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HWS453_4key

# HWS453_4key - University of Washington Department of...

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Unformatted text preview: University of Washington Department of Chemistry Chemistry 453 Winter Quarter 2010 Homework Assignment 4; Due at 5p.m. on 2/01/10 1) We learned that the Hamiltonian for the quantized harmonic oscillator is 2 2 2 2 ˆ 2 2 d x H dx κ µ = − + = . You can obtain the general expression for the energy, i.e. ( ) 1 2 n E n ω = + = , for n=0,1,2,… by solving Schrodinger’s equation ( ) ( ) ˆ H x E x ψ ψ = . Alternatively, one can guess the form of a trial wave function, and use the energy equation to obtain a refined wave function from the trial wave function and also determine the energy. a) Using a trial wave function of the form ( ) 2 1/2 x x e α α ψ π − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ , where α is a constant, calculate the energy E H d x dx = −∞ ∞ −∞ ∞ z z ψ ψ ψ 2 ¡ for the harmonic oscillator. Solution: ( ) ( ) 2 2 2 2 2 2 2 1/2 1/2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 8 2 ˆ 2 2 2 2 1 2 2 2 2 8 2 2 x x x x x x x d x e e d x H dx dx E dx e dx d x x e e dx x e dx dx e d x α α α α α α π α α α κ α ψ ψ µ π π α ψ π κ κ α α µ µ α κ π ∞ ∞ − − −∞ −∞ ∞ ∞ − −∞ −∞ ∞ ∞ − − − ∞ − ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ − + ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ = = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎡ ⎤ ⎡ ⎤ − + − − + ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ = = = − ∫ ∫ ∫ ∫ ∫ ∫ ∫ = = = = 2 2 2 2 2 2 2 2 2 2 2 8 2 1 2 8 2 8 2 1 2 4 x x e d x α α α α κ α π α π µ µ π µ α α µ α κ α α µ α µ ∞ − ⎡ ⎤ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ + = − + ⎢ ⎥ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎛ ⎞ = − + ⎢ ⎥ ⎜ ⎟ ⎝ ⎠ ⎣ ⎦ ∫ = = = = = b) Determine the value of α that minimizes the energy in part a. That is, using your expression for E from part a, set E α ∂ = ∂ and solve for α . Solution: 2 2 2 2 2 2 2 2 2 2 2 1 2 4 8 2 1 4 8 2 2 E E κ α α κ α α µ α µ α µ µ µκ κ α α α µ µ α κ µ ⎡ ⎤ ⎛ ⎞ = − + = − + ⎢ ⎥ ⎜ ⎟ ⎝ ⎠ ⎣ ⎦ ∂ = = − − + ⇒ = ⇒ = ∂ = = = = = = = = c) Using your result from part b, calculate the ground state energy of the harmonic oscillator and the ground state wave function. Compare these expressions to the expressions given in the text and in the notes obtained by solving Schroedinger’s equation. Solution: ( ) 2 2 1/2 1/2 2 1 2 x x x e e µκ α µκ α ψ π π − − ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = = The function is close to what is in the text but it is not normalized. The text has the normalized wave function: ( ) 2 1/2 /2 x x e α α ψ π − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ where µκ α = = . d) Repeat this procedure using the same trial wave function applied to the quartic oscillator which has the Hamiltonian 2 2 4 2 ˆ 2 2 d x H dx κ µ = − + = . Note the quartic oscillator differs from the harmonic oscillator by having the potential energy depend upon x 4 instead of x 2 ....
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HWS453_4key - University of Washington Department of...

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