By Gerry C, Knight P,

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**Example text**

5 51 Assume that cf (t) = eXt , and plug it into the differential equation we obtain the following quadratic equation X 2 − i∆(n + 1) + λ2 (n + 1) = 0, whose solutions are 1 X = (i∆ ± −∆2 − 4λ2 (n + 1)) 2 i = (∆ ± ∆2 + 4λ2 (n + 1)). 2 The general solution then is i cf (t) = e 2 ∆t AeiΩn t + Be−iΩn t , 2 where Ωn = ∆4 + λ2 (n + 1). From initial conditions, we have B = −A, so i cf (t) = Ae 2 ∆t eiΩn t − e−iΩn t i = i2Ae 2 ∆t sin (Ωn t) i = A e 2 ∆t sin (Ωn t) , where A is just a constant. 2 √ ci (t) = (ic˙f (t) + ∆cf (t))/ λ n + 1 A ei∆t/2 = √ λ n+1 A ei∆t/2 = √ λ n+1 ∆ sin(Ωn t) + Ωn cos(Ωn t) + ∆ sin(Ωn t) 2 ∆ sin(Ωn t) + Ωn cos(Ωn t) 2 − Using the second initial condition, ci (0) = 1, we obtain √ λ n+1 .

1) by identifying 1 x = α − β, y = −(α∗ − β ∗ ), and z = . 2) For |Ψ = |N Using Eq. 128a) we have ˆ CW (λ) = N |D(λ)|N 2 /2 † = e−|λ| N |eλˆa e−λˆa |N N =e N −|λ|2 /2 2 /2 = e−|λ| λn a ˆ†n (−1)n λn a ˆn N| |N n! n! n! (−1)n |λ|2n N! n! (N − n)! 3) where we have used the Laguerre polynomials expansion. 4) 40 CHAPTER 3. 5) we compute the integral in Eq. 4 2 2 W (α) = (−1)N e−2|α| LN (4|α|2 ). 13 a. For the state |ψ = N (|β + | − β ) 2 ψ|ψ = 1 = |N |2 [ β|β + −β| − β + −β|β + β| − β ] = |N |2 [2 + 2e−2|β| ] 2 For large β, e−2|β| ≈ 0 so this state is normalized for: 1 N =√ .

8 35 b. ∞ 2 2 d z|z z| = d z 1 − |z| ∞ ∞ 1 = 1 = 2π n=0 dφ 1 − |z|2 |z|2(n+n ) eiφ(n−n ) |n n | dr (1 − r) r(n+n )/2 2πδn,n |n n | 0 n=0 n =0 ∞ = 2π n=0 n =0 2π 0 1 = n=0 ∞ d|z|2 z n z n |n n | 0 n=0 n =0 ∞ ∞ ∞ 2 dr rn − rn+1 |n n| 0 1 |n n|, (n + 1)(n + 2) It does not resolve unity. c. We have proved that the state is not normalized for |z| < 1. Thus we drop the normalization constant and we write z = eiφ and we obtain the phase states |φ of Eq. 221). Obviously the the last states resolve unity as in Eq.