PS5sol - Solutions to Problem Set 5 Physics 341 by: Callum...

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Unformatted text preview: Solutions to Problem Set 5 Physics 341 by: Callum Quigley 1 Completeness Relation for Coherent States For this problem, Ill use the normalized coherent states from the last homework, namely: | i = e-| | 2 / 2 e a | i = e-| | 2 / 2 n ( n / n !) | n i . Writing = x + iy = re i we can compute Z d x d y | ih | = Z d x d y X m,n e-| | 2 m ( * ) n m ! n ! | m ih n | = X m,n Z r d r d e- r 2 r m + n e i ( m- n ) m ! n ! | m ih n | = 2 X n 1 n ! Z d re- r 2 r 2 n +1 | n ih n | (1.1) = X n | n ih n | = 1 where Ive used the facts that R d e i ( m- n ) = 2 n,m and R d rr 2 n +1 e- r 2 = 1 2 ( n + 1) = 1 2 n !. And so, 1 = Z d x d yf ( x,y ) | ih | f ( x,y ) = 1 (1.2) Note: If Id used the unnormalized coherent states | i = e a | i as theyre given in Shankar then wed have f ( , * ) = e-| | 2 / 2 Properties of Coherent States Note: Well now switch to Shankars conventions for coherent states | z i = e za | i , which are normalized as h z | z i = e | z | 2 . Using the facts that a | z i = z | z i and a = X + ( i/ 2 ~ ) P where 2 = m/ 2 ~ , we have z z ( x ) = z h x | z i = h x | a | z i = x z ( x ) + 1 2 x z ( x ) (2.1) which we can integrate to find z ( x ) = z (0) e- 2 x 2 +2 zx . (2.2) 1 To determine z (0) well consider the norm e | z | 2 = h z | z i = Z d x | z ( x ) | 2 = | z (0) | 2 Z d xe- 2 2 x 2 +2 x ( z + z * ) = 1 2 | z (0) | 2 e 1 2 ( z + z * ) 2 Z - d y e- y 2 / 2 = 2 2 | z (0) | 2 e | z | 2 e ( z 2 + z * 2 ) / 2 (2.3) z (0) = s 2 2 e- z 2 / 2 , where in the second line I switch integration variables to y = 2 x- z- z * . Putting this all together, z ( x ) = m ~ 1 4 e- z 2 / 2 e- ( m/ 2 ~ ) x 2 + 2 m/ ~ zx . (2.4) If we focus on the x dependence of the wavefunction z ( x ) e- ( x- z ) 2 e- 2 ( x- Re( z ) / ) 2 e 2 i Im( z ) x (2.5) then, comparing to (9.3.7) in Shankar, we can read off the expectation values h X i z = Re( z ) / h P i z = 2 ~ Im( z )...
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PS5sol - Solutions to Problem Set 5 Physics 341 by: Callum...

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