\(|F|=\psi\) --- (*)
where the Newtonian force (not force density) is found to be numerically equal to \(\psi\). This implies that when we place a particle in the field of another, where both energy densities are equal, the respective forces on each particle are also equal.
Unfortunately, we have lost directional information in expression (*). But we know that if,
\(\psi_A\gt\psi_B\)
then
\(F_{AB}\gt F_{BA}\)
the force from particle \(A\) on particle \(B\) is greater than the force from particle \(B\) on particle \(A\). Because, \(F_{\rho}\) force density and \(F\) force, share the same direction and \(F_{\rho}\) points toward lesser \(\psi\),
\(F_{\rho}=-\cfrac{\partial\psi}{\partial x}\)
The greater force pushes the particle along a radial line into the region of lesser \(\psi_A\) where both \(\psi\)s can be equal. We can switch the role of \(A\) and \(B\), and see that a particle in a region of lesser \(\psi\) will experience a force that pushes it along a radial line in the direction of greater \(\psi\).
So the point where \(\psi_A=\psi_B\) is a stable point. Any small displacement from this point, \(\delta r\) generates a negative force that returns the particle to the stable point. The particle can be oscillatory at the stable point.
We have forces in equilibrium. We have one particle pushing another towards lower \(\psi\). But we have action and reaction pair only if \(\psi_A=\psi_B\).
Around a particle, \(\psi\) was graphed as,
we see that for a given value of \(\psi\) there can be two stable points, \(x_v\) and \(x_c\). There can be a transition from \(x_v\to x_c\) by first, an absorption of an energy packet and then, emission of the same energy packet. After this transition, the state of the system changed \(x_v\to x_c\). But as neither particles gained nor lost energy, the energy state of the system is unchanged.
The reverse transition, \(x_c\to x_v\) involves the same process, an absorption of a photon to reach \(\psi_{max}\) and an emission of a photon to return to \(\psi_{o}\).
Is \(\psi_{max}\) a valid energy state for the particle in transition? Must be, otherwise the packet of energy would not be absorbed in the first place.
After emission, it is possible that the particle returns to its original position along \(x\), before the absorption of a photon. In this case the state of the system does not change. This is wrong please refer to the post "Photon Emission After Absorption" dated 15 Jul 2015.
\(\psi_{max}\) is a constraint presented by the particle not in transition. As long as the particle in transition has valid energy state,
\(\psi_n\le\psi_{max}\)
\(n=1,2...\)
It can absorb a photon to reach \(\psi_n\), emit the photon and return to its original position on \(x\), either \(x_v\) or \(x_c\).
For the transition \(x_v\to x_c\) and reverse, \(\psi_{n_{max}}\) must be sufficiently close to \(\psi_{max}\). Not all transit particle and non transit particle pairs can achieve a \(x_v\to x_c\) or \(x_c\to x_v\) transition, although in all cases, \(x_v\) and \(x_c\) are both valid.
The absorption and emission of energy packets result in the absorption and emission spectra lines respectively. \(\psi_{max}\) limits the emission spectra and truncates all spectra lines of the particle in transition above \(n_{max}\), ie
\(\psi_n\gt\psi_{max}\).
\(n\gt n_{max}\)
The emission spectra lines carry information from both the transit particle and the non-transit particle.
If the particle absorb enough energy such that \(\psi_n\gt\psi_{max}\), this particle will find itself in the region around the non transit particle where greater \(\psi\) is in the reverse \(x\) direction. The particle in transit will experience a retarding force. It is expelled from the field of the non-transit particle only if it has non-zero momentum when it reaches \(x_a\), where \(\psi(x=x_a)=0\). This is the equivalent of photoelectric effect.
When a transit particle receive a photon and make the transition \(\psi_o\to\psi_n\), where \(\psi_n\gt\psi_{max}\), the resultant force on the particle at \(x=x_v\) is,
\(F=\psi_{n}-\psi_o=m\cfrac{dv}{dt}\)
where \(\psi_o\) is the \(\psi\) around the non-transit particle at \(x_v\), \(\psi(x_v)=\psi_o\), that is also the ground energy of the transit particle. As the particle move,
\(F=\psi_{n}-\psi(x)=m\cfrac{dv}{dt}=m\cfrac{d^2x}{dt^2}\)
\(\cfrac{d^2x}{dt^2}+\cfrac{1}{m}\psi(x)=\cfrac{1}{m}\psi_{n}\)
We are interested in an expression for \(v_{x_a}\), the velocity of the transit particle at \(x=x_a\) given \(\psi_n\gt\psi_{max}\). \(v_{x_a}\) is velocity of the particle when it is ejected.
\(\psi_{max}\) Is New. In the case where the transit particle is an electron, \(x_c\) might corresponds to an orbit in the conduction band. \(\psi_{max}\) can prevent electrons from ever reaching the conduction band ie,
\(\psi_n\ne\psi_{max}\)
for all \(n\).
and the material remains a non conductor. However, the existence of \(x_c\) irrespective, allows for electrons to be introduced at \(x_c\), after the electron at \(x_v\) has been ejected (heat treatment). Such a non-conductive material is then made conductive.
When the transit particle is an electron, the condition, \(\psi_n\approx\psi_{max}\) in order for a \(x_v\to x_c\) transition answers the question: why some material are conductor and some are non-conductors although both solutions \(x_v\) and \(x_c\) for orbital radii are valid all the time?
\(\psi_{max}\) also explains why the emission spectra lines are truncated and refines the conditions for ejecting the particle from within \(\psi\) of the non-transit particle. \(\psi_{max}\) marks the point beyond which the particle experiences a retarding force. If the particle stops within the \(\psi\) of the non-transit particle, it will oscillate about \(x=x_a/2\) where \(\psi(x_a/2)=\psi_{max}\) and not be ejected. We have encountered such oscillations along a radial line before.
