## Stephen Barnett and Paul Radmore

Print publication date: 2002

Print ISBN-13: 9780198563617

Published to Oxford Scholarship Online: January 2010

DOI: 10.1093/acprof:oso/9780198563617.001.0001

Show Summary Details
Page of

PRINTED FROM OXFORD SCHOLARSHIP ONLINE (www.oxfordscholarship.com). (c) Copyright Oxford University Press, 2017. All Rights Reserved. Under the terms of the licence agreement, an individual user may print out a PDF of a single chapter of a monograph in OSO for personal use (for details see http://www.oxfordscholarship.com/page/privacy-policy). Subscriber: null; date: 25 February 2017

# (p.245) Appendix 8 CONTOUR INTEGRALS

Source:
Methods in Theoretical Quantum Optics
Publisher:
Oxford University Press

The Cauchy residue theorem states that the integral anticlockwise around a closed contour in the complex plane of any function f(z) which is analytic apart from poles within the contour is equal to 2πi multiplied by the sum of the residues of f(z) at these poles. We use this theorem to evaluate some of the integrals required in Chapters 5 and 6.

First we consider the integral

(A8.1)
required in deriving (5.5.11) and (6.4.40). As written, this is the form required in Chapter 6; for those integrals in (5.5.10), we identify ω with Δ, W with W 0, and note that t in (A8.1) can be set equal to zero. We choose a contour consisting of the part of the real axis from −R to +R, closed by the semicircle R exp(iθ) in the lower half plane, as shown in Fig. A8.1. The lower half plane is chosen to ensure that the integrand tends to zero on the semicircle as R → ∞. The simple pole of the integrand in the lower half plane is at and the residue there is Letting R → ∞ gives
(A8.2)
where the minus sign arises because the contour is traversed in the clockwise direction. Note that if the integrand contained exp(i ωt) rather than exp(−i ωt), then we could close the contour in the upper half plane and obtain the same result (A8.2).

The second integral

(A8.3)
is required in deriving (6.4.44). We choose a contour consisting of the part of the real axis from −R to +R, indented by constructing a small semicircle of radius ρ

Fig. A8.1 A semicircular contour closed in the lower half plane.

(p.246)

Fig. A8.2 An indented semicircular contour closed in the upper half plane.

in the upper half plane, closed by the semicircle R exp(iθ) also in the upper half plane, as shown in Fig. A8.2. The simple pole of the integrand in the upper half plane is at Δ = iγ and the residue there is . The integral around the large semicircle tends to zero as R → ∞ while the integral around the small semicircle as ρ → 0 is
(A8.4)
Hence
(A8.5)

The third integral we evaluate is that in (6.5.21), required to establish the completeness of the dressed states . Consider

(A8.6)
where
(A8.7)
We proceed by separating the integrand of I 3 into partial fractions, so that
(A8.8)
The principal part integral F(ω) is defined for ω real. In order to use contour integration to evaluate I 3, we need to extend the definition of F(ω), and hence of , to include complex ω by analytic continuation. We formally evaluate the integral (A8.7) by closing a contour in the upper half plane, as for I 2, giving
(A8.9)
(p.247) where S(ω) is the sum of the residues of at the poles of W 2(Δ) in the upper half plane. Consider the function G(ω) defined, for values of ω in the upper half plane, by
(A8.10)
Evaluating this by contour integration in the upper half plane gives
(A8.11)
Comparing this with F(ω) given in (A8.9), we see that F(ω) becomes G(ω) + πiW 2(ω) in the upper half plane. Similarly in the lower half plane, G(ω) = 2πiS(ω) and F(ω) becomes G(ω) = πiW 2(ω). The integrand of I 3 in (A8.8) analytically continued into the upper half plane is therefore given by
(A8.12)
The second term in this expression has no poles in the upper half plane because ω − G(ω) has no zeros there, as we now show. Writing ω = x + iy and setting ω − G(ω) = 0 leads to
(A8.13)
in which the real and imaginary parts must each be zero. However, the imaginary part is strictly positive if y > 0, corresponding to the upper half plane, and hence there are no solutions. We now evaluate I 3 by contour integration closing the contour, as before, with a large semicircle of radius R in the upper half plane. We add to I 3 the integral of (A8.12) around the semicircle, this additional integral being identically zero in the limit R → ∞. The second terms of I 3 and of the integral of (A8.12) combine to give a zero contribution because has no poles in the upper half plane. Hence I 3 is given by the remaining integrals, as follows:
(A8.14)
(p.248) where we note that W 2(R exp(iθ)) must tend to a finite value and G(R exp(iθ)) must tend to zero as R → ∞ in order for F(w) to be finite. Since I 3 is real, the second term of (A8.14) must be zero, as it is purely imaginary. The first term is just I 3/2 while the last is simply 1/2. Hence I 3 = 1 as required.

The Cauchy residue theorem can also be used to sum particular infinite series. Suppose f(z) is a function which is analytic at the integers ,… and tends to zero at least as fast as as . Then

(A8.15)
In Chapter 6 we require the sum
(A8.16)
It appears that the terms in this sum do not tend to zero sufficiently fast as n increases to allow use of (A8.15). However, this can be overcome by writing S 1 as
(A8.17)
the terms of which are of order n −2 for large n and therefore tend to zero sufficiently fast as n → ∞ to allow the application of (A8.15). The simple poles of are at z = ±ω/ω0. The residues of at these points are both . Hence
(A8.18)
We also require the sum
(A8.19)
(p.249) Again, the terms are of order n −2 for large n as required. The double poles of are at . The residue of at either pole may be calculated, using the formula for double poles, as
(A8.20)
From (A8.15), summing the residues, we have
(A8.21)