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Dirac Notation in building Path Integrals

  1. Dec 23, 2011 #1
    Alright, so I was wondering if anyone could help me figure out from one step to the next...
    So we have defined |qt>=exp(iHt/[itex]\hbar[/itex])|q>
    and we divide some interval up into pieces of duration τ

    Then we consider
    <[itex]q_{j+1}[/itex][itex]t_{j+1}[/itex]|[itex]q_{j}[/itex][itex]t_{j}[/itex]>
    =<[itex]q_{j+1}[/itex]|e-iHτ/[itex]\hbar[/itex]|[itex]q_{j}[/itex]>
    =<[itex]q_{j+1}[/itex]|1-(i/[itex]\hbar[/itex])Hτ+O(τ2)|[itex]q_{j}[/itex]>
    =[itex]\delta[/itex]([itex]q_{j+1}[/itex]-[itex]q_{j}[/itex])-(iτ/[itex]\hbar[/itex])<[itex]q_{j+1}[/itex]|H|[itex]q_{j}[/itex]>

    Dumb question:
    Where did the delta function come from? I know where the second term comes from and I'm assuming that the higher order terms are being tossed since the τ2 and higher order terms have been deemed sufficiently small...but where does this delta function come from?

    Is this something like <[itex]q_{j+1}[/itex]|1|[itex]q_{j}[/itex]> = 1 iff [itex]q_{j+1}[/itex]=[itex]q_{j}[/itex], or something having to do with independence? But...then I don't see why after this we have:

    (2[itex]\pi[/itex][itex]\hbar[/itex])-1[itex]\int[/itex]dp e(i/[itex]\hbar[/itex])p([itex]q_{j+1}[/itex]-[itex]q_{j}[/itex])
    coming from the delta function term .-. what?
     
  2. jcsd
  3. Dec 24, 2011 #2

    Bill_K

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    Science Advisor

    Your last question... because that's what the Fourier transform of a delta function is:

    δ(x) ≡ (1/2π) ∫eixy dy

    and put x = qj+1 - qj, y = p/ħ
     
  4. Dec 24, 2011 #3
    your 1st Q;
    it is the normalization of qi, qj
     
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