Photoelectron counting in quantum optics
Photoelectron counting in quantum optics
Photoelectron counting in quantum optics
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Solve ∂ ∂t P s = [ −yz∂ z − y ∗ z ∗ ∂ z∗ + γ(1 + |z| 2 (s − 1)) ] P sArbitrary s:P s (z, t) = e γt P (0) ( ze i(Ω−iγ/2)t) exp{−|z| 2 (s − 1)(1 − e γt )}Now TrĜ(s, t) ≡ ∑ ∞m=0 sm Trρ (m)tt .distribution p m (t) ≡ Trρ (m)TrĜ(s, t) ==, read off photoelectron <strong>count<strong>in</strong>g</strong>∫∫d 2 zP s (z, t) = d 2 zP (0) (z)e −|z|2 (s−1)(e −γt −1)∞∑∫(|z|s m d 2 zP (0) 2 ) mη t(z) e −|z|2 η t, η t ≡ 1 − e −γt .m!m=0Use normal order<strong>in</strong>g property of P-representation,p m (t) = Trρ(0) : (a† aη t) mm!e −a† aη t:, η t ≡ 1 − e −γt (11)