Chemistry Reference
In-Depth Information
Section IV for particular examples. The general solution is very complex and can
only be determined numerically.
To describe the dynamics of the system on the two spheres, we introduce two
sets of spherical coordinates (
θ
i
,ϕ
i
) and (
θ
f
,ϕ
f
), such that
⎧
⎨
X
1
=
sin
θ
i
cos
ϕ
i
X
3
=
cos
θ
i
⎩
X
5
=
sin
θ
i
sin
ϕ
i
and
⎧
⎨
X
2
=
sin
θ
f
cos
ϕ
f
X
4
=
cos
θ
f
⎩
X
6
=
sin
θ
f
sin
ϕ
f
The solution of the optimal control problem on each sphere corresponds to the one
of the Grusin model on the sphere.
We consider four different qualitative cases of control. We recall that the initial
and target state belong, respectively, to
S
i
and
S
f
.
•
Case (a): Passage from
X
3
=
1to
X
4
=
|
to state
i
|
1, that is, from state
2
2
|
(modification of the phase of the state
2
). The measurement operator is not
fixed but allows us to pass from
S
i
to
S
f
.
Case (b): Passage from
X
3
=
1to
X
2
=
1. We assume that the operator
Q
is
•
of the form
i
√
3
(
α
Q
=
|
1
1
|+
β
|
2
2
|+
γ
|
3
3
|
)
where
α
,
β
, and
γ
are real constants.
Case (c): Passage from a state of
S
i
to the state (
θ
f
=
π/
2
,ϕ
f
=
α
), where
•
α
∈
[0
,
2
π
]. The states associated to the measurement are given by (
θ
f
=
π/
4
,ϕ
f
π/
2). The angle
α
is chosen such that the cost to reach the target state from one of the three states
of the measured observable is the same. In this symmetric case, the optimal
trajectory does not depend on the initial state.
=
0), (
θ
f
=
3
π/
4
,ϕ
f
=
0), and (
θ
f
=
π/
2
,ϕ
f
=
i
√
2
e
iπ/
4
/
2
•
Case
(d):
Passage
from
|
ψ
i
=|
1
to
|
ψ
f
=
|
1
+
|
2
+
e
−
iπ/
4
/
2
|
3
, that is, (
θ
f
=
π/
4
,ϕ
f
=
π/
4). The three states associated with
the measurement are (
θ
f
=
0
,ϕ
f
=
0), (
θ
f
=
π/
2
,ϕ
f
=
0), and (
θ
f
=
π/
2
,ϕ
f
=
π/
2).
