HW3-sol - PHYSICS 4455 QUANTUM MECHANICS Problem Set 3...

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PHYSICS 4455 — QUANTUM MECHANICS Problem Set 3 — Solutions. 1. Getting more familiar with δ functions. Liboff Problem 3.6 Remember to regard these equalities in the following sense. To show blah = BLAH , you need to demonstrate that blah f ( y ) dy = BLAH f ( y ) dy for any (infinitely differentiable) function f . Specifically, follow closely Liboff in (3.31), “which establishes (3.30)” and part (h) of this problem! (a) - see part (d), setting a =− 1. (b) Here, blah ( y ) ,and BLAH =−δ (− y ) . We start with the left hand side (LHS) and integrate by parts: δ ( y ) f ( y ) dy = d dy [δ( y ) f ( y )] dy δ( y ) f ( y ) dy = 0 f ( 0 )=− f ( 0 ) Now the RHS: Before integrating by parts, we substitute t y (watch the limits of the integral!): −∞ +∞ δ (− y ) f ( y ) dy +∞ −∞ δ ( t ) f (− t ) d (− t −∞ +∞ δ ( t ) f (− t ) dt −∞ +∞ d dt [δ( t ) f (− t dt −∞ +∞ δ( t ) d dt [ f (− t dt = 0 + −∞ +∞ δ( t )[− f (− t dt f ( 0 ) (c) y δ( y ) f ( y ) dy = δ( y )[ yf ( y dy = [ yf ( y | y = 0 = 0 (d) Substitute t = ay whence dt = ady : First, when a > 0, the limits of integration remain unchanged: a > 0: −∞ +∞ δ( ay ) f ( y ) dy = −∞ +∞ δ( t ) f ( t / a ) dt a = a 1 f ( 0 )= a 1 −∞ +∞ δ( y ) f ( y ) dy Now, when a < 0, the limits of integration reverse: a < −∞ +∞ δ( ay ) f ( y ) dy = +∞ −∞ δ( t ) f ( t / a ) dt a a 1 +∞ −∞ δ( t ) f ( t / a ) dt a 1 f ( 0 )= | a | 1 f ( 0 a | 1 −∞ +∞ δ( y ) f ( y ) dy (e) First of all, note that y 2 a 2 has two zeros: one at y = a and a second at y a .So ,we get “hit” twice by this δ -function. To make sure that we pick up both contributions, we recall that the δ -function vanishes away from the zeros of its argument. Therefore, we may write −∞ +∞ δ( y 2 a 2 ) f ( y ) dy = a −ε a δ( y 2 a 2 ) f ( y ) dy + a −ε a δ( y 2 a 2 ) f ( y ) dy

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where ε> 0 is a small (infinitesimal) quantity (technically, we should consider the limit ε→ 0). Written in this way, the first (second) integral “feels” only the singularity at y =− a ( y =+ a ). Let’s consider the second integral first: In the vicinity of y a we may write g ( y )≡ y 2 a 2 = g ( a )+ g ( a )( y a 1 2 g  ( a )( y a ) 2 = 0 + 2 a ( y a 1 2 × 2 ×( y a ) 2 2 a ( y a O 2 ) Note that this is the Taylor expansion of the function g ( y y 2 a 2 near y a : g ( y )= g ( a g ( a )( y a 1 2 g  ( a )( y a ) 2 Since | y a |≤ε , the last term is O 2 ) andmuchsmallerthanthefirstterm ,as 0. Note also that the Taylor expansion terminates after the second derivative here, since g ( y ) is just a simple second-degree polynomial. Returning to the δ -function, we now have a −ε a δ( y 2 a 2 ) f ( y ) dy a −ε a δ( 2 a ( y a )) f ( y ) dy = 1 | 2 a | a −ε a δ( y a ) f ( y ) dy = f ( a ) | 2 a | Note that we have used part (d) of this problem. Finally, we perform the remaining integral (near y a ) in a similar manner. With y 2 a 2 =( y a )( y + a ) 2 a ( y + a O 2 ) ,we find a −ε a δ( y 2 a 2 ) f ( y ) dy = a −ε a δ(− 2 a ( y + a )) f ( y ) dy = 1 |− 2 a | a −ε a
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HW3-sol - PHYSICS 4455 QUANTUM MECHANICS Problem Set 3...

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