\(|\psi_n|=A_n\)
The following diagram shows two energy states, at \(a_{\psi\,i}\) and \(a_{\psi\,f}\).
As the particle gain energy, \(A_f\) increases, and
\(r^2=a^2_{\psi\,f} + A^2_f\) --- (*)
\(r\) increases.
Could it be that when the perimeters are equal,
\(2\pi \sqrt { \cfrac { ({ a^2_{\psi\,f} }+r^{ 2 }) }{ 2 } }=2\pi a_{\psi\,i}\)
a transit of energy states \(n_f\rightarrow n_i\) occurs; ie when
\(r^{ 2 }=2a^2_{\psi\,i}-a^2_{\psi\,f} \)
By substituting \(r\) into (*)
\(A_f=\sqrt{2}\sqrt{a^2_{\psi\,i}-a^2_{\psi\,f}}\)
\(A_i=0\)
Just on transition, \(A_i\) is small. This can explain a lower \(a_\psi\) to a higher \(a_\psi\) transition, but what about the reverse?
The reverse occurs when \(a_{\psi\,i}\) collapses back to \(a_{\psi\,f}\) but the amplitude, \(A_i\rightarrow A_f\) remains small.
The excess energy as a result of a reduced \(A_f\) at \(a_{\psi\,f}\) is emitted as a photon. In the limit when \(A_i\rightarrow 0\), this excess energy is just the rms value of \(A_f\) (not \(A^2_f\) as \(\psi\) is already energy),
\(E_{\Delta n}=\cfrac{1}{\sqrt{2}}A_f=\sqrt{a^2_{\psi\,i}-a^2_{\psi\,f}}\)
\(A_f\) is measured in meters (m), so more correctly,
\(E_{\Delta n}=E_o\sqrt{a^2_{\psi\,i}-a^2_{\psi\,f}}\)
where \(E_o\) (Jm-1) maps meters (m) to Joules (J) in the diagram above.
Is this the same result as before?
\(E_{ \Delta n }=E_{ o }\sqrt { a^{ 2 }_{ \psi \, i }-a^{ 2 }_{ \psi \, f } } \)
\( E_{ \Delta n }=E_{ o }\sqrt { (a_{ \psi \, i }-a_{ \psi \, f })(a_{ \psi \, i}+a_{ \psi \, f }) } \)
If \(E_o\) is,
\( E_{ o }=-mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \sqrt { \cfrac { (a_{ \psi \, i}-a_{ \psi \, f }) }{ (a_{ \psi \, i }+a_{ \psi \, f }) } } \)
then,
\( E_{ \Delta n }=-mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \sqrt { \cfrac { (a_{ \psi \, i }-a_{ \psi \, f }) }{ (a_{ \psi \,i }+a_{ \psi \,f }) } }\sqrt { (a_{ \psi \, i}-a_{ \psi \, f })(a_{ \psi \, i }+a_{ \psi \,f }) } \)
\( E_{ \Delta n }=mc^{ 2 }(\cfrac { a_{ \psi \, f } }{ a_{ \psi \, i } } -1)\)
Since,
\(a_{\psi\,n}=e^{ -\cfrac { Aa_{\psi\,n}+C }{ n^{ 2 } }}\)
\(E_{\Delta n}=mc^2\left\{e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i } }-1\right\}\)
which is the same expression for \(E_{\Delta n}\) in the post "What Does It Mean to Be Excited".
But why is,
\( E_{ o }=-mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \sqrt { \cfrac { (a_{ \psi \, i }-a_{ \psi \, f }) }{ (a_{ \psi \,i}+a_{ \psi \, f }) } } \) (Jm-1)???
Unless \(E_o\) is justified, this scenario is just probable.
Note: The perimeter of a ellipse is used here,
\(P=2\pi\sqrt{\cfrac{a^2+b^2}{2}}\)
where \(a\) and \(b\) are the radii of the ellipse.
