Introductory quantum optics: solutions manual by Gerry C.C., Knight P.L.

By Gerry C.C., Knight P.L.

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N αn √ √ n|m m + 1|n n! n ! m=0 n,n ∞ 2 = αe−|α| n=0 |α|2n √ n! n + 1 1 α| Cˆ |α = α| Eˆ + Eˆ † |α 2 1 α| Eˆ |α + α| Eˆ † |α = 2 1 = α| Eˆ |α + α| Eˆ |α ∗ 2 ∞ 1 |α|2n ∗ −|α|2 √ = (α + α ) e 2 n! 1) |α|2n √ n! n + 1 2 n=0 |α|2n+1 √ n! 7. 7 31 1 α| Eˆ − Eˆ † |α 2i 1 = α| Eˆ |α − α| Eˆ † |α 2i 1 = α| Eˆ |α − α| Eˆ |α ∗ 2i ∞ 2 |α|2n 1 √ = (α − α∗ ) e−|α| 2i n! n + 1 n=0 α| Sˆ |α = = ∞ −|α|2 (α)e n=0 ∞ |α|2n √ n! n + 1 −|α|2 = sin(θ)e n=0 |α|2n+1 √ n! n + 1 1 ˆ Cˆ 2 = E + Eˆ † Eˆ + Eˆ † 4 1 ˆ 2 ˆ †2 ˆ ˆ † ˆ † ˆ = E + E + EE + E E 4 −1 ˆ Sˆ2 = E − Eˆ † Eˆ − Eˆ † 4 −1 ˆ 2 ˆ †2 ˆ ˆ † ˆ † ˆ = E + E − EE − E E 4 ∞ ∞ Eˆ 2 = |n n + 1|m m + 1| n=0 m=0 ∞ = |n n + 2|, n=0 ∞ Eˆ †2 = |n + 2 n| n=0 Eˆ Eˆ † = 1, Eˆ † Eˆ = 1 − |0 0| Eˆ Eˆ † + Eˆ † Eˆ = 2 − |0 0| 32 CHAPTER 3.

N ¯ −1 a. Here we consider the following Hamiltonian ˆ = 1 ω0 σ H ˆ3 + ωˆ a† a ˆ + λˆ a† a ˆ(ˆ σ+ + σ ˆ− ). 2 Also we define the following “bare” states |ψ1n = |e |n |ψ2n = |g |n . Clearly ψ1n |ψ2n = 0. Using these basis we obtain the matrix elements of ˆ H. ˆ 1n = 1 ω0 |e |n + ωn|e |n + λn|g |n , H|ψ 2 ˆ 2n = − 1 ω0 |g |n + ωn|g |n + λn|e |n H|ψ 2 54CHAPTER 4. EMISSION AND ABSORPTION OF RADIATION BY ATOMS 1 ω0 + nω , 2 1 − ω0 + nω , 2 ˆ 1n = ψ1n |H|ψ ˆ 2n = ψ2n |H|ψ ˆ 1n = nλ, ψ2n |H|ψ ˆ 2n = nλ.

3) For a general case where we have the sum of n-photon inversions of Eq. 3 weighted with photon number distribution of the initial fields state we have ∞ |cn |2 W (t) = n=0 1 Ω2n ∆ sin(Ωn t) + Ωn cos(Ωn t) 2 2 − λ2 (n + 1) sin2 (Ωn t) . 4) Notice that the last equation is in agreement with Eq. 123) for ∆ = 0. 6 For an atom initially in the excited state and the cavity field initially in a thermal state the atomic inversion is 1 W (t) = 1+n ¯ Let ∞ n=0 n ¯ 1+n ¯ n √ cos(2λt n + 1) √ Ω(n) = 2λ n + 1.

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