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Unformatted text preview: CHAPTER 6 19. (a) We have Aa> = aa> It follows that < a  A  a > = a < a  a> = a if the eigenstate of A corresponding to the eigenvalue a is normalized to unity. The complex conjugate of this equation is < a  A  a >* = < a  A +  a > = a * If A + = A , then it follows that a = a *, so that a is real. 13. We have 〈 ψ  ( AB ) +  ψ 〉 = 〈 ( AB ) ψ  ψ 〉 = 〈 B ψ  A +  ψ 〉 = 〈 ψ  B + A +  ψ 〉 This is true for every ψ , so that ( AB ) + = B + A + 2. TrAB = 〈 n  AB  n 〉 = 〈 n  A 1 B  n 〉 n ∑ n ∑ = 〈 n  A  m 〉〈 m  B  n 〉 = m ∑ n ∑ 〈 m  B  n 〉〈 n  A  m 〉 m ∑ n ∑ = 〈 m  B 1 A  m 〉 = m ∑ 〈 m  BA  m 〉 = m ∑ TrBA 3. We start with the definition of  n > as  n 〉 = 1 n ! ( A + ) n  0 〉 We now take Eq. (647) from the text to see that A  n 〉 = 1 n ! A ( A + ) n  0 〉 = n n ! ( A + ) n 1  0 〉 = n ( n 1)! ( A + ) n 1  0 〉 = n  n 1 〉 4. Let f ( A + ) = C n n = 1 N ∑ ( A + ) n . We then use Eq. (647) to obtain Af ( A + )  0 〉 = A C n n = 1 N ∑ ( A + ) n  0 〉 = nC n ( A + ) n 1 n = 1 N ∑  0 〉 = d dA + C n n = 1 N ∑ ( A + ) n  0 〉 = df ( A + ) dA +  0 〉 5. We use the fact that Eq. (636) leads to x = h 2 m ϖ ( A + A + ) p = i m ϖ h 2 ( A + A ) We can now calculate 〈 k  x  n 〉 = h 2 m ϖ 〈 k  A + A +  n 〉 = h 2 m ϖ n 〈 k  n 1 〉 + k 〈 k 1  n 〉 ( 29 = h 2 m ϖ n δ k , n 1 + n + 1 δ k , n + 1 ( 29 which shows that k = n ± 1. 6. In exactly the same way we show that 〈 k  p  n 〉 = i m ϖ h 2 〈 k  A + A  n 〉 = i m ϖ h 2 ( n + 1 δ k , n + 1 n δ k , n 1 ) 7. Let us now calculate 〈 k  px  n 〉 = 〈 k  p 1 x  n 〉 = 〈 k  p  q 〉〈 q  x  n 〉 q ∑ We may now use the results of problems 5 and 6. We get for the above i h 2 ( k q ∑ δ k 1, q k + 1 δ k + 1, q )( n δ q , n 1 + n + 1 δ q , n + 1 ) = i h 2 ( kn δ kn ( k + 1) n δ k + 1, n 1 + k ( n + 1) δ k 1, n + 1 ( k + 1)( n + 1) δ k + 1, n + 1 ) = i h 2 ( δ kn ( k + 1)( k + 2) δ k + 2, n + n ( n + 2) δ k , n + 2 ) To calculate 〈 k  xp  n 〉...
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This note was uploaded on 03/09/2008 for the course PHYS 401 taught by Professor Mokhtari during the Spring '05 term at UCLA.
 Spring '05
 mokhtari

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