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Delta function and dirac notation

  1. Jan 15, 2016 #1
    Hello there !
    I found this discussion http://physics.stackexchange.com/qu...a-delta-function-position-space-wave-function about dirac notation and delta function .
    The one that answers to the problem says that ##<a|a>=1## and ##<a|-a>=0## .
    As far as i know:
    1) ##<a|-a>=δ(a-(-a))## which in this case is zero because delta function is zero everywhere except ##x=-a## where it goes to infinity. So i understand why this is zero.
    2) ##<a|a>=δ(a-a) → ∞ ## and not 1 . Am I wrong and this ##<a|a>## is obviously 1. Can you explain me please?
     
  2. jcsd
  3. Jan 15, 2016 #2

    bhobba

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    You are correct - it is infinity. Strictly speaking its undefined, but is usually taken as infinity. The issue with infinity and why its better to have it undefined is c∂(t) is also zero when t is not zero and infinity when t = 0 for any positive c. So, naively, c'∂(t) = c∂(t) for any positive c or c'. But it acts differently when integrated.

    Thanks
    Bill
     
    Last edited: Jan 15, 2016
  4. Jan 15, 2016 #3
    So it is wrong to normalize the function with the two delta functions that he gives using ##<a|a>=1## because it actually goes to infinity.
    So how could we normalize ##ψ(x)=δ(x-a)+δ(x-a)## in a correct way ?
     
  5. Jan 15, 2016 #4

    bhobba

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    They are distributions - normalisation makes no sense. The square of a delta function doesn't exist.

    I suggest getting hold of the following:
    https://www.amazon.com/The-Theory-Distributions-Nontechnical-Introduction/dp/0521558905

    It will help in many areas of applied math not just QM.

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
  6. Jan 15, 2016 #5
    Thank you for your response :)
     
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