ph2a_hw9_soln

# ph2a_hw9_soln - Quiz Problem 5 Solution Problem 3 Solution...

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Problem 3 Solution a) The condition that the wavefunction is normalized is: 1 = Z 0 | ψ ( x ) | 2 dx = Z 0 | A | 2 ± x x o ² 2 b e - 2 x/x o dx = | A | 2 x o 2 b Z 0 x 2 b e - 2 x/x o dx Using the hint (with a = 2 /x o ), this is equal to: = | A | 2 x o 2 b (2 b )! (2 /x o ) 2 b +1 = | A | 2 (2 b )! x o 2 2 b +1 For this to equal one, we must take (picking the phase of A so it’s real and positive): A = s 2 2 b +1 (2 b )! x o b) By deﬁnition: < x m > = Z 0 x m | ψ ( x ) | 2 dx = Z 0 x m | A | 2 ± x x o ² 2 b e - 2 x/x o dx = | A | 2 x o 2 b Z 0 x 2 b + m e - 2 x/x o dx = | A | 2 x o 2 b (2 b + m )! (2 /x o ) 2 b + m +1 Using the value of A we calculated above: 1

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= 2 2 b +1 (2 b )! x o 1 x o 2 b (2 b + m )! (2 /x o ) 2 b + m +1 = (2 b + m )! (2 b )! ± x o 2 ² m c) We have: Δ x = p < x 2 > - < x > 2 Using the result from the last part: = s (2 b + 2)! (2 b )! ± x o 2 ² 2 - ³ (2 b + 1)! (2 b )! x o 2 ´ 2 = x o 2 p (2 b + 2)(2 b + 1) - (2 b + 1) 2 = x o 2 p (2 b + 1) ((2 b + 2) - (2 b + 1)) = x o 2 2 b + 1 d) The time-independent Schrodinger equation is: - ¯ h 2 2 m d 2 ψ dx 2 ( x ) + V ( x ) ψ ( x ) = ( x ) Rearranging: ¯ h 2 2 m d 2 ψ dx 2 ( x ) = ( V ( x ) - E ) ψ ( x ) In this case, it’s easy to compute: d 2 ψ dx 2 ( x ) = A ³ x x o ´ b ³ b ( b - 1) x 2 - 2 b xx o + 1 x o 2 ´
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## This note was uploaded on 01/28/2011 for the course PH 2 taught by Professor Dudko during the Spring '09 term at UCSD.

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ph2a_hw9_soln - Quiz Problem 5 Solution Problem 3 Solution...

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