Chem 113A HW3 Solutions - Chem 113A Homework 3 Solutions...

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Chem 113A Homework 3 Solutions Question 1 Consider a particle trapped in a 1-D box. Show mathematically that the uncertainty in the position of the particle is always less than the length of the box, for any value of n . Why does this make physical sense for a particle in a 1-D box? Solution For a 1-D particle in a box we have the following: h x i = L 2 and h x 2 i = L 2 3 - L 2 2 n 2 π 2 The uncertainty in the position, σ x , is then given by: σ x = p h x 2 i - h x i 2 = L 2 3 - L 2 2 n 2 π 2 - L 2 4 1 / 2 = L 1 12 - 1 2 π 2 n 2 1 / 2 We assume, to the contrary, that σ x L , then L 1 12 - 1 2 π 2 n 2 1 / 2 L 1 12 - 1 2 π 2 n 2 1 / 2 1 π 2 n 2 - 6 12 π 2 n 2 1 Rearranging terms we have that - 6 11 π 2 n 2 But this is a contradiction, since the RHS of the above inequality is always greater than zero for all possible values of n . Thus our initial assumption that σ x L must be wrong, whence, σ x < L for all n . This makes physical sense since the particle is confined, by the boundary conditions, to be within the box length L . Thus, an uncertainty greater than the length of the box would mean a possibility to find the particle outside the box, which we know to be wrong. 1
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Question 2 A classical moving particle trapped in a 1-D box is equally likely to be found at any point along the length of the box of length L. In other words, its probability distribution is given by p ( x ) dx = 1 L dx 0 x L Determine the uncertainty in the position, σ x , , of a classical particle in a box. Compare this to σ x for a quantum mechanical particle in a box. Illustrate the correspondence principle by showing that that these uncertainties agree in the limit that n → ∞ . Solution For the classical particle, we have: h x i = Z L 0 xp ( x ) dx = Z L 0 x 1 L dx = L 2 h x 2 i = Z L 0 x 2 p ( x ) dx = Z L 0 x 2 1 L dx = L 2 3 Thus, σ 2 = h x 2 i - h x i 2 = L 2 3 - L 2 4 = L 2 12 For the quantum particle in a box, we have: σ 2 = L 2 12 1 - 6 n 2 π 2 We now take the limit as n → ∞ to obtain: lim n →∞ σ 2 = lim n →∞ L 2 12 1 - 6 n 2 π 2 = L 2 12 We see that classical mechanics emerges from quantum mechanics at large values of the quantum number n . Question 3 A fundamental property of the eigenfunctions of quantum mechanical operators is that these eigen- functions are orthogonal. In one dimension, this means: Z all space ψ * m ( x ) ψ n ( x ) dx = 0 for m 6 = n Using the trigonometric identity: sin( αx ) sin( βx ) = 1 2 cos[( α - β ) x ] - 1 2 cos[( α + β ) x ] Show explicitly that the wavefunctions of a particle in a 1-D box are orthogonal. What happens when m = n ? 2
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Solution For a 1-D particle in a box, the wavefunction, ψ n ( x ) is given by ψ n ( x ) = r 2 L sin nπx L Thus, the orthogonality relation is: Z L 0 ψ * m ( x ) ψ n ( x ) dx = 2 L Z L 0 sin mπx L sin nπx L dx = 2 L 1 2 Z L 0 cos ( m - n ) πx L dx - 2 L 1 2 Z L 0 cos ( m + n ) πx L dx = 1 L L π ( m - n ) sin ( m - n ) πx L - L π ( m + n ) sin ( m + n ) πx L L 0 = 1 π ( m - n ) sin [( m - n ) π ] - 1 π ( m + n ) sin [( m + n ) π ] = 0 Since m, n Z , it follows that m - n, m + n Z , thus, sin[( m - n ) π ] = sin[( m + n ) π ] = 0 for all m, n Z . If m = n , the integral is equal to 1 and the wavefunctions are said to be orthonormal.
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