Schrödinger's Equation

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.