Scattering, Perturbation and asymptotic states in LSZ reduction formula

Doubt 1: You cannot simply put $t=pminfty$ 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.

[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

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 $|lambda krangle$ 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."

$$|lambda krangle_W equiv int d^3{mathbf{k}'} f_W(mathbf{k} - mathbf{k}') |lambda krangle$$

We will come back to these.

Now, the free vacuum $|0rangle$ of $hat H_0$ and the true vacuum $|Omegarangle$ 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) $|Psi_1rangle$ 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

$$psi_i(x) equiv langle 0| hat phi(x) |Psi_irangle$$

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:

$$langle Omega| hat a_1(t) |Omegarangle = i int d^3{x}langle Omega| big( psi_1^*(t, vec x) partial_t hat phi (t, vec x) - partial_t psi_1^*(t, vec x) hat phi(t, vec x) big) |Omegarangle$$ $$= i int d^3{x} big( psi_1^*(t, vec x) partial_t langle Omega| hat phi(t, vec x) |Omegarangle - partial_t psi^*_1(t, vec x) langle Omega| hat phi(t, vec x) |Omegarangle big)$$ $$= i langle Omega| hat phi(t, vec x) |Omegarangle int d^3{x} (- partial_t psi_1^*(t, vec x))$$

The fact that $partial_t langle Omega| hat phi(t, vec x) |Omegarangle = 0$ follows directly from the fact that the vacuum state has zero energy, so $e^{- i hat H t} |Omegarangle = |Omegarangle$. Now as we want $langle Omega| hat a_1(t) |Omegarangle = 0$ for any $psi_1$, we can see that this is achieved if and only if $langle Omega| hat phi(x) |Omegarangle = langle Omega| hat phi(0) |Omegarangle = 0$. We will assume this is the case.

In the free theory, $langle 0| hat a_1(t) hat a_2^dagger(t) |0rangle = langle Psi_1 | Psi_2rangle = (psi_1, psi_2)_{KG}$. In an interacting theory, for any $hat a_1$ and state $|Psi_2rangle$ (not just a single particle state) we have

$$langle Omega| hat a_1(t) |Psi_2rangle = langle Omega| big( psi_1(t, cdot) , hat phi(t, cdot) big)_{KG}|Psi_2rangle$$ $$= big( psi_1(t, cdot), langle Omega| hat phi(t, cdot) |Psi_2ranglebig)_{KG} $$ $$= big( psi_1(t, cdot), psi_2(t, cdot) big)_{KG}$$

$$langlePsi_2| hat a_1(t) |Omegarangle = big( psi_1(t, cdot) , psi_2^*(t, cdot) big)_{KG}$$

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.

$$langle Omega| hat phi(x) |lambda krangle = C_lambda e^{-ikx}$$

where $C_lambda$ is a constant that depends on $lambda$.

We now want to see what our states $hat a^dagger_1|Omegarangle$ 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_2rangle_W = big( psi_{ k_1}(t, cdot), langle Omega |hat phi(t, cdot)|lambda_2 k_2rangle_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) |Omegarangle = big( psi_{ k_1}(t, cdot), {}_W langle lambda_2 k_2 |hat phi(t, cdot) |Omegarangle 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 $hat a^dagger_{ k_1}(t) |Omegarangle$ would overlap with would be $|lambda_1 k_1rangle$, 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) |Omegarangle = 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) |Omegarangle = 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

$$hat a^dagger_{lambda k}(pm infty) |Omegarangle = -C_lambda big( e^{ikx}, hat phi( x)big)_{KG} |Omegarangle vert_{t = pm infty} = | lambda k rangle$$ where we have chosen the normalization $langlelambda k | lambda' k' rangle = C_{lambda}^* C_{lambda'} (2 pi)^3 (2 omega_{lambda k}) delta_{lambda lambda'} delta^3(mathbf{k} - mathbf{k}')$. From the second equation, we can immediately see that

$$langlePsi | hat a_{lambda k}(pm infty) |Omegarangle = 0 hspace{0.15 cm} text{ for all } langlePsi | hspace{0.5 cm} Longrightarrow hspace{0.5 cm} hat a_{lambda k}(pm infty) |Omegarangle = 0.$$ 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.

$$ |lambda_1 k_1, ldots, lambda_n k_nrangle_{rm in} equiv hat a^dagger_{lambda_1 k_1}(-infty) ldots hat a^dagger_{lambda_n k_n}(-infty) |Omegarangle |lambda_1 k_1, ldots, lambda_n k_nrangle_{rm out} equiv hat a^dagger_{lambda_1 k_1}(+infty) ldots hat a^dagger_{lambda_n k_n}(+infty) |Omegarangle$$

The four-momenta $k_i$ will have masses $k_i^2 = m_{lambda_i}^2$ and no $|lambda_i k_irangle$ 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 $|Omegarangle$.