Photon Momentum

Momentum of a photon,

$\Delta\,p=\cfrac{2}{c}\int ^{ x_{z} }_{ 0 }{ \psi } d\, x$

Is $a_\psi=x_z$?  What does it mean when,

$2πx=i.nλ$

as $\psi$ oscillates at $f=\cfrac{c}{\lambda}$,

and the variation of $\psi$ around a particle given by,

$\psi=-i{ 2{ mc^{ 2 } } }\,ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }(x-x_z)))+c$

$c=i{ 2{ mc^{ 2 } } }\,ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }x_z))$

both of which is from the same wave equation,

$\cfrac{\partial^2\psi}{\partial\,t^2_c}=ic\cfrac{\partial^2\psi}{\partial\,x\partial\,t_c}$

where one space dimension has been replaced with a time dimension.

$a_\psi\ne x_z$.  Instead, in a 2D view, $\psi$ shape like a torus of radius $a_\psi$ and has extend from $a_\psi-x_z$ to $a_\psi+x_z$, the maximum energy density occurs on the perimeter of the circle with radius $a_\psi$.

In which case we have a bound on $x_z$,

$x_z\le a_\psi$

And,

$\Delta\,p\le\cfrac{2}{c}\int ^{ a_{\psi} }_{ 0 }{ \psi } d\, x$ --- (*)

The other restriction on $x_z$ is,

$m_{ \rho \, p }c^{ 2 }=m_{ \rho  }c^{ 2 }-\int _{ 0 }^{ 2x_{ z} }{ \psi  } dx=m_{ \rho  }c^{ 2 }-2\int _{ 0 }^{ x_{ z} }{ \psi  } dx\ge0$

So,

$\Delta\,p\le m_{ \rho  }c$

as we set $x_z=a_{\psi}$ and substitute into (*).  $m_{ \rho  }$ is the mass density of the particle fully expressed as mass.