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homework2_q2

# homework2_q2 - How do they apply(a dx|y(x)|2 = |A|2 dx...

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Sheet1 Page 1 Solution problem 2: An electron is moving freely in the x-direction. At t = 0 the electron is described by the wave function (neglect spin) y(x) = Aexp{-x2/2b2}exp{ip0x/(h/2p )}. (a) Compute the constant A such that - + dx|y(x)|2 = 1. � � � (b) Compute Dx at t = 0. (c) Compute Dp at t = 0 and show that DxDp = h/2 for the electron. (d) Assume that the electron has a position uncertainty of Dx = 10-10m. Compute its velocity uncertainty compared to the speed of light. Solution: Concepts, principles, relations that apply to the problem: The mean value and the root mean square deviation of an observable Why do they apply? The expression for the mean value of an observable A in the normalized state |y> is <A> = <y|A|y>. If |y> is not normalized th The root mean square deviation DA characterizes the dispersion of the measurement around <A>.
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Unformatted text preview: How do they apply? (a) -+ dx|y(x)|2 = |A|2 -+ dx exp{-x2/b2} = |A|2bp1/2. |A|2 = 1/(bp1/2). (b) Dx = (<x2> - <x>2))1/2. <x> = 0 from symmetry. <x2> = -+ x2|y(x)|2dx = -+ x2exp{-x2/b2} dx = b2/p1/2-+ x'2exp{-x'2} dx' = b2/2. Dx = b/21/2. Details of the calculation: (c) Dp = (<p2> - <p>2))1/2. p2y(x) = [(p0 + ihx/b2)2 + h2/b2]y(x). <p> = -+ dx(p0 + ihx/b2)|y(x)|2 = p0. <p2> = -+ dx[(p0 + ihx/b2)2 + h2/b2]|y(x)|2 Err:510 Dp = h/(21/2b), DxDp = (b/21/2) h/(21/2b) = h/2. (d) Dv/c = Dp/(mec) = h/(2mecDx) = 1.05*10-34/[2*9.1*10-31*3*108*10-10] = 1.9*10-3. (me = 9.1h10-31kg, h =1.05h10-34J-s, c = 3h108m/s). DA h (<(A - <A>2)>)1/2 = (<A2> - <A>2))1/2. py(x) =(h/i)(h/hx)y(x) = (p0 + ihx/b2)y(x)....
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