# Separation of Klein-Gordon-/Dirac-equation (Bohmian-mechanics)

With the function $$R{ e }^{ \frac { i }{ \hbar } S }$$ one can separate the Schrödinger equation $$i \hbar \frac{\partial \psi}{\partial t}=\left(-\frac{\hbar^{2}}{2 m} \nabla^{2}+V\right) \psi$$ into

\begin{aligned} &\rightarrow \frac { \partial \rho }{ \partial t } +\nabla \cdot (\rho v)=0\qquad\qquad\qquad \left(R={ \rho }^{ 2 },\quad v=\frac{1}{m} \nabla S\right)&\\ &\rightarrow \frac{\partial S}{\partial t}=-\left[\frac{|\nabla S|^{2}}{2 m}+V+Q\right]\qquad\qquad\left(Q=-\frac{\hbar^{2}}{2 m} \frac{\nabla^{2} R}{R}\right)& \end{aligned} My question is:

• Is it possible to separate the Klein-Gordon-/Dirac equation with the same function or is there a mathematical or physical reason why it's not possible?
• Is there another function or way to separate these equations to get a better feeling for the real and imaginary part (or the phase an absolut value)?

I tried to separate the Klein-Gordon-equation

$$$$\partial_{t}^{2} \psi-\nabla^{2} \psi+m^{2} \psi=0$$$$

with the function $$R{ e }^{ iS }$$ but I am stuck with

$$$$R\left[ \left( i{ \partial }_{ t }^{ 2 }{ S }-{ \left( { \partial }_{ t }{ S } \right) }^{ 2 } \right) -i\left( i\left( { S }_{ x }^{ 2 }+{ S }_{ y }^{ 2 }+{ S }_{ z }^{ 2 } \right) +{ \nabla }^{ 2 }S \right) \right] +{ \nabla }^{ 2 }R+{ m }^{ 2 }R+{ \partial }_{ t }^{ 2 }R+2i\cdot \left( { \partial }_{ t }{ S }{ \cdot \partial }_{ t }{ R-{ \nabla }S\cdot { \nabla }R } \right) =0$$$$

Edit: The equation above leads to: $$$$i\left[ R\left( { \partial }_{ t }^{ 2 }S-{ \nabla }^{ 2 }S \right) +2\cdot \left( { \partial }_{ t }{ S }{ \cdot \partial }_{ t }{ R-{ \nabla }S\cdot { \nabla }R } \right) \right] -R\left( { \left( { \partial }_{ t }{ S } \right) }^{ 2 }+{ \left( { \nabla }S \right) }^{ 2 } \right) +{ \nabla }^{ 2 }R+{ m }^{ 2 }R+{ \partial }_{ t }^{ 2 }R=0$$$$

Because $$S,R$$ are real one gets the following equations:

\begin{aligned} &\rightarrow 2\cdot \left( { \partial }_{ t }{ S }{ \cdot \partial }_{ t }{ R-{ \nabla }S\cdot { \nabla }R } \right) =R\left( { { \nabla }^{ 2 }S-\partial }_{ t }^{ 2 }S \right) &\rightarrow R\left( { \left( { \partial }_{ t }{ S } \right) }^{ 2 }+{ \left( { \nabla }S \right) }^{ 2 } \right) ={ \nabla }^{ 2 }R+{ m }^{ 2 }R+{ \partial }_{ t }^{ 2 }R \end{aligned}

The left equation yields to

$$$$2{ \partial }_{ \mu \\ }S{ \partial }^{ \mu }R=R\Box S$$$$

• I'd guess many people had tried but failed. Because this is a natural direction once Schrödinger's is separated, but I have not seen one (caution: I am quite ignorant in this field). I'd guess it is quite close to "not possible". Feb 26, 2020 at 14:38
• Note that your change of variables didn't separate the equation. The variables are still pritty much coupled, in a very non-linear way. Feb 26, 2020 at 22:26
• @AccidentalFourierTransform I know but when you 'separate' the SE S and R are also nonlinearly coupled but the two equations you get have a very well known interpretation and I am wondering if one can do the same thing with the KGE. Feb 26, 2020 at 23:15