\(\psi_T=\psi_1+\psi_2\)
\(\psi_T=\cfrac { mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ (x+a_e)^{ 3 } }+\cfrac { mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ (x-a_e)^{ 3 } } \)
where \(2a_e\) is the orbital separation. \(\psi_1\) and \(\psi_2\) are due to the two orbiting electrons individually.
We formulate
\(\psi_T \ge \psi_{min}\)
\(\cfrac { mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ (x+a_e)^{ 3 } }+\cfrac { mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ (x-a_e)^{ 3 } }\ge\psi_{min}\)
\(\cfrac { mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ (x+a_e)^{ 3 } }+\cfrac { mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ (x-a_e)^{ 3 } }\ge \cfrac{mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ (x_{min})^{ 3 } }\)
and so,
\(\cfrac { 1 }{ (x+a_{ e })^{ 3 } } +\cfrac { 1 }{ (x-a_{ e })^{ 3 } } \ge \cfrac { 1 }{ x_{ min }^{ 3 } } \)
Consider \(x \gt a_e\),
\(\cfrac { 1 }{ (x+a_{ e })^{ 3 } } +\cfrac { 1 }{ (x-a_{ e })^{ 3 } } \ge \cfrac { 1 }{ x_{ min }^{ 3 } } \)
\( \cfrac { (x-a_{ e })^{ 3 }+(x+a_{ e })^{ 3 } }{ (x+a_{ e })^{ 3 }(x-a_{ e })^{ 3 } } \ge \cfrac { 1 }{ x_{ min }^{ 3 } } \)
Since \( (x-a_{ e })^{ 3 }\gt 0 \),
\(\cfrac { (x+a_{ e })^{ 3 }(x-a_{ e })^{ 3 } }{ (x-a_{ e })^{ 3 }+(x+a_{ e })^{ 3 } }\le x^3_{ min } \)
This implies that given a \(\psi_{min}\) and a corresponding \(x_{min}\), the expression above for \(x\) must be smaller or equal to \(x_{min}\).
The constrain on \(x\) due to \(x_{min}\) is plotted in red. Given this solution for \(x\), with reference to the diagram below, we will find \(r\) such that \(|\psi|\gt|\psi_{min}|\).
\(\psi=\cfrac{1}{2\mu_o}B^2=-\cfrac{1}{2}\varepsilon_oE^2=\cfrac { mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ x^{ 3 } }\)
We shall consider the B field,
\(B=\sqrt{2\mu_o\psi}=\left( \cfrac { \mu_omc^{ 2 } }{ 6\pi } \right) ^{ 1/2 }\cfrac { 1 }{ x^{ 3/2 } } \)
In this diagram, the electrons in parallel orbits are travelling out of the paper at \(\psi_{-}\) and \(\psi_{+}\). Their respective \(B\) fields are circular with centers at \(\psi_{-}\) and \(\psi_{+}\). Without lost of generality we will set the angle on the side of \(B_{+}\) at 90o and plot for values of \(r\ge a_e\), given \(x\),
\(B_{//}=B_{+\,//}+B_{-\,//}\)
\(B_{//}=B_{+}sin(\theta)+B_{-}sin(\phi)\)
\(B_{//}=\left( \cfrac { \mu_omc^{ 2 } }{ 6\pi } \right) ^{ 1/2 }\left( \cfrac{\sqrt { r^{ 2 }-x^{ 2 } } }{r^{5/2}}+\cfrac{\sqrt { r^{ 2 }-x^{ 2 } } }{[(2a_e+x)^2+(r^2-x^2)]^{5/4}}\right)\)
and
\(B_{\bot}=B_{+\,\bot}+B_{-\,\bot}\)
\(B_{\bot}=B_{+}cos(\theta)+B_{-}cos(\phi)\)
\(B_{\bot}=\left( \cfrac { \mu_omc^{ 2 } }{ 6\pi } \right) ^{ 1/2 }\left( \cfrac { x }{ r^{ 5/2 } } +\cfrac {2a_e+x }{ [(2a_{ e }+x)^{ 2 }+(r^{ 2 }-x^{ 2 })]^{ 5/4 } } \right) \)
Therefore,
\(B^2=\cfrac { \mu_omc^{ 2 } }{ 6\pi } \left[\left( \cfrac{\sqrt { r^{ 2 }-x^{ 2 } } }{r^{5/2}}+\cfrac{\sqrt { r^{ 2 }-x^{ 2 } } }{[(2a_e+x)^2+(r^2-x^2)]^{5/4}}\right)^2+\\\left( \cfrac { x }{ r^{ 5/2 } } +\cfrac {2a_e+x }{ [(2a_{ e }+x)^{ 2 }+(r^{ 2 }-x^{ 2 })]^{ 5/4 } }\right)^2 \right]\)
If \(2a_e\lt\lt x\),
\(B^2=\cfrac { \mu_omc^{ 2 } }{ 6\pi } \left[ \left( \cfrac { \sqrt { r^{ 2 }-x^{ 2 } } }{ r^{ 5/2 } } +\cfrac { \sqrt { r^{ 2 }-x^{ 2 } } }{ r^{ 5/2} } \right) ^{ 2 }+\left( \cfrac { x }{ r^{ 5/2} } +\cfrac { x }{ r^{ 5/2 } } \right) ^{ 2 } \right] \)
\(B^2=\cfrac { \mu_omc^{ 2 } }{ 6\pi } \cfrac { 4 }{ r^{3 } } \)
So, we have
\(\cfrac{1}{2\mu_o}B^2= \cfrac{1}{2\mu_o}\cfrac { \mu_omc^{ 2 } }{ 6\pi } \cfrac { 4 }{ r^{ 3 } }\ge|\psi_{min}| \)
The LHS is just the expression for \(\psi\) of two electron charges at a distance \(r\).
\(\cfrac { 4 }{ r^{ 3 } }\ge\cfrac { 1 }{ x_{ min }^{ 3 } }\)
\(r\le\sqrt[3]{4}\,x_{ min }\)
Since, \(x_{min}\) is from the expression,
\(\psi_{min}=\cfrac{mc^{ 2 } }{ 12\pi } \cfrac { 1 }{ (x_{min})^{ 3 } }\)
around a single particle, so provided that \(x_{min}\gt\gt a_e\), we have approximately,
The shape of the graph is not actually what we expected. The approximation to \(\psi\) at \(r\) may need improvement.
