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Why should the uncertainty in measurement of two conjugate variables, say $p$ and $x$, be of the order of Planck's constant or higher and not lower? What is so sacrosanct about $h$? I always think $h$ shows up in discretization of energy in quantum theory. And even electrical engineering people use uncertainty relations without quantifying them. Should we necessarily quantify it the way we do?

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    $\begingroup$ This is not a choice. It follows from the commutation relations; in the simplest case $[x,p]=i\hbar$ so $\hbar$ comes up quite naturally. See en.wikipedia.org/wiki/… $\endgroup$ – ZeroTheHero Jan 8 '18 at 17:01
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There's nothing sacrosanct about $h$ (or $\hbar$).

Quite generally, given two self-adjoint operators $\hat{A}$ and $\hat{B}$ (associated with observables $A$ and $B$) such that $$\left [ \hat{A}, \hat{B} \right ]\neq 0,$$ you can prove the following inequality (Generalized Uncertainty Principle, GUP): $$\Delta\hat{A}\Delta\hat{B}\geq \frac{1}{2}\left | \left\langle \psi \right|\left [ \hat{A}, \hat{B} \right ]\left| \psi \right\rangle \right |,$$ where $\Delta\hat{A}$ ($\Delta\hat{B}$) is the standard deviation of $A$ ($B$) on the state $\left|\psi\right\rangle$.

If you identify $\hat{A}\equiv\hat{x}$ and $\hat{B}\equiv\hat{p}$, knowing the fundamental commutation relation $$\left [ \hat{x}, \hat{p} \right ] = i\hbar,$$ you get $$\Delta\hat{x}\Delta\hat{p}\geq \frac{\hbar}{2}=\frac{h}{4\pi}.$$ So, $h$ doesn't directly come from the uncertainity principle.

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    $\begingroup$ Well, h comes directly from the black body radiation , a measured constant to fit the data and dependent on the quantization hypothesis, so yes, it is the commutation relations that tie it up with the uncertainty principle. en.wikipedia.org/wiki/Uncertainty_principle#History $\endgroup$ – anna v Jan 8 '18 at 20:12
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To amplify @Vincenzo 's answer, the uncertainty principle is a generic property of Fourier analysis, and has little to say about dimensionfull conversion factors such as h, a mere conversion constant taking your from de Broglie momenta to wavelengths, $$ p=\frac{h}{\lambda}=\frac{2\pi \hbar}{\lambda}. $$
So, for your archetypal QM wave, $e^{ipx/\hbar}=e^{i2\pi x/\lambda}$, the uncertainty principle is $$ \Delta x ~\Delta \frac{1}{\lambda} ~\geq \frac{1}{4\pi}, $$ just like the signal processing Gabor limit you mentioned, $$ \Delta \nu ~\Delta t \geq \frac{1}{4\pi} ~. $$ The $\hbar$ in the popular version of the UP is strictly an artifact of your choice of (dimensions/units of) the momentum you substitute in it.

The dimensional Planck constant h quantifies the phase-space area of the fundamental quantum cell and the book of nature (Galileo) does it for us--we don't get to write it ourselves: we are just the book's avid readers. Its size is some basic resolution ability of our measurements. But, when we are dealing with actions, angular momenta, and, yes, energies integrated over time, which are much larger than it, we ignore it in a classical haze. It only becomes apparent and significant for such quantities and terrains comparable to it, basically atomic and subatomic physics, with some notable macroscopic extensions of quantum physics.

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Of course we need to quantify it.

I don't know how uncertainty principles are applicable in electrical engineering, but in Physics, it needs to be quantified so that we get an idea of the "amount" of uncertainty in either position or momentum,when one is known with sufficient accuracy. Heisenberg's principle is like a concept on which the whole of Quantum Mechanics and its branches rests.

The $h$ comes into account due to experiment and theoretical calculations. Mathematical calculations show that the uncertainty can be evaluated in terms of constants $h$ and $\pi$.

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    $\begingroup$ I think the OP means $\Delta k\Delta x\approx 1$ and similar Fourier-type relations. $\endgroup$ – ZeroTheHero Jan 8 '18 at 17:02
  • $\begingroup$ Even then, quantification needs to be done, isn't it? $\endgroup$ – Wrichik Basu Jan 8 '18 at 17:03
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    $\begingroup$ Yeah but the Fourier-type ones don't have $\hbar$ in them and follow from more classical arguments. They are quite different as they not related to incompatibility of non-commuting operators. $\endgroup$ – ZeroTheHero Jan 8 '18 at 17:05

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