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I Question about the deBroglie hypothesis

  1. Dec 2, 2016 #1
    Hi, I am a new member here : )
    I have always wondered about this. It seems that the answer is easy but I missed an important lecture.
    De Broglie proposed that just as light has both wave-like and particle-like properties, electrons also have wave-like properties. Why is that? I know the relation between matter and energy but I don't Understand what made him think that photons and localized particles should be the same in this regard. Please explain this point to me. Thank you all and please excuse my poor writing skills.
     
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  3. Dec 2, 2016 #2

    vanhees71

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    Welcome to the club :-).

    The answer is that de Broglie provided an important heuristic step in the development of quantum theory, but it's all outdated since 1926, when modern quantum theory was discovered/developed.

    The answer is that all his arguments are nowadays abandoned by the majority of physicists. First of all there's no wave-particle duality anymore, i.e., the interpretation of Schrödingers waves in non-relativistic quantum theory is now understood in the sense of Born's Rule, i.e., that ##|\psi(\vec{x})|^2## provides the probability density for finding the described particle at position ##\vec{x}##, no more, no less, but the particle (or better say "quantum") itself has neither classical particle nor classical wave properties, but it behaves as described by the formalism of quantum theory.

    Second de Broglie's ideas become even less accurate in the relativistic realm, where single-particle wave functions make sense only for free particles provided they have a mass. For interacting particles, the single-particle interpretation in the sense of Schrödinger's wave-mechanics formulation of non-relativistic QM doesn't hold anymore. Nowadays that's very clear, because we know that in reactions at relativistic energies and momenta particles are created and destroyed all the time, i.e., a single-particle description (or a description where the number of particles is conserved) is not sufficient anymore, and we always deal with a many-body system with non-fixed particle number (what's conserved are certain charges like electric charge, lepton number, etc. but not the particle number itself). That's why in relativistic QT one uses the formulation as local microcausal relativistic quantum field theory (QFT) which automatically provides the mathematical formalism to describe the creation and annihilation processes taking place all the time when particles interact at relativistic energies.

    Third, the photon is an extreme case in the sense that it is always relativistic, because it's a massless quantum (here I don't dare to use the words "particle" or "wave" at all!) with spin 1. Then it's not even sensible to define a position observable for free photons. In other words photons have no position you could measure. Of course, quantum electrcodynamics, the paradigmatic example for a relativistic QFT, describes photons (i.e., electromagnetic radiation) and electromagnetic interactions very well and with an astonishing accuracy (some predictions as the anomalous magnetic moment of the electron or the lamb shift of hydrogen spectral lines are accurate in agreement with experiment by 12 or more significant decimal places). QED provides you with the detection probability for a photon with the photo detector set up at a given place, but registering a photon at this place doesn't mean that you can say it had a certain position, because position of a photon cannot be defined as an observable at all!
     
  4. Dec 3, 2016 #3
    Thank you vanhees71
    your answer helped me a lot.
     
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