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: () {} 2 2 2 ,e x p e x p 22 mm a x td x x x a x it ππ ⎧⎫ ′′ Ψ= ⎨⎬ ⎩⎭ == (2) This integral, of course, could be calculated out explicitly: () {} 2 2 2 2 , exp exp 21 exp 2 2 1 1 am m x x x x a x aa x ia t t m m ⎪⎪ = + + = = (3) And the corresponding 2 , x t Ψ is: 1/2 2 2 2 2 2 x p 2 2 1 1 x xt at at m m + + = = (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 at m →= + ± = Then equation (4) changes into 2 2 2 , exp 2 a x ta x ± ± (5) Comparing equation (5) with the 2 ,0 x Ψ obtained from equation (3): 2 2 2 exp 2 a x ax (6) you will find that the only change is ± .
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So after a very large time interval, aa << ± 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 >0 cases: () 2 2 22 ,0 1 , 4 xt d x xt x d x x t x a = = ± (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 td x x t x t ix 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 ia t m ω = + = , then we have aa ωω = ± and 2 += . {} 2 2 2 2 2 2 2 2 2 2 ,4 2 4, 2 , 1 2 2 2 p t dx x dx x t x dxx x t dx x t a a a a =− Ψ Ψ Ψ Ψ + Ψ + ⎛⎞ ⎜⎟ ⎝⎠ = ∫∫ = = == ± = ± = = (9) So from equation (7), (8) and (9) we have: 2 2 2 2 2 2 2 14 4 2 x p ma t xx am a pp a σ + = = = = ± = (10)
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(e): since we have: 22 2 2 2 4 44 2 xp ma t m aa am am σσ + => = == (11) So we see that the uncertainty principle holds and the system is closest to the uncertainty limit when t =0. Question 2: (a): Suppose the wave function of this particle at =0 is ( ) ,0 x Ψ , then it must vanish outside the potential well since the potential well is infinitely deep. From the normalization we have: () 2 0 1 a x dx Ψ= (1) Because the particle is equally likely to be found at any point of that region, so we have: 0 00 const for x a x for x or x a ≤≤ <> (2) So from equation (1) and (2) it is obvious: [] 1 0, x for x a a (3) (b): One could expand the wave function given above using the eigenstates of the particle in the infinite potential well: nn n xc Φ where 2 sin n nx π Φ= with corresponding eigen value 222 2 2 n n E ma = = .
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This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.

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

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