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Schrödinger's equation and line width

  1. Oct 23, 2011 #1
    is the spectral linewidth something that is only explained by the uncertainty principle or do you also get this from schrödinger's equation? cause i would say, that schrödinger gives us discrete energy levels if we talk about atoms and therefore there should appear no linewidth.

    or do i need to take quantum electrodynamics into account to make proper predictions of spectral lines?
     
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  3. Oct 23, 2011 #2

    vanhees71

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    I vote for the latter explanation: Emission and absorption of photons is described by quantum field theory. In fact it has been invented by Dirac in 1927 to explain precisely this emission and absorption processes.
     
  4. Oct 23, 2011 #3
    okay, thank you...
     
  5. Oct 23, 2011 #4

    Bill_K

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    I disagree, and vote for the first one. While it's true that the emission of photons is ultimately described by QED, this is not necessary to explain the linewidth. The natural linewidth of atomic spectral lines is just one example of a general feature of a decaying stationary state, independent of the interaction that causes it, first elaborated by Weisskopf and Wigner in the context of nuclear physics. If the state has a time dependence e-iEt = e-(Γ/2)t e-iE0t where Γ is related to the lifetime of the state by τ= 1/Γ, the line shape for the emitted radiation is proportional to 1/((E - E0) + Γ2/4). This is known as a Lorentzian or Breit-Wigner form.

    QED becomes necessary if you want to calculate Γ itself, or higher order radiative corrections to the line shape.
     
  6. Oct 23, 2011 #5

    Ken G

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    I'd say your both right, because what vanhees71 is saying gives the explanation of why the Gamma that Bill_K is talking about is not zero! But if one is willing to accept its value as given by experiment, then one can proceed to an understanding of linewidth directly from the HUP, or from the (time-dependent) Schroedinger equation that is the source of the HUP. It sounds like Gavroy might be asking essentially "why do transitions happen at all" given the time-independent Schroedinger equation, but we are saying that you have to use the time-dependent Schroedinger equation, either because QED says the Hamiltonian is perturbed, or because experiment tells us there is a decay time there.
     
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