Do we have a potential (quantum) in Bohm quantum mechanics that is associated with the phase factor $S$?

Question

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?

Answer 1

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.