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?
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1Unfortunately 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 ($\Delta x\cdot\Delta p\geq\hbar/2$) – BioPhysicist Jun 27 '19 at 18:19
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4look atmy answer here https://physics.stackexchange.com/questions/479475/are-bubble-chamber-tracks-inconsistent-with-quantum-mechanics/479694#479694 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 '19 at 18:45
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@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-DoomsdayClockIsAt-90 Jun 28 '19 at 03:34
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@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. – BioPhysicist Jun 28 '19 at 09:58
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@GiorgioP Also, you can derive uncertainty relations even for sets of commuting operators that depend on your initial state if your initial state is not an eigenvector common to both operators. – BioPhysicist Jun 28 '19 at 10:06
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@AaronStevens In your comment you wrote that it is possible a measurement of the position and momentum of a particle at the same time. I do not understand from your reply how you think that such a simultaneous measurement can be made consistent with the interpretation of a measurement as a projection on an eigenvector of the operator representing the measured quantity. – GiorgioP-DoomsdayClockIsAt-90 Jun 28 '19 at 13:59
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@GiorgioP Oh ok, I see what you are saying. I misunderstood originally. My original comment was more to address the misunderstanding of the HUP, and I see now that I was not very precise in my language. – BioPhysicist Jun 28 '19 at 14:09
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@GiorgioP I suppose it would be better to say, then, that the HUP is a consequence of the inability to simultaneously measure position and momentum, not a result (as the OP says in the first sentence). That is what I was trying to get it. – BioPhysicist Jun 28 '19 at 16:48
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1@AaronStevens I fully agree with such a point of view. – GiorgioP-DoomsdayClockIsAt-90 Jun 28 '19 at 23:03
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I know a "kind of" answer.
A "velocity selector" AKA "Wein filter" will pass particles with a narrow fixed range of velocities. If we know the species (and therefore the mass) we have measured the momentum of all particles passing the filter without measuring position in the direction of travel (but we have measured position transverse to the direction of travel).
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.
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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? – BioPhysicist Jun 27 '19 at 23:38
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In "AKA Wein filter" as mentioned, is it mass & velocity pre known? Even in that case I think point of time of measurement also needed to comply HUP. My understanding is that how can we measure any physical parametre without observing the object. Can't we say that just mere observation itself is the act of knowing the position? in HUP what I read was that our eye observation needs light i.e. photon which when hits, particle will automatically alter its position and hence uncertainty. But my doubt is before and after our observation also there is light exists in space and hitting of photons. – Siva Kumar Valiveti Jun 28 '19 at 17:53
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@SivaKumarValiveti Your understanding of the HUP in terms of photon detection is somewhat false in that it assumes the particle has a well defined position and momentum the entire time. It's like you are thinking "the HUP just makes us uncertain about the true position and momentum of the particle." The "uncertainties" in the HUP really are standard deviations of probability distributions. There is no "true" position or momentum that we are "uncertain" about. – BioPhysicist Jun 29 '19 at 15:14
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@SivaKumarValiveti Generally the earliest stages of creating a particle beam give you at best roughly directionallity and a spread of velocities. By engineering things right you will know that a non-trivial number of particles in the beam have the right velocity to pass the selector, but you don't know the velocity of any given particle. Whether you know the particle species (and therefore mass) in advance or not depends a lot on the situation. – dmckee --- ex-moderator kitten Jun 30 '19 at 04:46
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@AaronStevens I'd like to quote few words from Ch 4 of the book Brief history of time. " In order to predict the future one has to be able to measure its present position and velocity accurately. The obvious way to do this is to shine light on the particle.Some of the waves of light will be scattered by the particle and this will indicate its position. More accurately one measures the position, the shorter the wavelength of the light that one needs and hence the higher the energy of a single quantum. So the velocity of particle will be disturbed by a larger amount." Could you pls explain this? – Siva Kumar Valiveti Jul 01 '19 at 15:44
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@SivaKumarValiveti I do not mean to invalidate that. I just mean to say that this is not the reason for the HUP. The HUP does not depend on the method of measurement. I think that serves as a nice analogy. The HUP is more concerned with the spread of measurements one would obtain upon measurements of similarly prepared systems from the inherently probabilistic nature of QM. It does not depend on the method of measurement. – BioPhysicist Jul 01 '19 at 16:25