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Problem Set 5 Solutions

# Problem Set 5 Solutions - Chemistry 120A Spring 2009...

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Chemistry 120A, Spring 2009 Problem Set 5 Solutions Due March 11, 2009 Problems 1. Variational method applied to the particle in a box. Use the variational trial function, ψ = c 1 x 2 ( a - x )+ c 2 x ( a - x ) 2 , where c 1 and c 2 are the variational parameters, a is the box length, and let m be the particle mass. (a) Obtain an expression for the energy in terms of the variational parameters and obtain two approximate eigen- values. (b) Find and plot the corresponding eigenvectors. What states do they best approximate? 2. Discuss why (in terms of e.g. boundary conditions, smoothness properties, etc.) each of the following functions is or is not an appropriate (admissible) trial funciton for the particle in a box problem of question 1 (where k is a variational parameter). (a) The boundary conditions for particle in a box are ψ (0) = ψ ( a ) = 0. In addition, the function must be smooth and continuous. ψ = x ( a - x ) + kx 2 ( a - x ) 2 ψ (0) = 0 ( a - 0) + k 0 2 ( a - 0) 2 = 0 ψ ( a ) = a ( a - a ) + ka 2 ( a - a ) 2 = a · 0 + k 0 2 (0) 2 = 0 This function also has no sharp discontinuities or spikes, so it is a good trial wavefunction. (b) ψ = e k ( x - a 2 ) - e - ka 2 ψ (0) = e k ( 0 - a 2 ) - e - ka 2 = e - ka/ 2 - e - ka/ 2 = 0 ψ ( a ) = e k ( x - a 2 ) - e - ka 2 = e ka/ 2 - e - ka/s 6 = 0 Therefore, this is not a valid trial wavefunction, as it is not zero at ψ ( x )=a (c) The final proposed trial wavefunctions have a different form on the left and right sides of the box. Therefore, we have a new boundary condition: the wavefunctions must be equal at a 2 ψ = e k ( x - a 2 ) - e - ka 2 x 0 , a 2 ψ = e - k ( x - a 2 ) - e - ka 2 x a 2 , a ψ (0) = e k ( 0 - a 2 ) - e - ka 2 = e - ka/ 2 - e - ka/ 2 = 0 ψ ( a ) = e - k ( a - a 2 ) - e - ka 2 = e - ka/ 2 - e - ka/ 2 = 0 1

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However, this function must also have both sides match at a 2 ψ a 2 left = e k · 0 - e - ka/ 2 = 1 - e - ka/ 2 ψ a 2 right = e - k cot 0 - e - ka/ 2 = 1 - e - ka/ 2 So, they also meet in the center of the box, and are thus continuous. ∂ψ a/ 2 ∂x left = ke k ( a/ 2 - a/ 2) == ke 0 = k ∂ψ a/ 2 ∂x right = - ke - k ( a/ 2 - a/ 2 = - ke 0 = - k But, the first derivative is not continuous at the boundary, and therefore the wavefunction is not smooth, and thus not a good choice for a trial wavefunction. 3. Linear variational method applied to the hydrogen atom. Use the variational trial function: | ψ i = c 1 | ω 1 i + c 2 | ω 2 i To calculate the energy of the hydrogen atom via the following steps: (a) In atomic units, our basis functions are | ω 1 i = N 1 e - 3 r 2 / 2 and | ω 2 i = N 2 e - r 2 / 2 . Work out the values of N 1 and N 2 , and the overlap matrix. Normalize | ω 1 i 1 = h ω 1 | ω 1 i 1 = Z 0 r 2 drN 1 e - 3 r 2 / 2 · N 1 e - 3 r 2 / 2 1 = N 2 1 Z 0 drr 2 e - 3 r 2 1 = N 2 1 1 12 r π 3 N 1 = 2 3 3 / 4 π 1 / 4 ! Normalize | ω 2 i 1 = h ω 2 | ω 2 i 1 = Z 0 r 2 drN 2 e - r 2 / 2 · N 2 e - r 2 / 2 1 = N 2 2 Z 0 drr 2 e - r 2 1 = N 2 2 π 4 N 2 = 2 π 1 / 4 2
The S matrix (overlap matrix) is then given by: S = h ω 1 | ω 1 i h ω 1 | ω 2 i h ω 2 | ω 1 i h ω 2 | ω 2 i We already know h ω 1 | ω 1 i = h ω 2 | ω 2 i = 1, becuase we just normalized them. Also, the diagonal elements h ω 2 | ω 1 i = h 2 | ω 1 i , because there are no imaginary components in either | ω 1 i or | ω 2 i .

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