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Unformatted text preview: 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 cos 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 2 = c 2 λ 2 so that λ 2 d ν d λ = c 2 νλ = c 1 ( ν / ν ) 2 3. We may use the formula for v g derived above for ν = 2 π T ρ λ 3/2 to calculate 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 1030 kg and w (0) = 106 m, the calculation yields w (1) = 1.7 x (1) = 1....
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 Spring '05
 mokhtari
 Schrodinger Equation, wave function, Somewhat different choices

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