ch02 - CHAPTER 2 1 We have(x dkA(k)e ikx dk N k 2 2 e ikx...

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CHAPTER 2 1. We have ψ ( x ) = dkA ( k ) e ikx -∞ = dk N k 2 + α 2 e ikx -∞ = dk N k 2 + α 2 co s kx -∞ because only the even part of e ikx = cos kx + i sin kx contributes to the integral. The integral can be looked up. It yields ψ ( x ) = N π α e - α | x | so that | ψ ( x )| 2 = N 2 π 2 α 2 e - 2 α | x | If we look at | A ( k ) 2 we see that this function drops to 1/4 of its peak value at k α .. We may therefore estimate the width to be k = 2 α . The square of the wave function drops to about 1/3 of its value when x =± 1/2 α. This choice then gives us k x = 1. Somewhat different choices will give slightly different numbers, but in all cases the product of the widths is independent of α . 2. the definition of the group velocity is v g = d ϖ dk = 2 π d ν 2 π d (1 / λ ) = d ν d (1 / λ ) =- λ 2 d ν d λ The relation between wavelength and frequency may be rewritten in the form ν 2 - ν 0 2 = c 2 λ 2 so that - λ 2 d ν d λ = c 2 νλ = c 1 - ( ν 0 / ν ) 2 3. We may use the formula for v g derived above for ν = 2 π T ρ λ - 3/2 to calculate

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v g =- λ 2 d ν d λ = 3 2 2 π T ρλ 4. For deep gravity waves, ν = g /2 π λ - 1 /2 from which we get, in exactly the same way v g = 1 2 λ g 2 π . 5. With ϖ = k 2 /2 m , β = / m and with the original width of the packet w(0) = 2 α , we have w ( t ) w (0 ) = 1 + β 2 t 2 2 α 2 = 1 + h 2 t 2 2 m 2 α 2 = 1 + 2 h 2 t 2 m 2 w 4 (0 ) (a) With t = 1 s, m = 0.9 x 10 -30 kg and w (0) = 10 -6 m, the calculation yields w (1) = 1.7 x 10 2 m With w (0) = 10 -10 m, the calculation yields w (1) = 1.7 x 10 6 m.
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