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Let's have two non commuting operators $\hat A$ & $\hat B$ corresponding to the physical quantities $A$ & $B$. Also let's assume we are given a wave function $\Psi(\vec{r})$. Because $\hat A \hat B \ne \hat B \hat A$ the two operators don't have the same set of orthonormal eigenvectors/eigenfunctions ($\{|a_i\rangle\}\neq\{|b_i\rangle\}$).

Now thinking in terms of the standard copenhagen interpretation we can conclude the following. If we measure $A$ the wave function will collapse into one of the eigenvectors of $\hat A$: ($|\Psi\rangle \xrightarrow{A}|a_k\rangle$) and we will measure its corresponding eigenvalue $\alpha_k$. However since $\{|a_i\rangle\}\neq\{|b_i\rangle\}$, the wave function now will have the form:

$$|a_k\rangle=\sum_{i=1}^\infty c_i|b_i\rangle$$

So we will be completely uncertain about the value of $B$. Now if we measure $B$ right after that, then by the same logic we will be uncertain about the value of $A$.

Now let's take a look at the pilot wave interpretation or Bohmian mechanics. In this view we start with initial conditions which give us the probability distribution of a particle is. This probability distribution evolves according to the Schrödinger equation. We think of $\Psi$ as a wave which guides the particle (wherever it is in the distribution). At each point in space a hypothetical particle in that position is "pushed" by the wave in such a way that satisfies the guiding equation:

$$\vec{p}=\hbar \ \mathfrak{Im}\left(\frac{\vec{\nabla}\Psi}{\Psi}\right)$$

Now here comes my confusion. Imagine we have some wave function of a particle $\Psi(\vec{r})$ and we decide to measure its position. By measuring it we conclude that it is located in point $\vec{r}_0$. This means that just at the moment of measuring the particle was pushed by the wave and had momentum:

$$\vec{p}_0=\hbar \ \mathfrak{Im}\left(\frac{\vec{\nabla}\Psi(\vec{r}_0)}{\Psi(\vec{r}_0)}\right)$$

So we would have measured both the position and the momentum of the particle with an arbitrary precision even though the position and momentum operators don't commute ($[\hat x, \hat p]\ne\hat 0$).

Where is the lapse in my logic?

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I think your confusion might stem from not understanding that the effective wave function of the particle still collapses in Bohmian mechanics; Bohmian mechanics just provides a solid theoretical framework for understanding how the effective wave function of a subsystem is related to the universal wave function and therefore provides a mechanism for how it collapses.

So if you actually perform a measurement that gives you the precise position of the particle, then you have caused the wave function to collapse to a delta function, and so the momentum of the actual particle is undefined; such measurements never actually happen. If you perform a measurement of the position that gives only the approximate position of the particle (as all actual measurements do), then you are unsure about what momentum the particle has because you don't know where exactly in the collapsed wave function the particle is. So the Heisenberg uncertainty principle still holds.

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  • $\begingroup$ But doesn't Bohmian mechanics try to get rid of this concept of measurements affecting the behaviour of the system? I thought the collapse of the wave function was something special to the copenhagen interpretation. $\endgroup$ Commented Nov 29 at 2:08
  • $\begingroup$ Also by calculating $\vec{p}_0=\hbar \ \mathfrak{Im}\left(\frac{\vec{\nabla}\Psi(\vec{r}_0)}{\Psi(\vec{r}_0)}\right)$ that would have been the momentum of the particle just before measurement. And if at the instant of the measurement that momentum jumps, then the particle would have to experience instantaneous infinite force. How do you explain that? $\endgroup$ Commented Nov 29 at 2:12
  • $\begingroup$ Bohmian mechanics gets rid of collapse at the fundamental level, i.e., the level of the universal wave function. At the level of subsystems, collapse still occurs, it's just that now it's not a postulate, it's an explainable process. And there is no such thing as instantaneous measurement. $\endgroup$
    – Travis
    Commented Nov 29 at 5:02
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Pilot wave theory is often called an interpretation of quantum theory. This is misleading since it adds particles on top of states and observables. As a result the predictions of pilot wave theory may differ from those of quantum theory and violate the Born rule in some circumstances, including violating the uncertainty principle. For a discussion of such proposed differences see this paper and references therein

https://arxiv.org/abs/2408.05403

I should note that decoherence explains the absence of quantum interference in everyday life so it is unclear what problem pilot wave theory is supposed to solve that isn't solved by quantum theory without collapse

https://arxiv.org/abs/quant-ph/0403094

Also pilot wave theory can't currently reproduce the predictions of relativistic quantum theories, i.e. - the vast bulk of experimentally tested predictions of quantum theory

https://arxiv.org/abs/2205.00568

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    $\begingroup$ It's not true that pilot wave theory can't reproduce the predictions of relativistic quantum theories. Here's an example for QED: arxiv.org/pdf/0707.3487 $\endgroup$
    – Travis
    Commented Nov 29 at 16:37
  • $\begingroup$ @Travis No. See Section 4 of arxiv.org/abs/2205.00568 the third paragraph on p.12 in particular for criticisms of the reference you gave. $\endgroup$
    – alanf
    Commented Nov 29 at 19:02

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