1081_PartUniversity Physics Solution

1081_PartUniversity Physics Solution - 39-6 Chapter 39...

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39-6 Chapter 39 (b) The probability of finding the particle is zero where 2 0, ψ = which occurs where sin 0 kx = and 2 / , 0, 1, 2, kx x n n πλ π == = / 2, 0, 1, 2, so 0, / 2, , 3 / 2, xn n x λλ λ = …… EVALUATE: The situation is analogous to a standing wave, with the probability analogous to the square of the amplitude of the standing wave. 39.30. sin , ψω t ∗∗ Ψ= so 22 ** 2 2 sin sin ψψ ω t t Ψ=ΨΨ= = . 2 Ψ is not time-independent, so Ψ is not the wavefunction for a stationary state. 39.31. IDENTIFY: To describe a real situation, a wave function must be normalizable. SET UP: | | 2 dV is the probability that the particle is found in volume dV . Since the particle must be somewhere , must have the property that | | 2 dV = 1 when the integral is taken over all space. EXECUTE: (a) In one dimension, as we have here, the integral discussed above is of the form 2 |( ) | 1 xd x −∞ = . (b) Using the result from part (a), we have () 2 2 2 2 ax ax ax e ed x x a ∞∞ = ∫∫ . Hence this wave function cannot be normalized and therefore cannot be a valid wave function. (c) We only need to integrate this wave function of 0 to because it is zero for x < 0. For normalization we have 2 1| | dx = = 2 2 00 0 bx bx bx Ae A Ae dx A e dx bb −− = , which gives 2 1 2 A b = , so 2 Ab = . EVALUATE: If b were positive, the given wave function could not be normalized, so it would not be allowable. 39.32. (a) The uncertainty in the particle position is proportional to the width of ψ x , and is inversely proportional to α . This can be seen by either plotting the function for different values of , finding the expectation value 2 x ψ xdx = for the normalized wave function or by finding the full width at half-maximum. The particle’s uncertainty in position decreases with increasing . The dependence of the expectation value 2 x ⟨⟩ on may be found by considering 2 2 2 2 x x xe dx x x −∞ −∞ = = 2 2 1 ln 2 x x −∞ 2 11 1 ln , 24 2 u u αα −∞ =− = ⎢⎥ ⎣⎦ where the substitution ux = has been made. (b) Since the uncertainty in position decreases, the uncertainty in momentum must increase. 39.33. * (,) a n d (,) xi y y fxy f xy y y ⎛⎞ −+ ⎜⎟ +− ⎝⎠ 2 * 1. y xi y ff f y = 39.34. The same. 2 * (,,) (,,)(,,) ψ xyz ψ ψ = 2 * ( , , ) ( ( , , )) ( ( , , ii i ψ xyze ψ ψ φφ φ = * . ψ ψ = The complex conjugate means convert all i ’s to– i ’s and vice-versa. ee ⋅= 39.35. IDENTIFY: To describe a real situation, a wave function must be normalizable. SET UP: | | 2 dV is the probability that the particle is found in volume dV . Since the particle must be somewhere , must have the property that | | 2 dV = 1 when the integral is taken over all space. EXECUTE: (a) For normalization of the one-dimensional wave function, we have 2 | dx = = ( ) 2 2 bx bx bx bx Ae dx Ae dx A e dx A e dx += + . 0
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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1081_PartUniversity Physics Solution - 39-6 Chapter 39...

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