "annihilate" the vacuum, even though we need to keep ${}_W \langle \lambda k |$ on the left. Defining $\omega_{\lambda k} \equiv (m^2_{\lambda} + \mathbf{k}^2)^\frac{1}{2}$, we have
\begin{equation*}
\begin{split}
\big(\psi_{ k_1}(t, \cdot), \psi_{ k_2 }(t, \cdot) \big)_{KG} = (2 \pi)^3 \int d^3{\mathbf{k}} f_W(\mathbf{k}_1 - \mathbf{k}) f_W(\mathbf{k}_2 - \mathbf{k}) (\omega_{\lambda_1 k} + \omega_{\lambda_2 k})e^{it(\omega_{\lambda_1 k} - \omega_{\lambda_2 k})} \\
\big(\psi_{ k_1}(t, \cdot), \psi_{ k_2 }^*(t, \cdot) \big)_{KG} = (2 \pi)^3 \int d^3{\mathbf{k}} f_W(\mathbf{k}_1 - \mathbf{k}) f_W(\mathbf{k}_2 + \mathbf{k}) (\omega_{\lambda_1 k} - \omega_{\lambda_2 k})e^{it(\omega_{\lambda_1 k} + \omega_{\lambda_2 k})}. \\
\end{split}
\end{equation*}
The top expression is not $\propto \delta_{\lambda_1 \lambda_2} \delta^3_W(\mathbf{k}_1 - \mathbf{k}_2)$ and the bottom expression is not $0$. However, if we take $\kappa \ll |\mathbf{k}_1 - \mathbf{k}_2|$ and also take $t \to \pm \infty$, they are! This hinges on our assumption that $m_{\lambda_1} \neq m_{\lambda_2}$ if $\lambda_1 \neq \lambda_2$. The $e^{it (\ldots)}$ term will oscillate wildly in both integrals if $\lambda_1 \neq \lambda_2$, causing them to be 0. In the top integral, this oscillation does not occur when $\lambda_1 = \lambda_2$. Furthermore, the top integral will be negligible unless $\mathbf{k}_1 = \mathbf{k}_2$. Taking the $f_W(\mathbf{k}) \to \delta^3(\mathbf{k})$ and $t \to \pm \infty$ limit, we can now write
\begin{equation*}
\begin{split}
\langle \lambda_2 k_2 | \hat a_{\lambda_1 k_1}^\dagger (\pm \infty) |\Omega\rangle = C_{\lambda_2} (2 \pi)^3 2 \omega_{\lambda_2 k_2} \delta_{\lambda_1 \lambda_2} \delta^3(\mathbf{k}_1 - \mathbf{k}_2) \\
\langle \lambda_2 k_2 | \hat a_{\lambda_1 k_1} (\pm \infty) |\Omega\rangle = 0.
\end{split}
\end{equation*}
These properties are even more important than I let on. This is because the states $| \lambda k \rangle$ are so generally defined: they are just momentum eigenstates with all the extra necessary data stuffed into $\lambda$. As they diagonalize the momentum operator, they form a basis of our entire state space! Therefore, we can immediately see from the first equation that
$$\hat a^\dagger_{\lambda k}(\pm \infty) |\Omega\rangle = -C_\lambda \big( e^{ikx}, \hat \phi( x)\big)_{KG} |\Omega\rangle \vert_{t = \pm \infty} = | \lambda k \rangle$$
where we have chosen the normalization $\langle\lambda k | \lambda' k' \rangle = C_{\lambda}^* C_{\lambda'} (2 \pi)^3 (2 \omega_{\lambda k}) \delta_{\lambda \lambda'} \delta^3(\mathbf{k} - \mathbf{k}')$. From the second equation, we can immediately see that
$$\langle\Psi | \hat a_{\lambda k}(\pm \infty) |\Omega\rangle = 0 \hspace{0.15 cm} \text{ for all } \langle\Psi | \hspace{0.5 cm} \Longrightarrow \hspace{0.5 cm} \hat a_{\lambda k}(\pm \infty) |\Omega\rangle = 0.$$
Apparently our asymptotic creation and annihilation operators behave almost exactly like our good old creation and annihilation operators from the free theory!
There's another important property I must mention, which is that two creation/annihilation operators that have different $\lambda k$ data will commute. This is a direct consequence of the fact that our creation/annihilation operators are spacial integrals weighted by wave packets that are spatially separated at large times. (For operators with the same $k$ but different $\lambda$, as $m_\lambda$ is different the wave packets will propagate at different speeds and will still succeed to separate.) Note that spatial separation is a property of wave packets but not of plane waves. This is another place where it is necessary to view plane waves as a limit of wave packets in order to properly understand your theory. In fact, the operators will not commute unless they are defined with this limiting procedure.
We are finally ready to define our incoming and outgoing multi-particle states. As our asymptotic creation operators only change the ground state in localized spatial regions and each spatial excitation is justifiably called a "particle state" we can say that acting with a few of them on the ground state will create a perfectly good multi-particle state. We will now define our incoming (created at $t = -\infty$) and outgoing (created at $t = + \infty$) multi-particle asymptotic states.
$$
|\lambda_1 k_1, \ldots, \lambda_n k_n\rangle_{\rm in} \equiv \hat a^\dagger_{\lambda_1 k_1}(-\infty) \ldots \hat a^\dagger_{\lambda_n k_n}(-\infty) |\Omega\rangle \\
|\lambda_1 k_1, \ldots, \lambda_n k_n\rangle_{\rm out} \equiv \hat a^\dagger_{\lambda_1 k_1}(+\infty) \ldots \hat a^\dagger_{\lambda_n k_n}(+\infty) |\Omega\rangle$$
The four-momenta $k_i$ will have masses $k_i^2 = m_{\lambda_i}^2$ and no $|\lambda_i k_i\rangle$ is allowed to equal another. Some people prefer to rescale $\hat \phi$ in order to hide those $C_\lambda$ prefactors but I will not. The nature of these prefactors will be explored much later. It is important to note that the total momenta of these states are approximately the sum of all $\mathbf{k}_i$, and the energy is approximately the sum of all $\omega_{\lambda_i k_i}$. This lends more credence to the notion that these are "multi-particle" states.
Now that we have successfully defined our incoming and outgoing asymptotic multi particle states and derived some important properties of our newly constructed asymptotic creation and annihilation operators, we have completed the framework necessary to derive the LSZ reduction formula. Using the properties defined here, you should be able to justifiably go through the steps as outlined in Srednicki.
To answer your doubt 2: In order to get our states to have the right properties, we needed these to be wave-packets that are widely separated in the distant past and future. Therefore, these states are only approximately momentum and energy eigenstates (although you can get as close as you want). As they're not perfect energy eigenstates, some time evolution will occur. You particles will start far apart, come together, interact, then (different) particles will leave.
TLDR: If you define creation and annihilation operators properly, using the Klein Gordon inner product with widely separated wave packets in the far past/future, you will get your actual particle states when acting with these operators on the true vacuum $|\Omega\rangle$.
Best Answer
I assume that we are working with a fixed order in perturbation theory, so that the set of all possible Feynman graphs is finite. I also assume that the theory has been regularized such that all graphs evaluate to finite amplitudes.
We call a bubble graph a Feynman diagram with no external legs.
We call a connected graph a Feynman diagram such that among its connected components there are no bubbles. Important: note that this definition, despite being used heavily in many QFT textbooks, is somewhat counterintuitive. Usually a connected graph is a graph with a single connected component. Not so here!
Consider an arbitrary Feynman graph $G$. We can naturally split it into the connected subgraph $G_0$ (in the sense defined above) and the bubble graph $G_b$, which in the general case is a union of bubbles. It can be obtained directly from the Feynman rules that
$$ \text{Amp}(G) = \text{Amp}(G_0) \cdot \text{Amp}(G_b). $$
Now consider a sum of all possible Feynman graphs with some fixed set of external legs. This sum is over the Cartesian product of connected graphs (in the sense defined above) and bubble graphs. Thus the sum factors out as
$$ \sum \text{graphs} = \sum \text{connected graphs} \cdot \sum \text{bubble graphs}. $$
This result is referred to as "bubble factorization formula".
Now note that the denominator in your formula is just a special case of the numerator where there's no field operators before $U(t,-t) = T \exp \{-i \int_{-t}^t dt V(t) \}$, or in our diagrammatic terminology, where there's no external legs. There's only one connected graph with no external legs which is the empty graph $\emptyset$ with no vertices and no edges, for which (again, directly from Feynman rules)
$$ \text{Amp} (\emptyset) = 1. $$
Thus we conclude that the denominator is just a sum of bubble graphs, and from the bubble factorization formula we conclude that the fraction is equal to the sum of connected graphs.
All this reasoning can be made precise if we introduce an upper order of perturbation theory and a regularization scheme, as it was assumed in my answer.