Chapter 6 Problem Solutions

# We thus expand our wavefunction in the given a free

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Unformatted text preview: 2 ‰- x ë 2 a 2 pa The free particle Hamiltonian is given by ` H= `2 px 2m Printed by Mathematica for Students so its eigenstates are just the momentum eigenstates, with energy eigenvalues E p = p2 2m . We thus expand our wavefunction in the Given a free particle of mass m initially in the state 2 1 yHxL = Xx y\ = Ch6ProblemSetKey.nb 2 ‰- x ë 2 a 2 pa The free particle Hamiltonian is given by ` H= `2 px 2m so its eigenstates are just the momentum eigenstates, with energy eigenvalues E p = p2 2m . We thus expand our wavefunction in the momentum basis as X p x\ = X p y\ = Ÿ „ x X p x\ Xx y\ ¬༼ 1 2p— Xx y\ = yHxL = ‰ -Â px — 1 pa è yH pL = ¶ Ÿ -¶ 1 p —a 2p „x‰ px -Â 2 ‰- x ë 2 a — 2 This is the Fourier transform of a Gaussian, and thus we have è yH pL = a — ‰ p - a2 p2 2 —2 Thus, the time evolution of the particle state in the position basis is given by Xx ‰ Â` - Ht — ¶ y \ = Ÿ -¶ „ p X x ‰ ¶ yHx, tL = Ÿ -¶ „ p ‰ 1 = 1 — Â t — 2m p2 — 2m — p\ X p y\ è Xx p\ yH pL t ¶ a ÿ 2p— yHx, tL = - `2 px Â - Ÿ -¶ „ p ‰ p ¶ a 2p Ÿ -¶ „ p ‰ p Â Â p2 — 2m 1 px ‰ — - a2 2 —2 -K t + ‰ Â Ât m— px — ‰ - a2 p2 2 —2 O p2 This is the inverse Fourier transform of a Gaussian. Therefore x2 1 yHx, tL = 2 a2 1 + ‰ p Ba + J Â—t Â—t m a2 NF ma and x2 1 - a-J yHx, tL = Â—t ma p Ba2 + J yHx, tL 2 Â—t m a2 —t 2 a2 B 1 + ‰ ma —t 2 2 x2 - N ‰ 2 a2 B 1 + ma 2p a2 2 2 F NF 1 = —t...
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