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A problem with the wave packet concept?

  1. Aug 16, 2012 #1
    A wave packet is formed when a multitude of sine waves of different periods are linearly added.
    This localizes the pdf, and hence creates certainty in a particle's position.


    However, basic sinusoidal functions are cyclical, hence the additton of two sinusoidal functions are always cyclical.

    Doesnt this imply that with the addition of multiple sine waves, we would expect to see wave packets at regular intervals of space? How do they produce a "lone" wave packet?
     
  2. jcsd
  3. Aug 16, 2012 #2
    This is not true, if by "cyclical" you mean "periodic". For example, the function f(x) = sin(x) + sin(sqrt(2)x) is not periodic.

    The way we create localized wave packets is that we superpose infinitely many sine waves so that they interfere destructively everywhere except in a small region of space. Doing this requires sine waves of *all* frequencies. Including infinitely many frequencies is how we can accomplish this amazing cancellation.

    You should read about gaussian wave packets; they are the prototype for localized "lone" wave packets.
     
  4. Aug 16, 2012 #3

    jtbell

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    The key to getting a wave packet that is localized in space and does not "repeat" elsewhere is to superpose an uncountably infinite number of waves with a continuous spectrum of wavelength and frequency, via a Fourier integral; as opposed to superposing a finite number of waves, or even a countably infinite number of them via a Fourier series.
     
  5. Aug 16, 2012 #4
    like countable and uncountable from Cantor's principles?

    Still cant get my head around the idea. If a wave packet is formed, will it be located near the origin? (on a graph)
     
  6. Aug 16, 2012 #5
    can i demonstrate the creation of a packet using my ti-84?
     
  7. Aug 16, 2012 #6

    Matterwave

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    It depends on what you mean by "demonstrate". Since a fourier integral is an integral and not a series (and therefore there are an uncountably infinite number of additions), one cannot see a progressive animation (e.g. adding the first, second, third, terms and so-forth) one usually sees for a fourier series converging into the correct wave-form.
     
  8. Aug 16, 2012 #7

    jtbell

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  9. Aug 16, 2012 #8
    This isn't very rigorous, but here's one way to sort of get an intuitive grasp on this. You have an uncountably infinite set of sinusoid waves, one for every possible wavelength. That means that no matter what wavelength you pick, no matter how big or small it is, there's a sine wave in the set somewhere that has that wavelength.

    Now say that you align the phases of every single one of these infinitely many sine waves such that they have a peak at the origin. So now, regardless of the exact details of how you do the integral, the resulting waveform is clearly going to have a peak at the origin. It has to--all of the waves are positive, and none are negative, so when the add up they're going to form a positive peak.

    Now, as you move further away from the origin, the phases start to get out of alignment, and the total sum decreases in some way, meaning the peak falls off. This is the normal behavior for a wavepacket, regardless of whether it's made up of a finite or infinite set of sinusoids.

    Now here comes the trick. You're worried that all of these infinite sinusoids are going to sync back up again at some other point and form another peak some distance away from the origin, let's call it X. But this is impossible, because our set of waves contains waves with every wavelength. So no matter what X is, there's a wave in the set with a wavelength of X/2, meaning it's at a trough at that point. In fact, any wave with a wavelength that divides that distance will have a trough at that point, so there are a whole bunch of waves in the set interfering destructively at point X.

    So whatever else is happening at point X, we know that it isn't getting 100% constructive interference the way the origin is. So the total sum of the waves must be at least a little bit smaller at point X than it is at the origin. Since X was arbitrary, we've just shown that there's no other point anywhere along the real line that is getting as big a peak as the origin is. In other words, the origin is special--it's the only point where every single wave has a peak simultaneously. Therefore, the waveform can't be periodic.

    The reason we were able to show this is that we knew that the set contained a wave with every conceivable wavelength, which can only happen if there's an uncountably infinite number of them. If it were finite, or countably infinite, then there would be some wavelengths that we wouldn't have. This means that there would be some values of X for which we couldn't guarantee that there is a destructively interfering wave, and the proof falls apart.
     
  10. Aug 16, 2012 #9
    Also, in response to your question in #4, note that we built the wave packet by carefully picking the phases of all of the waves so that they all simultaneously had a peak at the origin. We could just as easily have chosen to sync them up at some other place, so it's possible to build a wave packet at any point on the real line, just by choosing the phases of all of the individual waves correctly.
     
  11. Aug 17, 2012 #10

    Ben Niehoff

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    Actually, you don't need a continuum of wavelengths. You only need a set that is dense in the reals. The rationals will do.
     
  12. Aug 20, 2012 #11

    But imagine each of these sinusoids had the same amplitude. The peak at the origin would have an infinite amplitude.

    Unless the amplitude of each successive sinusoid follows a converging series.

    If that is the case, even with an infinite no. of sinusoids, it would be possible for constructive interference at some other point, as the amplitudes of all the other destructive sinusoids would be insignificant?

    I'm kinda confused on this.
     
  13. Aug 21, 2012 #12
    But the amplitudes don't matter. It's the phase cancellations that cause peaks and troughs, so unless every single cancelling wavelength happens to have an amplitude of 0, you're going to have at least some destructive interference. That doesn't prove that the wave doesn't have some kind of peak there, but it does prove that the overall waveform can't be perfectly periodic, which was your original concern.

    Really, at the end of the day, the easiest way to see this is just to watch the math work out. Imagine we have some function [itex]f(x)[/itex] that's localized to one region in space. For simplicity, let's make it a Gaussian: [itex]f(x) = e^{-\frac{1}{2}ax^2}[/itex]. Now, what we want to show is that this function can be made up of a continuum of sinusoids, i.e. [itex]f(x) = \int_{-\infty}^\infty{\frac{dk}{2\pi}\:e^{ikx}\tilde{f}(k)}[/itex]. So if we can find a function [itex]\tilde{f}(k)[/itex] which satisfies this equation, we're done.

    Of course, this function is just the Fourier transform of [itex]f(x)[/itex]. So we have:
    [tex]\tilde{f}(k) = \int_{-\infty}^\infty{dx\:e^{-ikx}f(x)} = \int_{-\infty}^\infty{dx\:e^{-ikx}e^{-\frac{1}{2}ax^2}}[/tex]

    Complete the square, shift, and integrate:
    [tex]\tilde{f}(k) = \int_{-\infty}^{\infty}{dx\:e^{-\frac{1}{2}a(x+\frac{ik}{a})^2-\frac{k^2}{2a}}}\\
    = e^{-\frac{1}{2a}k^2}\int_{-\infty}^{\infty}{d\bar{x}\:e^{-\frac{1}{2}a\bar{x}^2}}\qquad\left(\bar{x} = x + \frac{ik}{a}\right)\\
    =\sqrt{\frac{2\pi}{a}}e^{-\frac{1}{2a}k^2}[/tex]

    So we've found the Fourier transform of our Gaussian, and it's just another Gaussian. Therefore, we've proven if you want to make a peaked waveform, all you need to do is integrate together a bunch of sine waves whose amplitudes are themselves clustered around one value in "momentum space", and the result will be a wave with a peak at one spot, which falls off to 0 as you move away from it in either direction. You can make [itex]a[/itex] as big as you want, which will make [itex]f(x)[/itex] more and more sharply localized, until the wave packet is as tight as you want it to be. Due to the [itex]\frac{1}{a}[/itex] factor in the transformed equation, this also has the effect of making [itex]\tilde{f}(k)[/itex] less and less localized, which is the Heisenberg Uncertainty Principle in action.
     
    Last edited: Aug 21, 2012
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