3502_HWkey5 - Chem 3502/5502 Physical Chemistry II(Quantum...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Answers to Homework Set 5 From lecture 18: On the interval 0 to 1, what is the square modulus (in terms of a and b ) of the trial wave function in eq. 18-15? Eq. 18-15 was ! x ; a , b ( ) = x a 1 " x b ( ) (18-15) By analogy to eq. 18-5, we find the square modulus as " x ; a , b ( ) " x ; a , b ( ) = x a 1 # x b ( ) 0 1 $ x a 1 # x b ( ) dx = x 2 a 0 1 $ dx # 2 x 2 a + b 0 1 $ dx + x 2 a + 2 b 0 1 $ dx = x 2 a + 1 2 a + 1 0 1 # 2 x 2 a + b + 1 2 a + b + 1 0 1 + x 2 a + 2 b + 1 2 a + 2 b + 1 0 1 = 1 2 a + 1 # 2 2 a + b + 1 + 1 2 a + 2 b + 1 = 2 b 2 2 a + 1 ( ) 2 a + b + 1 ( ) 2 a + 2 b + 1 ( ) From lecture 19: Today’s in-class homework solution showed that the expectation value of the momentum operator for any real wave function must be zero. Why doesn’t the same proof hold for any complex wave function? Consider any complex wave function of one dimension f . f x ( ) p x f x ( ) = f * x ( ) " i h d dx # $ % & ( a b ) f x ( ) dx = " i h f * x ( ) df x ( ) dx # $ % & ( a b ) dx
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HW5-2 where we have assumed without loss of generality that f is normalized over the integration interval. We can try to solve the integral using integration by parts. If we use
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This note was uploaded on 12/18/2010 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.

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3502_HWkey5 - Chem 3502/5502 Physical Chemistry II(Quantum...

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