A helix.
In which case, \(a_\psi\) is related directly to the frequency of the oscillation of \(\psi\) by,
\(f=n\cfrac{c}{2\pi a_\psi}\)
\(n=1,2,3...\)
\(\omega=n\cfrac{c}{a_\psi}\)
But for a oscillator,
\(E=\cfrac{1}{2}kx^2\), \(\omega=\sqrt{\cfrac{k}{m}}\),
In this case,
\(x=a_\psi\)
So,
\(E=\cfrac{1}{2}m\omega^2a^2_\psi\)
\(E=n^2\cfrac{1}{2}mc^2=n^2KE\)
In the case of a photon,
\(E=n^2\cfrac{1}{2}m_pc^2\)
where \(m_p\) is the hypothetical mass of a photon and \(n=1,2,3...\)
A photon is quantized in \(n^2\) not \(n\) of its \(KE\) at \(c^2\).
In general, a particle has energy that is quantized not to \(n\) but \(n^2\). This assumes that the mass of the particle is in the center of \(\psi\) and \(E\) is the \(KE\) of the mass in addition to \(\psi\). (Note: this last statement may be wrong, \(\psi\) may just be the \(KE\) of the mass in circular motion. In which case, \(\psi\) and \(m\) are two views of the same phenomenon, and \(KE\) is not considered separately. \(KE=\psi\))
The fact that \(m\) is in circular motion and not stationary requires that the transmutation of \(\psi\) to mass via,
\(m_{ \rho \, p }c^{ 2 }=m_{ \rho }c^{ 2 }-\int _{ 0 }^{ 2x_{ z} }{ \psi } dx\)
be reconsidered. The \(KE\) of the mass that materialize as \(\psi\) decreases must be accounted for. (Note: Not unless \(\psi=KE\))
