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Can Heisenberg's Uncertainty principle be rewritten in terms of energy density

writing $$\Delta E \Delta T \geqslant \hbar/2$$ in factors of energy density $\Delta \sigma \text{ }= \frac{3\Delta E}{4\pi r^3}$

I get $$\Delta \sigma \frac{3\text{$\Delta $T}}{4\pi r^3} \geqslant \hbar/2$$

does $$\frac{3\text{$\Delta $T}}{4\pi r^3}$$ represent something?

Or does it make more sense to write $$\text{$\Delta $x} \frac{\text{$\Delta $E}}{c^2} \geqslant \frac{\hbar}{2}$$

EDIT after noting Willie's comment I've corrected the form to $\frac{\hbar}{2}$

However, now I see c is constant and can be moved to the right side - is it mathematical correct to write $$\text{$\Delta $x} \text{$\Delta $E} \geqslant \frac{\hbar c^2}{2}$$ or does this violate the Cauchy–Schwarz inequality? This is the question I was trying to understand at the beginning.

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Your algebra's not right. The correct expression should be $\frac{4}3 \pi r^3 \Delta\sigma\Delta T \geq h/2$. –  Willie Wong Jul 23 '11 at 1:18
    
Really? Ok - but I was using Kennard's "Zur Quantenmechanik einfacher Bewegungstypen" where planck's unit is modified by h/4Pi. Sorry, wenn mein Englisch ist neunzehnten Jahrhunderts –  metzgeer Jul 23 '11 at 3:56
    
""Sorry, wenn mein Englisch ist neunzehnten Jahrhunderts "" So, congratulation to more than 111 years of age! :=) –  Georg Jul 23 '11 at 9:52
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Please fix the question! The factor of c should not be squared, it should be $\Delta X \Delta E \approx \hbar c$, the density form is also wrong, as Willy Wong said. –  Ron Maimon Sep 18 '11 at 8:14
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1 Answer

It seems that you are missing what $\Delta T$ is here. It is not an interval of time. It is time uncertainty. Which is not what usually called $\Delta T$. The most standard application for this uncertainty principle is a linewidth. From this inequality you know that lifetime in upper state limits lower bound of energy dispersion. The larger lifetime the more precise your energy measurement (may be).

As a consequence, there is no use to convert this $\Delta T$ to $\Delta x$. It will mean not what you expect.

The same holds for $\Delta E$. It is not a real energy of some object.

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