Question about uncertainty principle and attempts at simultaneous measurement of position and momentum 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?
 A: 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.
A: 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.
A: "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.
