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Uncertainty principle for position and momentum: $$ \Delta x \Delta p \ge \frac{h}{4\pi}$$

So suppose we have a particle... and we have 2 different measuring devices. The first measuring device measures position. The second measuring device measures momentum. The two devices act simultaneously on the particle. What will happen? Will we get a definite value on both measuring devices?

I'm not asking about the practical impossibility of simultaneous measurements. I'm asking, what does the QM formalism say will happen in this situation when these 2 measuring devices act simultaneously on the particle. Or is such a situation impossible for some theoretical reason? If so, what is that reason?

Thanks.

EDIT: I don't think I'm communicating what I want to ask properly. Let's suppose I'm a scientist pre-QM. I want to construct an experimental setup that simultaneously measures position and momentum. Meaning in a single instant I get position and momentum measurement with arbitrary precision. Is such a setup possible in classical physics? What would the same setup actually do when taking QM into account?

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    $\begingroup$ What would happen if you tried to measure an electron's favorite movie? $\endgroup$
    – WillO
    Commented Dec 10, 2022 at 20:53
  • $\begingroup$ Has it been tried that there is absolutely no way to insert a measuring instrument, eg. a traveling photon, to insert this ad hoc into the wavefunction of the system and then checking how this affects the evolution of this system, hopefully resulting in a clean spike "collapse" due to this photon hitting the system? $\endgroup$
    – James
    Commented Dec 10, 2022 at 21:26
  • $\begingroup$ @josephh, so what would the measuring devices display? $\endgroup$
    – Ameet Sharma
    Commented Dec 11, 2022 at 1:58
  • $\begingroup$ What we can be sure of is that the device will never yield precise values for the particle’s position and momentum. What we do know is that we will get values constrained by the uncertainty relation. I know that is not the answer you were looking for, but when you simultaneously measure position and momentum, the relation $\Delta x\Delta p\ge\frac{1}{2}\hbar$ always holds. There is no way around this and will always hold no matter what the setup. Like stated above, this is a fundamental property of nature. $\endgroup$
    – joseph h
    Commented Dec 11, 2022 at 3:20
  • $\begingroup$ How do you get 2 different devices to measure the same target simultaneously? $\endgroup$
    – BowlOfRed
    Commented Dec 11, 2022 at 9:25

3 Answers 3

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Reading the question and comments, it seems there is a slight misunderstanding of the Heisenberg uncertainty principle.

This principle is a statistical law. What this means is that for an ensemble of particles that are prepared in the same way, the relationship between the standard deviations in the momentum and position measurements will be such that $$\sigma_x\sigma_p\ge\frac{\hbar}{2}$$

This means that the variance in the results of position measurements and the variance in that for momentum measurements cannot both be arbitrarily small. Your experiment has one device that measures position and the other device, momentum. What this means is you need to confine the particle as much as possible. But the more you confine it, the greater will be the variance in its momentum.

No matter how you do the experiment, and no matter how sophisticated your equipment is, a particle will never be found to exist such that it has precise position and momentum measurements, simultaneously.

What we do know is that in any experiment, regardless how it's performed, we will get values constrained by the uncertainty relation $$\Delta x\Delta p\ge\frac{1}{2}\hbar$$ which always holds. There is no way around this and will always hold no matter what the setup. Like stated above, this is a fundamental property of nature.

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  • $\begingroup$ Yes, I understand. The particle doesn't have simultaneous position and momentum. But my question is a bit different. What happens to the two measuring devices when they simultaneously attempt to measure position and momentum? The physics must give some kind of answer to this situation. Is it that both measuring devices will give values, but both those values are meaningless... Will one give a reliable value, and the other unreliable? Is it a 50/50 random choice which is reliable? Do the dials stay at 0? Do the devices explode? Something must happen. $\endgroup$
    – Ameet Sharma
    Commented Dec 11, 2022 at 4:19
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    $\begingroup$ See the edits Ameet. There is not much more that can be said without knowing more details of your experimental setup. Thanks. $\endgroup$
    – joseph h
    Commented Dec 11, 2022 at 5:44
  • $\begingroup$ The passage from the statistical uncertainty relations to what can be said about a single experiment is not so simple. See, for instance, nature.com/articles/srep02221 $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Dec 11, 2022 at 7:27
  • $\begingroup$ @GiorgioP Yeah. I have made an edit. Thanks. $\endgroup$
    – joseph h
    Commented Dec 11, 2022 at 9:12
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The question you are asking simply does not make any sense in Quantum Mechanics. Quantum Mechanics says that, upon a position measurement, the particle becomes a position eigenstate. And upon a momentum measurement, the particle becomes a momentum eigenstate.

There is no state that is simultaneously both position and a momentum eigenstate. So there is no state that the particle can take after the measurement that you're proposing. Quantum Mechanics says that particles must always be described by some state in the Hilbert space.

Hence, this question cannot be answered in Quantum Mechanics. If Quantum Mechanics is right (which we have no evidence against so far), then this question is meaningless because such a measurement does not exist.

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  • $\begingroup$ Let me ask the question a different way. Suppose we were going by classical physics. 1. Within classical physics, is there an experimental setup that could simultaneously measure position and momentum? 2. What exactly would the same experimental setup do when taking QM into account? I am asking about what the measuring device will do... I'm not asking about what state the particle is in. $\endgroup$
    – Ameet Sharma
    Commented Dec 11, 2022 at 13:21
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    $\begingroup$ Just to add a few words to this excellent answer: A measurement of position is by definition an action that leaves the particle in a position eigenstate, and likewise for momentum. So a "simultaneous measurement of position and momentum" cannot exist for the same reason that a married bachelor cannot exist. $\endgroup$
    – WillO
    Commented Dec 11, 2022 at 13:22
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    $\begingroup$ @AmeetSharma In the everyday world, we measure velocities by looking at the trajectory of the classical particle, and then calculating the ratio of displacement over time. Quantum Particles at least don't have a visible trajectory, so this method won't work. There may be an invisible trajectory but those theories are either disproved or incomplete. Quantum Measurements are very indirect in nature, for example, you measure energy levels off spectral lines. $\endgroup$
    – Ryder Rude
    Commented Dec 11, 2022 at 14:06
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    $\begingroup$ @AmeetSharma ... which makes it even more impressive that Quantum Mechanics captures the results of a variety of measurement techniques in elegant mathematics, without even needing the details of the experiment as input. For hidden trajectory theories, you should see Bell's Inequality and Kochen Specker theorem for the disproved ones. Bohmian Mechanics is a standing hidden variable theory but it can't explain Quantum Field Theory yet, and is hence incomplete. Also, Bohmian mechanics agrees with non-relativistic QM and hence does not allow for simultaneous position and momentum measurements. $\endgroup$
    – Ryder Rude
    Commented Dec 11, 2022 at 14:10
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"I'm not asking about the practical impossibility of simultaneous measurements."

Ah, but they are easily possible. Consider a diffraction grating: it takes light and directs it at different angles depending on wavelength. For a spectrometer, measuring the angle yields the wavelength. The theoretical resolution of such a spectrometer depends on how large the grating is: the more grooves or slits it has relative to a wavelength, the fussier it is about directing the light.

That's the wave picture. Now consider the particle picture. You detect a photon coming from the grating at a particular angle. Since momentum of a particle is inversely proportional to wavelength, that's a momentum measurement. But detecting the photon is also a position measurement, since the photon must have reflected from the grating, which is of a certain size at a certain location. The uncertainty principle tells you that the smaller the size, the more uncertain the momentum.

But that's the same thing the wave picture says. And, if you work through the math, it's quantitatively about the same as the Uncertainty Principle. But the Uncertainty Principle is a crude estimate: the wave picture can tell you precisely what the pattern of intensity/probability on your detector is.

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  • $\begingroup$ So with the diffraction grating then... let's say it's arbitrarily thin so uncertainty in position is arbirarily small. We know the position and we know the angle. So the "uncertainty" in "uncertainty principle" is only a statistical law saying how wide the difference in measurements can be over repeated experiments. It isn't literally saying when position is perfectly defined, momentum doesn't exist? So among the comments/ answers, I'm seeing two interpretations of uncertainty. 1) statistical variance. 2) literal non-existence of simultaneous position/momentum. 1) and 2) seem very different. $\endgroup$
    – Ameet Sharma
    Commented Dec 11, 2022 at 22:28
  • $\begingroup$ @AmeetSharma We don't know the lateral position, and for the lateral deflection that's what matters. As to what it means , well, when you do an experiment and correctly apply the theory, experiment and theory agree. There is no deeper meaning, and there cannot be, since everything we know about physics comes from experiments and observations. $\endgroup$
    – John Doty
    Commented Dec 11, 2022 at 22:39

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