Solution to Klein-Gordon equation: real field condition and other questions Sorry for the lengthy question, pretty much the whole text is the standard derivation of the solution of the KG equation which I included to illustrate my doubts, and some questions are at the end. 
The Klein-Gordon equation is
$$ (\partial^2+m^2)\phi(x)=0\tag{1}$$
where $\partial^2=\partial_\mu\partial^\mu$.
Taking a Fourier transform of $\phi$
$$\phi(x)=\int \frac{d^4 k}{(2\pi)^4}\phi(k)e^{ikx} \tag{2}$$
and inserting it into the equation we have
$$ \int \frac{d^4 k}{(2\pi)^4}(k^2-m^2)\phi(k)e^{ikx}=0\tag{3}$$
which implies
$$(k^2-m^2)\phi(k)=0 \implies \phi(k)=2\pi f(k)\delta(k^2-m^2) \tag{4}$$
for some function $f$. Define $\omega=\sqrt{\mathbf{k}^2+m^2}$ where $k=(k_0, \mathbf{k}$), then $$\delta(k^2-m^2)=\frac{1}{2\omega}\left[\delta(k_0-\omega)+\delta(k_0+\omega)\right]\tag{5}$$
And our solution becomes
$$\phi(x)=\int \frac{d^4 k}{(2\pi)^3}\frac{1}{2\omega}\left[\delta(k_0-\omega)+\delta(k_0+\omega)\right]f(k)e^{ikx}\tag{6}$$
performing the integration on $k_0$
$$\phi(x)=\int \frac{d^3 k}{(2\pi)^3}\frac{1}{2\omega}(f(\omega,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}+i\omega t}+f(-\omega,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}-i\omega t})\tag{7}$$
In the second term, we can change variable $\mathbf{k}\rightarrow-\mathbf{k}$ and get
$$\phi(x)=\int \frac{d^3 k}{(2\pi)^3}\frac{1}{2\omega}(f(\omega,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}+i\omega t}+f(-\omega,-\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{x}-i\omega t})\tag{8}$$
Now: all the sources I can find go on and impose that the field must be real, and write something like
$$\phi(x)=\int \frac{d^3 k}{(2\pi)^3}\frac{1}{2\omega}(a(\omega,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}+i\omega t}+a^\dagger(\omega,\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{x}-i\omega t})\tag{9}$$
and this is presented as the "general solution to the Klein-Gordon equation"
Questions:


*

*Why should we impose that the field is real, i.e. $f(-\omega, -\mathbf{k})=f(\omega, \mathbf{k})^\dagger$? I see no mathematical reason to do so, and I don't see why a complex field would pose physical problems. I know about the harmonic oscillator interpretation, but I'd say that this interpretation is a consequence of the fact that the field is real and not vice versa. Why do we eliminate complex field solutions?

*Are the $2\pi$ factors important? In equation $(4)$ I introduced a factor $2\pi$ just so that the final solution would coincide with the one given in standard sources, but again, I don't see another reason to add it (actually the whole "the solution must be a function times delta" is a bit sketchy, how to see this?)

*Some give the solution with $\sqrt{\omega}$ instead of $\omega$ in the denominator (Peskin & Schroeder for instance) in analogy with the harmonic oscillator. To try and get this I thought of defining $a(\omega,k)=\sqrt{\omega}f(\omega,k)$ instead of $a(\omega,k)=f(\omega,k)$. Does this make sense? Do the two solutions have any physical difference? 
 A: Answering the three questions in order:
1) Both the real scalar field and the complex scalar field are important and physically useful, and refer to different classes of physical objects. Imposing that $\phi$ is real means that you are interested in the real scalar field solutions of the Klein-Gordon equation. For example, a real scalar field represents a spin-0 particle which is its own antiparticle, and a complex scalar field represents a spin-0 particle that is not its own antiparticle.
2) The factor of $2\pi$ comes from the fact that $x$-space and $k$-space are related by a Fourier transform. Where to put the factors of $2\pi$ is a matter of convention. You can either require:
$$f(x)=\int\frac{dk}{2\pi}e^{-ikx}\tilde{f}(k)$$
$$\tilde{f}(k)=\int dx\;e^{ikx}f(x)$$
or you can require:
$$f(x)=\int dk\;e^{-ikx}\tilde{f}(k)$$
$$\tilde{f}(k)=\int \frac{dx}{2\pi} e^{ikx}f(x)$$
or you can even require:
$$f(x)=\int\frac{dx}{\sqrt{2\pi}}e^{-ikx}\tilde{f}(k)$$
$$\tilde{f}(k)=\int\frac{dk}{\sqrt{2\pi}}e^{ikx}f(x)$$
There are actually an infinite number of choices for conventions for defining this Fourier transform. Mathematicians tend to like the third "symmetric" convention, while in QFT we tend to use the first one. For a more in-depth treatment of this, see Fourier transform standard practice for physics.
3) The choice of scaling is arbitrary, as the calculation of any observable will not be affected by this (again, as long as the choice of convention is consistent throughout the calculation).
