Wrong. All Wrong.

What is the result of oscillating around a radius of $a_\psi$ and travelling at light speed, $c$?

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 $\psi=KE$)