3502_08_sept28_CC

# 3502_08_sept28_CC - Chem 3502/4502 Physical Chemistry...

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Chem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 8, September 28, 2009 Guest Lecturer Prof. Christopher J. Cramer Solved Homework Evaluate < x > and < x 2 > for a particle-in-a-box wave function. These expectation values are, generically x = 2 L sin n ! x L " # \$ % & ( ) * + * x sin n ! x L " # \$ % dx 0 L , 2 = 2 L sin n ! x L " # \$ % & ( ) * + * x 2 sin n ! x L " # \$ % dx 0 L , Noting that the complex conjugate in this case is simply the sine function (since the function is real) and using the trigonometric identity given as a hint in the assignment, these simplify to x = 1 L x 1 ! cos 2 n " x L # \$ % & ( ) * + , dx 0 L - 2 = 1 L x 2 1 ! cos 2 n " x L # \$ % & ( ) * + , dx 0 L - Additional simplification leads to x = 1 L xdx 0 L ! " x cos 2 n # x L \$ % & ( ) dx 0 L ! * + , - . / = L 2 " 1 L x cos 2 n # x L \$ % & ( ) dx 0 L ! x 2 = 1 L x 2 dx 0 L ! " x 2 cos 2 n # x L \$ % & ( ) dx 0 L ! * + , - . / = L 2 3 " 1 L x 2 cos 2 n # x L \$ % & ( ) dx 0 L ! The remaining integrals can be solved by integration by parts (or by reference to a good integral table). To demonstrate the former, let us consider the first case

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8-2 x cos 2 n ! x L " # \$ % dx 0 L & Integration by parts uses the relationship udv ! = uv " vdu ! If we choose u = x v = L 2 n ! sin 2 n ! x L " # \$ % du = dx dv = cos 2 n ! x L " # \$ % we have x cos 2 n ! x L " # \$ % & dx 0 L ( = xL 2 n ! sin 2 n ! x L " # \$ % & 0 L ) L 2 n ! sin 2 n ! x L " # \$ % & dx 0 L ( = 0 ) 0 ) L 2 4 n 2 ! 2 cos 2 n ! x L " # \$ % & 0 L = ) L 2 4 n 2 ! 2 + L 2 4 n 2 ! 2 = 0 The final result, then, is that x = L 2 Happily, this is a completely intuitive result. The average position is the center of the box! If we look at the plotted wave functions from last lecture, this should be obvious, as all have square moduli that are symmetric about the box center. To solve the integral remaining in < x 2 > requires two successive integrations by parts. The calculus is straightforward, if tedious, and is not shown here. An integral table can also be used to find x 2 cos ax ( ) dx ! = 2 x cos ax a 2 + a 2 x 2 " 2 a 3 sin ax If we plug in the appropriate values for a and evaluate over the integration limits 0 to L we have the remaining contribution to < x 2 > and our final result is
8-3 x 2 = L 2 3 ! L 2 2 n 2 " 2 The classical result is the first term on the r.h.s., which makes sense (the probability is uniform across the entire box (that is, P ( x ) = 1/ L ) for the classical case, but not the quantum case—when n approaches infinity, the rapid oscillation of the wave function again makes the probability essentially uniform and, as it should be, we see that the classical limit is recovered). The appearance of n in this expectation value reflects the highly non -uniform probability distribution of the lower-energy particle-in-a-box wavefunctions, which have reduced expectation values for < x 2 >, since those wave functions have low probabilities near the walls and higher probabilities near the center. This effect is larger as L gets larger, since more and more of the distances far from the center are disadvantaged compared to distances near the center. [Note that these results would be slightly more intuitive if we had transformed our coordinate system by L /2, so that the center of the box was zero, but the physics doesn’t change.] Note that < x > 2 < x 2 > because the eigenfunctions of the Hamiltonian are not eigenfunctions of the position operator. Put differently (but equivalently) [

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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_08_sept28_CC - Chem 3502/4502 Physical Chemistry...

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