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ps06soln - 1 Physics 316 Cornell University Solution for...

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1 Physics 316 Solution for homework 6 Spring 2005 Cornell University I. EXERCISE 1 Compute the position uncertainty Δ x in the ground state Φ 0 ( x ) = 1 p πa e - ξ 2 2 , ξ = x a , a = r ¯ h 0 . (1.1) for the harmonic oscillator potential. Use Δ x = p h ( x - h x i ) 2 i = q h x 2 i - h x i 2 and the expectation values h x i = Z -∞ x | Φ 0 ( x ) | 2 dx , x 2 fi = Z -∞ x 2 | Φ 0 ( x ) | 2 dx . (1.2) (prepared by Steve Drasco) First we will compute the average position h x i h x i = Z -∞ dx x | Φ 0 | 2 = 0 , (1.3) since x | Φ 0 | 2 is an odd function, and the integration boundaries are symmetric about the origin. Here we have used a general rule which applies to any odd function f ( - x ) = - f ( x ) Z A - A dx f ( x ) = Z 0 - A dx f ( x ) + Z A 0 dx f ( x ) , = - Z 0 A dx f ( - x ) + Z A 0 dx f ( x ) , = - Z A 0 dx f ( x ) + Z A 0 dx f ( x ) , = 0 . (1.4) We now compute x 2 fi x 2 fi = Z -∞ dx x 2 | Φ 0 | 2 , = 1 a π Z -∞ dx x 2 e - Ax 2 , (1.5) where we have defined A = a - 2 . Now note that x 2 e - Ax 2 = - ∂A e - Ax 2 , (1.6) so that we have x 2 fi = - 1 a π ∂A Z -∞ dx e - Ax 2 , = - 1 a ∂A A - 1 / 2 , = a 2 2 , (1.7) 2 where we have used the known Gaussian integral Z -∞ dx e - Ax 2 = r π A . (1.8) To prove this relation let I = R -∞ e - Ax 2 . Then we have I 2 = Z -∞ dx Z -∞ dy e - A ( x 2 + y 2 ) , = Z 0 dr Z 2 π 0 rdφ e - Ar 2 , = 2 π Z 0 dr re - Ar 2 , = π/A, (1.9) where in the second line we have converted from cartesian coordinates with area element dxdy , to polar coordinates with area element rdφdr . These results, Eqs. (1.3) and (1.7), give Δ x = q h x 2 i - h x i 2 = a 2 0 . 7 a. (1.10) So on average, measurements should find the particle at the origin, and these measurements should fluctuate on a scale of about 70% of the ground state’s classical maximum extension a . II. EXERCISE 2 What is the expectation value h V i and the uncertainty Δ V = p h V 2 i - h V i 2 of the harmonic oscillator potential V = 1 2 2 0 x 2 for the ground state wave function? This corresponds to the fact that the energy of a classical harmonic oscillator is on average split in equal parts between kinetic an potential energy.
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