4604_Exam1_solutions_fa05

4604_Exam1_solutions_fa05 - PHY4604 Exam 1 Solutions PHY...

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PHY4604 Exam 1 Solutions Department of Physics Page 1 of 8 PHY 4604 Exam 1 Solutions Monday October 3, 2005 (Total Points = 100) Problem 1 (10 points): Circle true or false for following (1 point each). (a) (True or False) One of the “breakthroughs” that lead to quantum mechanics was the idea of associating differential operators with the dynamical variables. (b) (True or False) The wave function Ψ (x,t) must vanish in a region of infinite potential. (c) (True or False) It is possible for a free particle to have a definite energy. (d) (True or False) In quantum mechanics particles can enter the “classically forbidden” region where V 0 > E ( i.e. KE < 0). (e) (True or False) The operator (A op -A op ) is hermitian. Note is we let O = A op -A op then O = A op - A op = - (A op -A op )= -O . (f) (True or False) If A op and B op are hermitian then A op -B op is also hermitian. Note is we let O = A op -B op then O = A op -B op = A op -B op = O . (g) (True or False) If P op is the parity operator, P op ψ (x) = ψ (-x) , then P op 2 = 1 . (h) (True or False) Solutions of Schrödinger’s equation of the form ) ( ) ( ) , ( t x t x φ ψ = Ψ correspond to states with definite energy E. (i) (True or False) Solutions of Schrödinger’s equation of the form ) ( ) ( ) , ( t x t x = Ψ correspond to states in which the probability density 2 | ) , ( | ) , ( t x t x Ψ = ρ is independent of time. (j) (True or False) Schrödinger’s equation is valid for all velocities even when v c. Problem 2 (30 points): Consider an infinite square well defined by V(x) = 0 for –L/2 < x < +L/2 , V(x) = + otherwise. We look for stationary states of the form h / ) ( ) , ( t iE n n n e x t x = Ψ . Parity is a good quantum number in this problem since V(-x) = V(x) and hence the stationary state solutions are either even or odd under parity as follows: ) ( ) ( ) ( ) ( ) ( ) ( x x x P n n n + + + + + + = = ) ( ) ( ) ( ) ( ) ( ) ( x x x P n n n = = (a) (5 points). Calculate the (normalized) even parity wave functions ) ( ) ( x n + + and their corresponding energies, + n E . V = +infinity V = +infinity Infinite Square Well -L/2 +L/2 x
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PHY4604 Exam 1 Solutions Department of Physics Page 2 of 8 Answer: ) / ) 1 2 cos(( 2 ) ( ) ( L x n L x n π ψ = + + + , 2 2 2 2 2 2 ) 1 2 ( 2 2 = = + + + n mL m k E n n h h , where n + = 1, 2, 3, … Solution: We look for solutions of the time-independent Schrödinger equation ) ( ) ( ) ( ) ( 2 2 2 2 x E x x V dx x d m = + h or ) ( )) ( ( 2 ) ( 2 2 2 x x V E m dx x d = h with h / ) ( ) , ( iEt e x t x = Ψ . In the region –L/2 < x < L/2 for E > 0 and V(x) = 0 we have ) ( ) ( 2 ) ( 2 2 2 2 x k x mE dx x d = = h with 2 2 h mE k = and m k E 2 2 2 h = The most general solution is ikx ikx Be Ae x + + = ) ( . The even parity solution have ψ (-x) = ψ (x) and hence A = B and hence ) cos( 2 ) ( ) ( ) ( kx A e e A x ikx ikx = + = + + The boundary condition at x = L/2 is ) 2 / cos( 2 0 ) 2 / ( ) ( kL A L = = + , which implies that kL/2 = (n + -1/2) π , where n + = 1, 2, 3, … Thus, ) / ) 1 2 cos(( 2 ) ( ) ( L x n A x n = + + + . The normalization is determined by requiring that 1 2 4 ) 2 sin( 2 ) 1 2 ( 4 ) ( cos ) 1 2 ( 4 ) / ) 1 2 (( cos 4 | ) ( | 2 ) ( ) ( 2 ) ( ) ( 2 2 2 / 2 / 2 2 2 / 2 / 2 ) ( 2 1 2 1 2 1 2 1 = = + = = = + + + + + + + + + L A y y n L A dy y n L A dx L x n A dx x n n n n L L L L n .
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4604_Exam1_solutions_fa05 - PHY4604 Exam 1 Solutions PHY...

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