5
$\begingroup$

Look at http://www.bohmian-mechanics.net/whatisbm_pictures_hydrogen.html. It is mentioned that in the rest states of a bound electron, the position of the electron is stationary, since the wavefunction's phase is not changing. In the Bohmian picture where the electron DOES have an exact position, it should be supposed to stay at a certain point, since it has no Bohmian trajectory.

However, we KNOW that the electron has a kinetic energy of ~13.6eV, which make it have a velocity of ~2E6 m/s. So isn't this a contradiction in the Bohmian picture of the atom?

Improve this question
$\endgroup$
1
  • $\begingroup$ This answer seems relevant to your question. Relevant quote: "So what regular quantum mechanics thinks of as kinetic energy (i.e. not potential) is to dBB theory sometimes quantum potential energy and sometimes is the actual motion through configuration space." $\endgroup$
    – Alfred Centauri
    May 3 '19 at 11:10
7
$\begingroup$

So isn't this a contradiction in the Bohmian picture of the atom?

In the Bohmian view, what would be kinetic energy can be 'stored' in the guiding wave. Here's an extensive excerpt from Reflections on the deBroglie–Bohm Quantum Potential

Consider the simple case of a cubical box with no classical potential inside and infinite potential outside. In other words, the particle cannot escape while the box remains intact. If an initially free (spinless) particle is trapped in such a box of side length $L$, then its wavefunction would take on the stationary waveform:

$$\psi = (2L)^{3/2}|\sin(n_1\pi\,x/L)\sin(n_2\pi\,y/L)\sin(n_3\pi\,z/L)|e^{iE_nt/\hbar} = Re^{iS/\hbar}$$

with total energy $E=(n^2_1 + n^2_2 + n^2_3)(\pi^2\hbar^2/2mL^2)$; where $n_1,\,n_2,\,n_3$ are positive integers. In Orthodox Quantum Theory, the particle is assumed to have kinetic energy only and to be bouncing back and forth between the walls of the box. In deBroglie–Bohm Theory, the value of the quantum potential is given by: $Q = -(\hbar^2/2m)(\nabla^2R)/R = (n^2_1 + n^2_2 + n^2_3)(\pi^2\hbar^2/2mL^2)$: This is the same magnitude as the particle’s kinetic energy is assumed to have in Orthodox Quantum Theory. However, since $S = -Et,\,\nabla S = 0$; i.e., the particle has zero momentum and therefore zero kinetic energy. All the energy of the system is potential with the kinetic energy of the quantum particle having become stored in the wave field (Bohm1952, 184; Riggs1999, 3072). What’s more, this energy will be returned to the particle if the wave field’s stationary state is disturbed, e.g., if any side of the box is removed. Surprisingly, this explanation was originally suggested by David Bohm when he wrote:

... the kinetic energy of the particle will come from the $\psi$ field, which is able to store up even macroscopic orders of energy when its wave-length is small (1953, 14, italics mine).

Improve this answer
$\endgroup$
1
  • $\begingroup$ What determines the trajectory of the particle after trying to measure it (and so disturbing the wavefunction)? $\endgroup$
    – Ali Lavasani
    May 9 '19 at 18:52
2
$\begingroup$

This is indeed one of the many disadvantages of Bohmian mechanics: in the ground state the electron is predicted to hover still in midair forever. Besides being exceptionally strange for a theory that sells itself on being classically intuitive, it flies in the face of experimental measurements of the electron's velocity, momentum, or kinetic energy.

The Bohmian reply to this problem is that all such experimental measurements are wrong. In other words, any attempts to measure properties of the Bohmian particle (besides its position) are doomed to accidentally instead measure properties of the pilot wave. The electron really is hovering still in midair, but we just can't observe it doing that. Several tricks like this allow Bohmian mechanics to evade contradiction with the experimentally verified uncertainty principle. Whether this is a deal breaker is up to you.

On the plus side, for highly excited states, the Bohmian trajectories are somewhat intuitive. For example, for atomic states, they give concrete realization to the orbits postulated in the Bohr model, so it's not all bad.

Improve this answer
$\endgroup$
5
  • $\begingroup$ I've downvoted your answer for two reasons: (1) it doesn't seem to address the OP's question (this answer doesn't even mention kinetic energy), and (2) it seems (to me) to be more of a rant than an earnest attempt to answer the question of "isn't this a contradiction in the Bohmian picture of the atom". $\endgroup$
    – Alfred Centauri
    May 4 '19 at 0:35
  • $\begingroup$ @AlfredCentauri First off, I'm pretty sure I did mention kinetic energy, and explicitly address both the OP's questions. Second off, OP seems to have poured a lot of faith and energy in Bohmian mechanics, if his last 20 questions about it are any judge, and sometimes you just need to hear it straight. $\endgroup$
    – knzhou
    May 4 '19 at 0:40
  • $\begingroup$ On its face, it's obviously a contradiction, as OP was suspecting. And the only fix in Bohmian mechanics is to assume ad hoc that the theory's main achievement (the definite trajectory) is almost entirely hidden from observation. The closer you look at the trajectory, the more problems pop up. $\endgroup$
    – knzhou
    May 4 '19 at 0:43
  • $\begingroup$ @knzhou I don't have any faith in Bohmian mechanics, I'm just trying to understand it. Its trying to offer a classical picture actually annoys me. Anyway thanks for your attention to my question. $\endgroup$
    – Ali Lavasani
    May 10 '19 at 3:42
  • $\begingroup$ @knzhou What do you mean that the trajectories are entirely hidden? My understanding is, the electron is moving inside the atom, because Bohmian mechanics is non-local, and while its marginal wavefunction is stationary , its movement depends on the universal wavefunction. Since we don't know the universal wavefunction, the electron's movement looks like random trajectories to us, and this is where the Born rule emerges from Look at sciencedirect.com/science/article/pii/037596019190116P. $\endgroup$
    – Ali Lavasani
    May 10 '19 at 3:47
1
$\begingroup$

In Bohmian mechanics, kinetic energy splits into a contribution from the actual motion of the particle and a contribution from the quantum potential. In the situation you reference, all of the standard quantum mechanical kinetic energy is stored in the quantum potential.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Oh, see @AlfredCentauri's comment above. $\endgroup$
    – Will
    May 3 '19 at 14:35
  • $\begingroup$ Is the quantum potential of a particle measurable? $\endgroup$
    – Ali Lavasani
    May 3 '19 at 14:36
  • $\begingroup$ Not directly, although people have worked on trying to map it out by inference. The follow-up question is, I imagine: if the quantum potential is not measurable, how can it contain part (or all) of the kinetic energy? The best I can do to answer that is: in Bohmian mechanics the standard quantum mechanical kinetic energy operator no longer corresponds directly to a property of the particle. If you think this is super weird, I agree! Bear in mind, though, that it is not entirely surprising, since in Bohmian mechanics we separate the wavefunction itself from the "actual particle." $\endgroup$
    – Will
    May 3 '19 at 14:46
  • $\begingroup$ I should say, people have tried to map out the quantum potential by inference in experiments, as an attempt to verify Bohmian mechanics, but to the best of my knowledge no such experiments have succeeded. I'm not positive though -- that would be a good question to ask in itself! $\endgroup$
    – Will
    May 3 '19 at 14:48
  • $\begingroup$ What do you mean by mapping the quantum potential? As far as I know, in Bohmian mechanics, the wavefunction is defined on the configuration space, which is infinite dimensional, and in principle inaccessible for us in the 3D world. $\endgroup$
    – Ali Lavasani
    May 6 '19 at 18:08

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.