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To take an example,

Feynman Lectures Vol 3 13-1

Let's think of an electron which ban be in either one of two positions [...] There are two possible states of definite energy for the electron. Each state can be described by the amplitude for the electron to be in each of the two basic positions. In either of the definite-energy states, the magnitudes of these two amplitudes are constant in time, and the phases vary in time with the same frequency. On the other hand, if we start the electron in one position, it will later have moved to the other, and still later will swing back to the first position.

(Feynman derives this behaviour from first principles in chapter 8-6 by solving equations for the state of an ammonia molecule, but that's a nice summary).

This sounds like the uncertainty principle - that you can know energy or position but not both. Is it possible to derive the uncertainty principle from such an analysis? Or is the uncertainty principle somehow axiomatic in the way this behaviour is derived?

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The uncertainty principle is a formal property of the Fourier transform, which means that it's derived. –  WillO Feb 26 at 14:21
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For me it does not sound like this section is related to the uncertainty principle. The propagation of an electron through a crystal is viewed as the time evolution from a state "the electron is at atom number n" to another state "the electron is at atom number n+1" and so on. –  Wildcat Feb 26 at 15:18
    
After reading the whole section in the book, I have a feeling that you was confused by degeneracy in the system described. See my answer for further comments. –  Wildcat Feb 26 at 16:05
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4 Answers

up vote 7 down vote accepted

You may be confusing two things here. The uncertainty principle, which states that for any two observables $A$ and $B$ $$ \sigma_{A}\sigma_{B} \geq \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right| \, , $$ can be derived (see, for example, Proof of the Schrödinger uncertainty relation in Wikipedia). Canonical commutation relation $[x,p_x] = i \hbar$, which using the uncertainty principle leads to $$ \sigma_{x}\sigma_{p_x} \geq \frac{\hbar}{2} \, , $$ is, however, postulated.


And for me, as I already said in the comment, it does not sound like this section is related to the uncertainty principle. The propagation of an electron through a crystal is viewed as the time evolution from a state "the electron is at atom number n" to another state "the electron is at atom number n+1" and so on.

Yes, you have degeneracy in the system, which is defined as follows

an electron which can be in either one of two positions, in each of which it is in the same kind of environment

The energy in the state "the electron in the first position" is identical to energy in the state "the electron in the second position" (states are said to be degenerate in such case), but it is something to be expected due to the presence of symmetry in the system (explicitly mentioned in the definition of the system).

The fact that the knowledge of the energy of an electron in such system tells you nothing about its position is due to degeneracy: energy lonely does uniquely specify (or, does not fix, as we say) the state of such system. So you are uncertain about the position of the electron because you are uncertain about its state.

The uncertainty principle, on the other hand, says that even when you know the state of a system (not just its energy), you can not specify both position and momentum of the electron with any desired precision. There is a fundamental limit to the precision with which position and momentum can be known simultaneously.

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Are you saying, then that in the system I described, you can know either energy or position of the electron but not both - but this is not due to the uncertainty principle? –  Sideshow Bob Mar 4 at 22:02
    
@SideshowBob, first, given a state of a system it is impossible to know both $x$ and $p$ of each particle in the system precisely. That is what uncertainty principle tells you. And, of course, it works for the system in question: if I give you the state of such system, you won't be able to specify both $x$ and $p$ of any particle precisely. –  Wildcat Mar 5 at 11:23
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@SideshowBob, second, the situation, when the energy of a system is given, rather than its state, is totally different. In this case you don't have some uncertainty in $x$, rather, you do not even know $x$, since you do not know the state of a system. So you can not say anything about $x$ in this case, and not just specify it exactly. And, again, it has nothing to do with uncertainty principle. You simply do not know the state of a system, thus you do not know values of physical observables. Except energy, of course, since it was given to you by someone for some reason. –  Wildcat Mar 5 at 11:24
    
Thanks. Much clearer now :) –  Sideshow Bob Mar 5 at 14:32
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There are several mathematical proofs for the Uncertainty Principle, although it is also based out of intuition. A good, fundamental proof of the mathematical sort is found here: http://www.tjhsst.edu/~2011akessler/notes/hup.pdf.

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Interesting and synthetic proof, thanks –  Guillaume Feb 26 at 15:32
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The Uncertainty Principle (UP) is not only a quantum mechanics property. There's a way to describe UP in an empirical approach (the famous Heisenberg microscope) but, from a mathematical (and historical) point of view, UP is intrinsically connected to Fourier transform (FT) properties. In fact if you assume that you have a gaussian wave packet in real space (at fix time) and you transform it; in the conjugate space you obtain a function with a gaussian profile (in absolute value). Now if you take the width in real space times the width in the conjugate one, you obtain exactly the equality 1/2. In the end recall the relation between momentum and wave number you have the UP equality (that's a lower limit of UP: gaussian functions minimize UP). There's a quantum approach in Dirac notation completely different. Nevertheless I stress the importance of the fact that it's not a merely quantum property but it's a mathematical consequence of FT, and obviously FT is present in quantum mechanics. I don't want to discuss if it's a nature property, I only say that if you use FT (and wave mechanics in general) you meet this situation.

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Is that to say, then, that $E=\hbar \nu$ is the axiomatic bit, and combining this with Fourier analysis we get the uncertainty principle? –  Sideshow Bob Mar 4 at 21:59
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E=hν is an Einstein supposition (obviously correct) in order to describe photoelectric effect. It's not an axiom, it's a physical relation. UP is in the FT mathematically before physics. If you then associate a physical meaning to the mathematical variables (i.e. x position and k wave number) you obtain UP in wave physics. Otherwise as I already told tou there are a lot of different contexts where UP is used. –  LC7 Mar 5 at 15:26
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Although the Uncertainty Principle can be derived from other aspects of quantum mechanics, it is still regarded as a principle, rather than as a result or a relation, because it is an empirical principle of quantum mechanics.

There is a complete discussion in the Stanford Encyclopedia of Philopsophy.

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protected by Qmechanic Feb 27 at 16:16

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