Interaction Picture Ladder Operators vs Free Field Operators Consider the case of an interacting scalar field theory with bare mass $m_0$. After having derived the LSZ reduction formula, all that is left is to compute the time ordered products of the Heisenberg picture full interacting field operators $\phi (t, \vec{x})$ acting on the full interacting field vacuum.
$$\left< \Omega \right|T(\phi(x)\phi(y))\left| \Omega \right>$$
We would like to relate the quantities involved in this expression to free field operators and the vacuum of the free theory.
Consider the interaction picture operators
$$\Phi_I(t,\vec{x}) = e^{-iH_0(t-t_0)}\phi(t_0,\vec{x})e^{iH_0(t-t_0)}$$
$$\Pi_I(t, \vec{x}) = e^{-iH_0(t-t_0)}\pi(t_0,\vec{x})e^{iH_0(t-t_0)}$$
where $\phi(t_0, \vec{x})$ and $\pi(t_0,\vec{x})$ are the Heisenberg picture field and conjugate momentum operators respectively of the full interacting theory at some time $t_0$. We then show that $\Phi_I(t,\vec{x})$ satisfies the Klein-Gordon equation
$$(\square +m_0^2)\Phi_I(t,\vec{x})=0$$
We thus justify that $\Phi_I(t,\vec{x})$ admits the form of a mode expansion
$$\Phi_I(t,\vec{x})=\int \frac{d^3p}{(2\pi)^3}(a_{p,I}(t)e^{-ip.x}+a_{p,I}^{\dagger}(t)e^{ip.x})$$
for some operator $a_{p,I}(t)$.
Using the equal time commutation relation on the Heisenberg fields $[\phi(t_0, \vec{x}), \pi(t_0,\vec{y})] = i\delta^{(3)}(\vec{x}-\vec{y})$ and the fact that $\partial_t \Phi_I(t,\vec{x}) = \Pi_I(t, \vec{x})$ we deduce that $[a_{p,I}(t), a_{p',I}^{\dagger}(t)] = i \delta^{(3)}(\vec{p}-\vec{p'})$. We have thus shown that the interaction picture fully interacting field and conjugate momentum operators mimic the exact same algebraic relations as those of the free field and conjugate momentum operators in the Heisenberg picture. Needless to say, the fact that the two sets of operators satisfy the same algebraic relations among themselves does not imply that their action on the state kets $\left | \psi \right> \in H$ of the "Hilbert space" of wavefunctionals are identical. There is some highly nontrivial relation between the ladder operators $a_{p,I}(t)$ that appear in the mode expansion above and the $a_p$ operators that appear in the mode expansion of the free theory. The states annihilated by the $a_I$ operators need not coincide with those annihilated by $a_p$ in general.
Why does the author in this reference https://www.thphys.uni-heidelberg.de/courses/weigand/QFT1-12-13.pdf page 54 eq 2.79 conclude that the operator $a_{p,I}(t)$ annihilates the vacuum state of the free theory $\left |0 \right>$? Why should the state $\left |0 \right>$ annihilated by $H_0$ be also annihilated by $a_{p,I}$? $H_0$ is expressed in terms of $a_p$ of the free theory, not $a_{p,I}$.
I am aware of similar questions being asked before but they just go on about justifying the mode expansion above, no reference to the vacuum state misunderstanding presented here.
 A: I think your reference makes the following two assumptions (stated in the two sentences above (2.79)):
Assumption 1: A non-zero state $|\psi\rangle$ exists such that $H_0 |\psi\rangle = a_I(\vec{p}) |\psi\rangle = 0$ for all $\vec{p}$.
Assumption 2: $0$ is an eigenvalue of multiplicity 1 of the free field Hamiltonian $H_0$, i.e. the free field vacuum vector $|0\rangle$ is (up to a non-zero complex number) the only non-zero vector that is annihilated by $H_0$.
Taking these two assumptions into account, the conclusion is $|\psi\rangle = c|0\rangle$, $c\in\mathbb{C}\backslash\{0\}$, and $a_I(\vec{p})|0\rangle = 0$.
Ad Assumption 1: The free Hamiltonian $H_0\geq 0$ is bounded from below. Let $|\psi\rangle$ be a state of lowest energy (it is implicitly assumed that such a state exists), i.e. $H_0|\psi\rangle = 0$. Moreoever, we have the commutator identity $[H_0,a_I(\vec{p})] = -E_p a_I(\vec{p})$, $E_p = \sqrt{|\vec{p}|^2+m^2}$ (eq. (2.78)). Thus,
$$H_0 a_I(\vec{p})|\psi\rangle = [H_0,a_I(\vec{p})]|\psi\rangle + a_I(\vec{p}) H_0|\psi\rangle = -E_p a_I(\vec{p})|\psi\rangle.$$
This implies that $a_I(\vec{p})|\psi\rangle$ is a (generalised) eigenstate of $H_0$ with eigenvalue $-E_p < 0$, contradicting $H_0 \geq 0$ unless $a_I(\vec{p})|\psi\rangle = 0$.
