I would like to understand where I am wrong in my proof of second quantization for Klein-Gordon field.
This is what I have done:
I start with the K.G equation:
$$ (\Box +m^2) \phi=0.$$
I write its Fourier decomposition: $$\phi(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \phi_k(t) e^{i \vec{k} \vec{x}}.$$
Using K.G equation I end up with:
$$ (\partial_t^2 +\vec{k^2}+m^2) \phi_k(t)=0.$$
I write $$\omega_k=\sqrt{\vec{k^2}+m^2}.$$
From here I understand that $\phi_k(t)$ satisfies the equation of an harmonic oscillator at each $k$.
Thus I can use my Q.M background to know that I can write:
$$\phi_k(t) \rightarrow \widehat{\phi_k}(t) = \frac{1}{\sqrt{2 \omega}}(a_k(t)+a_k(t)^\dagger).$$ (the t dependance is because I decide to work in Heisenberg picture with my operators). From now I will not put the "hat" on operators, the $a_k$ are operators.
I can prove that $a_k(t)=e^{-i \omega_k t}a_k(0)=e^{-i \omega_k t}a_k$ (it just comes from the commutation with the Hamiltonian).
I plug it in the equation I had at the beginning:
$$\phi(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \phi_k(t) e^{i \vec{k} \vec{x}} \rightarrow \int \frac{d^3k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega}}(e^{-i \omega_k t}a_k+e^{+i \omega_k t}a_k^\dagger) e^{i \vec{k} \vec{x}} $$
$$\phi(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega}}(e^{-ik.x}a_k+a_k^\dagger e^{+i \omega_k t + i \vec{k} \vec{x}})$$
And here I have a problem: in the second term I don't have a scalar product of 4-vectors. I could do a change of variables but then I would avec $a^{\dagger}_{-k}e^{+ik.x}$ in the second part. And for me it is a problem, I can't just redefine $a^{\dagger}_{-k}$ by $a^{\dagger}_{k}$ because the commutation relations would be wrong in such case.
Indeed, everything commutes for different k so for me there would be a problem. This question has already been asked here Quantization of a free field: Klein-Gordon case but for those reasons I don't get the answer.
Second question: Imagine that I no longer have my problem with $a_{-k}^{\dagger}$, I would end up with a field like this:
$$\phi(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega}}(e^{-ik.x}a_k+a_k^\dagger e^{+i k.x})$$
which is a real field ($\phi(t,\vec{x})^{\dagger}=\phi(t,\vec{x}))$. But why should it be real? I don't see where we would have made such an assumption? I know we can write a lagrangian for complex field but for me there is no reason for having an hermitic operator at the end?
Extra question: When we deal with interacting fields, we write them as:
$$\phi_{int}(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega}}(e^{-ik.x}a_k(t)+a_k(t)^\dagger e^{+i k.x}).$$
When I first read it I thought we did an assumption to deal with such an expression. But in fact, isn't it just a general decomposition in Fourier basis of any scalar field? It is probably a basic question but I would like to be sure of it.