64
2 The Discrete Spectrum and the Continuum
The crossing relation of S (k) arises from Eqs. (2.7) and (2.182):
S (−k
∗ ) = S (k)
∗ .
(2.190)
Using (2.190) in Eq. (2.189) one obtains:
(S (k) − 1)e ikR (s) =
1
iπ
P
+∞
0
(S (k ) − 1)e ik R (s)
k − k
+
(S (k ) ∗ − 1)e −ik R (s)
−k − k
dk
− 2
n
Res
(S (k ) − 1)e ik R (s)
k − k
k =k n
.
(2.191)
Equation (2.191) can then be written as a function of real and imaginary parts of
S (k), which one will also express as an energy integral:
((S (E) − 1)e
ikR (s)
) =
1
π
P
+∞
0
(S (E ) − 1)e ik R (s)
E − E
dE
−2
n
Res
(S (k ) − 1)e ik R (s)
k − k
k =k n
. (2.192)
Let us consider the convergence properties of the integral of Eq. (2.192). One has
|S (E)| = 1 on the real E-axis, so that one only has to consider the convergence of
the integral of Eq. (2.192) for E → +∞. According to Eqs. (2.165) and (2.166):
S (E
) − 1 = O(ln(E
) E
) ,
(2.193)
so that integrand in Eq. (2.189) is equal O(ln(E ) E −3/2 ) for E → ±∞ and
is absolutely converging. However, the latter convergence might be too slow, as
experimentally accessible energies are only a few hundreds of MeV. Hence, it would
be convenient to accelerate the convergence of the integral in Eq. (2.192). For this,
following the method used in Ref. [44] in the context of the dispersion relations of
charged particles, one will consider Eq. (2.192) at two different energies E and E 0 ,
with the E 0 -dependent formula subtracted from the E-dependent equation:
((S (E) − 1)e
ikR (s) − (S (E 0 ) − 1)e
ik 0 R (s)
)
= (E − E 0 )
1
π
P
+∞
0
(S (E ) − 1)e ik R (s)
(E − E)(E − E 0 )
dE
−2(k − k 0 )
n
Res
(S (k ) − 1)e ik R (s)
(k − k)(k − k 0 )
k =k n
. (2.194)