Orthogonality of Functions and Uncertainty Principle in Quantum Mechanics - Prof. Sandeep , Quizzes of Chemistry

Download Orthogonality of Functions and Uncertainty Principle in Quantum Mechanics - Prof. Sandeep and more Quizzes Chemistry in PDF only on Docsity! Problem 1 (5 points). For each pair of functions given, demonstrate whether the pair is orthogonal: € ψ1 = A1 8x 3 α 3 − 12x α ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ exp -x 2 2α 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ψ2 = A2 16x 4 α 4 − 48x 2 α 2 +12 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ exp -x 2 2α 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (-∞ ≤ x ≤ ∞) Solution: To determine if the two functions are orthogonal to one another, we compute the following integral. The integrand is odd over the symmetric integration limits, thus allowing us to write the answer immediately. Since the integral is zero, the pair is orthogonal. One can expand the integrand to obtain a sum of integrals of odd functions; either way, the integral is zero and the pair is orthogonal. These functions are the eigenfunctions for the quantum harmonic oscillator. Problem 2 (5 Points) Can the momentum and potential energy of a quantum particle existing in one- dimensional space be known simultaneously? Please show your detailed work. Consider that the potential is not constant for all values of position, x. Solution: Let’s consider a generic, position-dependent potential, V(x) (one-dimensional). We’ll let the associated operator for this potential energy be € ˆ V (x) . In order to test whether the potential energy and linear momentum can be determined simultaneously and exactly, we evaluate the commutator operating on some general function, f(x). € ˆ p x, ˆ V (x)[ ] f (x) = −i ddx , ˆ V (x) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ f (x) = −i d dx V (x) f (x)( ) + iV (x) df (x) dx = −iV (x) df (x) dx − if (x) dV (x) dx + iV (x) df (x) dx = −if (x) dV (x) dx Thus, the commutator evaluates to: € ˆ p x, ˆ V (x)[ ] = −i dV (x)dx ≠ 0 Since the commutator is not zero, the properties cannot be determined simultaneously and exactly. ! 1= ? "1 *"2 dx = A1A2 exp -x 2 # 2 $ % & ' ( ) 8x 3 # 3 * 12x # $ % & ' ( ) 16x 4 # 4 * 48x 2 # 2 +12 $ % & ' ( ) dx -+ + , -+ + , = A1A2 (even)(odd)(even) dx = A1A2 (odd) dx -+ + , -+ + , = 0 CHEM444 Section 11/81 Spring 2011 (11S) QUIZ 6 April 7, 2011 NAME: Score ______/10