|ψn|=An
The following diagram shows two energy states, at aψi and aψf.
As the particle gain energy, Af increases, and
r2=a2ψf+A2f --- (*)
r increases.
Could it be that when the perimeters are equal,
2π√(a2ψf+r2)2=2πaψi
a transit of energy states nf→ni occurs; ie when
r2=2a2ψi−a2ψf
By substituting r into (*)
Af=√2√a2ψi−a2ψf
Ai=0
Just on transition, Ai is small. This can explain a lower aψ to a higher aψ transition, but what about the reverse?
The reverse occurs when aψi collapses back to aψf but the amplitude, Ai→Af remains small.
The excess energy as a result of a reduced Af at aψf is emitted as a photon. In the limit when Ai→0, this excess energy is just the rms value of Af (not A2f as ψ is already energy),
EΔn=1√2Af=√a2ψi−a2ψf
Af is measured in meters (m), so more correctly,
EΔn=Eo√a2ψi−a2ψf
where Eo (Jm-1) maps meters (m) to Joules (J) in the diagram above.
Is this the same result as before?
EΔn=Eo√a2ψi−a2ψf
EΔn=Eo√(aψi−aψf)(aψi+aψf)
If Eo is,
Eo=−mc21aψi√(aψi−aψf)(aψi+aψf)
then,
EΔn=−mc21aψi√(aψi−aψf)(aψi+aψf)√(aψi−aψf)(aψi+aψf)
EΔn=mc2(aψfaψi−1)
Since,
aψn=e−Aaψn+Cn2
EΔn=mc2{e−Aaψf+Cn2f+Aaψi+Cn2i−1}
which is the same expression for EΔn in the post "What Does It Mean to Be Excited".
But why is,
Eo=−mc21aψi√(aψi−aψf)(aψi+aψf) (Jm-1)???
Unless Eo is justified, this scenario is just probable.
Note: The perimeter of a ellipse is used here,
P=2π√a2+b22
where a and b are the radii of the ellipse.