Saturday, December 6, 2014

Quantized \(x\), Discretely

From the previous post "Nothing Probabilistic, No Dice",

\( \nabla ^{ 2 }\psi +\left( \cfrac { n }{ x }  \right) ^{ 2 }\psi =0\)

where \(n=1,2,3,4...\) and when \(x=0\), \(n=0\) such that

\(\lim\limits_{x\rightarrow0}{\left( \cfrac { n }{ x }  \right)}=1\)

\(\psi\) is the total energy of the system, quantized.  We have,

\(\cfrac{1}{\psi}\cfrac{\partial^2\psi}{\partial\,x^2}=- \left( \cfrac { n }{ x }  \right) ^{ 2 } \)

\(\int { \cfrac { 1 }{ \psi  } \cfrac { \partial  }{ \partial \, x } \left\{ \cfrac { \partial \, \psi  }{ \partial \, x }  \right\}  } \, d\, x=-\int { \left( \cfrac { n }{ x }  \right) ^{ 2 } } \, d\, x\)

Integration by parts,

\(\cfrac { 1 }{ \psi  } \cfrac { \partial \, \psi  }{ \partial \, x } -\int { \cfrac { \partial \, \psi  }{ \partial \, x } \cfrac { \partial  }{ \partial \, x } \left\{ \cfrac { 1 }{ \psi  }  \right\}  } d\, x=\cfrac { n^{ 2 } }{ x } +A\)

Integrate again along \(x\),

\( \int { \cfrac { 1 }{ \psi  } \cfrac { \partial \, \psi  }{ \partial \, x } d\, x+\int { \int { \cfrac { 1 }{ \psi ^{ 2 } } \cfrac { \partial \, \psi  }{ \partial \, x }  }  } d\, \psi  } \, d\,x=\int { \cfrac { n^{ 2 } }{ x } +A } \,d\, x\)

\( \int { \cfrac { 1 }{ \psi  }  } d\, \psi +\int { \int { \cfrac { 1 }{ \psi ^{ 2 } }  }  } \left( d\, \psi  \right) ^{ 2 }=n^{ 2 }ln(x)+Ax+C\)

Let \( B={ e }^{ \cfrac { C }{ n^{ 2 } }  }\),

\( ln(\psi )-\int { \cfrac { 1 }{ \psi  }  } d\, \psi =n^{ 2 }ln(Bx)+ln(e^{ { Ax } })\)

\(n^{ 2 }ln(Bx)+ln(e^{ { Ax } })=0\)

\(Bx=e^{ -\cfrac { Ax }{ n^{ 2 } }  }\)

\(x=e^{ -\cfrac { Ax+C }{ n^{ 2 } }  }\)

A solution plot of the above \(y=x\) and \(y=e^{ -\cfrac { Ax+C }{ n^{ 2 } }  }\) is given below,


A zoomed version of these discrete solutions to \(x\) is given below,


Discrete values of \(x\) bunches up in a characteristic way for high values of \(n\), but the dependence of \(x\) on \(n\) is exponential of the reciprocal squared; not \(\cfrac{1}{n}\).

\(x\) is not unbounded, for \(\psi\) to be positive,

\(x\le x_a\),

and

\(x\le2x_z\)

since, \(x_a=2x_z\).