Initial values of creation/annihilation operators I have a question about creation/annihilation operators. For example, if I have an evolution equation for annihilation operator of photon
$$ \frac{da_k}{dt} = -i \omega_k a_k$$
I obviously obtain
$$a_k(t) = a_k(0) e^{-i \omega_kt} $$
I not fully understand how to find initial value of $a_k$. Should we just find it from expression of canonical variables $P_k$ and $Q_k$  or maybe I should go to Schrodinger representation since $a_k(0)$ does not depend on time? 
Or there is another way? 
 A: $a$ is an operator. There isn't a specific value to it, and even if you do provide a certain expression in the matrix form – it won't give you much information, as the expression entirely depends on the choice of the basis.
One example would be the generalization of the standard matrix form of the oscillator's lowering operator in the energy eigenstate basis:
$$ a \left| n \right> = \sqrt{n} \left| n - 1 \right>, $$
or
$$ a=\left(\begin{array}{ccccc}
0 & 1 & 0 & 0 & \dots\\
0 & 0 & \sqrt{2} & 0 & \dots\\
0 & 0 & 0 & \sqrt{3} & \dots\\
0 & 0 & 0 & 0 & \dots\\
\dots & \dots & \dots & \dots & \dots
\end{array}\right). $$
I trust you to do the obvious QFT generalization of this.
But this explicit expression won't actually give you much. In fact, all information is already encoded in the algebra of $a_{\bf p}$ and $a^{\dagger}_{\bf p}$.
The reason is – there's the Stone-von Neumann theorem that guarantees that there's a single unique representation of the algebra on the Hilbert space. So specifying an explicit expression of $a$ is equivalent to specifying a basis on the Hilbert space.
That is almost true for the case of QFT – the caveat being that the vacuum $\left| 0 \right>$ must lie in the Hilbert space.
A: The problem is that you are not very specific here. Let me try to infer as much as I can, and then you need to correct me. You are working in the Heisenberg picture here. That means your state is independent of time but you time-evolve your observable instead. Your notation suggests that you are considering a periodic problem with a single band of frequency $\omega_k$ (as there is no band index). $a_k$ annihilates one oscillator mode and $a_k^{\dagger}$ creates one. $k$ could be an element of the Brillouin torus or of the continuum, it is not clear which. 
You mention that this is the evolution equation of a photon. That means you already have a universally accepted definition for creation and annihilation operators (one for each polarization in fact), and these satisfy commutation relations with a transverse delta function. So you are already given $a_k(0)$, and this is the thing that you plug in. In the absence of interactions, the number of photons is conserved, so the evolution equation will look like the one you suggested. Once you add interaction, then this will evidently be no longer true. 
You will be able to find more details in chapter 13 of Herbert Spohn's book Dynamics of Charged Particles and Their Radiation Field. 
A: $a_k(0)$ is set by initial conditions, independent of the differential equation you have provided. It could be any normal operator.
