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Unformatted text preview: Chem 5314 homework #3, out of 45 marks Problem 1 – 10 marks In class, we found that the stationary states of the 1d harmonic oscillator have the form ψ n = A n ( x n +? x n 2 +? x n 4 + ··· ) e αx 2 (1) where A n is a normalization constant, the polynomial in brackets ends with x or a constant if n is respectively odd or even, and where α = mω 2 ~ (2) We did not derive a general formula for the coefficients ( i.e. the ?’s) in the polynomial. These could , although its not very practical, be determined by orthogonality . For all of question 1, express all your answers and do all your work in terms of the parameter α only. a – 5 marks) In particular, the second excited state ψ 2 has the form ψ 2 = A 2 ( x 2 + c ) e αx 2 (3) Find the constant c by requiring that ψ 2 be orthogonal to the ground state ψ . Solution: 0 = Z ψ 2 ψ * = A A 2 Z ∞∞ ( x 2 + c ) e 2 αx 2 (4) which means that Z ∞∞ x 2 e 2 αx 2 = c Z ∞∞ e 2 αx 2 (5) Evaluating the integrals gives c = 1 4 α (6) b – 2 marks) ψ 2 is also orthogonal to the first excited state ψ 1 . Why? (hint: symmetry) Solution: The orthogonality integral is Z ∞∞ ψ * 1 ψ 2 = A 1 A 2 Z ∞∞ x ( x 2 + c ) e 2 αx 2 (7) 1 which is zero because the integrand is an odd function of x , the domain of integration is even, and the integral from 0 to ∞ converges. c – 3 marks) Determine the normalization constant A for the ground state. Solution: We require that 1 = Z ∞∞ A 2 e 2 αx 2 (8) Evaluating the integral, we find that A = 2 α π 1 / 4 (9) Problem 2 – 4 marks Consider the operator ˆ A = x d dx d dx x (10) What does this operator do to a function...
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This note was uploaded on 02/08/2012 for the course CHEM 5314 taught by Professor Nielsen during the Fall '11 term at University of Texas at Dallas, Richardson.
 Fall '11
 Nielsen
 Physical chemistry, pH

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