- #26

- 3,039

- 1,077

Nice dickforce, a more accurate derivation of what i originally posted. What we didnt notice is that if we look at the equation

[tex]

\begin{array}{rcl}

- i \, \left(\nabla \times \mathbf{X}\right) & = & \frac{1}{c} \, \frac{\partial \mathbf{X}}{\partial t} + \, \mathbf{J}_{X}

\end{array}

[/tex]

and we rewrite it by multiplying both sides by i and take the case of free space where the current density is zero we have :

[tex]

\nabla \times \mathbf{X}\ & = & i \, \frac{1}{c} \, \frac{\partial \mathbf{X}}{\partial t}

[/tex]

If we now replace the curl operator with the hamiltonian operator so that [tex] \mathbf{H} & = & c \hbar \nabla \times [/tex]

the equation becomes the schrodinger equation [tex] \mathbf{H} \mathbf{X}\ & = & i \, \hbar \frac{\partial \mathbf{X}}{\partial t} [/tex]

So schrodinger equation seems to be a straightforward generalization of the equation of electromagnetic field, with the hamiltonian operator replacing as a more generic operator the curl operator.

[tex]

\begin{array}{rcl}

- i \, \left(\nabla \times \mathbf{X}\right) & = & \frac{1}{c} \, \frac{\partial \mathbf{X}}{\partial t} + \, \mathbf{J}_{X}

\end{array}

[/tex]

and we rewrite it by multiplying both sides by i and take the case of free space where the current density is zero we have :

[tex]

\nabla \times \mathbf{X}\ & = & i \, \frac{1}{c} \, \frac{\partial \mathbf{X}}{\partial t}

[/tex]

If we now replace the curl operator with the hamiltonian operator so that [tex] \mathbf{H} & = & c \hbar \nabla \times [/tex]

the equation becomes the schrodinger equation [tex] \mathbf{H} \mathbf{X}\ & = & i \, \hbar \frac{\partial \mathbf{X}}{\partial t} [/tex]

So schrodinger equation seems to be a straightforward generalization of the equation of electromagnetic field, with the hamiltonian operator replacing as a more generic operator the curl operator.

Last edited: