# solutionhw3 - Solutions to Homework Set 3 Test functions...

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Solutions to Homework Set 3 Test functions and distributions : For part a) we take any test function ϕ ( x ) and look at ( ϕ, fδ + f δ ) -∞ ϕ ( x ) { f ( x ) δ ( x ) + f ( x ) δ ( x ) } dx = -∞ { [ - ϕ ( x ) f ( x ) - ϕ ( x ) f ( x )] δ ( x ) + ϕ ( x ) f ( x ) δ ( x ) } dx = - ϕ (0) f (0) , and compare it with ( ϕ, f (0) δ ) -∞ ϕ ( x ) { f (0) δ ( x ) } dx = - ϕ (0) f (0) . The results are the same, and so the distributions + f δ and f (0) δ are equal. For b) part i) we have ∂t P -∞ ϕ ( x ) ( x - t ) dx = ∂t t - -∞ + t + ϕ ( x ) ( x - t ) dx = ϕ ( t - ) - - ϕ ( t + ) + t - -∞ + t + ϕ ( x ) ( x - t ) 2 dx = - 2 ϕ ( t ) + t - -∞ + t + ϕ ( x ) ( x - t ) 2 dx + O ( ) = t - -∞ + t + ϕ ( x ) - ϕ ( t ) ( x - t ) 2 dx + O ( ) P -∞ ϕ ( x ) - ϕ ( t ) ( x - t ) 2 dx as 0 . The tricky part was recognizing that ϕ ( t + ) + ϕ ( t - ) = 2 ϕ ( t ) + O ( 2 ) , and that you can rewrite the resultant divergent bit 2 ϕ ( t ) / as 2 ϕ ( t ) = ϕ ( t ) t - -∞ + t + 1 ( x - t ) 2 dx = t - -∞ + t + ϕ ( t ) ( x - t ) 2 dx. For b) part ii) we have P -∞ ϕ ( x ) ( x - t ) dx = P -∞ ϕ ( y + t ) y dy. Taking the derivative, we get ∂t P -∞ ϕ ( x ) ( x - t ) dx = P -∞ ϕ ( y + t ) y dy 1

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= P -∞ ∂y ϕ ( y + t ) y + ϕ ( y + t ) y 2 dy = - 2 ϕ ( t ) + P -∞ ϕ ( y + t ) y 2 dy = P -∞ ϕ ( y + t ) - ϕ ( t ) y 2 dy = P -∞ ϕ ( x ) - ϕ ( t ) ( x - t ) 2 dx. 2) One-dimensional scattering theory : This problem exploits the fact that the Wron- skian of any pair of solutions of a Schr¨ odinger equation is independent of x . The four
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