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How can you relate standing wave to a corpuscle at rest?

  1. May 16, 2014 #1
    I was going through De Broglie's acceptence speech and I found he said a particle at rest can be associated with a stationary wave.
    Now what We know wavelength= h/p(momentum).
    So for a particle at rest we get wavelength = h/0. This is undefined. So does not it state that a particle at rest is not wave like.
    Please help me to understand. I am getting confused.
  2. jcsd
  3. May 16, 2014 #2


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    Staff: Mentor

    According to Heisenberg's uncertainty principle, there is no such thing as a particle (wavefunction) with an exact value of momentum and a single well-defined wavelength. Instead, we must use a wave packet: a superposition (sum) of waves with a range of momenta and wavelengths.

    A "stationary" particle corresponds to a wave packet built of waves traveling in opposite directions, corresponding to a narrow range of + and - values of momentum centered on zero.

    If you've studied waves, you may recall that adding two traveling waves with the same wavelength, traveling in opposite directions at the same speed, produces a standing wave.
  4. May 16, 2014 #3
    Thanks for the reply. But I did not understand how you could say that a particle at rest corresponds to a wave packet built of waves travelling in opppsite directions? Please clarify.
  5. May 16, 2014 #4
  6. May 16, 2014 #5


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    Staff: Mentor

    A particle "at rest" is really a particle with momentum 0 ± (some small number). Suppose for the sake of discussion that the "small number" is 10. (Let's not worry about units here.) Then the wave packet is a superposition (sum) of waves corresponding to momentum -10, -9, -8, ..., -2, -1, 0, +1, +2, ..., +8, +9, +10, and waves "filling in" the intermediate values; it's a continuous distribution of momentum values.

    The waves with momentum -10 and +10 travel in opposite directions with the same wavelength, and combine to form a standing wave. Likewise for the waves with momentum -9 and +9, etc.
  7. May 16, 2014 #6


    Staff: Mentor

    I suppose that's a diplomatic way of expressing one of the fundamental issues of De Broglies hypothesis. In De Broglies matter wave theory a particle at rest has an infinite wavelength and infinite phase velocity (the phase velocity is the velocity of the individual waves) which doesn't really make physical sense. Particles were supposed to be wave-packets, but what are they when at rest? And one can always go to a frame where such is the case. That a theory breaks down by a simple coordinate system change means it has issues.

    To get around it as Jtbell points out you have to be a bit sneaky. That's the problem with it - its a mish mash of classical and quantum.

    QM solves the problem because you cant have a particle at rest since that would mean violating Heisenbergs uncertainly relation and knowing both momentum and position exactly. Theoretically you can have a particle with an exact momentum but that would be a wave of infinite extent and not physically realizable, so in practice there is always some uncertainty in both.

    Basically all these early ideas went out the window when Dirac came up with his transformation theory in about 1927 which is basically QM as we know it today. They are historical curiosities and IMHO hurt understanding rather than illuminate if you take them literally.

    If you want to learn QM IMHO its best to start with its conceptual core which is a generalisation of probability theory:

    Basically QM is one of the two most reasonable probability models for modelling physical systems. They are ordinary probability theory and QM. QM allows continuous transformations between so called pure states which is what you need for physical systems.

    The argument goes something like this. Suppose we have a system in 2 states represented by the vectors [0,1] and [1,0]. These states are called pure. These can be randomly presented for observation and you get the vector [p1, p2] where p1 and p2 give the probabilities of observing the pure state. Such states are called mixed. Standard probability theory is basically a theory about such mixed states. However it has a problem with continuous transformations which in physics you more or less require. To see this consider the matrix A that say after 1 second transforms one pure state to another with rows [0, 1] and [1, 0]. But what happens when A is applied for half a second. Well that would be a matrix U^2 = A. You can work this out and low and behold U is complex. Apply it to a pure state and you get a complex vector. This is something new. Its not a mixed state - but you are forced to it if you want continuous transformations between pure states.

    Basically QM is the theory that makes sense of pure states that are complex. That's its rock bottom essence in a nutshell.

    Last edited: May 16, 2014
  8. May 17, 2014 #7
    Thanks bhobba,jilang.
    And special thanks to bill. It really helped.
    Finally understood.
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