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4604_Exam3_solutions_fa06

4604_Exam3_solutions_fa06 - PHY4604 Fall 2006 Final Exam...

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PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 1 of 25 PHY 4604 Final Exam Solutions Wednesday December 6, 2006 (Total Points = 100) Problem 1 (25 points): Consider the following one-dimensional potential: + = 0 0 ) ( V x V L x L x L x < < 0 Where V 0 is a positive (real) constant. (A) Suppose that particles with mass m and energy E > 0 enter from the left in region 1 and travel to the right and encounter the potential V(x) . Let 2 2 2 2 ε mL E h = and E V r 0 1 + = . (1) (5 points) Calculate the probability that particles entering from the left in region 1 will be reflected back ( i.e. calculate the ratio of the probability flux in region 1 traveling to the left to the incident flux in region 1 traveling to the right, 1 1 / j j P R r s = ). Express your answer in terms of ε and r. Answer: P R = 1 . Solution: We look for solutions of the time-independent Schrödinger equation ) ( ) ( ) ( ) ( 2 2 2 2 x E x x V dx x d m ψ ψ ψ = + h or ) ( )) ( ( 2 ) ( 2 2 2 x x V E m dx x d ψ ψ = h with h / ) ( ) , ( iEt e x t x = Ψ ψ . In the region x < 0 (region 1) for E > 0 and V(x) = 0 we have ) ( ) ( 2 ) ( 2 2 2 2 x k x mE dx x d ψ ψ ψ = = h with L mE k ε = = 2 2 h and 2 2 2 2 2 2 2 ε mL m k E h h = = The most general solution is ikx ikx Be Ae x + + = ) ( 1 ψ , In the region 0 < x < L (region 2) we have iqx iqx De Ce x + + = ) ( 2 ψ , with rk V E m q = + = 2 0 ) ( 2 h , where E V k q r 0 1 + = = . In the region x > L (region 3) we have 0 ) ( 3 = x ψ , The boundary conditions at x = L implies that 0 ) ( 2 = L ψ and hence 0 = + + iqL iqL De Ce or iqL Ce D 2 + = . Thus, ) ( ) ( 2 2 iqx iqL iqx e e e C x + + = ψ , At x = 0 the wave function must be continuous with a continuous derivative which yields x V(x) -V 0 V = +infinity L Region 1 Region 2 Region 3

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PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 2 of 25 ) 0 ( ) 0 ( 2 1 ψ ψ = and ) 1 ( 2 iqL e C B A + = + 0 2 0 1 ) ( ) ( = = = x x dx x d dx x d ψ ψ and ) 1 ( 2 iqL e Ciq ikB ikA + + = Thus, ) 1 ( 2 iqL e C B A + = + and ) 1 ( 2 iqL e Cr B A + + = And ) ) 1 ( ) 1 (( 2 2 iqL e r r C A + + + = and ) ) 1 ( ) 1 (( 2 2 iqL e r r C B + + + = Thus, iqL e r r A C 2 ) 1 ( ) 1 ( 2 + + + = and A e r r e r r B iqL iqL 2 2 ) 1 ( ) 1 ( ) 1 ( ) 1 ( + + + + + + = . The relfection probability is given by 1 ) 1 )( 1 ( ) 1 )( 1 ( ) 1 ( ) 1 ( ) 1 )( 1 ( ) 1 )( 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( | | | | 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 = + + + + + + + + + + + + = + + + + + + + + = + + + + = = + + + + + + iqL iqL iqL iqL iqL iqL iqL iqL iqL iqL m k m k R e r r e r r r r e r r e r r r r e r r e r r e r r e r r e r r e r r A B P h h (2) (5 points) Calculate the probability that particles traveling to the right in region 2 will be reflected back ( i.e. calculate the ratio of the probability flux in region 2 traveling to the left to flux in region 2 traveling to the right, 2 2 / j j R r s = ). Express your answer in terms of ε and r. Answer: R = 1 . Solution: We see that 1 | | | | 2 2 2 2 = = = + iqL m q m q e C D R h h . (3) (5 points) Calculate the ratio of the probability flux in region 2 traveling to the right to the incident flux in region 1 traveling to the right, 1 2 / j j P r r = ). Express your answer in terms of ε and r. What are the maximum and minimum values of P( ε ) (express your answer in terms of r)?
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