My textbook and the following extract from feynman's lectures present the same idea regarding wavetrains and uncertainty in their wavelengths. Why is it that a wavetrain confined to some space has an uncertainty in its wavelength or the wave number? Is not a confined wave-train equivalent to a burst of successive pulses which can have a definite wavelength according to their origin or nature of origin. Extract

Next, i understand that the De Broglie relation relates the uncertainty in wavelength to the uncertainty in momentum but what links the finiteness of the wave-train to the uncertainty in position?


To find out what frequencies a signal contains you Fourier transform it.

If you have an infinite plane wave, $\psi$, then when you Fourier transform it you get a delta function centred on the wave frequency i.e. it contains a single frequency.

However a wave packet of the type shown is the product of an infinite plane wave, $\psi$, and some envelope function, $E$, e.g. a Gaussian. When you Fourier transform this product the frequency distribution you get is the Fourier transform of $\psi$ convolved with the Fourier transform of $E$. $\psi$ tranforms to a delta function but, assuming it's a Gaussian, $E$ transforms to another Gaussian. The convolution of the two gives a Gaussian i.e. the packet contains a range of frequencies not just a single frequency.

It's because the packet contains a distribution of frequencies that you can't assign it a definite frequency/wavelength. The best you can do is assign an average frequency.

Re the uncertainty in position: the position is given by the envelope function, $E$, as described above. In effect $E$ tells you the probability of finding the particle at some position/time. To get a precise position you need to shrink $E$ to a delta function, but if you do that you find it's Fourier transform goes to a constant i.e. the packet now has an infinitely uncertain frequency and therefore momentum.


John is right. I just want to try to give a more intuitive explanation.

Suppose you plot the sine and cosine of a wave that has a wavelength of exactly 1. Then of course the peaks and zero-crossings would be exactly 1 unit apart.

Now suppose you put an envelope on that plot, such as multiplying it by $x$. Here's that plot:

enter image description here

Notice the zero-crossings of the sine curve, and notice the peaks of the cosine curve. They are not 1 unit apart. They are shorter, because of the slope of the envelope. It's not a big effect, for this simple envelope, but you can see it.

So if the envelope is not a straight line, but a curve, the wavelength also varies along the curve.

  • $\begingroup$ I'm sorry, I don't get this. Multiplying by x did not change the position of peaks and zero crossings, right? (Zero crossings for sine curve will still occur at 0,π,2π,3π and so on and peaks for cos curve at 0,2π,4π... irrespective of the multiplying factor ie change in amplitude) so why should changing amplitude effect wavelength at all? Am I missing something? $\endgroup$ – Mahathi Vempati Jun 3 '16 at 12:30
  • $\begingroup$ Or is the measurement not being made along x axis? $\endgroup$ – Mahathi Vempati Jun 3 '16 at 12:32
  • 1
    $\begingroup$ @user93868: Just look at the plot. The zero-crossing and peaks do not occur at exactly 1, 2, and 3. They occur short of it, as you can see at 3. It's about a millimeter short. $\endgroup$ – Mike Dunlavey Jun 3 '16 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.