PS2sol - Solutions to Problem Set 2 Physics 341 by: Callum...

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Unformatted text preview: Solutions to Problem Set 2 Physics 341 by: Callum Quigley 1 The Dirac -function To evaluate, ( ax ) consider the cases a > 0 and a < 0 separately. Define y = ax . Then for a > , 1 = Z - ( y )d y = a Z - ( ax )d x (1.1) and for a < , 1 = Z - ( y )d y = a Z- + ( ax )d x =- a Z - ( ax )d x. (1.2) Combining these, we find ( ax ) = ( x ) | a | as required. Also, it should be clear that ( x ) = lim a 1 a x 2 + a 2 = , x = 0 , x 6 = 0 (1.3) as we expect for a -function. With this definition, we can compute its integral over R : Z ( x )d x = lim a a Z - 1 x 2 + a 2 d x = lim a a a = 1 , (1.4) as required. There are numerous ways to perform the above integral. The simplest probably being direct integration in terms of arctan. Another particularly nice way is to use the method of residues: that is, we consider x C and extend the integration contour, C , to enclose the upper-half plane. Because the integrand decays sufficiently rapid as | x | , the contribution from this extension vanishes. This contour encloses a pole in the integrand at x = ia , and so we just pick up the residue Z - 1 x 2 + a 2 d x = I C 1 ( x + ia )( x- ia ) d x = 2 i Res x = ia 1 ( x + ia )( x- ia ) = a . (1.5) 2 Gram-Schmidt Procedure The only thing youll need to solve this problem is to recall how the Gram-Schmidt procedure works. Starting from a given set of basis vectors {| e i i} , we construct a new orthogonal basis {| u i i} , where the k-th vector is determined by the previous...
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PS2sol - Solutions to Problem Set 2 Physics 341 by: Callum...

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