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# HW 03(1) - EEE 352Spring 2013 Homework 3Due January 28 2.12...

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EEE 352—Spring 2013 Homework 3—Due January 28 2.12 When the uncertainty principle is considered, it Is not possible to locate a photon in space more precisely than about one wavelength. Consider a photon with wavelength λ = 1 μ m. What is the uncertainty in the photon’s momentum? From the problem definition, we take Δ x = 10 6 m and from the footnote on p. 30, we have Δ p = 2 Δ x = 5.27 × 10 29 kg m / s . 2.18 Consider the wave function Ψ ( x ) = A cos n π x ( ) e i ω t , 1 2 x 1 2 . Determine A such that the wave function is properly normalized . The normalization is given such that 1 = Ψ ( x ) 2 dx 1/2 1/2 = A 2 cos n π x ( ) e i ω t 2 dx 1/2 1/2 = A 2 cos 2 n π x ( ) dx 1/2 1/2 = A 2 2 1 + cos 2 n π x ( ) \$ % & ' dx 1/2 1/2 = A 2 2 x + 1 2 n π sin 2 n π x ( ) \$ % ( & ' ) 1/2 1/2 = A 2 2 A = 2 . 2.20 An electron is given by a wave function given by ψ ( x ) = 2 a cos π x a ! " # \$ % & , a 2 x a 2 . The wave function is zero elsewhere. Calculate the probability of finding the electron between (a) 0 < x < a/4, (b) a/4 < x < a/2, (c) –a/2 < x < a/2. The expectation value of x is given by

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x = 2 a cos π x a ! " # \$ % & x cos π x a ! " # \$ % & dx x 1 x 2 = 1 a 1 + cos 2 π x a ! " # \$ % & ( ) * + , - xdx x 1 x 2 = 1 a x x 1 x 2 x a a 2 π sin 2 π x a ! " # \$ % & x 1 x 2 + 1 2 π sin 2 π x a ! " # \$ % & dx x 1 x 2 = x 2 x 1 a 1 2 π x 2 sin 2 π x 2 a ! " # \$ % & x 1 sin 2 π x 1 a ! " # \$ % & ( ) * + , - + 1 2 π ( ) a 2 π cos 2 π x a ! " # \$ % & x 1 x 2 = x 2 x 1 a 1 2 π x 2 sin 2 π x 2 a ! " # \$ % & x 1 sin 2 π x 1 a ! " # \$ % & ( ) * + , - + a 2 π ( ) 2 cos 2 π x 2 a
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