Can measurement of the momentum of a particle can be done without observing its position?

Uncertainty principle says that one cannot measure exactly the position and momentum of a particle at same time. As per common understanding when we are measuring momentum of an object it is implicit that we aware of its position. My doubt is in Quantum world, how can we measure momentum of a particle without knowing its position. Are the two momentum and position are mutually exclusive?

• Unfortunately you are using somewhat layman's interpretation of the uncertainty principle. You can make a measurement of the position and momentum of a particle at the same time. What the uncertainty principle tells us is that If you were to run your experiment many times you would find a spread in your measurements of both position ($\Delta x$) and momentum ($\Delta p$), and that for any system you will never be able to get the product of these two below a certain amount ($\Delta x\cdot\Delta p\geq\hbar/2$) – Aaron Stevens Jun 27 at 18:19
• look atmy answer here physics.stackexchange.com/questions/479475/… where bubble chamber tracks are discussed, momentum measured by $Bqv=mv^2/r$ taking into account the ionisation loss,and the interaction point measured at the main vertex. – anna v Jun 27 at 18:45
• @AaronStevens I am not sure that your statement about the possibiiity of a measurement of the position and momentum of a particle at the same time is consistent with the request of QM that the effect of any measurement is to project the wavefunction onto the eigenstate corresponding to the measured eigenvalue. A really simultaneous measurement would imply a common eigenvecctor of two non-commuting operators. – GiorgioP Jun 28 at 3:34
• @GiorgioP No. Claiming that the results of your measurements were both eigenvectors of your initial state implies a common eigenvector of two non-commuting operators. Or if the measurements weren't simultaneous, claiming that the second measurement doesn't change the state that the system was in after the first measurement. Even though they are closely related, non-common eigenvalues and the HUP are not saying the exactly same thing. – Aaron Stevens Jun 28 at 9:58
• @AaronStevens I fully agree with such a point of view. – GiorgioP Jun 28 at 23:03

The reason this is interesting is that while the Heisenberg principle limits your precision in measuring both $$x$$ and $$p_x$$ at the same time it does not limit your precision in measuring $$y$$ and $$p_x$$ at the same time.
• Something that confuses me is the statement of "measuring $x$ and $p_x$ at the same time". Won't the HUP only manifest itself after multiple measurements of similar systems? Isn't the precision of a single measurement more about the detector? – Aaron Stevens Jun 27 at 23:38