Is this Fourier derivation of Heisenberg principle true? 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?
 A: 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.)
