Unformatted text preview: . Solutions of this form correspond to states with definite energy since H op  Ψ > = E Ψ > . Substituting Ψ (x,t) into the time dependent equation yields ) ( ) ( 2 1 ) ( 2 2 2 2 2 x E x Kx dx x d m = + − h and hence ) ( 2 2 ) ( 2 2 2 2 2 = − + x x mf mE dx x d h h where I set K = m(2 π f) 2 . Setting h / 2 mf α = and 2 / 2 h mE = β yields ) ( ) ( ) ( 2 2 2 2 = − + x x dx x d . We must find the allowed solutions ( i.e. ψ (x) must be “squareintegrable”) of this differential equation. The differential equation can be converted into the “Hermite Differential Equation” and the solutions are “Hermite Polynomials” . We will not solve for the energy levels in this way, instead we will solve the problem using operators!...
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 Spring '07
 FIELDS
 mechanics, Energy, Force, Kinetic Energy, Potential Energy, Planck

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