3502_10_octo02_CC

3502_10_octo02_CC - Chem 3502/4502 Physical Chemistry II...

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Chem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 10, October 2, 2009 Guest Lecturer Mr Adam Huss Solved Homework We are asked to find < x > and < x 2 > for the first two harmonic oscillator wave functions, which are ! 0 x ( ) = μ " h # $ % % & ( ( 1/ 4 e ) μ x 2 / 2 h and " 1 x ( ) = k μ 4 # h $ % & ( ) 1/4 2 k μ h $ % & ( ) 1/2 x * + , , - . / / e 0 k μ x 2 /2 h So, for < x > we have x n = 0 = ! 0 x ( ) x ! 0 x ( ) x n = 1 = ! 1 x ( ) x ! 1 x ( ) Before we do any heavy lifting with the integrals, note that Ψ 0 is an even function, Ψ 1 is an odd function, and x is an odd function. So, we have integrals of either even x odd x even or odd x odd x odd. Either case results in an odd function, so the expectation value in both cases is zero. Remember that in our derivation x (or r ) represents the displacement from the equilibrium bond length between the two balls. Thus, a value of zero implies the equilibrium distance (i.e., the bottom of the potential energy well). Since all harmonic oscillator wave functions are either even or odd, all of the wave functions have < x > = 0. That is, they are all symmetric about the equilibrium length, and the average of all of our experiments to measure bond length will be the equilibrium bond length irrespective of what vibrational state we measure . As for < x 2 >, now life is not so easy, since parity says the integrals need not be zero. Let’s start with Ψ 0 .
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10-2 x 2 n = 0 = μ ! h " # $ $ % & 1/ 4 e ( μ x 2 /2 h x 2 () ) * μ ! h " # $ $ % & e ( μ x 2 / 2 h dx = μ ! h " # $ $ % & 1/ 2 x 2 e ( μ x 2 / h dx () ) * An integral table provides x 2 e ! ax 2 dx !" " # = 1 2 a $ a % & ( ) * 1/2 If we plug in the appropriate value of a we have x 2 n = 0 = k μ ! h " # $ $ % & h 2 μ " # $ $ % & ! h k μ " # $ $ % & = h 2 μ " # $ $ % & Let’s do a quick sanity check on this answer. It says that as the force constant k gets larger, the average of the squared displacement from the equilibrium bond length will get smaller. That makes sense: stiffer spring, less displacement as it vibrates. It also says that as the reduced mass goes up, the average displacement will be smaller. That too makes intuitive sense. Kinetic energy goes up with mass, so in the same vibrational period we will have to slow down (and thus go less far) to maintain the same energy if the mass increases. Now, what about the case for Ψ 1 ? There we have x 2 n = 1 = k μ 4 ! h " # $ $ % & 2 k μ h " # $ $ % & x ( ) * * * + , - - - e . μ x 2 / 2 h x 2 ./ / 0 μ 4 ! h " # $ $ % & 1 2 k μ h " # $ $ % & x ( ) * * * + , - - - e . μ x 2 / 2 h dx = μ 4 ! h " # $ $ % & 4 μ h " # $ $ % & x 4 e . μ x 2 / h dx ./ / 0 An integral table provides x 4 e ! 2 dx !" " # = 3 4 a 2 $ a % & ( ) *
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10-3 If we plug in the appropriate value of a we have x 2 n = 1 = μ 4 ! h " # $ $ % & 1/ 2 4 μ h " # $ $ % & 3 h 2 4 k μ " # $ $ % & ! h μ " # $ $ % & = 3 h 2 μ " # $ $ % & So, the average of the squared displacement is three times larger for n = 1 than for n = 0. We will look more closely at the wave functions themselves below, and this result will be consistent with the appearance of the probability densities.
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This note was uploaded on 12/14/2010 for the course CHEM 3502 taught by Professor Staff during the Spring '08 term at Minnesota.

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3502_10_octo02_CC - Chem 3502/4502 Physical Chemistry II...

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