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Wave packet and uncertainty principle, a mathematical approach

  1. Mar 29, 2008 #1
    Hello. I started reading this little book by Heisenberg. It starts giving a mathematical relation of width of wave packet and range of wavelengths necessary to mathematically construct it, and then drops in wavalength-momentum relations to give a quick insight on the nature of the uncertainty principle. But I'm just trying to understand the wave packet math part, not even its relation to physical properties of wave-particles.

    So, basically, a wave packet (one dimension) is mathematically constructed by the addition of sine waves, all around some wavelength [tex]\lambda[/tex]. This wave packet has amplitude different to zero within a range in x, this range being denoted as [tex]\Delta[/tex]x. Now, if [tex]\lambda[/tex] is a few times smaller than [tex]\Delta[/tex]x, there will be a few troughs inside that range, this number "n" being [tex]\Delta[/tex]x/[tex]\lambda[/tex]. So far, so good.

    But then, just like that, he goes on to say that the wave packet is constructable ONLY if the wavelength range (denoted as [tex]\Delta[/tex][tex]\lambda[/tex]) has a lower limit of at least [tex]\Delta[/tex]x/(n+1). In other words, the sine component of lowest wavelength has to be at least small enough to fit n+1 troughs within the wave packet width [tex]\Delta[/tex]x. In this way there is a compromise between the width of the packet and the range of wavelengths of sine components that make it, assuming a main wavelength [tex]\lambda[/tex]. That's it, no further explanation. How does he know this for sure? What is the proof of this?

    This is very important because it is the mathematical origin of the inequality. Then it is simple algebra to arrive to the uncertainty principle, assuming empirically found relations between wavelength and momentum, and interpreting the packet width as uncertainty in position.
    Last edited: Mar 29, 2008
  2. jcsd
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