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3502_24_nove06_LG

3502_24_nove06_LG - Chem 3502/5502 Physical Chemistry...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 24, November 06, 2009 Solved Homework We are given a perturbing potential V = k ( 1 – x ) for a particle in an otherwise "normal" box of length 1. The first order correction to the ground-state energy is simply the expectation value of the perturbing operator for the exact eigenfunction that would exist in the absence of the perturbation. That is a 1 1 ( ) = ! 1 0 ( ) x ( ) k 1 " x ( ) ! 1 0 ( ) x ( ) where Ψ 1 is the ground state particle-in-a-box wave function ! 1 0 ( ) x ( ) = 2 sin " x ( ) 0 # x # 1 E 1 0 ( ) = " 2 2 = a 1 0 ( ) (note that although our formal derivations used subscript "0" to represent a ground state, for the particle in a box the ground state quantum number is n = 1). The energy of the ground state in the absence of the perturbation is the zero-order eigenvalue plus the first order correction. We may expand the expectation value above as a 1 1 ( ) = k ! 1 0 ( ) x ( ) ! 1 0 ( ) x ( ) " ! 1 0 ( ) x ( ) x ! 1 0 ( ) x ( ) # \$ % & = k 1 " 1 2 ( ) * = k 2 where in making the first set of simplifications we have made use of the normalization of the ground-state wave function (so the first integral is 1) and our prior knowledge that < x > for a particle in a box wave function is always L /2, which in this case is 1/2. Note that the result is completely intuitive. It says that, to first order, the ground state energy goes up by k /2. This is the value that the perturbing function has in the middle of the box, and it is what we would obtain as an exact answer if the bottom of the

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24-2 box had potential k /2 instead of the usual convention of zero. Since the perturbing
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3502_24_nove06_LG - Chem 3502/5502 Physical Chemistry...

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