Are the creation and annihilation operators time-dependent?

Question

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).

Answer 1

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!

Answer 2

First, even for the simple harmonic oscillator annihilator and creation operators depend on the position operator so coordinate dependence should not be a surprise. Second quantization is applies to a quantum field. The field configuration in space-time is the degree of freedom in the classical system and this is turned into an operator. This operator has to convey the temporal and spatial dependence of the field configuration at least at the expectation value level. In other words the expectation value of the field in a given state gives information on the likelihood of observing a given configuration.

I do not understand why the comparison of two completely distinct Hamiltonians, one seemingly for the free field, and the other for an interacting field (or field in the presence of some external force) would produce an apparent contradiction. The interaction will cause the system described by the operators to gain/lose energy and change the state. Presumably, if the interaction is due to another field with a corresponding set of operators then in the larger system something is conserved, energy, topological charge, etc. So all that is happening is that the overall state is changing in such a way as to preserve the conservation laws, as in a classical hard body collision.

As for representation, it looks like you are in momentum or occupation number representation and the 4-coordinate dependence is not apparent in k-space. This is somewhat of an advantage in describing the field dynamics, rather that trying to draw a picture that fills all of space and time we just draw the k-space representation of the same state, which for free fields in a coherent state is much easier. There is a field version of x and p. As the p operator is the derivative with respect to x, the field p operator in configuration representation would some type of derivative with respect to the field configurations (likely a variational derivative). This operator extends over all coordinate values.

As for time dependent operators the total time derivative of the operator equals the commutator of that operator with the Hamiltonian (which comes from classical mechanics Poison brackets) plus the partial derivative of the operator with respect to time. The partial measures "explicit" time dependence. the total time derivative of the system can be non-zero even without explicit time dependence due to implicit time dependence.