# ch06 - CHAPTER 6 19(a We have A|a> = a|a> It follows that...

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CHAPTER 6 19. (a) We have A|a> = a|a> 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. (6-47) 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. (6-47) to obtain

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A f ( A + )|0 ⟩= AC 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 ⟩= d f ( A + ) dA + |0 5. We use the fact that Eq. (6-36) 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 we may proceed in exactly the same way. It is also possible to abbreviate the calculation by noting that since x and p are hermitian operators, it follows that
k | xp | n ⟩= n | px | k * so that the desired quantity is obtained from what we obtained before by interchanging k and n and complex-conjugating. The latter only changes the overall

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• Spring '05
• mokhtari
• Trigraph, A+ A+, |B |m ∑|B|m, d α+

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