Prob-sheet9-08-solutions.pdf

# Prob-sheet9-08-solutions.pdf - PHYS 234 Quantum Physics...

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PHYS 234: Quantum Physics 1 (Fall 2008) Assignment 9 – Solutions Issued: November 14, 2008 Due: 12.00pm, November 21, 2008 1. Two possible eigenfunctions for a particle moving freely in a region of length a , but strictly confined to that region, are shown in the figure below. When the particle is in the state corresponding to the eigenfunction ψ I , its total energy is 4 eV. a) What is its total energy in the state corresponding to ψ II ? b) What is the lowest possible energy for the particle in this system ? solution a) In the lowest energy state n = 1 , ψ has no nodes. Hence ψ I must correspond to n = 2 , and ψ II to n = 3 . Since the energy of the n th state E n n 2 and E I = 4 eV, then E II E I = 3 2 2 2 ; E II = 9 eV b) By the same analysis, E 0 E I = 1 2 2 2 ; E II = 1 eV

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PHYS 234, Fall 2008 Assignment 9 2. It can be proved that, in general, the inner product of any two different eigenfunctions (of the same given potential) is always zero. This property is called orthogonality . The inner product is defined as, Z -∞ ψ m ( x ) * ψ n ( x ) dx = 0 m 6 = n. Form the inner product of the eigenfunctions for the n = 1 and n = 3 states of the infinite square well potential. Show that the inner product is equal to zero. In other words, show that Z -∞ ψ * 1 ( x ) ψ 3 ( x ) dx = 0 Hint: The following identities may be useful: cos( v ) cos( v ) = [cos( u + v ) + cos( u - v )] / 2) cos( u ) - cos( v ) = - 2 sin u + v 2 sin u - v 2 solution The inner product that is formed is analysed as follows Z + -∞ ψ * 1 ψ 3 dx = 2 a Z + a/ 2 - a/ 2 cos( πx a ) cos( 3 πx a ) dx The integral is simplified using the substitution u = πx a and v = 3 πx a and the relationship from the problem sheet cos( v ) cos( v ) = [cos( u + v ) + cos( u - v )] / 2 , hence Z + -∞ ψ * 1 ψ 3 dx = 1 a Z + a/ 2 - a/ 2 cos( 4
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