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Daily_014.SOLN

# Daily_014.SOLN - x 2/a 2 2 dx = A 2 Z ∞-∞ e-x 2/a 2 2...

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Phys 274 – c Eric B. Szarmes 18 February 2011 Daily Homework 14 SOLUTIONS Q6S.8 Graphs of the wavefunction and its square are shown below. The wavefunction is ψ ( x ) = A sin(3 πx/L ) for 0 x L , and ψ ( x ) = 0 for x < 0 and x > L . A ! (x) x 0 L A 2 | ! (x) | 2 x 0 L L L 1 3 2 3 Note that this wavefunction contains 3 half-cycles of the sine curve between x = 0 and x = L , and its square contains 3 lobes between those limits. By definition, if A is chosen so that the wavefunction is normalized, then the area under the | ψ ( x ) | 2 -curve between 0 and L is unity. But this means that the area under the middle lobe (which equals the probability to locate the quanton between L/ 3 and 2 L/ 3) is just 1/3: Prob { 1 3 L < x < 2 3 L } = 1 3 Q6S.9 a) We are given ψ ( x ) = A e - x 2 /a 2 . For this wavefunction to be normalized, we require Z -∞ | ψ ( x ) | 2 dx = 1 . (1) Therefore, 1 = Z -∞
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Unformatted text preview: x 2 /a 2 ) 2 dx = A 2 Z ∞-∞ ( e-x 2 /a 2 ) 2 dx = A 2 a r π 2 = A 2 r πa 2 2 (2) and we must then have A 2 = r 2 πa 2 , so that A = h 2 πa 2 i 1 / 4 . (3) b) The speciﬁed interval (from-0.1 nm to +0.1 nm) is much smaller than the range over which the exponential varies appreciably (which is on the order of a = 1.5 nm). Therefore, the wavefunction may be taken to be constant across that small interval (equal to its value at x = 0) and removed from the integral. We then have Probability = Z +0 . 1 nm-. 1 nm | ψ ( x ) | 2 dx ≈ | ψ (0) | 2 Z +0 . 1 nm-. 1 nm dx = A 2 × 2(0 . 1 nm) = r 2 πa 2 × 2(0 . 1 nm) = r 2 π 1 . 5 2 × 2(0 . 1) = 0 . 1064 . (4) 1...
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