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 ($[\hat x, \hat p]\ne[\hat p, \hat x]$).

Where is the lapse in my logic?

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?