# What is the significance of Planck's constant in De Broglie–Bohm theory or Pilot-wave theory?

In standard QM, Planck's constant seems to be a constant that describes the smallest quanta of energy in some way. Does De Broglie–Bohm theory have an alternate interpretation of that constant and what it means about space-time or particles (or both)?

• it probably doesn't need to be said, but I'll do it anyway so no one else has to do it: as far as we know, there is no such thing as a smallest quantum of energy – Christoph Jan 23 '15 at 12:16

Bohm's interpretation of the QM says that a particle moves along a trajectory, as in the classical mechanics, however, the potential in which this particle moves is supplemented by a "quantum potential". In all the equation of movement relevant for this question is

(1) $\frac {∂S}{∂t} + \frac {(\nabla S)^2}{2m} + \left[V(x) - \frac {\hbar ^2}{8m} \left(2\frac {\Delta P}{P} - \frac {(\nabla P)^2}{P^2} \right) \right] = 0$,

where $S$ and $P$ are defined by expressing the wave-function as

(2) $\psi = P \ e^{iS/ \hbar}$.

Here is what Bohm says about the constant $\hbar$ :

" ... in the classical limit $\hbar \to 0$ the above equations are subject to a very simple interpretation. The function $S(x)$ is a solution of the Hamilton-Jacobi equation."

However, as long as the limit $\hbar \to 0$ doesn't hold, the trajectory of a particle is influenced, besides $V(x)$, by a "quantum potential" generated by the wave-function,

$V^{quantum} = -\frac {\hbar ^2}{8m} \left(2\frac {\Delta P}{P} - \frac {(\nabla P)^2}{P^2} \right)$.

Thus, the magnitude of $\hbar$ relative to the considered problem separates between the classical and quantum behavior.

• Interesting. I'd like to understand your bolded point more concretely. The units of h are energy multiplied over a distance. Can you add some interpretation of those units in this context? – B T Jan 23 '15 at 19:10
• @BT , no $\hbar$ has dimensions of energy multiplied by *time*. $\hbar$ is action. The relevance of this quantity is inherited from classical mechanics - from the principle of least action, which says that a particle follows in space the trajectory on which the action is minimal. And what is that action? the Lagrangian (that has dimensions of energy) multiplied by time. – Sofia Jan 23 '15 at 19:58
• Ah right, I misspoke. So the ℏ has units of action. Didn't realize that was a type of unit. Sounds like you're saying that by using 0 as the value for h, we basically get classical mechanics, and that h is the value that is required to describe the "quantum potential". Could you elaborate on what the quantum potential is, and what ℏ means to it? Perhaps ℏ relates to how much mass is necessary to generate a given frequency and/or amplitude of particles' waves (purely speculating here)? – B T Jan 23 '15 at 20:58
• @BT : you ask me too many things at once. Please, one step at one time. What $\hbar$ means to the quantum potential you saw in my answer: let $\hbar \to 0$, and we have classical mechanics. But, worse that that : the uncertainty principle says $\Delta x \Delta p_x \ge \hbar/2$. But if $\hbar \to 0$, there is no interdiction to have $\Delta x \Delta p_x = 0$. So, no uncertainty principle - classical mechanics. Well, about quantum potential can you post a question? In the space of the comments it's difficult to answer. The quantum potential tells the particle where it can be and where not. – Sofia Jan 23 '15 at 21:24
• @BT : do you remember the 2slit experiment? Well, the quantum potential doesn't allow the Bohmian trajectories to pass through the forbidden fringes. – Sofia Jan 23 '15 at 21:26