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!...
View
Full
Document
 Spring '07
 FIELDS
 mechanics, Force

Click to edit the document details