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Normalising a wave function to dirac's delta

  1. Oct 19, 2013 #1
    Hi guys!

    There is something I fail to understand in normalising wave functions to dirac's delta.

    Let's take the free particle solution as an example.
    Please note that my question is not about using the solution as much as about undestanding the concept.

    -----
    The situation

    The general form of the free particle solution:
    [tex]\Psi(x,t) = A e^{i(kx-ωt)}
    [/tex]
    And the normalisation is done with an overlap integral of two different wave functions, [itex]\Psi_{k'}(x,t)[/itex] and [itex]\Psi_{k}(x,t)[/itex] relying on the following representation of dirac's delta:
    [tex]\delta (k-k') = \frac{1}{2 Pi}\int e^{ix(k-k')}dx[/tex]
    The normalised free particle then will have the form
    [tex]\Psi(x,t) = \frac{1}{\sqrt{2 Pi}} e^{i(kx-ωt)}[/tex]

    -----
    So here is what I don't undestand.

    The general definition of a normalized wave function is
    [tex]\int \Psi^{*} \Psi dx = 1[/tex]
    So, as I undestand it, the scalar product is taken using a function and its complex conjugate.
    When normalising to dirac's delta, however, this doesn't seem to be the idea. (The momentum of the functions used is different). Why is this, i.e. why is the normalization to dirac's delta done with two different wave functions? And what does it mean to define the normalization like this, using two different functions?

    Thank you in advance.
     
    Last edited: Oct 19, 2013
  2. jcsd
  3. Oct 20, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    You want to find an orthonormal basis of states for a quantum problem where the states are discrete you pick them so ##<\Psi_i,\Psi_j>=\delta_{ij}##. So in particular ##\int \Psi^*_i\Psi_i dx = 1##. For free particles you can't do that because they aren't normalizable in that sense. The integral diverges. The delta function normalization is a continuous analog of the discrete problem.
     
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