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I noticed that I can derive Heisenberg principle very easy.

$\begin{aligned}p=\dfrac{h}{\lambda }=\dfrac{h\nu }{\lambda \nu }=\dfrac{h\nu }{c}=h\nu \left( c=1\right) \\ Fourier\Delta v\cdot \Delta x >1\left( \cdot h\right) \\ \Delta h\nu \Delta x >h\\ \Delta p\cdot \Delta x >h\end{aligned}$

Though I never seen this stated in books?

Mercury
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    The uncertainty in Fourier IS the uncertainty of position and momentum in QM, and it actually appears in many book and videos. See for example https://www.youtube.com/watch?v=MBnnXbOM5S4 . However, the proof with the commutators generalized the concept to other pairs of physical quantities (like angular momentum components) – Ofek Gillon May 06 '22 at 11:02
  • I do remember, during my studies, seeing a proof that, for any pair of operators such that [a,b]=i, the following properties are verified: 1) their eigenvalues are real and 2) the dispersions of those eigenvalues satisfy Δa.Δb>1/2. There may be some assumptions hidden here that escaped me at the time, I'll post again if I find more details. – Miyase May 06 '22 at 11:10
  • @OfekGillon yes something is mentioned about in this utube but is like not official but a little bit laymann. There is no derivation also. I mean real QM books. – Mercury May 06 '22 at 17:24

1 Answers1

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The confusion here is between a mathematical fact and a physical principle. Uncertainty principle in Fourier analysis means that the width of the function $\Delta t$ and the width of its spectrum $\Delta \omega$ are related via $$\Delta t \Delta \omega \geq 1$$ (The constant can be other than $1$, depending on how the standard deviations $\Delta t, \Delta\omega$ are defined.)

Heisenberg uncertainty principle establishes a limit on the accuracy of simultaneous measurement of physical quantities. In some cases, e.g., when we are dealing with position $x$ and $momentum$, while working with wave functions (rather than, e.g., the matrix representation), the formal mathematical statement of the Heisenberg uncertainty principle would be just that of the Fourier analysis. However, the principle holds deeper physical meaning, it applies to variables that are not continuous, not necessarily canonically conjugate of each other, and does not depend on the representation that one is working with (wave mechanics, matrix mechanics, path integrals, etc.)

Roger V.
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  • My question is about the derivation not about meaning and other applications? Is there something wrong in the derivation that way? If not why it is not taught and shown that Heisenberg easy follows from Fourier theorem? I am worried by this c=1 is it ok? – Mercury May 06 '22 at 12:58
  • Using de Broglie relation and the Fourier relation I get the HUP in one stroke. – Mercury May 06 '22 at 13:02
  • It is not clear what function you apply the Fourier transform to. I explained in my answer why Heisenberg does not follow from the Fourier theorem - they are not the same thing: one is a physical principle, the other is a mathematical statement. Sometimes this physical principle is expressed in this mathematical form. – Roger V. May 06 '22 at 13:16
  • The function is the wavefunction psi(x). If it is localized in dx then the frequencies dv and dx obey Fourier dv.dx>1. (this is math). Physics - de Broglie p=h/l. Then HUP follows. This is exactly a relation between two physical quantities for simultaneos measurement dp.dx>1. How does this mean less that the same expression? – Mercury May 06 '22 at 15:31