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The closest thing I can think as to why is because of the wave-particle duality of matter given by λ=h/mv, which also explains itself. But I don't know exactly why.

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- #1

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The closest thing I can think as to why is because of the wave-particle duality of matter given by λ=h/mv, which also explains itself. But I don't know exactly why.

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blue_leaf77

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Could you please explain/elaborate on that?

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https://www.physicsforums.com/threa...ly-uncertainty-principle.811326/#post-5093474

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Nugatory

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That result falls out of the mathematical formalism. If a quantum system is in a state in which some observable has a definite value, then that state must be an eigenstate of that observable. Thus, if the system state is such that two different observables both have definite values, then the state must be an eigenstate of both observables. However, there is a basic theorem that says a state can be an eigenstate of two different observables only if they commute, so if the observables don't commute then there is no state in which they both have definite values.Could you please explain/elaborate on that?

Of course this answer just begs the question, WHY does the universe behave according to this particular set of mathematical rules? Physics has no really satisfactory answer to that question. The universe behaves the way it does, and it's up to us to discover rules that describe that behavior.

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Nugatory

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Ah - yes, zero eigenvalues do allow for a number of interesting special cases.

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It is not only zero eigenvalue that matters. What matters is whether or not the operators have a common eigenvector. This is not necessarily related to the eigenvalue being zero.Ah - yes, zero eigenvalues do allow for a number of interesting special cases.

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jtbell

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https://www.physicsforums.com/showthread.php?p=3295464#post3295464

https://www.physicsforums.com/showthread.php?p=2972106#post2972106

Any wave function (not just quantum-mechanical ones) has a similar uncertainty principle. In general, it involves the wavenumber k, which in QM is related to the momentum p. Electrical engineers see this uncertainty principle in electronic signal processing. It's part of Fourier analysis.

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blue_leaf77

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Sorry for late reply, I was trying to find a lecture note that derives this but couldn't find it in time. As Nugatory said, Heisenberg uncertainty principle is a purely mathematical consequence of the formulation of QM we are using. The derivation can be found in here (section 2), in particular the end result looks likeCould you please explain/elaborate on that?

$$

(\Delta A)^2 (\Delta B)^2 \geq \left( \langle\psi|\frac{1}{2i}[A,B]|\psi\rangle\right)^2

$$

If ##[A,B]=0## then the two observables would not obey the uncertainty principle.

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https://www.physicsforums.com/threa...ly-uncertainty-principle.811326/#post-5093474

This proof even admits the derivation of the very states for which the equality sign holds, and these are the coherent and squeezed states (i.e., Gaussians).

Of course the uncertainty principle also holds for ##[A,B]=0##. Only then it's not very exciting, because in this case it just states the obvious fact that the standard deviations of the two observables in any state are not negative :-).

- #13

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Heisenberg found that then he could use classical equations. At least that's what The Feynman Lectures says.why is it like that?

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It's called the spectral theorem, and the proof isn't particularly difficult.Sorry for late reply, I was trying to find a lecture note that derives this but couldn't find it in time. As Nugatory said, Heisenberg uncertainty principle is a purely mathematical consequence of the formulation of QM we are using. The derivation can be found in here (section 2), in particular the end result looks like

$$

(\Delta A)^2 (\Delta B)^2 \geq \left( \langle\psi|\frac{1}{2i}[A,B]|\psi\rangle\right)^2

$$

If ##[A,B]=0## then the two observables would not obey the uncertainty principle.

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