Two questions to Wick's theorem: particle scattering vs. free field & evaluating a contraction During a read in a self-study book about many-body physics, I came across Wick's theorem. There were two questions arising before I could start to grasp the idea of Wick's theorem. In the hope of finding some help, let me share these questions:
i) Wick's theorem only applies to free fields which can be found within the interaction picture. It was further introduced to chop down and thus evaluate a string of (fermionic) operators in the Dyson expansion for the scattering matrix (S-matrix), which is the time-evolution operator in the interaction picture. My question is since scattering is a form of interaction, why would we be able to apply Wick's theorem? Or in other words: To which extend does scattering imply the field to be free? Is it because scattering happens instantaneously in contrast to the time before and after scattering? I guess that in the process of particle scattering, there might be some intrinsic field evolving, such as a Coulomb field if we have a high concentration of (e. g.) electrons. Is the field still "free"?
ii) As an introduction to Wick's theorem, the following contraction of two fermionic field operators (creation operator $\hat{a}_l^\dagger(t_1)$ creates a particle in state $l$ at time $t_1$, annihilation operator $\hat{a}_k(t_2)$ destroys a particle in state $k$ at time $t_2$) is solved via
$\text{contr}[\hat{a}_k(t_2) \hat{a}^\dagger_l(t_1)] = \hat{a}_k(t_2) \hat{a}_l^\dagger(t_1) + \hat{a}_l^\dagger(t_1) \hat{a}_k(t_2)$
due to Wick's theorem for $t_2>t_1$ (so far so good). But now this equals
$\hat{a}_k(t_2) \hat{a}_l^\dagger(t_1) + \hat{a}_l^\dagger(t_1) \hat{a}_k(t_2) = (a_ka_l^\dagger + a_l^\dagger a_k)e^{-i(E_kt_2-E_1t_1)}$
and this I do not understand. Probably, we are in the interaction picture and hence operators and wavefunction both possess time-dependency. However, using the rules for going from one picture to the other, I am unable to find this solution. It starts with taking away the time-dependency of the operators, that is e. g. $\hat{a}_l^\dagger(t_1) = e^{iH_lt_1} a_l e^{-iH_lt_1}$, which can be done for all four operators. I cannot see if anything was done on the wavefunction as it is not expressed in the term above.
If you could give me any hint or detect some faulty reasoning with respect to the two questions i) and ii) I would be very thankful and can attack Wick's theorem.
Thanks in advance. Best, Gee.
 A: Wick's theorem applies to free fields, so how can it be used in interacting theories? The solution is the interaction picture.
Wick's theorem is about time-ordered products of Heisenberg picture operators in a free field theory. In the interaction picture, you write your Hamiltonian as a free part plus a small perturbation for the interactions: $H = H_0 + H_{\mathrm{int}}$. Then, starting from the Schrödinger picture operators $a$, you can form the interaction picture operators $a_I(t) = e^{iH_0 t} a e^{-iH_0 t}$ as well as the regular Heisenberg picture operators $a(t) = e^{iHt} a e^{-iHt}$. The $a_I(t)$ are thus defined to be precisely the Heisenberg picture operators of the free field theory. That is why you can use Wick's theorem on them. (The notation can be confusing: the subscript $I$ is sometimes omitted because the context of Wick's theorem always means that the fields are in the interaction picture.)
For your example, let me use bosonic fields because I am more familiar with them. I assume it would generalise to fermionic fields with the appropriate modifications. The free part of the Hamiltonian would be
$H_0 = \sum_k E_k\, a_k^\dagger a_k$.
The (Schrödinger picture) commutation relations are $[a_k, a_l^\dagger] = \delta_{kl}$. Thus
$$ [H_0, a_l] = \sum_k E_k\, \overbrace{[a_k^\dagger, a_l]}^{-\delta_{kl}} a_k = -E_l a_l $$
which we can write $[H_0, \cdot] a_l = -E_l a_l$.
Using the formula $e^A B e^{-A} = e^{[A,\cdot]} B$ (which is defined to be $B + [A, B] + \frac{1}{2} [A, [A, B]] + \cdots$), we find
$$ a_{lI}(t) = e^{iH_0 t} a_l e^{-iH_0 t} = e^{i[H_0, \cdot] t} a_l = e^{-iE_l t} a_l $$
which I think is the result you are looking for.

Here's the bigger picture: You see that working with the interaction picture fields is exactly the same as working in the free field theory, which is (relatively) easy. The price we pay for this is that the states evolve according to a complicated time evolution operator. We have $\left|{\psi_I(t_1)}\right> = U(t_1, t_0) \left|\psi_I(t_0)\right>$ with $U(t_1, t_0) = e^{iH_0 (t_1 - t_0)} e^{-iH (t_1 - t_0)}$ and the Dyson series says that, for $t_1 > t_0$, we can write
$$ U(t_1, t_0) = T \exp\left[-i \int_{t_0}^{t_1} dt\, H_I(t)\right] $$
with $T$ the time ordering symbol and $H_I(t) = e^{iH_0 t} H_{\mathrm{int}} e^{-iH_0 t}$. Expanding this gives you a perturbation series in terms of Feynman diagrams.
In comparison, working directly with the Heisenberg picture operators $a(t)$ in an interacting theory is generally very difficult as you need to solve the whole time evolution before you can even write them down.

Addendum: For fermionic operators, the commutation relations are replaced by anticommutation relations $\{a_k, a_l^\dagger\} = \{a_l^\dagger, a_k\} = \delta_{kl}$ (where $\{A,B\} = AB + BA$). Using the general identity $[AB,C] = A\{B,C\} - \{A,C\}B$, we then have
$$ [H_0, a_l] = \sum_k E_k\, [a_k^\dagger a_k, a_l] = \sum_k E_k\, \left(a_k^\dagger \overbrace{\{a_k, a_l\}}^0 - \overbrace{\{a_k^\dagger, a_l\}}^{\delta_{kl}} a_k\right) = -E_l a_l. $$
All other derivations are the same as in the bosonic case.
