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Lecture10

# Lecture10 - CHEM 356 Lecture 10 Fall 2009 1 Expectation...

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CHEM 356, Lecture 10, Fall 2009 1 Expectation Values for the Particle–in–a–box. Position : The expectation value for ˆ ? in eigenstate 𝜓 𝑛 of the translational energy operator can be calculated as ? 𝑛 = 𝐿 0 𝜓 𝑛 ( ? ) ˆ ?𝜓 𝑛 ( ? ) ?? = 2 𝐿 𝐿 0 ? sin 2 ( 𝑛𝜋? 𝐿 ) ?? = 1 𝐿 𝐿 0 ? [ 1 cos 2 𝑛𝜋? 𝐿 ] ?? = 1 𝐿 { ? 2 2 𝐿 0 𝐿 0 ? cos 2 𝜋𝑛? 𝐿 ?? } or ? 𝑛 = 𝐿 2 . Similarly, but with a few more mathematical steps, the expectation value of ˆ ? 2 can be obtained as ? 2 𝑛 = 2 𝐿 𝐿 0 ? 2 sin 2 𝑛𝜋? 𝐿 ?? = ( 𝐿 2 𝜋𝑛 ) 2 ( 4 𝜋 2 𝑛 2 3 2 ) As the variance equals the square of the uncertainty, let us use this relationship: ? ) 2 𝑛 = ? 2 𝑛 − ⟨ ? 2 𝑛 = ( 𝐿 2 𝜋𝑛 ) 2 [ 4 𝜋 2 𝑛 2 3 2 ] 𝐿 2 4 = ( 𝐿 2 𝜋𝑛 ) 2 [ 4 𝜋 2 𝑛 2 3 2 𝜋 2 𝑛 2 ] = ( 𝐿 2 𝜋𝑛 ) 2 [ 𝜋 2 𝑛 2 3 2 ]

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CHEM 356, Lecture 10, Fall 2009 2 or ? ) 𝑛 = 𝐿 2 𝜋𝑛 [ 𝜋 2 𝑛 2 3 2 ] 1 2 . Momentum : The expectation values for ˆ 𝑝 ? and ˆ 𝑝 2 ? are calculated in exactly the same manner. For ˆ 𝑝 ? we have 𝑝 ? = 2 𝐿 𝐿 0 sin 𝑛𝜋? 𝐿 ( 𝑖 ? ?? ) sin 𝑛𝜋? 𝐿 ?? = 2 𝐿 ( 𝑖 𝑛𝜋 𝐿 ) 𝐿 0 sin 𝑛𝜋? 𝐿 cos 𝑛𝜋? 𝐿 ?? , and hence 𝑝 ? 𝑛 = 0 , for every value of 𝑛 . Evaluation of 𝑝 2 ? is carried out similarly; 𝑝 2 ? 𝑛 = 2 𝐿 ( 𝑖 ) 2 𝐿 0 sin 𝑛𝜋? 𝐿 ? 2 ?? 2 sin 𝑛𝜋? 𝐿 ?? = ( 𝑖 ) 2 ( 1) ( 𝑛𝜋 𝐿 ) 2 2 𝐿 𝐿 0 sin 2 𝑛𝜋? 𝐿 ?? , and so we see that 𝑝 2 ? is given by 𝑝 2 ? 𝑛 = 2 ( 𝑛𝜋 𝐿 ) 2 .
CHEM 356, Lecture 10, Fall 2009 3 Because 𝑝 ? = 0, the uncertainty for ˆ 𝑝 ? is simply 𝑝 ? ) 𝑛 = 𝑝 2 ?

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