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

Physics Asked by NicAG on January 22, 2021

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

$$begin{aligned}
&rightarrow frac { partial rho }{ partial t } +nabla cdot (rho v)=0qquadqquadqquad left(R={ rho }^{ 2 },quad v=frac{1}{m} nabla Sright)&
&rightarrow frac{partial S}{partial t}=-left[frac{|nabla S|^{2}}{2 m}+V+Qright]qquadqquadleft(Q=-frac{hbar^{2}}{2 m} frac{nabla^{2} R}{R}right)&
end{aligned}$$

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}
Rleft[ left( i{ partial }_{ t }^{ 2 }{ S }-{ left( { partial }_{ t }{ S } right) }^{ 2 } right) -ileft( ileft( { 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+2icdot left( { partial }_{ t }{ S }{ cdot partial }_{ t }{ R-{ nabla }Scdot { nabla }R } right) =0
end{equation}

Edit:
The equation above leads to:
begin{equation}
ileft[ Rleft( { partial }_{ t }^{ 2 }S-{ nabla }^{ 2 }S right) +2cdot left( { partial }_{ t }{ S }{ cdot partial }_{ t }{ R-{ nabla }Scdot { nabla }R } right) right] -Rleft( { 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:

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

The left equation yields to

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

One Answer

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.

Answered by Damon on January 22, 2021

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