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3502_19_octo26_LG

3502_19_octo26_LG - Chem 3502/5502 Physical Chemistry...

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Unformatted text preview: Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 19, October 26, 2009 Solved Homework We seek ! x ; a ( ) p x ! x ; a ( ) = N 2 x a 1 " x ( ) " i h d dx # \$ % & ’ ( 1 ) x a 1 " x ( ) [ ] dx where N is the normalization constant on ξ . We could do this the usual way (take the derivative, solve all the integrals), but just for fun, let’s do something more general. Consider any real 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 where we have assumed without loss of generality that f is normalized over the integration interval. We can solve the integral using integration by parts. If we use u a b ! dv = uv a b " v a b ! du u = f x ( ) dv = df x ( ) dx dx du = df x ( ) dx dx v = f x ( ) then we may write f x ( ) df x ( ) dx ! " # \$ % & a b ’ dx = f x ( ) [ ] 2 a b ( f x ( ) df x ( ) dx ! " # \$ % & a b ’ dx or f x ( ) df x ( ) dx ! " # \$ % & a b ’ dx = 1 2 f x ( ) [ ] 2 a b 19-2 but, we know that a well-behaved wave function must go to zero at its integration endpoints, so the r.h.s. of the final equation is just 0. Thus, any real wave function has an expectation value of 0 for the momentum operator. We can think of this result as deriving from a superposition of left- and right-moving particle wave functions. If you think about this more deeply, it should now be clear why only the the m l = 0 wave functions of the spherical harmonics can be real: they have no z component angular momentum. Those with positive or negative values of m l are necessarily complex, so we can think of them as rotating, in some sense, either clockwise or counterclockwise, about appropriate axes. When we make linear combinations to get real wavefunctions, the expectation value of L z necessarily becomes zero—because the wave functions are real—and the linear combinations are no longer eigenfunctions of L z . Variational Calculations on the H Atom As we did with the particle in a box, it is often helpful to consider a system where the answer is known exactly in order to evaluate the utility of different variational protocols. Let us now consider the hydrogen atom. The question in which we are interested is, what if we were to replace the exact 1s wave function ! 100 r , " , # ( ) = Z 3/ 2 \$ e % Zr (19-1) with a different functional dependence on r , namely ! 1 s r , " , # ; \$ ( ) = Ne %\$ r 2 (19-2) where N is the normalization constant and α is a variational constant. We can find N quickly from ! 1 s r , " , # ; \$ ( ) ! 1 s r , " , # ; \$ ( ) = 1 = N 2 e % 2 \$ r 2 r 2 dr sin " d " d # 2 & ’ & ’ ( ’ = 4 & N 2 r 2 e % 2 \$ r 2 dr ( ’ = 4 & N 2 1 8 \$ & 2 \$ ) * + ,- ....
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3502_19_octo26_LG - Chem 3502/5502 Physical Chemistry...

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