IV. Planification temporelle : le système TLP-GP
2. Etat de l'art de la planification temporelle
2.3. Algorithmique des planificateurs temporels actuels
B.4.2
Evolu¸c˜ao com intera¸c˜ao ´atomo-campo
Passado o tempo τ , “a intera¸c˜ao ´e novamente ligada”, e a evolu¸c˜ao temporal passa a ser ditada pelo modelo Jaynes-Cummings ressonante. Como novamente o tempo dessa intera¸c˜ao ser´a muito curto com rela¸c˜ao a κ−1, vamos considerar
uma evolu¸c˜ao unit´aria por um tempo χ. Para obter essa evolu¸c˜ao vamos escrever o estado da eq. (B.44) em termos dos autoestados do Hamiltoniano Jaynes- Cummings ressonante: H0= 1 2~ωσz+ ~ω a†a +1 2 + ~Ω a†σ−+ σ+a, (B.45)
que tem estado fundamental |0i = |g, 0i com energia nula e os demais auto- estados s˜ao os dupletos |n, ±i = √1
2(|e, ni ± |g, n + 1i), com energias εn± =
~{ω (n + 1) ± Ωn}, onde Ωn= Ω√n + 1. As rela¸c˜oes acima s˜ao invertidas por
|g, 0i = |0i , |e, ni = √1 2(|n+i + |n−i) , |g, n + 1i = √1 2(|n+i − |n−i) . (B.46) Usando a defini¸c˜ao
eij = T−1|ii hj| = exp (α+K+) exp (α−K−)|ii hj| , (B.47)
para os casos de interesse, temos ejj = j X r=0 ∞ X s=0 (α−)r(α+)s j r j− r + s j− r |j − r + si hj − r + s| = ∞ X m=0 min{m,j}X l=0 (α−)j−l(α+)m−l j l m l |mi hm| , (B.48) ej,j+1 = j X r=0 ∞ X s=0 (α−)r(α+)s j r j− r + s j− r (j + 1)1 2(j + 1− r + s) 1 2 j + 1− r |j − r + si hj − r + s + 1| = ∞ X m=0 min{m,j}X l=0 (α−)j−l(α+)m−l j l m l (j + 1)1 2(m + 1) 1 2 l + 1 |mi hm + 1| , (B.49) ej+1,j = ∞ X m=0 min{m,j}X l=0 (α−)j−l(α+)m−l j l m l (j + 1)1 2(m + 1) 1 2 l + 1 |m + 1i hm| . (B.50) Passando ent˜ao para operadores do sistema ´atomo-campo, temos
|ei he| ⊗ ejj = 1 2 ∞ X m=0 min{m,j}X l=0 (α−)j−l(α+)m−l j l m l ×
{|m+i hm+| + |m−i hm−| + |m+i hm−| + |m−i hm+|} , |gi hg| ⊗ ejj = (α−)j|0i h0| +12 ∞ X m=0 min{m+1,j}X l=0 (α−)j−l(α+)m+1−l j l m + 1 l × {|m+i hm+| + |m−i hm−| − |m+i hm−| − |m−i hm+|} ,
|ei hg| ⊗ ej,j+1 = 1 2 ∞ X m=0 min{m,j}X l=0 (α−)j−l(α+)m−l j l m l (j + 1)1 2(m + 1) 1 2 l + 1 ×
{|m+i hm+| − |m−i hm−| − |m+i hm−| + |m−i hm+|} , |gi he| ⊗ ej+1,j = 1 2 ∞ X m=0 min{m,j}X l=0 (α−)j−l(α+)m−l j l m l (j + 1)1 2(m + 1) 1 2 l + 1 ×
A evolu¸c˜ao temporal por um tempo χ ditada pelo modelo Jaynes-Cummings ressonante pode ser escrito como o seguinte mapa:
|0i h0| 7−→ |0i h0| , |n+i hn+| 7−→ |n+i hn+| , |n−i hn−| 7−→ |n−i hn−| , |n+i hn−| 7−→ e−2iΩnχ|n+i hn−| ,
|n−i hn+| 7−→ e2iΩnχ|n−i hn+| .
(B.52)
Feita esta evolu¸c˜ao pelo tempo χ, podemos descrever o estado do sistema por ρ(τ + χ) = 12P∞ j=0 e−2jκτ αj− n ajj11 |0i h0| + 1 2 P∞ m=0 α−m−1− ajj11 j m+ 1 h
|m+i hm+| + |m−i hm−| − e−2iΩmχ |m+i hm−| − e2iΩmχ |m−i hm+|i +Pmin{m,j}l=0 αm+ α−α+l j l m l × ajj00 +ajj11m+1−lm+1 α+ + 2aj,j+101 e−κτ(j+1) 1 2 (m+1)12 l+1 cos φ |m+i hm+| + + ajj00 +ajj11m+1−lm+1 α+ − 2aj,j+101 e−κτ(j+1) 1 2 (m+1)12 l+1 cos φ |m−i hm−| + + ajj00 −ajj11m+1−lm+1 α+ − 2iaj,j+101 e−κτ(j+1) 1 2 (m+1)12 l+1 sin φ e−2iΩm χ|m+i hm−| + + ajj00 −ajj11m+1−lm+1 α+ + 2iaj,j+101 e−κτ(j+1) 1 2 (m+1)12 l+1 sin φ e2iΩm χ|m−i hm+| , (B.53) onde os coeficientes aij
kls˜ao dados pelas equa¸c˜oes (B.32,B.35,B.40,B.41). Passando para a base fatorada, obtemos ρ(τ + χ) = 12P∞j=0 e−2jκτ αj−najj11 |g,0i hg, 0| + P∞ m=0 α−m−1− ajj11 j m+ 1 h
sin2 (Ωmχ) |e, mi he, m| + cos2 (Ωmχ) |g, m + 1i hg, m + 1| + −i sin (Ωmχ) cos (Ωmχ) (|e, mi hg, m + 1| − |g, m + 1i he, m|)]
+Pmin{m,j}l=0 αm+ α−α+l j l m l ×
ajj00 cos2 (Ωmχ) + ajj11m+1−lm+1 α+ sin2 (Ωmχ) − aj,j+101 e−κτ(j+1) 1 2 (m+1)12 l+1 sin φ sin (2Ωmχ) |e, mi he, m| + +
ajj00 sin2 (Ωmχ) + ajj11m+1−lm+1 α+ cos2 (Ωmχ) + aj,j+101 e−κτ(j+1) 1 2 (m+1)12 l+1 sin φ sin (2Ωmχ) |g, m + 1i hg, m + 1| + + ajj00 −ajj11m+1−lm+1 α+
isin (Ωmχ) cos (Ωmχ) + aj,j+101 e−κτ (j+1)
1 2 (m+1)12
l+1 (cos φ + i sin φ cos (2Ωmχ)) |e, mi hg, m + 1| + +
−ajj00 −ajj11m+1−lm+1 α+isin (Ωmχ) cos (Ωmχ) + aj,j+101 e−κτ
(j+1)12 (m+1)12
l+1 (cos φ − i sin φ cos (2Ωmχ)) |g, m + 1i he, m| , (B.54)
Fazendo o tra¸co parcial, obtemos o estado atˆomico dado por ρat (τ + χ) = 12P∞j=0 e−2jκτ αj− n ajj11 |gi hg| + P∞ m=0 α−m−1− ajj11 j m+ 1 h
sin2 (Ωmχ) |ei he| + cos2 (Ωmχ) |gi hg|i+ Pmin{m,j} l=0 αm+ α−α+ l j l m l ×
ajj00 cos2 (Ωmχ) + ajj11m+1−lm+1 α+ sin2 (Ωmχ) − aj,j+101 e−κτ(j+1) 1 2 (m+1)12 l+1 sin φ sin (2Ωmχ) |ei he| + +
ajj00 sin2 (Ωmχ) + ajj11m+1−lm+1 α+ cos2 (Ωmχ) + aj,j+101 e−κτ(j+1) 1 2 (m+1)12 l+1 sin φ sin (2Ωmχ) |gi hg| , (B.55)
Ainda vamos testar o resultado e escrevˆe-lo com uma cara mais bonita... mas por ora ´e isso.
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