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)?

  • 1
    $\begingroup$ 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 $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – B T Jan 23 '15 at 19:10
  • $\begingroup$ @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. $\endgroup$ – Sofia Jan 23 '15 at 19:58
  • $\begingroup$ 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)? $\endgroup$ – B T Jan 23 '15 at 20:58
  • $\begingroup$ @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. $\endgroup$ – Sofia Jan 23 '15 at 21:24
  • $\begingroup$ @BT : do you remember the 2slit experiment? Well, the quantum potential doesn't allow the Bohmian trajectories to pass through the forbidden fringes. $\endgroup$ – Sofia Jan 23 '15 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.