PHYS08200605006 D.K. Hazra - Homi Bhabha National Institute

CHAPTER 5. THE SCALAR BI-SPECTRUM DURING PREHEATING
instance, Ref. [114]). The Mathieu equation possesses unstable solutions that are known
to grow rapidly when the values of the parameters fall in certain domains known as the
resonant bands. As discussed in detail in Refs. [60, 63], sinceq ≪ 1 in the situation of our
interest, one falls in the narrow resonance regime. In such a case, the first instability band
is delineated by the condition 1−q < A k < 1+q, which turns out to be equivalent to the
condition
0 < k a < √ 3Hm. (5.15)
It should be emphasized here that the time evolution of the quantities A k and q are such
that, once a mode has entered the resonance band, it remains inside it during the entire
oscillatory phase.
Note that, in Eq. (5.11), we can neglect the term involvingk 2 providedk 2 /(m 2 a 2 ) ≪ 1.
This condition can be rewritten as
( ) 2 k H 2
≪ 1. (5.16)
aH m2 On the other hand, the condition to fall in the first instability band, viz. Eq. (5.15), can be
expressed as [63]
( ) 2 k H
≪ 1. (5.17)
aH 3m
Given that, H < m immediately after inflation, it is evident that the first of the above two
conditions will be satisfied if the second is. In other words, being in the first instability
band implies that one can indeed neglect the k 2 term in Eq. (5.11). But, clearly, this is
completely equivalent to ignoring the k 2 term in the original equation (1.16). Therefore,
we can conclude that, provided we fall in the first instability band (which is the case for
the range of modes and parameters of our interest), it is perfectly valid to work with the
super-Hubble solution (5.9) even during the preheating phase.
The above conclusion can also supported by the following arguments. As discussed
in Ref. [63], in the first instability band, the Floquet index is given by µ = q/2. In such a
case, the mode V k behaves as V k ∝ e µσ . However, in the situation of our interest, since
we have a time dependent Floquet index, the corresponding solution can be written as
(∫
V k (η) ∝ exp
)
µdσ ∝ a 3/2 (5.18)
which, in turn, implies that v k = V k /a 1/2 ∝ a. Further, since, f k = v k /z and z ∝ a during
preheating (i.e. if one makes use of the fact that ǫ 1 = 3/2 on the average), we arrive at
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