244
Solutions
For small values of P max , W
2
2
Δω
2 , so to keep P max < 10
−4 we simply
need |W | < 10
−2
|Δω|. Since |Δω| = 100 cm
−1 , in atomic units we have W <
4.5 · 10
−6 , so Δt min = 220000 a.u. = 5.3 ps.
3.2 With constant V and Δε = 0, Eq. (3.35) yields
c 2 (t) = −
V
∗
Δε
e
iΔε t/
− 1
and
|c 2 (t)|
2
=
2|V |
2
Δε 2
1 − cos
Δε t
2
=
4|V |
2
Δε 2 sin
2
Δε t
2
.
The exact Rabi solution, Eq. (3.23), is
|c 2 (t)|
2
=
4|V |
2
Δε 2 + 4|V | 2 sin
2
Δε 2 + 4|V | 2 t
2
.
We see that, with |V | | Δε, the TDPT solution approximates the exact one, with a
slightly larger amplitude and a slightly smaller oscillation frequency:
4|V |
2
Δε 2 =
4|V |
2
Δε 2 + 4|V | 2
1 +
4|V |
2
Δε 2
and
Δε
2
=
Δε 2 + 4|V | 2
2
1 +
4|V |
2
Δε 2
−1/2
Δε 2 + 4|V | 2
2
1 −
2|V |
2
Δε 2
.
With Δε = 0, Eq. (3.35) yields
c 2 (t) = −
i V
∗ t
and
|c 2 (t)|
2
=
|V |
2 t
2
2 .
The exact Rabi solution, Eq. (3.23), in this case is simply
|c 2 (t)| 2 = sin 2
|V | t
=
|V | t
−
1
6
|V | t
3
+ O(t 5 )
2
=
|V | 2 t 2
2
1 −
|V | 2 t 2
3 2 + O(t 4 )
.
This solution is acceptably accurate if t /|V |.
3.3 The Fourier transform of A(t) as given by Eq. (3.89) is