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handin09

# handin09 - Physics 212 Classical and Modern Physics Spring...

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Physics 212 Classical and Modern Physics Spring 2008 Solutions for Hand-In 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 0 ) ψ ( x ) . (b) For the first trial solution we have ψ 1 ( x ) = Ax 2 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 0 ) 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 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 0 ) 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 0 ) . In a classically forbidden region, U 0 > E , so ( E - U 0 ) < 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 -→ 3 dx = - Cke - κx -→ d 2 ψ 1 dx 2 = + Ck 2 e - κx .

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