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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?

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  • $\begingroup$ Why not switch to the momentum representation? $\endgroup$
    – Roger V.
    Commented Mar 8, 2023 at 12:18
  • $\begingroup$ @RogerVadim Of course, that makes sense. But even then the issue of finding the correct initial condition remains. $\endgroup$
    – raeel
    Commented Mar 8, 2023 at 12:27
  • $\begingroup$ 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.) $\endgroup$
    – Roger V.
    Commented Mar 8, 2023 at 12:29
  • $\begingroup$ Your initial condition is underdetermined. Check out this solution to the Cauchy problem of KG. $\endgroup$
    – LPZ
    Commented Mar 8, 2023 at 13:50

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