252
Solutions
The condition to impose is q
· h
= 0. We obtain
q
2 cos(2θ) sin(2θ) − h
2 cos(2θ) sin(2θ) + q · h(cos
2
(2θ) − sin
2
(2θ)) = 0
which gives
tg(4θ) =
2q · h
h 2 − q 2
where q and h are the norms of q and h, respectively.
5.4 Using Eq. (2.66) and taking into account that g
(α)
ii = 0 for real wavefunctions,
we obtain immediately
t
(α)
11 = t
(α)
22 = −
g
(α)
12
2
and
t
(α)
12 =
∂g
(α)
12
∂ Q α
.
For the Landau–Zener model we have, using Eq. (5.29)
t 12 (Q) = −
β
3
ΔQ
(1 + β 2 ΔQ 2 ) 2 .
Close to a conical intersection, using Eq. (5.42) with q = h we obtain
t 12 =
x y
(x 2 + y 2 ) 2 ( ˆ
x − ˆ
y) .
According to (5.39) U 1 has a cusp at the conical intersection. For U
1 we have
U
1 = U 1 −
2
2
α
t
(α)
11
M α
= U 1 +
2
2
α
g
(α)
12
2
M α
so that U
1 is discontinuous (it diverges) at the intersection.
5.5 Equation (5.1) gives the adiabatic states in terms of the diabatic ones, with
tg(2θ) = −2H 12 /ΔH , according to (D.5). We have then
tg(2θ) = −
2x y
y 2 − x 2 = − tg(2γ ) .
In the last equation we switched to polar coordinates x = r cos γ and y = r sin γ .
We can therefore choose θ = −γ , obtaining
|ϕ 1 = |η 1 cos γ − |η 2 sin γ
|ϕ 2 = |η 1 sin γ + |η 2 cos γ .