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Unformatted text preview: 4.5. THE HYDROGEN MOLECULAR ION 55 Now evaluate the expectation energy: ( E ) = ( ax ( ℓ − x ) | H | ax ( ℓ − x ) ) = | a | 2 (Bigg x ( ℓ − x ) vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle − ¯ h 2 2 m ∂ 2 ∂x 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle x ( ℓ − x ) )Bigg You can substitute in the value of | a | 2 from the normalization requirement above and apply the Hamiltonian on the function to its right: ( E ) = 30 ℓ 5 ¯ h 2 m ( x ( ℓ − x ) | 1 ) The inner product is by definition the integral integraltext ℓ x ( ℓ − x ) d x , which was given to be ℓ 3 / 6. So the final expectation energy is ( E ) = ¯ h 2 10 2 mℓ 2 versus ¯ h 2 π 2 2 mℓ 2 exact. The error in the approximation is only 1.3%! That is a surprisingly good result, since the parabola ax ( ℓ − x ) and the sine a ′ sin( πx/ℓ ) are simply different functions. While they may have superficial resemblance, if you scale each to unit height by taking...
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This note was uploaded on 11/13/2011 for the course PHY 4458 taught by Professor Garvin during the Fall '11 term at University of Florida.
- Fall '11