Shankar CH1 Derivative of Dirac delta

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 2K views
LAHLH
Messages
405
Reaction score
2
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.
 
Physics news on Phys.org
[tex]\frac{d f(x')}{dx} = 0[/tex]

Because x and x' are independent variables.
 
But isn't 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?