4604_Exam1_solutions_fa06

# 4604_Exam1_solutions_fa06 - PHY4604 Fall 2006 R D Field PHY...

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PHY4604 Fall 2006 R. D. Field Department of Physics Page 1 of 12 Exam 1 Solutions PHY 4604 Exam 1 Solutions (Total Points = 100) Problem 1 (20 points): Circle true or false for following (2 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) Solutions of Schrödinger’s equation of the form ) ( ) ( ) , ( t x t x φ ψ = Ψ correspond to states with definite energy E . (c) (True or False) Solutions of Schrödinger’s equation of the form ) ( ) ( ) , ( t x t x φ ψ = Ψ correspond to states in which the probability density 2 | ) , ( | ) , ( t x t x Ψ = ρ is independent of time. (d) (True or False) The wave function Ψ (x,t) must vanish in a region of infinite potential. (e) (True or False) It is possible for a free particle to have a definite energy. (f) (True or False) In quantum mechanics particles can enter the “classically forbidden” region where V 0 > E ( i.e. KE < 0). (g) (True or False) The operator A op A op is hermitian. (h) (True or False) If A op and B op are hermitian then A op B op is also hermitian. (i) (True or False) The commutator operator [(p x ) op ,(x 2 ) op ] is equal to h i 2 . (j) (True or False) In position-space the commutator operator [(p x ) op ,sin(kx)] is equal to ) cos( kx k i h . Problem 2 (30 points): Consider an electron with mass m e confined within an infinite square well defined by V(x) = 0 for 0 < x < L , V(x) = + otherwise. (a) (2 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 , and normalize the eigenfunctions V = +infinity V = +infinity Infinite Square Well 0 L x

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PHY4604 Fall 2006 R. D. Field Department of Physics Page 2 of 12 Exam 1 Solutions so that the probability of finding the electron somewhere in the box is one. Answer: ) / sin( 2 ) ( L x n L x n π ψ = Solution: For the region outside of 0 < x < L 0 ) ( = x ψ and inside the region ) ( ) ( 2 2 2 2 x E x d x d m e ψ ψ = h or ) ( ) ( 2 2 2 x k x d x d ψ ψ = with e m k E 2 2 2 h = The most general solution is of the form ) cos( ) sin( ) ( kx B kx A x + = ψ . The boundary condition at x = 0 gives 0 ) 0 ( = = B ψ and the boundary condition at x = L gives 0 ) sin( ) ( = = kL A L ψ which implies that π n kL = with n = 1, 2, 3,… Thus, ) / sin( ) ( L x n A x n π ψ = with 2 2 2 2 2 mL n E n h π = . 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 = = = = = +∞ π π π π π ψ ψ Thus, L A / 2 = . These states are called stationary because the probability density and all the expectation values are independent of time. (b) (2 points) Show that the wavefunctions Ψ n (x,t) correspond to states with definite energy ( i.e. show that E = 0 ). Solution: We see that n n n n n n n op n n E dx t x t x E dx t t x t x i dx t x E t x E = Ψ Ψ = Ψ Ψ = Ψ Ψ = > < +∞ +∞ +∞ ) , ( ) , ( ) , ( ) , ( ) , ( ) , ( h and 2 2 2 2 2 2 2 ) , ( ) , ( ) , ( ) , ( ) , ( ) , ( n n n n n n n op n n E dx t x t x E dx t t x t x dx t x E t x E = Ψ Ψ = Ψ Ψ = Ψ Ψ = > < +∞
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