Hw6_ans - Chem 113A HW6 Solutions 1 The wavefunction for the ground state(n=0 for a quauntum harmonic oscillator is given by planckover2pi1 Ψ x =

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Unformatted text preview: Chem 113A HW6 Solutions 1. The wavefunction for the ground state (n=0) for a quauntum harmonic oscillator is given by planckover2pi1 : Ψ ( x ) = ( α π ) 1 4 e- 1 2 αx 2 For the average potential energy: < E potential > = integraltext Ψ * ( x ) ˆ E potential Ψ ( x ) dx = ( α π ) 1 2 integraltext ∞-∞ e- 1 2 αx 2 1 2 kx 2 e- 1 2 αx 2 dx = k 2 radicalbig α π integraltext ∞-∞ x 2 e- αx 2 dx = k radicalbig α π integraltext ∞ x 2 e- αx 2 dx From integration tables, we know the following integral: integraltext ∞ x 2 e- αx 2 dx = 1 4 α radicalbig π α Using this information, we find the average potential energy to be: < E potential > = k radicalbig α π 1 4 α radicalbig π α = k 4 α = 1 4 radicalBig k μ planckover2pi1 = 1 4 planckover2pi1 ω For the average kinetic energy: < E kinetic > = integraltext Ψ * ( x ) ˆ E kinetic Ψ ( x ) dx = radicalbig α π integraltext ∞-∞ e- 1 2 αx 2 ( − planckover2pi1 2 2 μ ∂ 2 ∂x 2 ) e- 1 2 αx 2 dx = − planckover2pi1 2 2 μ radicalbig α π integraltext ∞-∞ e- 1 2 αx 2 ∂ 2 ∂x 2 e- 1 2 αx 2 dx = − planckover2pi1 2 2 μ radicalbig α π integraltext ∞-∞ e- 1 2 αx 2 ∂ ∂x ( − αxe- 1 2 αx 2 ) dx = α planckover2pi1 2 2 μ radicalbig α π integraltext ∞-∞ e- 1 2 αx 2 ( e- 1 2 αx 2 − αx 2 e- 1 2 αx 2 ) dx = α planckover2pi1 2 μ radicalbig α π integraltext ∞ ( e- αx 2 − αx 2 e- αx 2 ) dx From integration tables, we know the following integral: integraltext ∞ x 2 e- αx 2 dx = 1 4 α radicalbig π α Using this and the previous integral, we find for the average kinetic energy < E kinetic > = α planckover2pi1 2 μ radicalbig α π ( 1 2 radicalbig π α − 1 4 radicalbig π α ) = 1 4 α planckover2pi1 2 μ = 1 4 radicalBig k μ planckover2pi1 = 1 4 planckover2pi1 ω 2. (a) integraltext Ψ * ( x, 0)Ψ( x, 0) dτ = 1 ⇒ integraltext [ c ( φ + φ 1 )] * [ c ( φ + φ 1 )] dτ = 1 c 2 integraltext ( φ * φ + φ * φ 1 + φ * 1 φ + φ * 1 φ 1 ) dτ = 1 ⇒ c 2 (1 + 0 + 0 + 1) = 1 ⇒ 2 c 2 = 1 ⇒ c = ± 1 √ 2 (b) For our system at t=0, there are two possible eigenstates based on the wavefunction given: the ground state and the first excited state. As a result, the corresponding energiesgiven: the ground state and the first excited state....
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This note was uploaded on 12/05/2011 for the course CHEM 113A taught by Professor Lin during the Fall '07 term at UCLA.

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Hw6_ans - Chem 113A HW6 Solutions 1 The wavefunction for the ground state(n=0 for a quauntum harmonic oscillator is given by planckover2pi1 Ψ x =

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