Problem Set 4 Solution

4 1 note that this result is exactly half of the

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Unformatted text preview: ȹ ȹ 2 ȹ π Ⱥ k = . 4α = V ∞ ∫−∞ x 2 e−α x 2 1/ 2 1 ȹ π ȹ ȹ ȹ 2α ȹ α Ⱥ This result can be simplified further if desired, using the definition of α, € ȹ m k ȹ1 / 2 α = ȹ 2 ȹ . ȹ Ⱥ € dx 9 10 6 . continued Then, V k 4α = 1 = 4 V 1/ 2 ȹ 2 ȹ1 / 2 1 ȹ k ȹ ȹ k ȹ = ȹ ȹ ȹ m k ȹ 4 ȹ m Ⱥ ȹ Ⱥ 1/ 2 h 1 ȹ k ȹ ȹ ȹ . 4 2π ȹ m Ⱥ = Using the definition of the harmonic frequency, € 1/ 2 1 ȹ k ȹ ȹ ȹ , 2π ȹ m Ⱥ ν0 = the expectation value becomes € 1/ 2 h 1 ȹ k ȹ ȹ ȹ 4 2π ȹ m Ⱥ 1 = hν 0 . 4 V = V 1 Note that this result is exactly half of the ground state eigenvalue, E0 = hν 0 . 2 € Kinetic Energy Expectation Value € For a normalized wavefunction, the expectation value T of the kinetic energy is T ∞ * ∫−∞ ψ0 ( x) Tˆ ψ0 ( x) dx . = € 2 d 2 ˆ Substituting the form of the wavefunction and using that T = − , 2m dx 2 € T 1/ 2 2 ȹ α ȹ ȹ ȹ 2 m ȹ π Ⱥ € =− ∞ ∫−∞ e−α x 2 /2 d2 dx 2 e−α x 2 /2 Evaluating the second derivative, €...
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This document was uploaded on 12/05/2013.

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