I am looking into the section of the book by Peskin and Schroeder in which they connect the $S$-matrix to probabilities.
They start by considering the in state, which is a two-particle state \begin{equation} |\phi_{A}\phi_{B}\rangle_{in} \equiv \int \frac{d^3 k_A}{(2\pi)^3}\frac{1}{\sqrt{2E_{k_A}}}\phi_A(k_A) \int \frac{d^3 k_B}{(2\pi)^3}\frac{1}{\sqrt{2E_{k_B}}}\phi_B(k_B)|k_Ak_B\rangle_{in}. \tag{4.68} \end{equation} As one can see, it is not a product of momentum eigenstates, but it is smeared over momentum space with some functions.
Instead, the out state is considered to be the state with definite momenta of particles \begin{equation} {}_{out}\langle p_1p_2 \dots p_n|. \end{equation} Eventually, they compute the probability density for the scattering process as \begin{equation} {\cal P}({A} {B}\to 1,2,\dots, n)=\prod_{f=1}^n \frac{d^3p_f}{(2\pi)^3}\frac{1}{2E_f}|{}_{out}\langle p_1p_2 \dots p_n| \phi_{A}\phi_{B}\rangle_{in}|^2. \tag{4.74} \end{equation}
Question. Why is this formula so asymmetric with respect to the change of in and out particles? Why do in particles feature as wave packets, while out particles enter as eigenstates of momentum? P$\&$S explain this by, essentially, saying that this is how the design of a collider works. That is, the incoming particles are the beam that hits the target, so they are wave packets, while detectors can measure particles with definite momentum. Is this a real reason? What are the conditions on the wave packets then except that they are normalised? It seems that P$\&$S suggest that these are simultaneously localised in space, at least in the transverse direction and, at the same time, have a sharp peak at a given momentum. It seems that these two things are incompatible.
My guess. I guess, that in reality the problem that one encounters is that unless the in states are integrated against test functions, then (4.74) has a product of momentum conserving delta functions. This is why amplitudes where integrated against test functions. I guess, the same could have been done with out particles, just this is not necessary, as the issue with delta functions can be resolved smearing in states only.
What is the actual reason?
Highlighting the mathematical issue. What I've said in words and what may need a clarification is the following. If we just consider the scattering probability for eigenstates of momenta, we will end up having a factor \begin{equation} |{\cal M}|^2\Big(\delta^4(\sum p_{in}- \sum p_{out})\Big)^2. \end{equation} So we get an issue of $(\delta(x))^2$ type. Having wave packets instead of eigenstates of momentum resolves it as the formula above gets replaces with something like \begin{equation} \int |{\cal M}|^2\delta^4(\sum p_{in}- \sum p_{out})\delta^4(\sum p'_{in}- \sum p_{out}) dp_{in} dp'_{in}\phi(p_{in})\phi^*(p'_{in}) \end{equation} in the sense that this expression becomes finite and well-defined.
What I, probably, want. I think, it makes sense to expect something symmetric in ingoing and outgoing particles. For example, can one write a formula of the form \begin{equation} {\cal P}({A} {B}\to 1,2,\dots, n)=\prod_{f=1}^n \frac{d^3p_f}{(2\pi)^3}\frac{1}{2E_f} \frac{d^3p_A}{(2\pi)^3}\frac{1}{2E_A} \frac{d^3p_B}{(2\pi)^3}\frac{1}{2E_B} |{}_{out}\langle p_1p_2 \dots p_n| p_{A}p_{B}\rangle_{in}|^2 \times (\dots) . \end{equation} The last formula definitely does not exist because of the issue with the Dirac delta squared. Is there anything reminiscent of it with test functions involved?
Edit 1.
Ok, I think it is clear why one integrates over final states, but not over initial states. The reason is simply that the initial state is assumed to be known and fixed, while the final state is known only probabilistically and it is this probability that we compute. It is the probability density not the probability because the space of potential final states is continuous/infinite-dimensional. If one considers the reverse process (with $p_1,\dots, p_n$ incoming and measures $p_{\cal A}$ and $p_{\cal B}$) then the probability density computed needs to be integrated over $p_{\cal A}$ and $p_{\cal B}$. So, the whole asymmetry on that part comes from the fact that the problem is asymmetric with respect to in and out states.
Still, the question why we actually need to smear the initial states as in (4.68) is not entirely settled for me.