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Unformatted text preview: Physics 212 Classical and Modern Physics Spring 2008 Solutions for HandIn Set #9 A62 — Uncertainty The uncertainty in the x position is given in the prob lem statement: Δ x = 10 12 m. This implies a mini mum uncertainty in the x component of the momentum given by the Heisenberg Uncertainty Principle: Δ x Δ p ≥ ¯ h ⇒ Δ p ≥ ¯ h Δ x . The mass is certain, so this can be rewritten as m Δ v ≥ ¯ h Δ x , or Δ v ≥ ¯ h m Δ x ≥ 1 . 06 × 10 34 J · s (0 . 04 kg)(10 12 m) ≥ 2 . 6 × 10 21 m/s. A65 — Schr¨odinger equation for classically forbidden situation (a) Rearranging the Schr¨ odinger equation gives d 2 ψ ( x ) dx 2 = 2 m ¯ h 2 ( E U ) ψ ( x ) . (b) For the first trial solution we have ψ 1 ( x ) = Ax 2 dψ 1 dx = 2 Ax d 2 ψ 1 dx 2 = 2 A. Plugging these into the rearranged Schr¨ odinger equa tion gives 2 A ? = 2 m ¯ h 2 ( E U ) Ax 2 . where I have used the symbol ? = to emphasize that were are tying to see if the function ψ 1 “works,” i.e., makes the equality true for all values of x . In this case ψ 1 is not a soluton. The lefthand side is a positive constant, and the righthand side is quadratic in x ; these clearly can’t be equal for all x . The function ψ 1 ( x ) is not a solution. (c) For the second trial solution we have ψ 2 ( x ) = B sin kx dψ 2 dx = Bk cos kx d 2 ψ 1 dx 2 = Bk 2 sin kx. Plugging these into the rearranged Schr¨ odinger equa tion gives Bk 2 sin kx ? = 2 m ¯ h 2 ( E U ) B sin kx. To see if ψ 2 is a soluton, first cancel all common factors, leading to k 2 ? = 2 m ¯ h 2 ( E U ) . In a classically forbidden region, U > E , so ( E U ) < 0. This means that the righthand side is a positive con stant. But the lefthand side is a negative constant (for real k ), and the two sides can’t be equal. Therefore the function ψ 2 is not a solution. (You could argue that ψ 2 is a solution if k is an imaginary number. This actu ally leads to solutions like ψ 3 . If you’re interested in pursuing this further, talk to your instructor.) (d) For the third trial solution we have ψ 3 ( x ) = Ce κx→ dψ 3 dx = Cke κx→ d 2 ψ 1 dx 2 = + Ck 2 e κx ....
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This homework help was uploaded on 04/11/2008 for the course PHYS 212 taught by Professor Ladd during the Spring '08 term at Bucknell.
 Spring '08
 Ladd
 Physics

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