Chemistry Reference
In-Depth Information
(
θ
f
=
0
,ϕ
f
=
0), a result of case (a) leads to
C
1
=
π/
4. By symmetry, we also
deduce that the cost from (
θ
f
=
π/
2
,ϕ
f
=
0) and (
θ
f
=
π/
2
,ϕ
f
=
π/
2) is the
same (i.e.,
C
2
=
C
3
=
C
). We determine this time numerically. For the cost
C
,we
have obtained
m
f
1
/
1
.
126 and a corresponding time
C
equal to 1.422. Using
Eq. (28), it can be shown that
1
⎛
⎞
m
f
+
2
π
2
1
1
⎝
⎠
C
=
−
1
arcsin
√
2
m
f
m
f
+
+
To compute
C
, we have used the fact that for the extremal trajectory
θ
is not
a
monotonous function of time and that the minimum of
θ
is arccos(1
/
1
m
f
).
Figure 16 represents the three optimal trajectories reaching the target state. A
straightforward calculation then gives the total cost
C
T
that can be written as
follows:
+
π
4
cos
2
θ
i
+
C
(sin
2
θ
i
cos
2
ϕ
i
+
sin
2
θ
i
sin
2
ϕ
i
)
C
T
=
C
0
+
π
4
cos
2
θ
i
+
C
sin
2
θ
i
=
C
0
+
2
1.5
1
0.5
0
0
0.4
0.8
1.2
φ
Figure 16.
Optimal trajectories for the time-minimum problem starting from the different states
associated with the measurement.
