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Kronecker Delta and Dirac Delta

  1. Jan 7, 2012 #1
    Hello PF,

    When I was studying Quantum mechanics, I realized that this equality should be true,

    [tex] <{\psi}_{n} \mid {\psi}_{m}>=\int {\psi}_{m}^*{\psi}_{n}dx={\delta }_{mn}[/tex]

    So [itex] {\psi}_{m}^*{\psi}_{n}[/itex] must be equal to dirac delta function so that we provide the kronecker delta as a solution of the integral.

    Therefore, this equation must be true, mustn't it?

    [tex]\int \delta (x-x')dx={\delta }_{mn}[/tex]

    Or, if it is wrong, what is the expression [itex]{\psi}_{m}^*{\psi}_{n}[/itex] equal to?

    Thanks for your opinions and helps.
     
    Last edited: Jan 7, 2012
  2. jcsd
  3. Jan 7, 2012 #2

    Pengwuino

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    Gold Member

    No, it's equal to the Kronecker delta. The Dirac delta does not enter in at all here unless you know something about the actual functions [itex]\Psi[/itex]

    [itex] \int \delta(x - x')dx = 1[/itex] only if you integrate over x = x' within your integration. That's all you can say about that equation.
     
  4. Jan 7, 2012 #3
    Sorry, I mean this [itex]{\psi}_{m}^*{\psi}_{n}[/itex] must be equal to dirac delta. I corrected it now.
     
  5. Jan 7, 2012 #4

    Pengwuino

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    Gold Member

    Not quite.

    [itex]\int_{All Space} \psi^*_m\psi_n dx = \delta_{mn}[/itex] where the integration is over the whole space that you're looking at. Nowhere in this does a Dirac delta come into play and it's only valid when you look at the integration over the whole space. Look at a specific, easy example like the simple harmonic oscillator with say, n = 0 and m = 1. The product [itex]\psi^*_{m = 1} \psi_{n = 0}[/itex] will clearly not be 0, but the integration will be.
     
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