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Sakurai: Page 56

  1. Apr 23, 2013 #1
    From "Modern Quantum Mechanics, revised edition" by J.J. Sakurai, page 56.

    In equation (1.7.31) it is given,
    [tex]\begin{eqnarray}
    \delta(x' - x'') & = & | N |^2 \int dp' \exp \left[ \frac{ip'(x'-x'')}{\hbar} \right] \\
    & = & 2 \pi \hbar | N |^2 \delta(x' - x'' )
    \end{eqnarray}[/tex]
    How does the right side happen. Is this a definition of the delta function?
     
  2. jcsd
  3. Apr 23, 2013 #2
    It's a fourier transform.

    To see this the fourier transformation is given by
    [tex]\mathcal{F}[\delta (x)] = \int \delta (x)\exp \left(-i2\pi px\right) dx = \frac{1}{2\pi}\exp (0) = \frac{1}{2\pi}[/tex]
    Inverse transformation gives
    [tex]\delta(x)=\frac{1}{2\pi} \int \exp \left( ipx\right) dp[/tex]

    And thus [tex]\int \exp \left(ipx\right) dp = 2\pi \delta (x)[/tex]

    Can you see it now?
     
    Last edited: Apr 23, 2013
  4. Apr 23, 2013 #3

    Bill_K

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    Science Advisor

    Actually, δ(x) = (1/2π) ∫eipx dp
     
  5. Apr 23, 2013 #4
    Ok, my bad. Been 3 years since I actually 'performed' a Fourier transformation. Should've checked it
    I forgot the [tex]2\pi[/tex] factor in the exponential.
     
  6. Apr 23, 2013 #5
    OK, I see it now. Thanks.
     
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