PHYS4221(Fall2010)_Solution_2

PHYS4221(Fall2010)_Solution_2 - 1 PHYS4221 Suggested...

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Unformatted text preview: 1 PHYS4221 Suggested Solution of Homework 2 Question 1 Let C = A- B . From the question, we are given h u | C | u i = 0 for all | u i ∈ H . If A 6 = B , i.e. C 6 = 0, there must be two different states | a i , | b i ∈ H such that h a | C | b i 6 = 0 . We will see that this gives a contradiction. Let | u 1 i = N 1 ( | a i + | b i ), where N 1 is a normalization factor. Then 0 = h u 1 | C | u 1 i = | N 1 | 2 ( h a | C | a i + h a | C | b i + h b | C | a i + h b | C | b i ) = | N 1 | 2 ( h a | C | b i + h b | C | a i ) . Since N 1 6 = 0, we have h a | C | b i + h b | C | a i = 0 . (1) Similarly, letting | u 2 i = N 2 ( | a i + i | b i ), 0 = h u 2 | C | u 2 i = | N 2 | 2 ( h| a | - i h| b | ) C ( | a i + i | b i ) = i | N 2 | 2 ( h a | C | b i - h b | C | a i ) (Note the complex conjugation of i for the bra vector h u 2 | ). Again, N 2 6 = 0, therefore h a | C | b i - h b | C | a i = 0 . (2) Comparing (1) and (2) gives h a | C | b i = 0, which is a contradiction. Hence we must have C = 0 i.e. A = B . Question 2 Making use of the identity [ A n ,B ] = n- 1 X m =0 A m [ A,B ] A n- m- 1 (see e.g. lecture notes, no. 1), we have a n ,a † m = n- 1 X k =0 a k a,a † m a n- k- 1 = n- 1 X k =0 a k m- 1 X l =0 a † l a,a † a † m- l- 1 ! a n- k- 1 ([ A,B ] =- [ B,A ]) = n- 1 X k =0 a k m- 1 X l =0 a † m- 1 ! a n- k- 1 ( a,a † = 1 ) = m n- 1 X k =0 a k a † m- 1 a n- k- 1 . Question 3 First note that ˆ Df ( x ) = f ( x + η ) = f ( x ) + ηf ( x ) + η 2 2! f 00 ( x ) + ... = " ∞ X n =0 1 n ! η d dx n # f ( x ) ≡ exp η d dx f ( x ) . 2 Hence we may identify the operator ˆ D as ˆ D = exp η d dx . To find the eigenfunctions of ˆ D , it suffices to find the eigenfunction of the operator inside the exponential i.e. ( d/dx ) (A more ”professional” way of saying this is: the generator of the transformation ˆ D is ( d/dx )). Clearly d dx f β ( x ) = βf β ( x ) = ⇒ f β ( x ) = exp( βx ) , where β can be any complex number. So { f β ( x ) } forms a set of eigenfunctions of ˆ D , with eigenvalues exp( ηβ ). Note that there are degenerate eigenvalues because exp( ηβ ) = exp η β + i 2 nπ η , n = 0 , ± 1 , ± 2 ,......
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PHYS4221(Fall2010)_Solution_2 - 1 PHYS4221 Suggested...

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