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Unformatted text preview: Solutions to Problem Set 2 Physics 342 by: Callum Quigley 1 2 d DeltaFunction Potentials (i) We want to preserve the basic property that 1 = Z d 2 x 2 ( ~x ) = Z d r d r 2 ( ~x ) . (1.1) Thus, in polar coordinates the two dimensional deltafunction becomes 2 ( ~x ) = 1 r ( r ) ( ) . (1.2) Note that 1 2 r ( r ) would not work as a definition, since the delta function should also satisfy f (0 , 0) = Z d r d r 2 ( ~x ) f ( r, ) , (1.3) and this alternative would only work in the special cases where f = f ( r ). Then, recalling that L = i ~ , we can write the Hamiltonian as H = ~ 2 2 m 2 + g 2 ( ~x ) = ~ 2 2 m rr + 1 r r + L 2 2 mr 2 + g r ( r ) ( ) (1.4) (ii) If = 0 then = ( r ), and the Schr odinger equation reads 0 = ( H E ) ( r ) = ~ 2 2 m rr + 1 r r 2 mg ~ 2 r ( r ) ( ) + 2 mE ~ 2 ( r ) . (1.5) Note that E = E  for bound states. Defining = r p 2 m  E  / ~ , the Schrodinger equation can be written ( 2 +  2 ) ( ) = A ( ) ( ) ( ) . (1.6) where A = 2 mg/ ~ 2 . For 6 = 0, the RHS vanishes and we recognize what remains as the modified Bessel equation of order 0. 1 The general solution (at least away from = 0), is then given by ( ) = BI ( ) + CK ( ) (1.7) 1 Recall that this is related to the ordinary Bessel equation by 7 i , or equivalently by flipping the sign in the last term of the LHS. 1 for some coefficients B,C C . I ( ) ,K ( ) are the modified Bessel functions (of order 0), which are related to their standard counterparts via I ( ) = J ( i ). However, I is not normalizable since I ( ) e 2 , as = Z  I ( )  2 d (1.8) So I ( ) does not correspond to a bound state. On the other hand, K ( ) decays at infinity and so is normalizable. In particular, K ( ) e 2 ,  ln( / 2) , (1.9) is the EulerMascheroni constant, whose precise value doesnt matter here. Even thoughis the EulerMascheroni constant, whose precise value doesnt matter here....
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 Spring '10
 Sghy
 Physics, mechanics

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