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Topic: Quick Question: Heisenberg Uncertainty  (Read 1570 times)

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Offline pocantheses

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Quick Question: Heisenberg Uncertainty
« on: October 16, 2014, 05:13:40 PM »
In Heisenberg's Uncertainty Principle of the form:
[tex]\Delta x\cdot m\Delta v \geq \frac{\hbar}{4\pi }[/tex]

Why is there a '4Π' underneath Planck's constant? Why is this important and why does Π appear in there?





Offline Mitch

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Re: Quick Question: Heisenberg Uncertainty
« Reply #1 on: October 16, 2014, 06:32:20 PM »
In applications where it is natural to use the angular frequency (i.e. where the frequency is expressed in terms of radians per second instead of rotations per second or Hertz) it is often useful to absorb a factor of 2π into the Planck constant. The resulting constant is called the reduced Planck constant or Dirac constant. It is equal to the Planck constant divided by 2π, and is denoted ħ (pronounced "h-bar").
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From http://en.wikipedia.org/wiki/Planck_constant
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Offline mjc123

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Re: Quick Question: Heisenberg Uncertainty
« Reply #2 on: October 17, 2014, 04:52:22 AM »
Then the RHS of the inequality above should be h/4π, not ħ/4π.

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