Two Quantum Wells, Quantum Tunneling, $v_{min}$

From the post "Nature Has Orbits"

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$ --- (1)

$v$ is complex.  Where,

$u=ln(cosh(x- x_{ z }))$,  $x_z=\cfrac{1}{2}x_a$  and

$\psi_c=\psi_n-\psi_{max}$

A plot of which,

The scaling factor $\cfrac{\psi_c}{m}$ plays a minor role in the shape of $v^2$ vs $x$.  $v^2$ and so the orbiting particle $KE$, is confined to a narrow range about $\psi_{max}$, at $x=x_z$ of the particle.  At values of $x$ closer $x=x_z$, $v^2$ has a minimum, $v^2_{min}$.

On approach to $x_z$ from higher values of $x$, since,

$\lim\limits_{x\to x_z^{+}}\cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\to\infty$

This means $v$ increases rapidly as the particle fall from higher orbit but because of the speed limit $c$, at light speed, the particle does not reach $x=x_z$.

Also, on approach to $x_z$ from lower values of $x$,

$\lim\limits_{x\to x_z^{-}}\cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\to(-i)\infty$

As $v$ is already complex, $-i$ rotates $v$ back along the radial line and the particle is expelled beyond $x=x_z$ with great velocity.  The particle jumps to higher orbit, $x\gt x_z$ at light speed $c$ (assuming that it has reach light speed, $c$ before passing through $x=x_z$).  A plot of $v^2$ around $x_z$ is shown below,

To obtain $v_{min}$ consider this, when $x\to x_z\pm\delta$, where $\delta=small$,

$\lim\limits_{x\to x_z}\cfrac{d(v^2)}{dx}=\lim\limits_{x\to x_z}\cfrac{d(v^2)}{du}\cfrac{du}{dx}=\epsilon$

as

 $\delta\to small$, not necessarily zero,  $\epsilon\to0$.  Substitute equation (1) for $v^2$,

$\cfrac{d(v^2)}{dx}=-\cfrac{\psi_c}{m}\cfrac{d}{du}\left\{\cfrac{1}{u}\left(\psi_c+u\right)e^{u}\left(e^{2u}-1\right)^{1/2}\right\}\cfrac{du}{dx}=\epsilon$

Consider,

$\cfrac { e^{ u }\left\{\psi _{ c }(-u+e^{ 2u }(2u-1)+1)+u^{ 2 }(2e^{ 2u }-1)\right\} }{ u^2\sqrt { e^{ 2u }-1 }  } \cfrac{du}{dx}=\epsilon$

$\cfrac { e^{ u }\left\{u^{ 2 }(2e^{ 2u }-1)+\psi _{ c }(u(2e^{ 2u }-1)-(e^{ 2u }-1))\right\} }{ u^{ 2 }\sqrt { e^{ 2u }-1 }  } \cfrac { du }{ dx } =\epsilon$

$\cfrac { e^{ u }\left\{u(\psi _{ c }+u)(2e^{ 2u }-1)-\psi _{ c }(e^{ 2u }-1)\right\} }{ u^2\sqrt { e^{ 2u }-1 }  } \cfrac { du }{ dx } =\epsilon$

$e^{ u }\left\{ \cfrac { (\psi _{ c }+u) }{ u } e^{ u }-\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) }  \right\} \cfrac { du }{ dx } =\cfrac { e^{ u }(e^{ 2u }-1)^{ 1/2 } }{ (2e^{ 2u }-1) } \epsilon$

$\cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }=\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1)\cfrac { du }{ dx }  } \epsilon +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) }$

$\because \cfrac{\epsilon}{\cfrac { du }{ dx }}=\cfrac { dv^{ 2 } }{ du }$

$\cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }\cfrac { du }{ dx } =\cfrac { e^{ u }(e^{ 2u }-1) ^{1/2}}{ (2e^{ 2u }-1) } \cfrac { du }{ dx } \cfrac { dv^{ 2 } }{ du } +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1)}{ u^{ 2 }(2e^{ 2u }-1) } \cfrac { du }{ dx }$

$\cfrac { du }{ dx }=tanh(x)=\cfrac{(e^{2u}-1)^{1/2}}{e^u}$,

$\cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }\cfrac { (e^{ 2u }-1)^{ 1/2 } }{ e^{ u } } =\cfrac { e^{ u }(e^{ 2u }-1) ^{1/2}}{ (2e^{ 2u }-1) }  \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) } \cfrac { (e^{ 2u }-1)^{ 1/2 } }{ e^{ u } }$

$\cfrac { (\psi _{ c }+u) }{ u } e^{ u }(e^{ 2u }-1)^{ 1/2 }=\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) }  \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) }$

Wrong! Never divide by zero or the possibility of zero.  But $\cfrac { du }{ dx }$ was not actually divided with! 

So,

$v^2=-\cfrac{\psi_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) }  \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \right\}$

So, when $\delta$ is sufficiently small, from the diagram above, the velocity is minimum $v=v_{min}$ when,

$\cfrac{d(v^2)}{dx}=0$

which occurs before $x=x_z$.   This leads to,
$\require{cancel}$
$v^2_{min}=-\cfrac{\psi_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) }  \cancelto{0}{\cfrac { dv^{ 2 } }{ dx }} +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1) ^{3/2}}{ u^{ 2 }(2e^{ 2u }-1) } \right\}_{x=x_{min}}$

$v^2_{min}=-\cfrac{\psi^2_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \right\}_{x=x_{min}}$

Since,

$u=log(cosh(x-x_z))$

$\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) }=\cfrac { cosh(x-x_z)sinh^{ 3 }(x-x_z) }{ ln^{ 2 }(cosh(x-x_z))(2cosh^{ 2 }(x-x_z)-1) } =\cfrac { sinh(2(x-x_z))sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z))cosh(2(x-x_z)) } =tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) }$

All $v^2_{min}$ for $\psi_c\gt0$ lie on this line,

$y=tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) }$

For each elevation $\psi_c$ in energy, the particle orbits at different radius with different velocity.  The minimum velocities squared of all particles elevated to different $\psi_c$s are given by the curve above.  If energy above this minimum $KE$ represented by the curve is emitted, then this curve is the baseline of the emission spectrum.  This is emission across a continuous spectrum not in packets of discrete frequencies.

How??  In packets of $\psi$, at low speed, as the particle's photon equivalent, but across a continuous spectrum as v^2 is continuous.

The particle is in circular motion around with velocity, $v_{min}$ in a direction perpendicular to $x$.  Only when $x\to x_z^{-}$ does the particle switch direction and travel along $x$ with great velocity.

When $x\to  x_z^{+}$, the particle is in circular motion with radius, $x_{radius}=x$.  When $x\to x_z^{-}$, the particle travels along the radial line, $x$.

The steep values of $v$ on either side of $x=x_z$ restrict the location of the particle to small values around $x=x_z$.  This is a Quantum Well except that $x=x_z$ is also forbidden.  They are two Quantum Wells side by side.  It is however possible to escape from the lower Quantum Well in the region $x\lt x_z$ by approaching $x=x_z$.  At that point $iv$ is rotated by the $-i$ factor and the particle speed beyond $x=x_z$ along the radial line, in the direction of $x$ into the adjacent Quantum Well at very high speed.  This is quantum tunneling, but it is possible only from the well at lower region $x\lt x_z$, to the well at higher region $x\gt x_z$.

How is $iv$ applied as the particle approachs $x_z^{-}$?  Does the particle turn slowly transcribing a spiral or does it turn abruptly near $x_z^{-}$.

Note:

Do not try to take the square root of $-i$,  rotate $v$ by $i$ for the first negative sign, then rotate $v$ by $-i$ as the result of taking the limit to approach $x_z^{-}$.