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Amplitude of 0 for Waves

  1. Oct 20, 2013 #1
    When a wave for light or an electron is represented on a graph, it is shown with the y-axis representing amplitude. Now, these waves cross the x-axis (where amplitude=0) but what does this actually mean, for that instant? Is it that the wave is no longer observed for that instant?

    Also, since all objects have their own wavefunctions, what does the concept of 0 amplitude mean? Is the object not observed for that one instant as well?

    I'm new to the concept of the wave properties of matter and quantum physics as a whole, so any clarification on the above matters (and other related material, if you wish) would be greatly appreciated!

  2. jcsd
  3. Oct 20, 2013 #2


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    The modulus squared of the wave function is the probability (density) to find the particle at a point (region) in a position measurement. If the amplitude is zero, you won't find the particle at this point (region).
  4. Oct 20, 2013 #3
    Okay. So since all objects have a wavefunction, what exactly does it mean for the object (ex. couch) to have a 0 probability?
    Does it cease to exist at some infinitesimal point that we just never see?
  5. Oct 20, 2013 #4


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    Before you perform a measurement, quantum systems don't have a definite value for the physical quantity you measure. If the wave function is zero in some region, the particle is not in this region. But you can't say where it is with certainty. The probability to find it in a region with a large amplitude is simply higher than in a region with a low amplitude. So it is very unlikely to find the particle in the region where it's wave function has a node.
  6. Oct 25, 2013 #5

    Claude Bile

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    I think the obvious is being overlooked - just because a wave occupies a zero value at some point in time (or space) does not mean the amplitude is zero at that time (space).

    Amplitude is the magnitude of a complex number. The value of the field is the projection of the complex number onto the real axis.

  7. Oct 25, 2013 #6
    Nevertheless it does mean that the probability of locating a particle at that point, or probability of "particle" manifesting itself into a particle if measured at that point, is zero. Isn't it?
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