Physical Foundations of Cosmology

168 The very early universe

using the results (4.101), we find that

〈

φ^2 (x)

〉

=

1

2 π^2

∫

k^2

√

k^2 +m^2 φ(χ ̄)

(

1

2

+nk

)

dk. (4.102)

Vacuum contribution to VeffFirst we consider only vacuum fluctuations and set
nk= 0 .The integral in (4.102) is divergent ask→∞.To reveal the nature of
the divergence we regularize this integral by introducing the cut-off scalekc=M.
Taking into account thatm^2 φ(χ ̄)=V′′,we can rewrite the third term in (4.94) as

1

2

V′′′

〈

φ^2

〉reg

vac=

1

8 π^2

∂m^2 φ(χ ̄)

∂χ ̄

∫M

0

k^2 dk

√

k^2 +m^2 φ(χ ̄)

=

∂Vφ

∂χ ̄

, (4.103)

where

Vφ=

1

4 π^2

∫M

0

√

k^2 +m^2 φ(χ ̄)k^2 dk≡

I

(

mφ(χ ̄)

)

4 π^2

(4.104)

is simply equal to the energy density of the vacuum fluctuations.
Using (4.103), (4.94) becomes

χ ̄;α;α+Veff′(χ ̄)= 0 , (4.105)

whereVeff(χ ̄)=V+Vφis the one-loop effective potential. The integralIwhich
enters (4.104) can be calculated exactly:

I(m)=

1

8

[

M

(

2 M^2 +m^2

)√

M^2 +m^2 +m^4 ln

m

M+

√

M^2 +m^2

]

.

Taking the limitM→∞,we obtain the following expression for the effective
potential:

Veff=V+V∞+

m^4 φ(χ ̄)

64 π^2

ln

m^2 φ(χ ̄)

μ^2

, (4.106)

where the divergent term

V∞=

M^4

4 π^2

+

m^2 φ

16 π^2

M^2 −

m^4 φ

32 π^2

ln

2 M

e^3 /^4 μ

+O

(

1

M^2

)

can be absorbed by a redefinition of constants in the original potential. For instance,
in the case of the renormalizable quartic potential

V(χ ̄)=

λ 0

4

χ ̄^4 +

m^20

2

χ ̄^2 + 0 , (4.107)