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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

\begin{equation} \partial_{t}^{2} \psi-\nabla^{2} \psi+m^{2} \psi=0 \end{equation}

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

\begin{equation} 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 \end{equation}

Edit: The equation above leads to: \begin{equation} 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 \end{equation}

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

\begin{equation} 2{ \partial }_{ \mu \\ }S{ \partial }^{ \mu }R=R\Box S \end{equation}

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  • $\begingroup$ 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". $\endgroup$
    – verdelite
    Feb 26, 2020 at 14:38
  • $\begingroup$ Note that your change of variables didn't separate the equation. The variables are still pritty much coupled, in a very non-linear way. $\endgroup$ Feb 26, 2020 at 22:26
  • $\begingroup$ @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. $\endgroup$
    – NicAG
    Feb 26, 2020 at 23:15

1 Answer 1

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Under certain reasonable condition, your first equation reduces to the continuity equation (which I call pseudo continuity equation); and your second equation gives the energy equation. I have recently posted a paper about this here: https://web.ma.utexas.edu/mp_arc/c/20/20-94.pdf . You need to jump to section 8 to see the derivation.

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