What Does It Mean to Be Excited

From the post "de Broglie Per Person"

$2\pi a_{ \psi  }=n\lambda$

$h_n=2\pi a_{ \psi  }mc=mc.n\lambda$

with $n=1$.

$2\pi a_{ \psi  }=\lambda_1$

$h_1=2\pi a_{ \psi  }mc=mc.\lambda_{1}$

At $_na_\psi$,

$h_n=2\pi a_{\psi\,n}.mc=mc.\lambda_n$

When a particle is excited we have,

$h_{f}\rightarrow h_{i}$

and

$\lambda_{f}\rightarrow\lambda_{i}$

And when the particle returns from an excited state, $\lambda_{i}$ is applied to $h_{f}$ and we formulate,

$E_{i\rightarrow f}=h_{f}f_{i}=2\pi a_{ \psi\,_f  }.mc.f_{i}=\lambda_{f}mc.\cfrac{c}{\lambda_i}$

which is the energy of the particle at $h_f$.

Since,

$2\pi a_{\psi\,n}=2\pi e^{ -\cfrac { Aa_{\psi\,n}+C }{ n^{ 2 } }  }= \lambda_{n}$

$\cfrac{\lambda_{f}}{\lambda_{i}}= e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }$  --- (*)

We have,

$E_{i\rightarrow f}=mc^2.e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }=\psi_f+x.F$

The energy required at $n_f$ is $E_{i\rightarrow f}$, the energy in excess is $x.F$, the work function.

$E_{\Delta n}=E_{i\rightarrow f}-E_f=mc^2.e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }-mc^2=x.F$

$E_{\Delta n}=mc^2\left\{e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }-1\right\}=x.F$ --- (1)

where $E_{\Delta n}$ is the energy gained by the particle.  In this case the, the particle loses energy and $E_{\Delta n}$ is negative.

In this derivation, the Planck constant $h$, is formulated for each solution of $a_\psi$.  The resulting $h_i$ is applied at the appropriate level to obtain $\psi_i+x.F$, the work function.  $x.F$ is emitted as a photon.

Alternatively,

$\Delta E=E_{f}-E_{i}=mc^2-mc^2=0$

What happened?

When we apply $h_f$ to $\lambda_i$, we are suggesting that a wave at $\lambda_i$ is caught at energy level $h_f$.  $\lambda_i$ adjusts itself to $\lambda_f$ and there are energy changes in the process.  When $\lambda_i$ is due to a larger $a_{\psi\,n}$, its amplitude is also smaller  (cf. post "H Bar And No Bar"; this amplitude need to be quantified).  When $\lambda_i$ constrict to $\lambda_f$ at energy state $n=f$ it does so at smaller amplitude, it loses energy.

This lost in energy appears as the work function $x.F$ and is emitted as a photon.  The material cools.  The wave will regain its amplitude as heat is applied.  In the diagram above we considered only the fundamental frequency, when $2\pi a_{\psi\,n}=\lambda_n$.

In the post "Particle Spectrum", we took reference at $n=1$, where $h_o=h_1$.  The expressions for $E_{\Delta n}$ and $\lambda_p$ are valid when the final energy state is $n=1$.

Intuitively, $\psi$ has a longer path length at $n_i$, the difference in energy $E_{\Delta h}$ due to a change in $h_n$ as the wave move from $n_i$ to $n_f$ is,

$E_{\Delta h}=h_{f}f_i-h_{i}f_i=\cfrac{c}{\lambda_i}(h_{f}-h_{i})=\cfrac{c}{2\pi a_{\psi\,i}}(h_{f}-h_{i})$

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

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

$E_{\Delta h}=mc^2\left(e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }-1 \right)$

Which is the same expression as (1).  Both derivations are equivalent.

$\psi$ of a particle is excited to $n_i$ energy state, $a_\psi=a_{\psi\,i}$.  On its return to a lower $n_f$ energy state, $a_\psi=a_{\psi\,f}$ where $a_{\psi\,f}\lt a_{\psi\,i}$, it loses energy in the form of a photon.

This is consistent with the observation of spectra lines, but this particle is not orbiting around any nucleus.

To be excited is then $a_{\psi\,i}\rightarrow a_{\psi\,f}$ where $a_{\psi\,f}\gt a_{\psi\,i}$; to have a higher $a_\psi$.

Note: $E_{\Delta h}$ and $E_{\Delta n}$ are not the difference in energy as the particle move from $n_i$ to $n_f$ energy state.  But are energy changes as a result of $h_n$ changes between energy states.

In all cases, for,

$n_i\gt n_f$,    $f_i\gt f_f$

which is consistent with the post "Discreetly, Discrete λ And Discrete Frequency, f", unless the frequency profile is folded upwards.