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Unformatted text preview: CHEM 120A Problem Set 2 Solutions David Hoffman , Tara Yacovitch , Doran Bennett 2/5/2010 1. The solution can be found in the McQuarrie Solutions manual. 2. (a) Here we must show that the stationary states for system are orthonormal. The stationary states are quantized as follows, n ( ) = r 1 2 e in (1a) To prove they are orthonormal we must look at their overlap, Z 2 d * n ( ) m ( ) (1b) Z 2 d r 1 2 e in r 1 2 e im (1c) When n 6 = m Some intervening algebra yields, 1 2 i n m e i ( n m ) 2 (1d) Notice that n,m Z thus m n Z as well and finally we have, 1 2 i n m e i ( n m ) 2 = 0 (1e) On the other hand if n = m then, Z 2 d * n ( ) n ( ) = 1 (1f) Due to the fact that the wavefunctions are normalized. Equations ( 1c ) and ( 1f ) mean that Z 2 d * n ( ) n ( ) = n,m (1g) 1 QED 1 In class we were able to integrate the Schrodinger equation for the special case of stationary states. At first we thought that this was surely useless because stationary states are just that, stationary. But then we learned that we could create a superpo sition state of stationary states that evolved with time. In addition we learned that as long as we had a complete basis set (what does that mean?) we could create many different wavefunctions by projecting our chosen function onto the complete basis set. We will use these principles in the next two problems. (b) Here we can use the result proved in class that for a single stationary state the complete wavefunction is, n ( ,t ) = n ( ) e iE n t/ ~ (2a) Where is determined from the timeindependent Schr odinger equation. Then the general solution to the Schrodinger equation is a linear superposition of all the n s, ( ,t ) = X n = c n n ( ) e iE n t/ ~ where the c n s are subject to the constraint X n  c n  2 = 1 (2b) Plugging in our answers for n and E n from question 1 we arrive at our answer. (c) In this question we must project our initial function into the basis set of stationary states to determine the c n s. To do that we use the following formula, c n = Z 2 * n ( )( , 0) (3a) Using Eulers formula 2 and the fact that the basis functions are orthonormal (equation ( 1g )) we quickly find that only c 1 , c 1 , c 2 and c 2 are nonzero and our final wavefunction is, ( ,t ) = 1 2 2 e i i ~ t/ 2 I e i i ~ t/ 2 I e 2 i 2 i ~ t/I + e 2 i 2 i ~ t/I (3b) 1 The left hand side is the Kronecker delta which is defined as ij = ( 1 if i = j, 0 if i 6 = j If you need more information I would suggest looking at this website: http://en.wikipedia.org/wiki/ Kronecker_delta ....
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 Spring '10
 DAVIDCHANDLER

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