137A_HW04_solutions.pdf - Physics 137A Problem Set 4...

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Physics 137A: Problem Set 4 solutions February 18, 2017 Problem 1 (a) The momentum space wavefunction is given by φ ( k ) = 1 2 π Z -∞ ψ ( x ) e - ikx dx (1) For the infinite square well the energy eigenstates are ψ n ( x ) = r 2 L sin ( k n x ) where k n = L (2) Or writing the sine as complex exponential ψ n ( x ) = r 2 L 1 2 i h e ik n x - e - ik n x i (3) and so we have φ ( k = 1 2 π r 2 L 1 2 i Z -∞ h e ik n x - e - ik n x i e - ikx dx (4) φ ( k ) = 1 1 2 i Z -∞ h e i ( k n - k ) x - e - i ( k n + k ) x i dx (5) φ ( k ) = 1 1 2 i " e i ( k n - k ) x i ( k n - k ) - e - i ( k n + k ) x i ( k n + k ) # L 0 (6) φ ( k ) = - 1 e - ikx 2 e ik n x ( k n - k ) - e - ik n x ( k n + k ) L 0 (7) Getting a common denominator φ ( k ) = - 1 2 e - ikx ( k 2 n - k 2 ) h e ik n x ( k n + k ) - e - ik n x ( k n - k ) i L 0 (8) φ ( k ) = - 1 2 e - ikx ( k 2 n - k 2 ) [2 k n cos( k n x ) - 2 ik sin( k n x )] L 0 (9) 1
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From the boundary conditions in the infinite square well we have k n = nπ/L so that sin( k n L ) = 0 and cos( k n L ) = ( - 1) n φ ( k ) = 1 2 k n k 2 n - k 2 h 1 ± e - ikL i (10) where we have + if n = 1 , 3 , 5 , ... and - if n = 2 , 4 , 6 , ... (b) We have | φ ( k ) | 2 = 1 4 k 2 n ( k 2 n - k 2 ) 2 1 ± e - ikL 1 ± e ikL (11) | φ ( k ) | 2 = 1 4 k 2 n ( k 2 n - k 2 ) 2 2 ± ( e ikL + e - ikL ) (12) | φ ( k ) | 2 = 1 2 k 2 n ( k 2 n - k 2 ) 2 (1 ± cos( kL )) (13) (c) For n = 1, we have k n = π/L | φ ( k ) | 2 = πL 2( π 2 - k 2 L 2 ) 2 (1 + cos( kL )) (14) Using wolfram alpha online this plot (using a variable x = kL ) shows that the function peaks at k = 0 (d) The term (1 + cos( kL )) goes to zero at k = π/L . However, the denominator also goes to zero at k = π/L . The plot above shows that the function stays non-zero at k = π/L (as can be shown mathematically by L’Hopital’s rule) and the first zero is at k = 3 π/L . However, since we are just looking for a rough estimate here, it would be fine to take k = π/L as an estimate of the width. 2
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Making a rough measure that σ k 3 π/L (the distance between 0 and the first zero) we have the momentum uncertainty is ~ times this or σ p ~ 3 π/L . It would also be fine if we estimated the momentum as the distance between the zeros on both sides of the plot σ p 6 ~ π/L . The main
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