1081_PartUniversity Physics Solution

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online