Answer(1)

Answer(1) - PHY 4221 Quantum Mechanics I, Fall term of 2004...

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Preliminary Test, ( Answers ) Question 1 st : Method 1 st : It could be noticed that in the previous studies we got familiar with another simpler example of 1D rigid box problem: you have a 1D rigid box whose two boundaries are located at x=0 and x=a, then you know the wave function of the particle confined in this box is (in which n are any positive integers): ( 29 2 sin 0 0 0 n n x x a x a a x or x a π ψ < < ÷ = < . Now the question is similar, differing from the above case only in that the rigid box now is located at x L = ± . So you can use the results of the above case directly, with some mathematical transformation: Set y x L = + , and we have [ ] 0,2 y L as a new variable, and thus this question is changed into follows: a particle is confined in a rigid box located at [ ] 0,2 y L . So simply using the above result you know that the wave function in our question can be written as: ( 29 2 sin 0 2 2 2 0 0 2 n n y y L y L L y or y L < < ÷ = < . Changing this result back into x we obtain: ( 29 ( 29 2 sin 0 2 2 2 0 0 2 n n x L x L L y L L L x L or x L L + < + < ÷ - = + < + . This is, of course, equivalent to the following and our final result: ( 29 ( 29 1 sin 2 0 n n x L L x L x L L x L or x L + - < < ÷ = . Method 2 nd : You may also use the ordinary method as you do in any other cases, that is, solving
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Answer(1) - PHY 4221 Quantum Mechanics I, Fall term of 2004...

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