I'm working a lot on Bohmian mechanics related theories and just realised something that had escaped me until now but actually seems to throw a wrench in the core idea of the theory. I tried to clarify my doubt by reading Bohm's original paper but this did not really help, as I don't find his explanations satisfying. So, here's the problem.
In order for Bohmian mechanics to truly describe the same dynamics as regular quantum mechanics, we need the interactions to involve the wave functions, not the particles. For example, between two electrons, we would have something like:
$$ V_{ee} = \int\psi^*_1\psi_1\frac{1}{|\mathbf{r}_2-\mathbf{r}_1|}\psi^*_2\psi_2 d\mathbf{r}_{1}\mathbf{r}_{2} $$
All the particles do is coast along, following the guiding equation. Now, however, if the interactions are controlled by the wave function, so are the interactions between particles and measuring apparatus. Which means that if we had a wavefunction in a state of superposition, that superposition should still carry over to the apparatus. Sure, the actual Bohmian particles of the apparatus will slide into a definite, deterministic position. But that's not what we see. We see the wavefunction, and the wavefunction interacts with the EM field; the wavefunction reflects light that we see with our own eyes, light that is captured by our own retina's wavefunction, and so on. In all of this process, at no point comes a moment where the Bohmian particles finally step in and their deterministic outcome overrides the unitary quantum evolution of the wavefunction, causing the collapse. Because the interaction is one-way: the wavefunction drives the Bohmian particles, but the Bohmian particles never feed anything back. They're just along for the ride. Am I missing something or is this an actual problem?
EDIT: since I've received a comment that made me think my explanation wasn't clear, I'll try to express this with an example.
Suppose we have an experimental setup with two sensors, a "left" and a "right" sensor. Suppose also we prepare an electron in a way that its wavefunction is in a superposition between going left and going right:
$$ \psi = \frac{\sqrt{2}}{2}\psi_L+\frac{\sqrt{2}}{2}\psi_R $$
Now, the Bohmian particle, depending on its initial position, will actually ride only one of these two crests of the wavefunction - let's say the left one; the other is empty. Some time passes and the wavefunction hits the detector. It interacts by a coupling Hamiltonian:
$$ H_{meas} = H_L+H_R $$
that represents, for example, the Coulomb interaction of the electron with the atoms inside the detector, which will be bumped by it and thus measure the hit. Each term represents the interaction with one specific detector. Let's say the apparatus has a light that becomes red if the electron hits the left side, and blue if it hits the right. The overall wavefunction (instrument + electron) thus evolves into:
$$ \psi_{tot} = \psi_{det}\left(\frac{\sqrt{2}}{2}\psi_L+\frac{\sqrt{2}}{2}\psi_R\right) \rightarrow \frac{\sqrt{2}}{2}\psi_{red}\psi_L+\frac{\sqrt{2}}{2}\psi_{blue}\psi_R $$
At this point, we still have our measurement problem. Bohmian mechanics should solve it because the particle really is on the left. But how should the detector know? Measure isn't an abstract thing; it is the result of a specific series of interactions governed by the same quantum mechanical laws as everything else. And all interactions pass through an Hamiltonian, and affect the wavefunctions - not the Bohmian particles. If it weren't like this, then for example electron-electron interactions in a Helium atom would look very different! Coulomb interactions aren't described well between two point-like particles, and if they were, we would have a way to tell what the real position of the Bohmian particles is. Clearly, they aren't, and we don't.
In his 1952 paper [the second part, Phys. Rev. 85, 2 (1952) pp.180] Bohm makes a similar reasoning but then concludes that, basically, once the measurement is performed, our ignorance about which packet contains the particle disappears and we can then continue our following calculations by just restricting the wavefunction to that packet alone. But I'm not sure by which mechanism should our ignorance disappear. The theory doesn't really seem to provide any.