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In the uncertainty principle , whats the significance of the inequality sign ?
Why is it not equal to over there ?
Why do we say greater than equal to instead of just equal to ?

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  • $\begingroup$ The HUP just gives a lower bound in the uncertainty of $\Delta x$ and $\Delta p_x$, it is of course possible to measure both quantities with a higher uncertainty than given by the HUP, but not less. $\endgroup$
    – ahemmetter
    Mar 25 '15 at 11:27
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    $\begingroup$ Also, this is better moved to physics.SE $\endgroup$
    – ahemmetter
    Mar 25 '15 at 11:31
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    $\begingroup$ @andynitrox So the minimum uncertainty that will be there is given by it and uncertainty can be more than it too ? $\endgroup$
    – pikachu
    Mar 25 '15 at 11:45
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    $\begingroup$ physics.stackexchange.com/questions/170684/… $\endgroup$
    – Mithoron
    Mar 25 '15 at 11:46
  • $\begingroup$ @Mithoron Got it ! Thanks . Should I delete this post since it has an answer over there ? $\endgroup$
    – pikachu
    Mar 25 '15 at 11:54
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As andynitrox said in the comments :

The HUP just gives a lower bound in the uncertainty of Δx and Δpx, it is of course possible to measure both quantities with a higher uncertainty than given by the HUP, but not less.

The inequality exists to show that, the uncertainty is atleast this much and in real-life cases, its much larger than this . (source)(provided by Mithoron)

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    $\begingroup$ Refer this for more information. $\endgroup$
    – pikachu
    Mar 27 '15 at 18:21
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In addition to my comment further above I might add that the Heisenberg Uncertainty Principle has nothing to do with an error in a measurement. The HUP does not state that we cannot measure something more accurately than this lower bound, but instead that it is inherently not determined to a better degree.

To illustrate, the two properties of a one-dimensional wave packet , frequency and location, are similarily interrelated.

Imagine a sharp wave packet, a peak, something like this: enter image description here

You'll have no trouble locating 'the wave packet', simply because it is fairly sharp. Conversely, for a monochromatic (and therefore technically infinitely long wave 'packet'), it is very easy to figure out what frequency or wavelength it has:

enter image description here

However, trying to find out where this wave is, cannot really be answered. Keeping in mind that a wave packet is essentially made up of different monochromatic waves with different frequencies and amplitudes, you'll see that the sharper a packet is, the more frequencies it contains, so it will become nearly impossible to say what frequency it has.

enter image description here

We can give a lower bound for the accuracy of both measurements: the certainty in the measurement of the frequency and of the location of such a wave packet. We cannot obviously measure both at the same time with arbitrary accuracy. Note that we may very well see exactly what such a wave packet looks like, but we can still not pin-point one single frequency and one single location that desribes such a packet. That is basically the essence of the Heisenberg Uncertainty Principle.

Using the wave number $k$ and as location $x$, we can describe this relation in the case of a wave packet as $$\Delta k \cdot \Delta x \geq 2\pi$$

The case of the wave-particle duality is analog to this purely mathematical example (the factor $2\pi$ comes from the coefficient in the Fourier transform that is used to switch between the $k$-space and the $x$-space).

Images: http://www.newtonphysics.on.ca/heisenberg/image615.gif, https://skullsinthestars.files.wordpress.com/2007/12/sinesnapshot.gif and http://hyperphysics.phy-astr.gsu.edu/hbase/waves/imgwav/wpac5.gif

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  • $\begingroup$ Please leave a reference or a link to where the diagrams can be found. Fair use affords these kind of cases, but it's always best to give credit. $\endgroup$
    – jonsca
    Mar 25 '15 at 22:46
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    $\begingroup$ Thanks for the extra information. I am accepting your answer and not mine since my answer has been quoted from your comments and you won't get the credit if I do so. $\endgroup$
    – pikachu
    Mar 27 '15 at 18:20

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