Nothing Pops?

The expression $r=\sqrt{2a^2_{\psi\,i}-a^2_{\psi\,f}}$ also opens up the possibility that

$2a^2_{\psi\,i}-a^2_{\psi\,f}\lt 0$

$a_{\psi\,f}\gt \sqrt{2}a_{\psi\,i}$   both $a_{\psi\,i}$ and $a_{\psi\,f}$ being positive.

In this case, photon emission based on the equal perimeter criterion does not occur.  A state transition is likely to be when $r=a_i$.

It is tempting to set,

$\cfrac{1}{2}mc^2=E_{\Delta n}=mc^{ 2 }(\cfrac { a_{ \psi \, f } }{ a_{ \psi \, i } } -1)$

where $E_{\Delta n}$ as derived in the post "Amplitude, $A_n$" dated 23 Dec 2014, is the energy in $A_f$.  And $m$ is the mass density of the photon.  This simply states that the photon is at light speed along $A_f$.  Then,

$a_{ \psi \, i }=\cfrac{2}{3}a_{ \psi \, f }$

Of course,

$a_{ \psi \, f }\gt a_{ \psi \, i }$

Does this suggest that unless $a_{ \psi \, i }\lt\cfrac{2}{3}a_{ \psi \, f }$, we have no emission?  That the energy states denoted by $a_{ \psi \, i }$ and $a_{ \psi \, f }$ must first be sufficiently far apart for photon emissions to occur with a state transition between them?

Any energy state $a_{\psi\,f}$ above $a_{\psi\,i}$, where

$a_{\psi\,i}\lt a_{\psi\,f}\lt \cfrac{3}{2}a_{\psi\,i}$

is invisible (with no photon emission) when a state transition from $a_{\psi\,i}$ occurs.  Or does it simply mean that photons can be emitted with velocity less than light speed along $A_f$?

What happen to $A_f$ if not emitted?