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# HWII - PHY4211 Quantum Mechanics I Fall Term of 2005...

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PHY4211 Quantum Mechanics I Fall Term of 2005 Homework Set II (A Possible Set of Solution, October 09, 2005) Question 1: (a): Normalization is easy to take as: 1/ 4 2 a A π = (1) (b): Using the propagator of a free particle it could be shown that: ( ) ( ) { } 1/4 2 2 2 , exp exp 2 2 m m a x t dx x x ax i t i t π π ⎫⎛ Ψ = ⎭⎝ = = (2) This integral, of course, could be calculated out explicitly: ( ) ( ) { } 1/ 4 2 2 1/ 4 2 2 , exp exp 2 2 2 1 exp 2 2 1 1 a m m x t dx x x ax i t i t a ax i at i at m m π π π Ψ = = + + = = = = (3) And the corresponding ( ) 2 , x t Ψ is: ( ) 1/2 2 2 2 2 2 1 2 , exp 2 2 1 1 a ax x t at at m m π Ψ = + + = = (4) (c): the Sketching is ignored here, instead I will give some qualitative explanation. To consider this problem simply, let us do some transformation: 2 2 1 a a a at m = + ± = Then equation (4) changes into ( ) { } 1/2 2 2 2 , exp 2 a x t ax π Ψ = ± ± (5) Comparing equation (5) with the ( ) 2 ,0 x Ψ obtained from equation (3): ( ) { } 1/2 2 2 2 ,0 exp 2 a x ax π Ψ = (6) you will find that the only change is a a ± .

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So after a very large time interval, a a << ± and the wave function remains Gaussian type. Since that a ± becomes very small, however, at the final stage the Gaussian wave function has been largely dispersed. (d): since at time t =0 we have 0 x = and ( ) 2 1/ 4 x a = for Gaussian wave function, then it is direct to generalize this result to t >0 cases: ( ) ( ) ( ) ( ) 2 2 2 2 , 0 1 , 4 x t dx x t x x t dx x t x a = Ψ = = Ψ = ± (7) However, the calculation of ( ) p t and ( ) 2 p t cannot be generalized in such a way, so I will do it as following: For ( ) p t , you see from the below equation: ( ) ( ) ( ) , , p t dx x t x t i x = Ψ Ψ = (8) that, the integrand is an odd function, and as a result, the above integral must be equal to 0. For ( ) 2 p t , let us assume that: 2 1 a i at m ω = + = , then we have aa ω ω = ± and 2 a a ω ω + = . ( ) ( ) ( ) ( ) { } ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 , , , 4 2 4 , 2 , 1 2 2 2 p t dx x t x t x dx x t x dxx x t dx x t a a a a ω ω ω ω ω ω ω ω ω ω = − Ψ Ψ = − Ψ = − Ψ + Ψ = − + = = = = = = = = = ± = ± = = (9) So from equation (7), (8) and (9) we have: 2 2 2 2 2 2 2 2 2 1 4 4 2 x p m a t x x am a p p a σ σ + = = = = = = ± = (10)