Why is \(\psi\) all over the place but \(E=h.f\) take on discrete values?
\(\psi\) is measured as energy density (Jm-3). It is a standing wave wrap around the center of the particle. That is why it is all over the place. As a wave it can superimpose and create a bigger wave. Many \(\psi\) waves of the same type sum to create particle clouds. Earth itself is one big \(g^{-}\) particle, that is the superposition of many \(\psi\)s from many \(g^{-}\) particles.
Numerically \(\psi\) equals the Newtonian force. It is expected that under equilibrium when all forces sum to zero, that \(\psi\) of a particle is equal to the \(\psi\) of its neighboring particles. As a wave, the frequency, \(f\) of \(\psi\), measures its energy. This is consistent with Planck equation,
\(E=h.f\)
Planck equation, provides energy information of a particle as a whole. As a complete wave around the particle, \(\psi\) can only take on certain frequencies such that the wave are of integer multiple of wavelengths on the perimeter of a circle, center at the particle.
\(n.\lambda_{\psi}=2\pi a_\psi\)
and
\(n.\lambda_{\psi}mc=2\pi a_\psi mc=h.f=E\)
for \(n=0,1,2,3...\)
It is expected that as \(E\) changes, \(\psi\) around the particle changes correspondingly, given the total volume of \(\psi\). The energy of \(\psi\), \(E\) takes on discrete values.
\(\psi\) itself is all over the place.
