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Tuesday, December 23, 2014

Amplitude, An

Obviously,

|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 nfni occurs;  ie when

r2=2a2ψia2ψf

By substituting r into (*)

Af=2a2ψia2ψf

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, AiAf  remains small.

The excess energy as a result of a reduced Af at aψf is emitted as a photon.  In the limit when Ai0, this excess energy is just the rms value of Af (not A2f as ψ is already energy),

EΔn=12Af=a2ψia2ψf

Af is measured in meters (m), so more correctly,

EΔn=Eoa2ψia2ψf

where Eo (Jm-1) maps meters (m) to Joules (J) in the diagram above.

Is this the same result as before?

EΔn=Eoa2ψia2ψf

EΔn=Eo(aψiaψf)(aψi+aψf)

If Eo is,

Eo=mc21aψi(aψiaψf)(aψi+aψf)

then,

EΔn=mc21aψi(aψiaψf)(aψi+aψf)(aψiaψf)(aψi+aψf)

EΔn=mc2(aψfaψi1)

Since,

aψn=eAaψn+Cn2

EΔn=mc2{eAaψf+Cn2f+Aaψi+Cn2i1}

which is the same expression for EΔn in the post "What Does It Mean to Be Excited".

But why is,

Eo=mc21aψi(aψiaψf)(aψi+aψf)  (Jm-1)???

Unless Eo is justified, this scenario is just probable.

Note:  The perimeter of a ellipse is used here,

P=2πa2+b22

where a and b are the radii of the ellipse.