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4604_Exam2_solutions_fa05

# 4604_Exam2_solutions_fa05 - PHY4604 Exam 2 Solutions PHY...

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PHY4604 Exam 2 Solutions Department of Physics Page 1 of 7 PHY 4604 Exam 2 Solutions (Total Points = 100) Problem 1 (10 points): Circle true or false for following (1 point each). (a) (True or False) Bohr’s model of the atom assumes the orbital angular momentum is quantized according to h n L = , with n = 1, 2, 3, …, which is also true quantum mechanically. (b) (True or False) The eigenvalues of hermitian operators are real numbers. (c) (True or False) Eigenfunctions of hermitian operators corresponding to different eigenvalues are orthogonal. (d) (True or False) If an observable, O , commutes with the Hamiltonian and if O does not depend explicitly on time, then <O> is zero. (e) (True or False) If [A, B] = 0 , then A B 0 . (f) (True or False) All quantum operators can be expressed in terms of functions or differentials. (g) (True or False) If the three operators J 1 , J 2 , and J 3 , all commute with each other then they are said to form a “Lie Algebra”. (h) (True or False) SU(2) is the group of all unitary 2 × 2 matrices with determinant equal to one. (i) (True or False) The spin 2 ( i.e. s = 2) gravaton has five possible spin states. (j) (True or False) In SU(2), 2 × 2 = 4.

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PHY4604 Exam 2 Solutions Department of Physics Page 2 of 7 Problem 2 (30 points): Consider the ( one dimensional ) wave function at t = 0 given by 2 2 ) ( a x A x + = ψ , where A and a are real constants. (a) (5 points) Find the normalization constant A such that 1 | ) ( | 2 = +∞ dx x ψ and sketch the probability density . | ) ( | ) ( 2 x x ψ ρ = Answer: π 3 2 a A = Solution: We see that 1 4 2 ) ( 1 2 ) ( 1 | ) ( | 3 2 0 2 2 2 2 2 2 2 2 2 = = + = + = +∞ +∞ +∞ a A dx a x A dx a x A dx x π ψ and hence π 3 2 a A = , where I used ) / ( tan 2 1 ) ( 2 ) ( 1 1 3 2 2 2 2 2 2 a x a x a a x dx x a + + = + and 3 0 2 2 2 4 ) ( 1 a dx x a π = + . Thus, 2 2 2 3 2 ) ( 1 2 | ) ( | ) ( a x a x x + = = π ψ ρ (b) (5 points) Compute <x> and <x 2 > and x using the position space wave function ψ (x) . Answer: 0 >= < x , 2 2 a x >= < , and x = a Solution: We see that 0 ) ( | ) ( | 2 2 2 2 2 = + = >= < +∞ +∞ dx a x x A dx x x x ψ 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 ) ( 2 ) ( | ) ( | a a A dx a x x A dx a x x A dx x x x = = + = + = >= < +∞ +∞ +∞ π ψ , where I used ) / ( tan 2 1 ) ( 2 ) ( 1 2 2 2 2 2 2 a x a x a x dx x a x + + = + and a dx x a x 4 ) ( 0 2 2 2 2 π = + . We see that a x x x = > < > < = 2 2 . (c) (10 points). Find the momentum space wave function at t = 0 , where +∞ = dx e x p x ixp x h h / ) ( 2 1 ) ( ψ π φ and verify that it is properly normalized. Sketch the probability density . | ) ( | ) ( 2 x x p p φ ρ = ρ (x) -5.0 0.0 5.0
PHY4604 Exam 2 Solutions Department of Physics Page 3 of 7 Answer: h h / | | ) ( x p a x e a p = φ Solution: We see that h h h h h h h h h h h h / | | / | | 0 2 2 2 2 2 2 2 2 / 2 2 2 ) / cos( 2 2 ) / sin( 2 ) / cos( 2 2 ) ( x x x p a p a x x x ixp x e a e a A dx a x xp A dx a x xp A i dx a x xp A dx a x e A p + +∞

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