3063_Exam3_solutions_sp06

# 3063_Exam3_solutions_sp06 - PHY3063 Spring 2006 R D Field...

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PHY3063 Spring 2006 R. D. Field Final Exam Solutions Page 1 of 9 April 25, 2006 PHY 3063 Final Exam Solutions Problem 1 (35 points): Consider an particle with mass m confined within an infinite square well defined by V(x) = 0 for 0 < x < L , V(x) = + otherwise. (a) (5 points): Using Schrödinger’s equation calculate the allowed stationary state eigenfunctions ψ n (x) , where the complete wavefunctions are given by h / ) ( ) , ( t iE n n n e x t x = Ψ ψ . Normalize the eigenfunctions so that the probability of finding the particle somewhere in the box is one. Answer: ) / sin( 2 ) ( L x n L x n π = Solution: For the region outside of 0 < x < L 0 ) ( = x and inside the region ) ( ) ( 2 2 2 2 x E x d x d m e = h or ) ( ) ( 2 2 2 x k x d x d = with e m k E 2 2 2 h = The most general solution is of the form ) cos( ) sin( ) ( kx B kx A x + = . The boundary condition at x = 0 gives 0 ) 0 ( = = B and the boundary condition at x = L gives 0 ) sin( ) ( = = kL A L which implies that n kL = with n = 1, 2, 3,… Thus, ) / sin( ) ( L x n A x n = with 2 2 2 2 2 mL n E n h = . The normalization is arrived at by requiring that 2 4 ) 2 sin( 2 ) ( sin ) / ( sin 1 ) ( ) ( 2 0 2 0 2 2 0 2 2 LA y y n LA dy y n LA dx L x n A dx x x n n L n n = = = = = +∞ Thus, L A / 2 = . These states are called stationary because the probability density and all the expectation values are independent of time. (b) (5 points): Show that the allowed energy levels of the system are, E n = E 0 n 2 , where ) 2 /( 2 2 2 0 mL E h = is the ground state energy and n = 1, 2, 3, …. Why is n = 0 excluded as a possible energy level? Solution: We see from above that 0 2 2 2 2 2 2 E n mL n E n = = h with 2 2 2 0 1 2 mL E E h = = . The state with n = 0 correstions to ψ 0 (x) = 0 which is not normalizable. (c) (10 points): Consider the operator, O = (x) op (p x ) op ( i.e. the product of the position operator times the momentum operator). Is the operator O hermitian? Calculate the expectation value of the operator Q for the n th stationary state ( i.e. calculate < ψ n |O| ψ n >). Answer: O is not hermitian and 2 / h i xp n x = > < . Solution: We see that x x x x xp x p x p xp O = = = ) ( ) ( since (x) op and (p x ) op do not commute. Also, V = +infinity V = +infinity Infinite Square Well 0 L x

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PHY3063 Spring 2006 R. D. Field Final Exam Solutions Page 2 of 9 April 25, 2006 () 2 4 2 cos sin 4 1 ) sin( 2 2 2 ) / 2 sin( 2 2 ) / cos( ) / sin( 2 ) ( ) ( ) ( ) )( ( 2 0 2 0 2 0 0 h h h h h h h i n n i y y y n i dy y y n L L n L i dx L x n x L n L i dx L x n L x n x L n L i dx dx x d x x i dx x xp x xp n n L L n n n op x n n x =  − = = = = = = = > < +∞ +∞ π ψ where I used θ 2 sin cos sin 2 1 = and I let y = 2n π /L. (d) (15 points): Suppose the particle in this infinite square well has an initial wave function at t = 0 given by ) / ( sin ) 0 , ( 2 L x A x = Ψ .
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3063_Exam3_solutions_sp06 - PHY3063 Spring 2006 R D Field...

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