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I understand we cannot state with arbitrary precision the position and momentum of a micro-particle as we superpose infinite waves to create a wave packet at the exact position of the particle and hence cannot say which wave is corresponding to its momentum. This is theoretically said. Also some people say that measurement of one disturbs the other and some say it's reality.

Firstly, the theoretical explanation: why should I consider that a wave out of many superposed will give us the momentum or position? We might use another single wave which is not sine in nature and might get the results. Or why won't a single sine wave suffice? This all considers the micro-particle as wave because of de Broglie.

Secondly, the practical thing seems legit.

Thirdly, the electron must have some kinetic energy and position. It is who we are measuring and can't get it exact so why blame the electron as in theoretical concept. How we quantify the uncertainty, it must depend on the measuring device and various other factors.

I might need to edit this post based on your answer so that I can explain more clearly my thinking.

Edit: From here is what gets the measurement problem:

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 Text: Imagine that you're blind and over time you've developed a technique for determining how far away an object is by throwing a medicine ball at it. If you throw your medicine ball at a nearby stool, the ball will return quickly, and you'll know that it's close. If you throw the ball at something across the street from you, it'll take longer to return, and you'll know that the object is far away. The problem i­s that when you throw a ball -- especially a heavy one like a medicine ball -- at something like a stool, the ball will knock the stool across the room and may even have enough momentum to bounce back. You can say where the stool was, but not where it is now. What's more, you could calculate the velocity of the stool after you hit it with the ball, but you have no idea what its velocity was before you hit it. This is the problem revealed by Heisenberg's Uncertainty Principle. To know the velocity of a quark we must measure it, and to measure it, we are forced to affect it. The same goes for observing an object's position. Uncertainty about an object's position and velocity makes it difficult for a physicist to determine much about the object.

Edit: After seeing this answer one of my doubts is very well cleared, the other answers are also written and helpful:

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Text: There is a definine velocity and momentum, we just don't know it. Nope. There is no definite velocity--this was the older interpretation. The particle has all (possible) velocities at once;it is in a wavefunction, a superposition of all of these states. This can actually be verified by stuff like the double-slit experiment with one photon--we cannot explain single-photon-fringes unless we accept the fact that the photon is in "both slits at once".So, it's not a knowledge limit. The particle really has no definite position/whatever.

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    $\begingroup$ It is actually a mathematical law, for conjugate operators. For any two operators the following applies: $\endgroup$ – BeastRaban Oct 7 '14 at 15:29
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    $\begingroup$ @BeastRaban: Yeah, but OP is sort of getting at why that mathematical law applies to Nature. $\endgroup$ – DanielSank Oct 7 '14 at 15:34
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    $\begingroup$ The uncertainty principle tells us that the classical approximation to quantum reality yields a model too rich in nature. $\endgroup$ – Johannes Oct 7 '14 at 15:50
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    $\begingroup$ related: physics.stackexchange.com/q/35972 physics.stackexchange.com/q/57030 $\endgroup$ – user4552 Oct 7 '14 at 15:51
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/24068/2451 and links therein. $\endgroup$ – Qmechanic♦ Oct 7 '14 at 16:06
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It has obviously been experimentally tested many times, but it's also derivable theoretically from the mathematics of quantum mechanics. Two basic ideas are important here--the first is the notion of a Fourier series, which allows you to represent any arbitrary periodic function (a square wave, for example) as a potentially infinite sum of different sine and cosine functions of different wavelengths. This works with position wavefunctions too, so for example a wave packet, representing the wavefunction of a particle whose position has been narrowed down to some fairly narrow range, can be represented as a sum of wavefunctions that all vary sinusoidally and extend infinitely in all directions. The second is the idea that momentum eigenstates--wavefunctions which have an exact momentum rather than a range of possible momentums--themselves look like infinite sinusoidal wavefunctions with unique wavelengths when their position amplitudes are calculated, a fact related to the de Broglie relation which says that momentum is directly related to wavelength by $ p = \frac{h}{\lambda} $, which is also part of the basic mathematical structure of QM.

Putting these together, any localized position function like a wavepacket must be represented as a sum of different sinusoidal wavefunctions with different wavelengths, and thus different momenta; and something similar is true if you reverse the role of position and momentum, since when you calculate the amplitudes in momentum space of a wavefunction with an exact position, you get a sinusoidal wavefunction with a unique wavelength in momentum space. Viewed this way, even in classical physics there is a similar sort of "uncertainty relation" between the spread of positions a wave packet is confined to and the spread of wavelengths in its Fourier decomposition, although this wouldn't normally be described as an "uncertainty" since neither spread is interpreted as the probability of getting a particular measurement result as it is in QM.

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  • $\begingroup$ +1 for the uncertainty principle applies to classical waves too. It's really not that mysterious when viewed this way. $\endgroup$ – adipy Oct 7 '14 at 15:55
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Like everything of importance in physics it's an experimentally testable fact. If it wasn't, we wouldn't be talking about it. Secondly it is, of course, built into the theory, otherwise the theory wouldn't be correct. What it is NOT, is a measurement "problem".

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  • $\begingroup$ see my edit for info regarding measurement problem $\endgroup$ – RE60K Oct 7 '14 at 16:44
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    $\begingroup$ @Aditya: You are assigning too much importance to incorrect answers (probably because they are reinforcing your misconceptions about quantum mechanics). May I suggest to work trough a modern textbook on quantum mechanics for clarification? $\endgroup$ – CuriousOne Oct 7 '14 at 16:50
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    $\begingroup$ @Aditya: Who to believe? This isn't religion. The only instance that counts is nature, and the only way you can be sure that whatever someone says is true (in a scientific sense), is by doing the experiments yourself. In case of quantum mechanics some of the fundamental experiments (like the double slit experiment), can be done easily at home with the help of some rather common equipment (like a digital camera, for instance). When we teach physics students, it's very important that we make them repeat key experiments as part of their studies, it's the only way to learn about nature. $\endgroup$ – CuriousOne Oct 7 '14 at 17:11
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    $\begingroup$ Like everything of importance in physics it's an experimentally testable fact. If it wasn't, we wouldn't be talking about it. Many people consider interpretations of quantum mechanics to be of importance in physics, but they aren't testable. You could argue that they're philosophy rather than physics, but the people publishing on this stuff are generally physicists publishing papers in physics journals. I don't think the boundaries are as tidy as you seem to suggest. $\endgroup$ – user4552 Oct 7 '14 at 18:38
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    $\begingroup$ Not sure whether it is Occam, I think Popper and Hemple, together and separately, suggested a physical theory is one that is (at the very least theoretically) falsifiable. Quantum interpretations are to my knowledge completely equivalent in terms of results -> thus they constitute aesthetic preferences, rather the physical theories. (at least according to these venerable gentlemen) $\endgroup$ – BeastRaban Oct 7 '14 at 19:22
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It is actually a mathematical law, for conjugate operators.

For any two operators the following applies: $$ \langle (\Delta A)^2 \rangle \langle (\Delta B)^2 \rangle \geq \frac{1}{4}|\langle[A,B]\rangle| $$ So for conjugate coordinates this means the Heisenberg uncertainty principle - for more detail - see Sakurai Chap.

(supposed to be a remark but something happened and I can't edit it)

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