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Problem Set 4 Solution

# Problem Set 4 Solution - C hemistry 4 60 Dr Jean M Standard...

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Chemistry 460 Dr. Jean M. Standard Problem Set 4 Solutions 1. One of the important properties of Hermitian operators is that their eigenfunctions form a complete set. This means that any arbitrary function f x ( ) may be exactly expressed as a linear combination of eigenfunctions, f x ( ) = c n ψ n x ( ) n = 1 , where c n are the expansion coefficients and n x ( ) are the eigenfunctions. We showed in class that the expansion coefficients c n may be calculated using the relation c n = n * x ( ) f x ( ) −∞ dx . Assume that the function you wish to represent is a step function, shown in the figure below and defined by the relation f ( x ) = 0 C 0 x < 0 0 x L x > L Χ Ψ Ϊ Ω Ϊ Ϋ ά Ϊ έ Ϊ . x =0 x=L f(x)=C Here, C is a constant. Using the eigenfunctions of the particle in an infinite box, calculate analytically the first six linear expansion coefficients, c 1 , c 2 c 3 c 4 c 5 , and c 6 , for the step function. Recall that the eigenfunctions of the particle in an infinite box are n x ( ) = 2 L sin n π x L Λ Ν Μ Ξ Π Ο , 0 x L 0, x < 0, x > L Χ Ψ Ϊ Ϊ Ω Ϊ Ϊ Ϋ ά Ϊ Ϊ έ Ϊ Ϊ , where n is the quantum number and L is the width of the box. Using the six coefficients you determined, construct a representation of the step potential for the case in which L =2 and C =1. Note that since the expansion was truncated after only six terms, it is an approximation rather than an exact result.

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2 1. continued The coefficients c n to be calculated are c n = ψ n * x ( ) f x ( ) −∞ dx . Substituting the form of the function f x ( ) and the particle in a box wavefunctions yields c n = C 2 L sin n π x L Λ Ν Μ Ξ Π Ο 0 L dx . Here, since the particle in a box wavefunctions are 0 outside the range 0 x L , the limits of integration are reduced to 0 to L . Also, on that range, the function f x ( ) equals a constant C , which can be pulled out of the integral. The integral to be evaluated is sin n x L Λ Ν Μ Ξ Π Ο 0 L dx = L n cos n x L Λ Ν Μ Ξ Π Ο 0 L = L n cos n ( ) cos 0 ( ) [ ] sin n x L Λ Ν Μ Ξ Π Ο 0 L dx = L n cos n ( ) 1 [ ] . Note that this integral equals 0 for n even, and is non-zero for n odd, sin n x L Λ Ν Μ Ξ Π Ο 0 L dx = 0, n even 2 L n , n odd Χ Ψ Ϊ Ϊ Ω Ϊ Ϊ Ϋ ά Ϊ Ϊ έ Ϊ Ϊ . Therefore the coefficients are c n = 0, n even 2 LC n 2 L , n odd Χ Ψ Ϊ Ϊ Ω Ϊ Ϊ Ϋ ά Ϊ Ϊ έ Ϊ Ϊ .
3 1. continued The analytical forms of the first six coefficients in the expansion are given below, along with numerical results for the specific case L =2, C =1. n c n (analytic) c n (numerical) 1 2 LC π 2 L 1.2732 2 0 0 3 2 LC 3 2 L 0.4244 4 0 0 5 2 LC 5 2 L 0.2546 6 0 0 Using the expansion coefficients listed above, the step function f x ( ) with L =2 and C =1 may be approximately expanded as f x ( ) 1.2732 sin x 2 Λ Ν Μ Ξ Π Ο + 0.4244 sin 3 x 2 Λ Ν Μ Ξ Π Ο + 0.2546 sin 5 x 2 Λ Ν Μ Ξ Π Ο .

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Problem Set 4 Solution - C hemistry 4 60 Dr Jean M Standard...

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