Explanation Given Information: A particle is bound by a potential energy of the form. U ( x ) = { 0 | x | < 1 2 a ∞ | x | > 1 2 a Calculation: The given potential energy U ( x ) is U ( x ) = { 0 for | x | < 1 2 a ∞ for | x | > 1 2 a The wave function outside must be zero. The wave function inside is: ψ ( x ) = A sin ( k x ) + B cos ( k x ) The wave function is zero at both boundaries that is at + 1 2 a and − 1 2 a . The wave function must be zero at + 1 2 a . ψ ( + 1 2 a ) = 0 A sin ( k ( + 1 2 a ) ) + B cos ( k ( + 1 2 a ) ) = 0 A sin ( k a 2 ) + B cos ( k a 2 ) = 0.......... ( 1 ) The wave function must be zero at − 1 2 a ψ ( − 1 2 a ) = 0 A sin ( k ( − 1 2 a ) ) + B cos ( k ( − 1 2 a ) ) = 0 − A sin ( k a 2 ) + B cos ( k a 2 ) = 0 A sin ( k a 2 ) − B cos ( k a 2 ) = 0..... ( 2 ) Add the above two equations ( 1 ) and ( 2 ) 2 B cos ( k a 2 ) = 0 B cos ( k a 2 ) = 0 This equation is valid when k a 2 = ( 2 n + 1 ) π 2 n = 0 , 1 , 2 , 3 , ...... In general, this can be written as k a 2 = n π 2 ( n is odd ) From the above equation k = n π a ( n is odd ) In this case, the wave function is only in cosine function and it is valid for odd values of n . Hence, in this case, the wave function is ψ ( x ) = B cos ( k x ) Substitute for n π a for k . ψ ( x ) = B cos ( n π x a ) ( n is odd ) Apply normalization condition. ∫ − a 2 a 2 | ψ ( x ) | 2 d x = 1 ∫ − a 2 a 2 B 2 cos 2 ( n π x a ) d x = 1 B 2 ⋅ a 2 = 1 B = 2 a Then, for odd values of n the wave function is, ψ ( x ) = 2 a cos ( n π x a ) ( n is odd ) The energy E is, E = k 2 h 2 2 m Substitute n π a for k

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ISBN: 9780805303087

Author: Randy Harris

Publisher: Addison Wesley

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Chapter 5, Problem 30E

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The allowed energies and corresponding normalized wave function

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Chapter 5, Problem 31E

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