It would just be like Fourier Transformation where energy is conserved, ie the square of the coefficients of the decomposed series sums to the total energy on the left hand side.
We know from \(E=hf\),
\(2hf=2hf_0+2hf_1+2hf_2+...\)
where by symmetry both particles decompose in a similar manner,
\(f=f_0+f_1+f_2+...\)
assuming that no energy is lost as a result of the collision that decompose the particles.
We know from Fourier Transform that, when
\(f(x)=sin(ax)\)
then,
\(\hat{f}(\omega)=\sqrt{2\pi}.\cfrac{\delta(\omega-a)-\delta(\omega+a)}{2i}\). The sum of the square of the coefficients of the series,
\(RHS=\left(\cfrac{\sqrt{2\pi}}{2i}\right)^2+\left(\cfrac{\sqrt{2\pi}}{2i}\right)^2=-\pi\)
where both components at \(-a\) and \(a\) are considered. And we retain the negative sign without reasoning that energy must be positive and so only the absolute value is considered. Since, within the particle \(f(x)\) is in a circular motion, we apply a factor \(2\pi\),
\(2\pi*RHS=-2\pi^2=-\cfrac{1}{2}*\left(2\pi\right)^2\)
Because we considered both positive and negative frequency, we have split the energy between the two frequency components and have half the energy each. We will just keep this in mind.
We are trying to formulate,
$\require{AMScd}$
\begin{CD}
f@>{\text{collision Xform}}>>f_0+f_1+f_2+...
\end{CD}
where a type of particle decomposes into component particles after a collision with a similar type of particle.
til' next time...
