Description of a Classical Klein-Gordon Field with Momentum Distribution

So if I want to solve the free KG equation, the solution is of the form $$\phi(x) = \int_{p\in \mathbb{R}^4}\frac{1}{2\omega_\vec{p}} \left( a(\vec{p})e^{-ipx} +b(\vec{p})e^{ipx} \right),$$ where the mode coefficients $$a$$ and $$b$$ should depend on the initial conditions. Now if I want to implement some momentum distribution $$f(\vec{p})$$, along which the modes are distributed, how would I go about doing that?

I suppose one would need to choose the correct initial condition $$\phi(0,\vec{x})$$?

In the special case I am interested in, the situation is even a bit simpler, the field is only evaluated at $$x=0$$, meaning I have the representation $$\phi(t) = \int_{p^0} \frac{1}{2\omega_\vec{p}} \left( a(\vec{p})e^{-p^0t} +b(\vec{p})e^{ip^0t} \right),$$ and the initial condition $$\phi_0 = \int_{p^0} \frac{1}{2\omega_\vec{p}} \left( a(\vec{p})+b(\vec{p}) \right) = \dots$$ Am I correct in assuming that the functions $$a,b$$ evaluated at $$\vec{p}$$ should then correspond to the amplitude of the momentum mode $$\vec{p}$$, given the distribution $$f$$?
I would imagine this amplitude to be some sort of occupation number, i.e. $$\sqrt{f(p)\text{d}p}$$.

What would be the correct way to formulate this? Or is there literature that goes more in depth on that?

• Why not switch to the momentum representation? Commented Mar 8, 2023 at 12:18
• @RogerVadim Of course, that makes sense. But even then the issue of finding the correct initial condition remains. Commented Mar 8, 2023 at 12:27
• Think of a classical wave packet: you transform it in momentum space, and you know $f(p)$. Now you can go back into the position representation and find your initial condition, if needed. (I assume that $f(p)$ is known, so it contains all the information.) Commented Mar 8, 2023 at 12:29
• Your initial condition is underdetermined. Check out this solution to the Cauchy problem of KG.
– LPZ
Commented Mar 8, 2023 at 13:50