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Unformatted text preview: 1 Physics 316 Solution for homework 10 Spring 2005 Cornell University I. EXERCISE 1 A. A particle in an infinite square well extending between x = 0 and x = L has the wave function ( x, t ) = A 2 sin x L e i E 1 h t + sin 2 x L e i E 2 h t , E n = n 2 h 2 2 2 mL 2 . (1.1) Choose A so that the wave function is normalized to 1. Lets first find A such that ( x, t ) is normalized. We know (first question of Quiz no. 2) that it suffices to show that ( x, 0) is normalized. 1 = h  i = " A * r L 2 (2 h 1  + h 2  ) #" A r L 2 (2  1 i +  2 i ) # =  A  2 L 2 (4 + 1) , (1.2) and we get A = q 2 5 L (up to an arbitrary phase factor). Hence, the normalized state is:  i =  1 i 2 5 e i E 1 h t +  2 i 1 5 e i E 2 h t . (1.3) B. If a measurement of the energy is made, what are the possible results of the measurement, and what is the probability associated with each? Recall that for a system described by a state vector of the form  i = X n  n i A n , (1.4) the probability P n of finding the system in state  n i is given by P n =  A n  2 . (1.5) From Eq. (1.3), we then see that the two possible outcomes of an energy measurement are E 1 or E 2 , such that P 1 = fl fl fl fl 2 5 e i E 1 h t fl fl fl fl 2 = 4 5 , P 2 = fl fl fl fl 1 5 e i E 2 h t fl fl fl fl 2 = 1 5 , (1.6) and P n = 0, for all other values of n . C. When the energy is measured for many particles, each having been prepared in this state, what would the average energy be. This average is called the expectation value. When a quantity X can take on a discrete set of values X n , each with probability P n , its expectation value is defined to be h X i = n P n X n . Here we have h E i = 4 5 E 1 + 1 5 E 2 = 8 5 E 1 , (1.7) since, because E n n 2 for the infinite well, we have E 2 = 4 E 1 . 2 II. EXERCISE 2 A. In classical mechanics, we have put dx/dt = p x /m . In quantum mechanics, this is replaced by a corresponding relation between expectation values: d dt h x i = h p x i m . Verify this relation by using the definition of h x i , the Schroedinger equation, and the fact that you can write i h x for p x . Lets show that d dt h x i = h p x i m . To do so, we only need to use the timedependent Schrdinger equation and its complex conjugate: i h t ( x, t ) = h 2 2 m 2 x 2 ( x, t ) + V ( x )( x, t ) i h t * ( x, t ) = h 2 2 m 2 x 2 * ( x, t ) + V ( x ) * ( x, t ) ....
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This note was uploaded on 09/28/2008 for the course PHYS 316 taught by Professor Hoffstaetter during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 HOFFSTAETTER
 Physics, Work

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