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In Bohm quantum mechanics when we express the wavefunction as $\psi= R \exp(iS/\hbar)$ a quantum potential that depends on R is found, is there another potential that depends on the phase function S?

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No, there's only the classical (V), and supplementary quantum (Q) potentials found in Bohmian mechanics (BM). Giving the derivation will hopefully illuminate where this comes from (you may also find Bohm's 1952 papers here and here good to read in this regard), although we do come across other familiar formula involving $S$ in this process:

Starting with the SE,

\begin{equation} i \hbar \frac{\partial\Psi}{\partial t} = - \frac{\hbar^2}{2m} \nabla^2{\Psi} + V\Psi, \end{equation}

and the wave function in polar form of amplitude, $R$, and phase, $S/\hbar$,

\begin{equation} \Psi = R \, e^{iS/\hbar}\,, \qquad R,\, S \in \mathbb{R}\,. \end{equation}

Substituting this into the SE and separating the real and imaginary parts then yields

$$ \frac{\partial S}{\partial t} = \frac{\hbar^2}{2m}\frac{\nabla^2 R}{R} - \frac{(\nabla{S})^2}{2m} - V \tag{1} $$ and $$ \frac{\partial R}{\partial t} = -\frac{1}{2m}\left( R\,\nabla^2 S + 2\,\nabla{R}\,\nabla{S} \right) \tag{2} $$

respectively, where (1), in the classical limit $\hbar\to0$, is the Hamilton-Jacobi equation,

\begin{equation} \frac{\partial S}{\partial t} = -H, \end{equation} and the null term,

\begin{equation} Q = -\frac{\hbar^2}{2m}\frac{\nabla^2 R}{R}, \end{equation}

is the quantum potential (which can also be expressed using $R = \left|\Psi\right|$).

Now, from (2), using $\rho = R^2$ (Born's rule*), we may also obtain the continuity equation,

\begin{equation} \frac{\partial\rho}{\partial t}+\nabla\cdot{\left( \rho \frac{\nabla{S}}{m} \right)} = 0\,, \end{equation}

where the probability current,

$$ \vec j = \rho \frac{\nabla{S}}{m}\,, $$

is now apparent. Velocity is therefore

$$ \vec v = \frac{\nabla{S}}{m}\,, $$

and momentum

$$ \vec p = \nabla{S}\,. $$

*You may notice that Born's rule here was assumed, and used in order to make the results of BM line up with standard quantum mechanics - this is why under BM it's instead known as the quantum equilibrium hypothesis (QEH) - it isn't required by BM to always hold true, but is assumed to do so typically.

Anyway, I hope that clears things up for you.

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  • $\begingroup$ Thank you my dear for your elucidating answer. I however found an article that says there is an addition potential. Here is the link belo $\endgroup$ – Maxwell Sep 12 '18 at 9:31
  • $\begingroup$ Dear Toby Hawkins, Thank you very much for your elucidating comments. I also came across this article that assumes another potential to emerge. It is in the arXiv arxiv.org/abs/1710.01256 $\endgroup$ – Maxwell Sep 12 '18 at 9:44
  • $\begingroup$ @Maxwell You're welcome. That paper seems interesting, but it's uniquely proposing a new 'spin potential' through an extension of Bohm's derivation. It's not a typical feature of BM. BM, as it stands, treats particle positions separate to everything else; spin is still part of the wave function (as with regular QM). It (like any QM interpretation) isn't complete though, so it's good people are exploring new ideas. $\endgroup$ – Toby Hawkins Sep 12 '18 at 23:35

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