Something that always confused me when first hearing about second quantization were the dependencies of the creation and annihilation operators.
On the one hand I have seen expressions such as
$$ \hat{H} = \sum_k \epsilon_k\hat{a}^\dagger_k\hat{a}_k $$
which just represent the static energy of a system and on the other hand one can describe electron-electron scattering using the hamiltonian
$$ \hat{H} = \sum_{k,k',q}V_q \hat{a}^\dagger_{k+q}\hat{a}^\dagger_{k'-q} \hat{a}_{k}\hat{a}_{k'}$$
which seems to be a time dependent process. Furthermore, I am confused about the picture these operators are used in since one can "always" calculate there equation of motion using the Heisenberg equation of motion (for example for operator combinations such as $\hat{a}_{k}^\dagger\hat{a}_{k'}$)
$$ \frac{\partial}{\partial t} \hat{a}_{k}^\dagger\hat{a}_{k'} = \frac{i}{\hbar} \left[\hat{H}, \hat{a}_{k}^\dagger\hat{a}_{k'}\right]$$
I cannot wrap my head around, why this time dependence seems to arise from a seemingly "static" hamiltonian.
More confusion arises from the fact that operator combinations such as $\hat{a}_{k}^\dagger\hat{a}_{k'}$ seem to be "instantaneous", but can lead to complicated behaviour of the system (e.g. oszillations in driven two-level-systems).
Best Answer
In Schrodinger picture, as long as you don't have any time-varying external fields, the creation and annihilation operators are time-independent like every other operator. In Heisenberg picture, they should be time-dependent, like every other operator.
This is physically intuitive. The creation operator at time $t$ is $$a^\dagger(t) = e^{- i H t} a^\dagger(0) e^{i H t}$$ which means that it creates a particle at a time $t$ in the past, rather than a particle right now. For example, in a harmonic oscillator, the states just evolve by phases, so raising at an earlier time just changes the result you get by a phase; that's why $a^\dagger(t)$ is just a phase times $a^\dagger(0)$ there. In an interacting field theory, where the particle you create can decay or interact with other particles, $a^\dagger(t)$ would be a very complicated combination of creation and annihilation operators at time $t = 0$. So dramatic time dependence of the $a^\dagger(t)$ is not physically surprising; it means things can happen.
There's one more confusing point: in free relativistic quantum field theory, the creation and annihilation operators in momentum space just evolve via phases. It's conventional to pull out these phases in the definition of the quantum field, $$\phi(x) = \int \frac{d\mathbf{p}}{(2\pi)^3 \sqrt{2 E_p}} (a_{\mathbf{p}} e^{-ipx} + a_{\mathbf{p}}^\dagger e^{ipx})$$ so that we get factors of $e^{\pm ipx}$, where $px = p^\mu x_\mu$ is manifestly Lorentz invariant, and the creation and annihilation operators here are just constant. This remains true even for an interacting theory because we typically work in the interaction picture, when all operators behave as they do in the free theory. However, I don't think most nonrelativistic field theory (i.e. condensed matter) books do this, so watch out!