Shankar CH1 Derivative of Dirac delta

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SUMMARY

The discussion focuses on the derivative of the Dirac delta function as presented in Shankar's "Principles of Quantum Mechanics." Specifically, it examines the equality involving the derivative of the delta function and a test function, where the integral of the product is differentiated. The participant questions the validity of pulling the derivative out of the integral, suggesting that the product rule should apply. The conclusion drawn is that the second term indeed simplifies to the derivative of the test function evaluated at the point where the delta function is centered, confirming the correctness of Shankar's approach.

PREREQUISITES
  • Understanding of Dirac delta functions
  • Familiarity with integral calculus and differentiation
  • Knowledge of the sifting property of delta functions
  • Basic concepts of quantum mechanics as outlined in Shankar's text
NEXT STEPS
  • Study the properties of the Dirac delta function in detail
  • Learn about differentiation under the integral sign
  • Explore the product rule in the context of distributions
  • Review examples of applying the sifting property in various integrals
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Students and professionals in physics, particularly those studying quantum mechanics, as well as mathematicians interested in distribution theory and integral calculus.

LAHLH
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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.
 
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[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?
 

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