From the post "Nothing Probabilistic, No Dice",
\(\nabla ^{ 2 }\psi +\left( \cfrac { n }{ x } \right) ^{ 2 }\psi =0\)
we consider the fundamental frequency, \(n=1\) and \(x=a_\psi\)
\(\nabla ^{ 2 }\psi +\left( \cfrac { 1 }{ a_{ \psi } } \right) ^{ 2 }\psi =0\)
From the post "H Bar And No Bar",
\( \hbar =a_{ \psi }.mv\)
\( \nabla ^{ 2 }\psi +\left( \cfrac { mv }{ \hbar } \right) ^{ 2 }\psi =0\)
\( \hbar ^{ 2 }\nabla ^{ 2 }\psi +\left( mv \right) ^{ 2 }\psi =0\)
But,
\( \cfrac { 1 }{ 2 } mv^{ 2 }=E -V\)
So,
\( \cfrac { \hbar ^{ 2 } }{ 2m } \nabla ^{ 2 }\psi +\left( E-V \right) \psi =0\)
\( -\cfrac { \hbar ^{ 2 } }{ 2m } \nabla ^{ 2 }\psi +V\psi =E\psi \) --- (*)
And we almost have Schrödinger's Equation. Consider,
\( \psi=|\psi |e^{ iwt }\)
\(\dot { \psi } =i\omega |\psi |e^{ iwt }=i\omega\psi\)
\(\hbar\dot { \psi } =i\hbar\omega\psi=iE\psi\)
as \(E=hf=\hbar\omega\)
\(-i\hbar\dot { \psi }=E\psi\)
Substitute the above into (*),
\( -i\hbar\dot { \psi }= -\cfrac { \hbar ^{ 2 } }{ 2m } \nabla ^{ 2 }\psi +V\psi \)
this is the conjugated Schrödinger's Equation.
And we conjugate both sides and replace \(\psi^{*}\) with \(\psi\) (symbolically)
\( i\hbar\dot { \psi }= -\cfrac { \hbar ^{ 2 } }{ 2m } \nabla ^{ 2 }\psi +V\psi \)
where Euler's identity \(e^{i\pi}=-1\) leads to \(e^{i\pi/2}=i\); and \(i\) rotates between orthogonal dimensions.
\(\pi/2\equiv\) orthogonal
And \(\psi\) is clearly energy density. Thank you very much.
