Scattering, Perturbation and asymptotic states in LSZ reduction formula

Question

I was following Schwarz's book on quantum field theory. There he defines the asymptotic momentum eigenstates $|i\rangle\equiv |k_1 k_2\rangle$ and $|f\rangle\equiv |k_3 k_4\rangle$ in the S-matrix element $\langle f|S|i\rangle$ as the eigenstates of the full Hamiltonian i.e., $H=H_0+H_{int}$. Therefore, the states $|i\rangle=|k_1 k_2\rangle$ is defined as

$$|k_1 k_2\rangle=a_{k_1}^{\dagger}(-\infty) a_{k_2}^{\dagger}(-\infty)|\Omega\rangle$$

where $|\Omega\rangle$ is the vacuum of the full interacting theory. Then the LSZ reduction formula connects the S-matrix element $\langle f|S|i\rangle$ to the Green' functions of the interaction theory defined as $$G^{(n)}(x_1,x_2,...x_n)=\langle \Omega|T[\phi(x_1)\phi(x_2)...\phi(x_n)]|\Omega\rangle.$$ Here are a few doubts.

Doubt 1 When the particles are far away, the interaction can be considered to be adiabatically switched off. Therefore, at $t=\pm\infty$ the states are really free particle states and should have been written as

$$|k_1 k_2\rangle=a_{k_1}^{\dagger}(-\infty) a_{k_2}^{\dagger}(-\infty)|0\rangle$$

and $$|k_3 k_4\rangle=a_{k_3}^{\dagger}(+\infty) a_{k_4}^{\dagger}(+\infty)|0\rangle$$ where $|0\rangle$ is the vacuum of the free theory. I do not understand why these the states $|i\rangle$ and $|f\rangle$ are derived from $|\Omega\rangle$ instead of $|0\rangle$.

Doubt 2 The initial and final states were derived from the vacuum of the interacting theory $|\Omega\rangle$. According to my understanding, this suggests that the states $|i\rangle\equiv |k_1 k_2\rangle$ and $|f\rangle\equiv |k_3 k_4\rangle$ are eigenstates of the full Hamiltonian $H$. Since then there is no perturbation, there should not be any scattering or transition at all.

---

More references Even Peskin & Schroeder, Bjorken & Drell, Srednicki take the same approach as Schwartz; they too define the external momentum eigenstates to the eigenstate of the full Hamiltonian $H$. However, if the system was initially in a stationary state why should it undergo a transition in absence of any perturbation?

Answer 1

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 $|\Omega\rangle$.

Answer 2

Doubt 1: You cannot simply put $t=\pm\infty$ since all formulas become meaningless unless the limit is carefully done. The proof of the LSZ theorem by Haag and Ruelle shows that one needs the interacting vacuum. You can understand the relativistic situation by first looking at the simpler nonrelativistic situation, where a paper by Sandhas [1] gives the nonrelativistic analogue of the treatment by Haag and Ruelle.

Doubt 2: The asymptotic single-particle states are eigenstates, but these don't scatter. You need more than one particle for nontrivial scattering, and the product states are no longer eigenstates.

In Thirring's Course in Mathematical Physics, Vol. 3, there is a clear discussion of asymptotic states, again in the nonrelativistic situation. They are not eigenstates of the Hamiltonian: The (generalized) eigenstates are not the asymptotic plane waves but the the solutions of the Lippmann-Schwinger equations. (Take 2 particles and view them in the center of mass frame, to see the connection.) This makes your Doubt 2 moot.

[added 17.2.2024] The Haag-Ruelle scattering theory is treated in detail in Secition XI.16 of Volume III of the treatise by Reed and Simon. Sandhas works in second quantized quantum mechanics, which is nonrelativistic quantum field theory. For the relations to Haag-Ruelle theory, see the Remarks on Section XI.16 in Reed and Simon III, p. 379. The asymptotic states are built fron the vacuum of the asymptotic Hamiltonian, which is free and related to the dynamical Hamiltonian by a unitary transformation with the Möller operator (also called wave operator). If there are no bound states than the asymptotic Hamiltonian acts on the same space as the dynamical Hamiltonian, but if there are bound states the asymptotic state space is structurally different.

[1]: W. Sandhas, Definition and Existence of Multichannel Scattering States, Communications in Mathematical Physics 3.5 (1966): 358-374. https://projecteuclid.org/download/pdf_1/euclid.cmp/1103839514

Answer 3

When I first faced the issue of in-out asymptotic states reading Bjorken-Drell I've had your same doubt 2) and I've worked out some details trying to disentangle it. Before writing them down, I should mention that other answers specify that product state are not eigenstate of the complete (*) four-momentum $P^\mu$ (of which $P^0$ is the interacting Hamiltonian) and advocate to more advanced mathematical treatment of scattering theory in the context of axiomatic QFT. Instead my answer only refers to (and thus is valid only at the level of rigor of) the treatment of asymptotic states/evolution given in Bjorken-Drell type/level books (and only the case of a single scalar field involved). The (non-normalizable) asymptotic states are in fact (generalized) eigenstates of the complete four-momentum $P^\mu$, as it clearly is stated in Bjorken-Drell section 16.3, formulas 16.10 and 16.11, i.e

$[P^\mu, a_{in}(k)]=-k^\mu a_{in}(k)\;$ and $\;[P^\mu, a^\dagger_{in}(k)]=k^\mu a^\dagger_{in}(k)\;$ (16.10)

$P^\mu |k_1,...,k_n,in>=(\sum_{i=1}^n k_i) |k_1,...,k_n,in>\;$ (16.11)

where $|k_1,...,k_n, in>= a^\dagger_{in}(k_1)...a^\dagger_{in}(k_n)|0>$, $\;a^\dagger_{in}(k)\;$ and $\;a_{in}(k)$ are creation/destruction operators for in single particle states with four momentum $k$ and $|0>$ is the vacuum of the complete hamiltonian $H=H_0+H_i$. The same is valid for out asymptotic states $\;|k_1,...,k_m, out>= a^\dagger_{out}(k_1)...a^\dagger_{out}(k_m)|0>$, i.e they are eigenstate of the four-momentum $P^\mu$ but they are not the same set of eigenstate as the in-eigenstates $|k_1,...,k_n, in>$ (unless $m=n=1$, i.e $|k, in>=|k, out>$ only for one particle states).

This is the crucial point to understand and the help comes from section 16.6 of Bjorken-Drell's book. In this section the authors introduced the formalism of S-matrix operator, specifying definition and basic properties. The definition of $S$ implies that one can map in-eigenstates $|\alpha, in>=|k_1,...,k_n, in>$ to out-eigenstates $|\beta,out>=|k_1,...,k_m, out>$ through the $S$ operator, i.e $|\beta,out>=S^\dagger|\alpha, in>$ and the other significant property is that $S$ is unitary: this means that $S$ is nothing more than a unitary change of basis between the complete orthonormal set of in-eigenstates $|\alpha, in>$ and the complete orthonormal set of out-eigenstates $|\beta,out>$. The main question mark with this interpretation is that in fact the two complete orthonormal set of in/out-states must be eigenstates of the same 4-momentum operator $P^\mu$: the only possible eventuality when we can have two different set of eigenstates for the same operator is if this operator is degenerate (at least one of it's eigenspaces have dimension greater than one). Then the question is: Is $P^\mu$ degenerate? The answer is yes and it's easy to see it: if we apply $P^\mu$ to $|k_1,...,k_n, in>$, we get $P^\mu |k_1,...,k_n,in>=(\sum_{i=1}^n k_i) |k_1,...,k_n,in>$ and the same eigenvalue $\sum_{i=1}^n k_i$ of $P^\mu$ is associated with every possible state $|k'_1,...,{k'}_{n'},in>$ which satisfy $\sum_{i=1}^{n'} k'_i=\sum_{i=1}^n k_i$ (thus total momentum conservation is automatically implied in the fact that, through the S-matrix, we can only transform the basis in the degenerate eigenspace of fixed total momentum to make it possible that in and out state are eigenstates of the same operator $P^\mu$). I hope this clarify the issue.

Edit (*): I wrote "interacting four momentum" but must have written "complete four-momentum", because this can generate confusion with the interacting picture 4-momentum, instead the momentum used by B&D in chapter 16 is the complete 4-momentum $P^\mu=P^\mu_0+P^\mu_i$ of the theory (free+interaction part).

Answer 4

As a simple counterpart of switching on interactions in QFT one can consider a harmonic oscillator to which one adds extra terms to the Hamiltonian, e.g. the $x^4$ term. In the latter case, while doing so, not only the new vacuum state is different from the old one -- the same refers to all eigenstates of the Hamiltonian. The counterpart of one-particle states in QFT in the case of the oscillator is the second lowest-energy eigenstate (after the vacuum). The explicit connection between old and new eigenstates in most cases cannot be written out exactly -- only as a series in perturbation theory.

Let us now discuss a bit more what the single-particle asymptotic states in the interacting QFT are. As I mentioned before, these are the exact eigenstates of the complete Hamiltonian. Physically, these correspond to a single particle traveling trough an empty universe. It has nothing to interact with, so it has the dynamics of a free particle.

It is important to keep in mind that, despite this particle is free in the sense I explained above, it is not the same as a free particle in the free theory. Besides the analogy with the oscillator, for which the eigenstates deform, one can keep in mind the analogy in the classical field theory. For example, let us consider a small sphere of bare mass $m_0$ as a model of the electron. Let us now allow it to have non-vanishing electric charge, which amounts to switching on the interaction with the electro-magnetic field. Then, the sphere creates electric field. The complete energy of a system considered as a whole includes not only the bare mass $m_0$, but also the potential energy of a sphere interacting with its own E-M field $E_U$ and the energy of the electric field $E_{E^2}$ in the space around the sphere. We can look at this system from other reference frames and it gets clear, that the system as a whole behaves as it is a free particle with an effective mass $m=m_0+E_U+E_{E^2}$. Thus, by switching on the interaction with the E-M field, the effective mass of the electron has changed. Experimentally, one always sees the electron, which is dressed with this E-M cloud and has mass $m$, not the bare mass $m_0$ that appears in the Lagrangian. So, any computation, which is set to compute the scattering of electrons -- which is exactly what the S-matrix does -- as the input and the output should feature these dressed electrons, not the bare ones.

The story with multi-particle asymptotic states is a bit more tricky. It is clear intuitively that if a particle in the interacting theory is far away from anything else than it evolves as if it was a single particle in the universe. In the same way, we can consider several particles, which are spatially separated from each other and, thus, their interactions can be ignored. Their evolution is then well approximated by a free evolution in interacting theory. One way to phrase it is that the associated multi-particle asymptotic state is a tensor product (in the sense of representations of the Poincare algebra) of single-particle states. This, in particular, means that the total energy for such a state is the sum of energies of single-particle states. Rephrasing this in terms of the Hamiltonian, we have $H_{total}=H_1+\dots + H_n$, that is the total Hamiltonian is the sum of individual Hamiltonians. This highlights the fact that these particles are non-interacting: each evolves with its own Hamiltonian and there are no cross terms.

However, this is only true once the particles are spatially separated and one can ignore interactions between them. When particles get closer to each other, interactions are no longer negligible and they undergo the scattering. This is how the scattering is set experimentally and this is what the $S$-matrix computes.

No, we go back to your questions.

Doubt 1.

In the scattering setup one does not assume that interaction terms in the action are adiabatically switched off. Instead, one says that when particles are far away from each other, interactions between them can be neglected, so these are approximately free particles. But these are free particles in the interacting theory, which are very different from free particles in the free theory, as I explained above. This is why, the story is not at all about switching off the interactions adiabatically.

In the free theory one has, indeed, \begin{equation} |p_1,p_2\rangle_0 = a^\dagger_{p_1} a^\dagger_{p_2}|0\rangle. \end{equation} In interacting theory \begin{equation} |p_1,p_2\rangle \ne a^\dagger_{p_1} a^\dagger_{p_2}|0\rangle, \qquad |p_1,p_2\rangle \ne a^\dagger_{p_1} a^\dagger_{p_2}|\Omega\rangle \end{equation} if $a^\dagger$ is understood as in the free theory. Instead $|p_1,p_2\rangle$ is some complex state made of multi-particle states in the free theory. Given the complexity of this relation, one needs to design procedures like the Kallen-Lehmann representation and the LSZ prescription to see how states like $|p_1,p_2\rangle$ can be accessed and how the S-matrix, featuring these states, can be computed.

Doubt 2.

As I explained above, the multi-particle asymptotic states are eigenstates of the complete Hamiltonian only as long as particles are spatially separated and interactions between them can be ignored. Thus, strictly speaking, they are exact eigenstates of the complete Hamiltonian only for one-particle asymptotic state. For more than one particle, asymptotic states are only approximate eigenstates of the complete Hamiltonian and when particles get closer to each other, they do scatter.