- #1
eljose
- 492
- 0
let,s suppose Bohmian mechanics was true then we would have trajectories in the form:
[tex]m\frac{d^{2}x}{dt^2}=-\nabla(V+U_{b}) [/tex] (1) [tex]U_{b}=-\frac{\hbar^2}{2m}\nabla^{2}\psi [/tex] being psi the solution to schroedinguer equation...but the trajectories in (1) comes from the Hamiltonian..
[tex]H=H_0+U_b [/tex] with this we could form the new Schroedinguer equation with function [tex]\psi_{1}(x)[/tex],with new trajectories...proceeding this infinite times we would have that the real trajectories..(after quantizying infinite times) are:
[tex]m\frac{d^{2}x}{dt^2}=-\nabla(V+U_{total}) [/tex]
with U total [tex]U_{total}=-\frac{\hbar^2}{2m}\nabla(\sum_{n=0}^{\infty}\psi_{n}(x)) [/tex]
[tex]m\frac{d^{2}x}{dt^2}=-\nabla(V+U_{b}) [/tex] (1) [tex]U_{b}=-\frac{\hbar^2}{2m}\nabla^{2}\psi [/tex] being psi the solution to schroedinguer equation...but the trajectories in (1) comes from the Hamiltonian..
[tex]H=H_0+U_b [/tex] with this we could form the new Schroedinguer equation with function [tex]\psi_{1}(x)[/tex],with new trajectories...proceeding this infinite times we would have that the real trajectories..(after quantizying infinite times) are:
[tex]m\frac{d^{2}x}{dt^2}=-\nabla(V+U_{total}) [/tex]
with U total [tex]U_{total}=-\frac{\hbar^2}{2m}\nabla(\sum_{n=0}^{\infty}\psi_{n}(x)) [/tex]