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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 $\vec F(x) = q(E(x) + \vec v \times \vec B(x))$ ?

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The question confuses classical and quantum mechanical frameworks. To get a predictive model one has to stick to the one that pertains to the problem. Photons and particles belong to the quantum mechanical frame.

In this frame both the photon and the particle are point particles as defined by the standard model of particle physics. The calculations of this model are done using quantum field theory, with the use of Feynman diagrams.This link shows the connection of classical forces to the Feynman diagram level in the quantum framework.

I always "imagine" that the particle is split into infinite pieces in a way that at each point x in space

This is very wrong. The particle in space is a point particle . It is the probability, Π, of finding it when measuring that covers a space region with the wavefunction, $P=ψ*ψ.$ . See this answer which describes individual photons interactions .

since a photon is an electromagnetic wave,

Wrong, the photon is an elementary particle of the standard model. Not an electromagnetic wave. An electromagnetic wave is built up by a superposition of photons, the way a building is made of bricks, but a brick is not a building. The photon wavefunction, a complex numbers function, has the E and B field information so the superposition of many photons builds up the classical formula. An example of how this happens is shown using quantum field theory ,the classical fields are emergent from the underlying quantum level.

First of all, is there any chance that the photon does not interact with the particle

Depending on the boundary conditions: distances , x,y,z of photon direction of momentum of photon, the probabilities become infinitessimal for large distances. When they do interact, they follow the Feynman diagram point interactions between particles . In highe energy physics experiments with two colliding beams , the volumes are squeezed so the probability of interaction is large enough to measure interactions, but the interest is in the fourvectors energy and momentum, which characterize uniquely the interaction, and not space variables , so the sum of the results can be calculated in the standard model.

after the interaction, how will the magnitude of the interaction be determined

Since all model predictions are probabilistic, a large number of same boundary condition strikes have to accumulated , and then the probability distribution is known and can be compared with the QFT calcualtions. At the moment the standard model is very well validated with high energy experiments.

For low energies and quantum mechanical space solutions please read up the link I gave for the double slit experiment. Here is another answer.

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