Lets have a particle in vacuum. If we do not observe, the Quantum Theory & Copenhagen interpretation says that the particle is nowhere and at the same time it is somewhere ( i.e the quantum state $\psi$ will determine its position with probability $\phi^2 (x)$ at $x$ ... ), so in order to comprehend this idea, I always "imagine" that the particle is split into infinite pieces in a way that at each point $x$ in space, there is $\% [\psi^2 (x) * 100]$ of the particle at that point.
However, lets say that we wanted to observe our particle and send a photon.
[since a photon is an electromagnetic wave, we do not need to know the exact location of the particle to send the photon to that point ( we are not sniping the particle )]
First of all, is there any chance that the photon does not interact with the particle ? I mean after all, before the interaction, the wave function haven't collapsed, so the particle is "nowhere", so how can the photon interact with the particle ?
Secondly, (assuming that the photon interacts the particle for certain) since the position of the particle will be determined by $\phi^2 (x)$ after the interaction, how will the magnitude of the interaction be determined ? After all, the particle does not have a definite position, so depending on the collapsed wave function, the force on the particle will be determined by $\vec F(x) = q(E(x) + \vec v \times \vec B(x))$ ?