Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Shankar CH1 Derivative of Dirac delta

  1. Mar 30, 2010 #1
    Hi,

    On p67 of shankar Principles of QM, he considers the delta functions derivative. He says:

    [tex] \int \delta'(x-x')f(x')dx'= \int \frac{d\delta(x-x')}{dx}f(x')dx'= \frac{d}{dx}\int \delta(x-x') f(x')dx'=\frac{df(x)}{dx} [/tex]

    I don't understand how the second equality follows, how can the derivative just be pulled out like that here? I'm not sure if differentiating under the integral vs externally changes things from what I may have expected. But I thought some kind of product rule of the form:

    [tex] \frac{d}{dx}\int \delta(x-x') f(x')dx'=\int \left[ \frac{d\delta(x-x')}{dx} f(x')+\frac{d f(x')}{dx} \delta(x-x') \right] dx' [/tex]

    would be in operation.
     
  2. jcsd
  3. Mar 30, 2010 #2
    [tex]\frac{d f(x')}{dx} = 0[/tex]

    Because x and x' are independent variables.
     
  4. Mar 30, 2010 #3
    But isnt the second term:

    [tex]
    \int \frac{d f(x')}{dx} \delta(x-x') \right] dx'=\frac{d f(x)}{dx}
    [/tex]

    by using the sifting property of the delta function?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Shankar CH1 Derivative of Dirac delta
  1. Dirac delta (Replies: 6)

  2. Dirac delta (Replies: 24)

  3. Dirac delta (Replies: 8)

Loading...