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I Planck's constant and quantization of energy

  1. Sep 25, 2016 #1
    Given:
    ##\textbf{E}=\hbar \textbf{k}##
    where ##\textbf{k} = [\vec{k}_1, \vec{k}_2,\vec{k}_3, i c \omega]##
    If ##\textbf{k}## can vary continuously, how does the equation imply that energy is quantized?

    For example, ##y = m x +b## where ##m = \hbar## does not imply quantized ##y##.
    For ##\textbf{E}## to be quantized mustn't ##\textbf{k}## be quantized?

    And why should ##\hbar## be considered anything other than a unit conversion?
     
    Last edited: Sep 25, 2016
  2. jcsd
  3. Sep 25, 2016 #2

    mfb

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    Staff: Mentor

    Right. For bound states it is.
    You can work in units where it is equal to 1. Yes, it is just a unit conversion - but the fact that this conversion is possible is not trivial.
     
  4. Sep 25, 2016 #3

    Nugatory

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    It doesn't. Quantization of energy appears when you solve Schrodinger's equation for bound states. The simplest example is the one-dimensional infinite square well; in the solutions to Schrodinger's equation for that potential ##k## can only take on discrete values.
     
  5. Sep 26, 2016 #4

    vanhees71

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    ...and please don't use the awful ##\mathrm{i} c t## convention of the SRT pseudometric. Particularly when it comes to QFT, with that you'll confuse yourself even more than the subject itself can ever do when done in the real-time formalism ;-).
     
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