Is the uncertainty principle a technical difficulty in measurement or is it an intrinsic concept in quantum mechanics irrelevant of any measurement?
Everyone knows the thought experiment of measuring the position of an electron. One can detect electron's position by hitting it with a photon, due to Compton scattering the collision of the photon with electron will change electron's momentum. This experiment is used to explain uncertainty principle to layman, but it is over simplified, isn't it? It also gives an impression that if there was a better suited experimental method the uncertainty principle becomes irrelevant.
I personally think it is intrinsic as it arises from the non-zero commutator of position and momentum operators irrespective of the measurement process. Am I right?
EDIT: My question is similar to certain extent to this question and this question. The answers there are nice but they focus on explaining basics of quantum mechanics more than they comment on the technical difficulty part. In the answers of question 2 there are statements like "So, it's not a knowledge limit" and "you're sort of correct when you say it's an observational limit" without further comments
To summarize, assume hypothetically we managed to find a way in the future where we can have a look at an electron without disturbing it by measurement or causing its wave function to collapse, would the uncertainty principle still hold in such a case?? Why/Why not?
Answer
Yes, the experiment is oversimplified, because the uncertainty principle is not about "disturbance through measurement". Although that's what Heisenberg said (one of the things he said), it turned out you can't interpret it that way in a very rigorous sense.
Whether there is something like "disturbance through measurement" that gives rise to an uncertainty relation is currently under heated debate in the quantum foundations community (see the work of Ozawa and recently some collaborators on the one hand and the work of Busch, Lahti and Werner on the other, if you want, I can look up some references).
That said, your opinion is correct in the sense that this is exactly how you derive the uncertainty relation. With position and momentum, one could ask the question "but why don't position and momentum commute" and then one can turn to the Fourier transformation and remark that the uncertainty relation is something that holds for any wave (water, electromagnetic, etc.), because the Fourier transform tells us that a small wave packet must consist of a lot of frequencies and a wave with only one frequency is infinite in space, etc. Now, since we have wave functions, we have this phenomenon in quantum mechanics as well. This means that indeed the uncertainty principle in our formalism does not require any measurement, it is an intrinsic property of the wave function in phase space.
EDIT: Even with your extended question, assuming everything else would hold, the uncertainty principle should still be there. It just tells you that the product of the variances of momentum and position are lower bounded, which comes from the wave-function itself. There is no reference to any measurement in the uncertainty principle other than that you need to measure to actually compare anything.
Being more concrete, I would say the following: Given a state of an electron (i.e. a preparation scheme that prepares the exact same physical electron over and over again), you can measure momentum and you will obtain a probability distribution according to the wave function (repeating the experiment multiple times). Then, assuming you have no disturbance in measurement, you measure the position of the exact same state. In that case, this will also have some probability distribution according to the wave function. These two measured probability distributions have variances whose product is lower bounded. That's what the uncertainty principle tells you.
The question of whether or not you could have information without disturbance (at least asymptotically) is still a matter of debate...
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