Transition amplitude for QED+QFD+QCD interactions

Question

As I understood, Feynman diagrams are nothing more than pictures for the transition amplitudes (up to some orders). For this we introduce a interaction vacuum state $|\Omega\rangle$ then we are able to calculate: $$\langle\Omega|T\{\phi(x_1)...\phi(x_n)\}|\Omega\rangle$$ I thought this means the creation of some particle at $x_n$ and annihilation at some other space time point.

But if I like to have QED/QFD/QCD interactions in one diagram, do I need a common interaction vacuum to write such transition amplitudes (to create for example leptons, W-Bosons or other hadrons in one process)? Is there a common state for QED, QFD and QED or better for the standard model? Or are they different? But how can I interpret these processes in this case?

Answer 1

At the strict mathematical level, it is yet unknown whether such theories can be defined at all.

At the level of perturbation theory, which is commonly used by particle physicists to calculate measurable predictions, the state $\left| \Omega \right>$ (interacting vacuum) can be evaluated assuming the adiabatic hypothesis, which is: interactions are too negligible to influence the states of elementary particles in the far past and future, where the distances between particles were/will be too large for them to interact. An example of the derivation is given in my answer to this PSE question.

Spoiler alert: we account for the change in the vacuum state from $\left| 0 \right>$ to $\left| \Omega \right>$ by excluding the diagrams with disconnected bubble subgraphs (bubble graphs are those which don't have external legs).

Answer 2

Any theory with a Hamiltonian with a spectrum bounded from below has a unique vacuum state $|\Omega\rangle$, barring any spontaneous symmetry breaking which would give degenerate vacua. That statement is true essentially by definition; a vacuum state is the state with lowest energy, conventionally taken to be zero.