How do particles interact in Bohmian mechanics / pilot wave theory / de Broglie–Bohm theory?

Question

I've read that in the de Broglie–Bohm interpretation of QM, the particle directed by its wavefunction has a trajectory (meaning both position and velocity) and that these are the only properties possessed by the particle. Other properties, like spin and mass, are attributed to the wavefunction

But if a Bohmian particle only has position and velocity, how does it interact with other Bohmian particles to transfer momentum? I would think the particle must have some momentum to transfer to another particle when they collide.

Answer 1

In pilot wave theory, there is a wave and a particle. In truth, both are in configuration space, the wave is a function like $\Psi(\vec{r}_1, \vec{r}_2,\vec{r}_3, \dots, \vec{r}_n,t)$ (a function of time and configuration space) and the particle too, is really just a dynamically changing position in configuration space $Q(t)=(\vec{r}_1, \vec{r}_2,\vec{r}_3, \dots , \vec{r}_n)$. Spin, phase, everything else is by rights part of the wavefunction. And the particle exerts no effect whatsoever on the wavefunction which evolves by Schrödinger (or Schrödinger-Pauli, or one of those with relativistic corrections) and nothing else.

Even velocity you want to be careful about because the time derivative of $Q$ is not the same thing as the the velocity operator from regular quantum mechanics, and if you weight the configuration space position $Q$ with the masses, it is not the same as the momentum operator from regular quantum mechanics.

If you look at dBB theory (de Broglie-Bohm theory), there is a quantum potential energy (which is determined buy the wavefunction an there is a classical potential. Both together guide the particle through configuration space. 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. Do not confuse the motion of the particle(s) through configuration space (and the associated energy) with the regular kinetic energy operator from quantum mechanics. The regular kinetic energy operator from regular quantum mechanics contains two terms.

So when a regular quantum person talks about momentum transfer, they could be talking about an initial momentum situation where all the kinetic energy is (in dBB theory) in quantum potential energy. So you can stick to energy transfer and note that there can be quantum potential (depending on the wavefunction and the position) energy and classical potential (depending just on position) energy. And then you can see that both are really functions of time, the wavefunction and the position in configuration space.

Since the particle doesn't affect the wave, the dynamics are all really in the wave, the particle just tells you which region of the wave is occupied if you succeed at breaking the wave into disjoint regions that will never interact (overlap) again. So what a regular quantum person would call momentum transfer is when the wave separates into different regions, regions a regular quantum person identifies with different momentum eigenvalues. Whether or not that corresponds to any particular motion of the particle is pretty much a side issue.

The short story is that to Copenhagen, no measurement is a measurement of a preexisting property (unless maybe it was in the eigenstate of that operator before the measurement), and dBB is the same except for position which is the sole thing that had a preexisting value that wasn't just a property of the wave. Even the mass times the velocity of the particle is not the same as the eigenvalue of the momentum operator as applied to the wavefunction, which is a property of the wave anyway, not of the particle.

edit

Please accept that in regular quantum mechanics there is classical potential energy and "all other energy" and that a Copenhagenist will say that that energy is all kinetic, and hence that there must be momentum in a situation where dBB theory might have no motion whatsoever. The two theories radically disagree on when there is momentum in the system. So you can't just describe a double slit experiment as an experiment to "transfer momentum" when talking about two theories that radically disagree about who has momentum and when.

So how is a particle detected in a double slit?

You have something somewhere that moves differently based on whether the screen/hitter interact in one place versus another place. The wavepacket for those different options start to separate and eventually will never again overlap in configuration space, the one that has the world particle is what happened, the other packets are the empty packets. It's is always, I repeat always about the separation of wavepackets in the dBB it is never, absolutely never about anything else in the dBB theory.

To be fair, it's actually the same in Copenhagen too. To get real predictions and avoid the quantum zeno effect making things never move you have to be objective about when, where, and how measurements happen, and once you get honest and detailed about that, then everyone agrees again. Whether dBB, Transactional, Ithaca, MWI, MIW, or Copenhagen they all literally turn to the exact same picture and setup about identifying that the Schrödinger equation always holds and that a time comes where it is practical to ignore the effects of the other parts because the different parts never overlap so don't affect the ratios of outcomes anymore.

People agree that that's why and when it is meanful to say an experimental result has occured. They simply disagree about the words they use and the stories they tell. Which is why it is not helpful to use words from Copenhagen that don't have exact cognates in dBB, it confuses the issue about what you are talking about.

And a double slit is clear. You have something somewhere that moves differently based on whether the screen-and-traveller interact in one place on the screen versus another place on the screen. It could be air molecules near the screen, it could be your eyeball, it could be the ink on your lab notebook or the parts of your hard drive or the parts of the screen itself, it could be lots of things. The wavepacket for those different options start to separate and eventually will never again overlap in configuration space, the part of the wavefunction that has the world particle is what happened, the other packets are the empty packets, and you can ignore them now because they no longer do anything measureable or detectable. Whether the dBB theory says that motion came from a transfer of kinetic energy from the traveller to the screen or from a transfer of quantum potential energy from the traveller to the screen doesn't matter and ... it is just not at all obvious. It could depend on whether the particle was in the leading edge of the wavepacket or was a straggler near the tail of the wavepacket. The theory will tell you, but you can't just assume a result because you want to assume it.

People assuming that at hidden variable theory works the way they want to assume it does rather than learning the theory is why people think so poorly of them. And it happens to Relativity too, and its just not fair to bring your assumptions and preconceptions to a theory and then blame the theory.

Answer 2

No, it's not the Bohmian particle 1 that interacts with another Bohmian particle, 2. It's the wave-packet 1 interacting with the wave-packet 2. In the book of D. Dürr "Bohmian Mechanics" one can find a section on the scattering theory.

But I believe that for your question suits more the section 15.1.2, "Asymptotic Velocity and the Momentum Operator", (unfortunately it's very much mathematics). The author speaks of the velocity far from the interaction region. Then, after a not simple proof he obtains

$\hat V _{\infty} = \frac {1}{i} \nabla \psi $

I also quote from this section

"The asymptotic velocity is experimentally an easily accessible quantity, and it is therefore convenient to introduce the corresponding self-adjoint velocity operator $\hat V _{\infty}$ (or the momentum operator $\hat P_{\infty} = m \hat V _{\infty}$ )".

Of the energy conservation takes care the Schrodinger equation. Of the linear momentum conservation I don't recall some special material because the wave-function of two colliding quantum objects has to obey this this conservation.

Answer 3

The BI says that the initial positions of the particles (the "hidden variables") plus the Schrödinger equation (the "pilot wave") result in deterministic trajectories for all the particles, and hence account for all experimental measurements. So if some of the particles interact (due to the wave function, which reflects the geometry and properties of the particles and of the entire experimental apparatus), we will see the results of that interaction appear, deterministically. The situation appears to be this simple.

So the transfer of momentum (and other interactions) is seen to be a classical explanation, evolving out of the simplicity of the BI and the fact that the wave function scales up to include classical mechanics, if we use a classical Hamiltonian or if we apply statistics as in electric current, temperature, or pressure, all of which are classical statistical properties.