Confusion about creation/annihilation operators with interaction

[gasgas]

I'm currently going through Peskin and Schroeder's Intro to QFT textbook, chapter 4.2 "Perturbation Expansion of Correlation Functions".

So we use the evolution of an operator in the Heisenberg picture from some time t0 to an arbitrary time t>t0 assuming we know how the field looks at t0. Then we only keep the dominant case with no interactions, so this field is then the standard field in the interaction picture with a diagonalizible Hamiltonian (in the book it is a real scalar field with phi-4 interaction).

I am now confused because my professor commented in lecture that in the Fourier expansion of the field at t0 the operators a and a† do not have to be the usual annihilation and creation operators of the free field, but they have to be in order to explicitly calculate the field in the interaction picture. Do we then assume that t0 is far away enough in the past where there is no interaction and can use the results for a free theory?

 

[Demystifier]

This question can be answered at two levels, abstract theoretical and practical.

At the abstract theoretical level $t_0$ can be arbitrary. The interaction picture is a unitary transformation that interpolates between the Heisenberg picture (where only operators depend on $t$) and Schrödinger picture (where only states depend on $t$). In the interaction picture both depend on $t$, the operators evolve by the free Hamiltonian and states by the interaction Hamiltonian. The $t$-dependence of matrix elements of operators does not depend on the picture.

At the practical level, ultimately the goal is to compute scattering and decay amplitudes, so eventually $t_0$ is put far into the past where particles are "free". We do that because that makes the computations easier and happens to be sufficient for computing things that can actually be measured in most real experiments, at least in particle physics.

Speaking of the $t$-dependence of $a$ and $a^{\dagger}$ operators, it depends on the picture you use. In the Heisenberg picture they have a complicated dependence on $t$, so they are not the usual free destruction and creation operators. By contrast, in the interaction picture, they are the usual $t$-independent free destruction and creation operators.