Problem in understanding notation of scattering amplitude The Schrodinger equation:  $$
i \frac{d}{d t}\left|\psi_{t}\right\rangle=H\left|\psi_{t}\right\rangle
$$
and solutions are  given by
$$\left|\psi_{t}\right\rangle=U(t)|\psi_\text{in}\rangle \equiv e^{-\imath H t}|\psi_\text{in}\rangle$$
In scattering experiment we are interested in solution with the following asymptotic condition
\begin{gathered}
U(t)|\psi\rangle \underset{t \rightarrow-\infty}{\longrightarrow} U^{0}(t)\left|\alpha\right\rangle \\
\qquad \qquad \underset{t \rightarrow+\infty}{\longrightarrow} U^{0}(t)\left|\beta\right\rangle 
\end{gathered}
where $ U^0(t)=e^{-\imath H_0 t}$ with $H_0$ the Hamiltonian of a free theory.
From the expression above we can show that
$$
|\psi_\text{in}\rangle=\lim _{t \rightarrow-\infty} U(t)^{\dagger} U^{0}(t)\left|\alpha\right\rangle \equiv \Omega_{-}\left|\alpha \right\rangle
$$
and
$$
|\psi_\text{in}\rangle=\lim _{t \rightarrow+\infty} U(t)^{\dagger} U^{0}(t)\left|\beta\right\rangle \equiv \Omega_{+}\left|\beta\right\rangle
$$
We are intereste in the probability that a particle that entered the collision with in asymptote $|\phi\rangle$ will be observed to emerge with out asymptote $|\chi\rangle .$ To evaluate this probability we note that the actual state at $t=0$, which will evolve from the in asymptote $|\phi\rangle$ is $|\phi_\text{in}\rangle=\Omega_{-}|\phi\rangle$, while the actual state at $t=0$, which would evolve into the out asymptote $|\chi\rangle$ is $|\chi_ \text{out} \rangle=\Omega_{+}|\chi\rangle .$
This probability is given by
\begin{aligned}
w(\chi \leftarrow \phi) &=|\langle\chi(t) \mid \phi(t)\rangle|^{2} \\
&=|\langle\chi_\text{out}\mid \phi_\text{in}\rangle|^{2} \\
&=\left|\left\langle\chi\left|\Omega_{-}^{\dagger} \Omega_{+}\right| \phi\right\rangle\right|^{2} \\
&=|\langle\chi|S| \phi\rangle|^{2}
\end{aligned}
Now in  this notes Quantum Field Theory by Mark Srednicki at page 51 they have that the scattering Amplitude from a initial state $\mid i\rangle$ to a final state $\mid f\rangle$ is given by
$$\langle f \mid i\rangle=\left\langle 0\left|a_{1^{\prime}}(+\infty) a_{2^{\prime}}(+\infty) a_{1}^{\dagger}(-\infty) a_{2}^{\dagger}(-\infty)\right| 0\right\rangle \tag{5.13}$$
From expression  $(5.13)$ for example if the initial state is orthogonal to the final state than the scattering amplitude is zero.
My question is shouldn't this scattering amplitude be given by $\langle f \mid S\mid i\rangle$ where $S$ is the  $S$ matrix?
 A: The transformation from Interaction picture to Heisenberg picture is given by
$$
O_{H}(t)=U^{\dagger}(t, 0) O_{\mathrm{ip}}(t) U(t, 0)
$$
with $U(t, 0)=e^{i H_{0} t} e^{-i H t} .$
Suppose we have $\left|\psi_{t}\right\rangle=U(t)|\alpha\rangle_\text{in} \equiv e^{-\imath H t}|\alpha\rangle _\text{in}$ with asymptote  $|\phi\rangle$ so we have
$$|\alpha\rangle_{\text {in }}=\lim _{t \rightarrow-\infty} U^{\dagger}(t, 0)|\alpha\rangle$$
so \begin{aligned}
\lim _{t \rightarrow-\infty} a_{H}^{\dagger}(k ; t)|\alpha\rangle_{\text {in }} &=\lim _{t \rightarrow-\infty} U^{\dagger}(t, 0) a^{\dagger}(k) U(t, 0)|\alpha\rangle_{\text {in }} \\
&=\lim _{t \rightarrow-\infty} U^{\dagger}(t, 0)|k, \alpha\rangle \\
&=|k, \alpha\rangle_{\text {in }} \\
&=a_{\text {in }}^{\dagger}(k)|\alpha\rangle_{\text {in }}
\end{aligned}
From above we have that
$$\lim _{t \rightarrow-\infty} a_{H}^{\dagger}(k ; t)=a_\text{in}^{\dagger}(k)$$
Similarly we can show that $$\lim _{t \rightarrow +\infty} a_{H}(k ; t)=a_\text{out}(k)$$
Now since $$\mid i\rangle=a_\text{in}^{\dagger}(k_1)a_\text{in}^{\dagger}(k_2)\mid 0\rangle$$
and
$$\mid f \rangle=a_\text{out}^{\dagger}(k'_1)a_\text{out}^{\dagger}(k'_2)\mid 0\rangle$$
we obtain the result.
