
Quantum Vacuum Energy and the Casimir Effect Guim Aguad´e Gorgori´o
step of the damping function regularization presented is
on the use of the Abel-Plana formula
∞
X
n=0
f(n)−
∞
Z
0
f(t)−1
2f(0) = i
∞
Z
0
f(it)−f(−it)dt
e2πt −1
(30)
where we identify f(x) = F(x)e−δω(x).
The convergence of the right-hand side integral lets us
apply the limit δ→0 directly. What we need to un-
derstand is evaluation of F(it)−F(−it) by the means
of complex calculus. The square roots will have branch-
ing points ±iy to be rounded, and so F(it)−F(−it) =
2ipt2−y2.
The intermediate step of equation (16) runs through
(x= 2πt). What remains is the evaluation of a integral
by parts and a well known integral to be evaluated by
means of complex calculus or the gamma function and
the Riemann zeta function.
∞
Z
0
dxx3
ex−1= Γ(4)ζ(4) = π4
15 .(31)
VI.C The actual expression used by Casimir was
∞
X
(0)1
F(n)dn −
∞
Z
0
F(x)dx =−1
12F0(0) + 1
720F000(0) + ...
(32)
where the subindex (0) accounts for the non polariza-
tion of the n= 0 mode, and so contains the first term
(1/2)F(0). Casimir took into account the space integrals
and so used an already integrated expression for F
F(u) =
∞
Z
0
dx(x+u2)(1/2)f(n2π/akm) (33)
with kmbeing in fact the inverse regularization pa-
rameter. With this, the only non-vanishing derivative
is F000(0) = −4, as higher derivatives are in terms of
(π/akm), and go to zero as akm=a/δ >> 1. From here,
obtaining the Casimir force (17) is direct.
VI.D The first step in the zeta function regulariza-
tion is to rewrite (20) in order to obtain ζ(2s−3). We
use again the polar coordinates (k⊥, ϕk), and the change
k⊥=y(πn/a). We may see
k2
⊥+πn
a2!
1
2−s
=y2+ 1
1
2−sπn
a1−2s.(34)
The other (πn/a)2come from dk1dk2=k⊥dk⊥=
(πn/a)y(πn/a)dy. The demonstration of the functional
or reflection equation (23) is not trivial nor short [6]. The
key observation of the extension of ζto the whole com-
plex plane is the observation that in 0 < Re(s)<1, ζ(s)
has a symmetry in relation with ζ(1 −s), and together
with the expansion of Γ in the complex plane and many
other developments we obtain (23).
Once implemented the ζR(−3) value, the sum returns
(π/a)3(1/120). The remaining integral converges under
the limit as
lim
s→0
∞
Z
0
dyy(y2+ 1)(1/2)−s=−1
3(35)
and so the final result for the energy is (16), as ex-
pected.
Acknowledgments
Foremost, I would like to express my gratitude to my
advisor Joan Sol`a for his valuable time and patience on
my doubts and ignorance, together with his respect and
understanding of my personal situation. I also thank
my university colleagues for everyday supporting and my
family for being always by my side.
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