Consider a transition from \(n_i\rightarrow n_f\),
\(E_{ \Delta h 1}=mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } (a_{ \psi \, f }-a_{ \psi \, i })\)
Consider another transition from \(a_{ \psi \, f }+a_{ \psi \, i } \rightarrow a_{ \psi \, f }\), (This is an extrapolation, it is not proven that \(a_{ \psi \, f }+a_{ \psi \, i } \) is a solution.)
\( E_{ \Delta h 2}=\cfrac { mc^{ 2 } }{ a_{ \psi \, f }+a_{ \psi \, i } } (a_{ \psi \, f }-(a_{ \psi \, f }+a_{ \psi \, i } ))=\cfrac { mc^{ 2 } }{ a_{ \psi \, f }+a_{ \psi \, i } } (-a_{ \psi \, i })\)
such that, \(a_{\psi\,i}\) is in between the paths of \(E_{ \Delta h 1}\) and \(E_{ \Delta h 2}\).
If we take the geometric mean of the two,
\(\bar{E}_{\Delta h}=\sqrt{E_{ \Delta h 1}.E_{ \Delta h 2}}=mc^2\sqrt{\cfrac{a_{ \psi \, i }-a_{ \psi \, f }}{a_{ \psi \, f }+a_{ \psi \, i } }}\)
and let
\(E_o=-\cfrac{\bar{E}_{\Delta h}}{a_{\psi\,i}}=-mc^2\cfrac{1}{a_{\psi\,i}}\sqrt{\cfrac{a_{ \psi \, i }-a_{ \psi \, f }}{a_{ \psi \, f }+a_{ \psi \, i } }}\) Jm-1
We have an expression for \(E_o\). A negative sign is necessary because a change from a higher value of \(a_\psi\) to a lower value of \(a_\psi\) (ie. a positive value of \((a_{ \psi \, i }-a_{ \psi \, f })\) ) results in a loss of energy.
Note: If,
\(E_{ \Delta h\,f}=2\pi a_{\psi\,f}mc.f\)
\(E_{ \Delta h\,i}=2\pi a_{\psi\,i}mc.f\)
And,
\(E_{ \Delta h\,i}+E_{ \Delta h\,f}=2\pi a_{\psi\,i}mc.f+2\pi a_{\psi\,f}mc.f\)
\(2\pi (a_{\psi\,i}+a_{\psi\,f})mc.f=E_{ \Delta h\,i+f}\)
And,
\(E_{ \Delta h\,i+f}=\cfrac { mc^{ 2 } }{ a_{ \psi \, f }+a_{ \psi \, i } } (a_{ \psi \, f }-(a_{ \psi \, f }+a_{ \psi \, i } ))\) --- (*)
This does not proof that \(a_{ \psi \, f }+a_{ \psi \, i } \) is a solution to \(x\) for \(\psi\) but it shows that (*) is a valid expression.
