B.2 Double-Well Potential
411
B.2.2 Above the Barrier—(E o > 0)
The momentum for an untrapped trajectory can be written
p = ±
2m(E o + 2Bx 2 − x 4 ) = ±
2m(h 2 − x 2 )(x 2 + g 2 ),
(B.26)
where
h
2
= B +
(B 2 + E o ) and g
2
= −B +
(B 2 + E o ).
(B.27)
The turning points are now given by x ± = ±h. The action is given by
J =
1
2π
pdx =
√
2m
π
h
−h
dx
(h 2 − x 2 )(x 2 + g 2 )
=
2
3
√
2m
π
h
κ
3
[κ
2 (K(κ) − E(κ)) + κ
2 E(κ)],
(B.28)
where κ 2 = (1−κ 2 ) and the modulus, κ, is defined as κ 2 =
h 2
h 2 +g 2 . From Eq. (B.28),
we obtain
˙
=
πh
√
2m κ K(κ)
.
(B.29)
and thus
(t) =
πh
√
2m κ K(κ)
t + (0).
(B.30)
The canonical transformation is obtained in the usual manner. Since p = m ˙
x, we
can write
h
x
dx
(h 2 − x 2 )(x 2 + g 2 )
=
2
m
t =
2κK(κ))
eπ
.
(B.31)
We therefore obtain
x = h cn
2K(κ))
π
, κ
.
(B.32)
If we substitute Eq. (B.32) into Eq. (B.26), we find