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Unformatted text preview: PHY 4221 Quantum Mechanics Fall Term Of 2004 Problem Set No.1 (Answer) October 1, 2004 Question 1: You could expand ( 29 ( 29 exp exp A B A ξ ξ as following: ( 29 ( 29 ( 29 1 exp exp ! ! n n n m m n m A B A A B A n m ξ ξ ξ ξ ∞ ∞ = = = × × ∑ ∑ We will pick out the terms containing k ξ and thus the above equation could be rewritten as: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 exp exp ! ! 1 ! ! 1 ! ! ! ! ! ! ! ! ! ! n n n m m n m n k k n k n n n k n n k k n k n k n k k n k n n k n k k n n n k k n k A B A A B A n m A B A n k n A B A n k n k A B A k n k n k A B A A k n k n k ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ∞ ∞ = = ∞ = = ∞ = = ∞ = = ∞ = = = × × = × × = × × = × ×  = × ×  = ∑ ∑ ∑∑ ∑∑ ∑ ∑ ∑ ∑ ( 29 { } ! k n n n n k k k n C A B A A ∞ = = × × ∑ ∑ (1) The following is to prove that ( 29 { } [ ] , , , , , k n n n n k k n total number of A is k C A B A A B A A A A =  × × = ××× ∑ 1 4 4 442 4 4 4 43 (2) To achieve this, we firstly notice that when k =1 and 0, this relation obviously holds. Then we assume that when k = t , the relation still holds, and try to see what happen when k = t +1. Now we have ( 29 { } [ ] , , , , , t n n n n t t n total number of A is t C A B A A B A A A A =  × × = ××× ∑ 1 4 4 442 4 4 4 43 and when k = t +1, the right hand side becomes: [ ] [ ] [ ] 1 , , , , , , , , , , , , , , , total number of A is t total number of A is t total number of A is t B A A A A B A A A A A A B A A A A + ××× = ××× ×××...
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 Spring '11
 CFLO
 Work, total number, Hyperbolic function, CTN, ×B ×A−

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