Question

A1. A few useful properties of the Dirac delta function The Dirac δ(z) functions is defined by δ(z) = 0 if |メ0, δ(z) = oo ifx-0 but the integral of the function over any interval containing the zero of the argument is unity, Equivalent, if f(x) is continuous at the origin One should treat the Dirac δ function as a limit of a sequence of functions peaked at x-0 As the limit is approached, the height of the peak increases and the width shrinks in such a way that the area under the curve remains unity. There are many different representations for the δ function. For example Another useful representation is a) By multipying both sides of the following equations by a differentiable function f(z) and integrating over r, verify the following equations: 1 181)--5(z) b) Prove the following relations 9 c) Prove that the δ(z) function represented by (2) has the property: δ(z)-0 if! 0. Demonstrate that δ(x) function represented by (2) is properly normalized, ie.

Answer 1

a) delta (x) - lim_elementof rightarrow 0 1/pi - elementof/elementof^2 + x^2 delta(-x) = lim_elementof rightarrow 0 1/pi elementof/elementof^2 + (-x)^2 = lim_elementof rightarrow 0 1/pi elementof/ elementof^2 + x^2 Hence delta (x) = delta (-x) Proved (3) To prove d/ dx delta (x) = d/ dx d (-x) since delta (x) = delta (-x) d/ dx delta (x) d/ dx d(-x) (4) Since delta (x) = lim_elementof rightarrow 0 1/pi elementof/elementof^2 + x^2 Since x delta (x) since integral^infinity_- infinity f(x) delta (x) dx = f(0) f(x) = x f(0) = 0 integral^infinity_- infinity x delta (x) dx = 0 Then x delta (x) = 0 (5) To prove x d/dx delta (x) = -delta (x) Since x delta (x) = 0 (now diff with respect of x) delta (x) + x d/d(x) d (x) = 0 x d/dx delta (x) = -delta (x) \r\n b) Prove the following integral^infinity_- infinity detla (a - x)