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[ ]
1 0
K ϕa (∆t) =
0 e −∆t/2Tϕ
[ 0 0
K ϕb (∆t) =
0 √ 1 − e −∆t/T ϕ
]
(3.76)
(3.77)
K 1a and K 1b capture the energy relaxation process, while K ϕa and K ϕb capture
dephasing. Since both energy relaxation and dephasing are “non-unitary” process,
they need to be broken up mathematically into two steps, yielding the following
decoherence operations:
ρ ′ (t + ∆t) = K 1a (∆t) ρ(t + ∆t)K † 1a(∆t) + K 1b (∆t) ρ(t + ∆t)K † 1b
(∆t) (3.78)
ρ ′′ (t + ∆t) = K ϕa (∆t) ρ ′ (t + ∆t)K † ϕa(∆t) + K ϕb (∆t) ρ ′ (t + ∆t)K † ϕb
(∆t) (3.79)
Intuitively, K 1b transfers some of the population from the excited state into the
ground state while K 1a ensures that the resulting state is still normalized. K ϕb
reduces the phase information of the state while K ϕa , again, ensures normalization.
For accurate results, the decay operation needs to be interleaved into the
stream of rotation and coupling operations fairly frequently. ∆t should be chosen
to be much smaller than any timescales of rotation or coupling, i.e.:
√
1
∆t ≪ min
Xn 2 + Yn 2 + Zn
2
and ∆t ≪ min 1
C nm
(3.80)
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