From quantum ﬂuctuations to seed perturbationsAs the 4-by-4 symmetric tensor, the pertur-bations in metric tensor can have 10 independent degrees of freedoms: 1( g00)+3( g0i)+6(

Chapter 7

Cosmological linear perturbation theoryFrom quantum fluctuations to seed perturbations

So far, in this course, we have studied the expansion history and properties of the FRW universe: homo-geneous and isotropic expanding universe. Although accurately confirmed on large scales, our Universeis far from the FRW universe on scales smaller than a few tens of Mega-parsec. On these scales, theUniverse reveals the large-scale structure (LSS) that we can measure from the distribution of galaxies,or fluctuations of temperature and polarizations of CMB.

Studying the origin and evolution of the large-scale structure is at the forefont of the modern cosmologyresearch not only becasue it is directly related to studying our cosmic origin, but also because it encodeswealthy cosmological information. To unveil the cosmic mysteries of dark matter and dark energy, tomeasure the mass of neutrino mass, to decipher the physical conditions at the very first moment of theUniverse, we study the large-scale structure of the Universe.

Just like the case for the homogeneous, isotropic background universe, the basic theory describingthe generation and evolution of LSS is the general theory of relativity. In particular, focusing on theformation and evolution of large-scale structure at high redshifts (before the last-scattering time of theCMB), we can treat the large-scale structure as the small perturbations on top of the FRW backgrounduniverse. It is verified by the fact that the amplitude of CMB anisotropies is of 10s of micro-Kelvin,which yields δT ≡∆T/T ' 10−5.

In this chapter, we shall study the perturbed Einstein equation in the inflationary universe with singlescalar field. First, we study the left-hand side of the Einstein’s equation which must be applicable to thegeneral cases. We then study the right-hand side of the Einstein equation with perturbations generatedfrom single scalar field. This is the case for the inflationary universe, and we end this chapter by derivingthe initial conditions for the LSS that inflation predicts.

Here, we use the Latin letters for the space indices and Greek letters for the space-time indices.

7.1 Perturbations in space-time geometry

In this section, we focus our attention to the Einstein equation including small perturbations.

1

2 CHAPTER 7. PERTURBATION THEORY

7.1.1 Classification of perturbations

The space-time geometry is encoded in the metric tensor. For the background universe, we have shownthat the FRW metric

ds2 = −d t2 + a2(t) gi j(x)d x id x j (7.1)

uniquely describes the homogeneous, isotropic and expanding universe. For the discussion of this chap-ter, to avoid the notational clutter, we only focus on the spatially flat background universe:

ds2 = −d t2 + a2(t)δi jd x id x j . (7.2)

That is, we raise and lower the space indices by using δi j . As the 4-by-4 symmetric tensor, the pertur-bations in metric tensor can have 10 independent degrees of freedoms:

1(δg00) + 3(δg0i) + 6(δgi j) = 10 . (7.3)

Of course, all 10 perturbations are in general a function of spacetime coordinate xµ. In perturbationtheory, we find it convenient to decompose these 10 components into scalar-, vector-, and tensor-types.

The time-time component δg00(xµ) is a scalar function, and let’s call this A(xµ):

δg00(xµ) = 2A(xµ) . (7.4)

Next, the time-space components δg0i(xµ) form a spatial vector. Note that any spatial vectors V can befurther deomposed by the sum of the curl-free mode (V‖) and the divergence-free mode (V⊥)

V= V‖ +V⊥ ≡∇v +∇×ω , (7.5)

where the vector calculus identities∇×(∇v) = 0 and∇·(∇×ω) = 0 guarantee the curl- and divergence-free conditions. In Fourier analysis, we call curl-free modes and divergence-free modes, respectively,longitudinal and transverse modes, because the respective Fourier components are parallel and perpen-dicular to the wavevector k. We therefore, further decompose the time-space components as

δg0i(xµ) = a [∂iB(x

µ) + Ci(xµ)] , (7.6)

with one scalar function B(xµ) and the transverse vector C(xµ) satisfying ∇ ·C= 0.

Then, we are left with a spatial tensor: δgi j(xµ), that we shall decompose as two scalar modes (D(xµ)and E(xµ)) and one transverse vector function F i (∂i F

i = 0) and one traceless, transverse symmetrictensor function Hi j (Hi j = H ji and H i

i = Hi j, j = 0) as

δgi j(xµ) = a2

�

2D(xµ)δi j + 2�

∂i∂ j −13δi j∇2

�

E(xµ) + 2∂{i F j}(xµ) +Hi j(x

µ)�

. (7.7)

Here, we introduce

∂{i F j}(xµ)≡

12

�

∂i F j(xµ) + ∂ j Fi(x

µ)�

(7.8)

to ensure the symmetry of δgi j . Note that Hi j(xµ) has left with only two degrees of freedom: 6 (sym-metric) - 1 (traceless) - 3 (transverse) = 2.

To summarize, we have four scalar (A, B, D, E), four vector (Ci , Fi) and two tensor (Hi j) degrees offreedom to describe total 10 components of the metric perturbations. The final form of the perturbedmetric is

ds2 = −(1−2A)d t2+2a [∂iB + Ci] d td x i+a2�

(1+ 2D)δi j +�

∂i∂ j −13δi j∇2

�

2E + 2∂{i F j} +Hi j

�

d x id x j

(7.9)

7.1. PERTURBATIONS IN SPACE-TIME GEOMETRY 3

As we shall see later, in linear perturbation theory, the scalar-, vector-, tensor-type perturbations evolveindependently, and we can study them separately. The four vector modes correspond to the vorticitiesthat decay on all scales proportional to 1/a2(t), so we often neglect these contributions. The tensormode corresponds to the gravitational waves.

7.1.2 Choice of coordinate: Gauge transformation

When describing the background FRW universe, we can define the constant-time hypersurface by usingthe physical quantities such as temperature or density. It is because the temperature and density evolvesin time, and the homogeneity of the Universe demands that they must equal at equal time. This is notany more true for the Universe with perturbations.

At each point in space-time, we have a value for physical quantities, say ρ(t,x), and it is our discretionto divide the quantities to its background value (ρ) and perturbation (δρ):

ρ(t,x) = ρ(t) +δρ(t,x) (7.10)

In fact, there are infinitely many ways to make such a division, depending on how one defines theconstant-time hypersurface, the process that we call slicing. In addition to the choice of slicing, one’schoice of how to charting the constant-time hypersurface can also change the definition of perturbations,which is is called threading.

Such an ambiguity of discerning the perturbation from the backround is genuine in general relativity; itis one of the core idea of the general relativity that no observer occupies the special place of interpretingthe physics, and the viewpoint of all observers must be equivalent. We call this phenomena that themeaning of perturbation depends on the choice of coordinate the guage ambiguity, and the transforma-tion between different coordinate choices is called gauge transformation. In this section, we shall spellout the transformation between two different choice of coordinate system.

Let’s consider two coordinate systems xµ and yµ, differ by an infinitesimal displacement spacetimevector ξµ:

yµ = xµ + ξµ (7.11)

Let us further denote the displacement vector by its scalar components (ξ0 and ξ) and vector component(ζi , with ∂iζ

i = 0) so thatξi = ∂ iξ+ ζi . (7.12)

Scalar perturbations

For a scalar quantity S(xµ), we have one value at one space-time point. That is,

S(xµ) = S(yµ) , (7.13)

orS(x0) +δS(xµ) = S(y0) +δS(yµ) . (7.14)

Arranging the perturbation terms, we find that

δS(yµ) = δS(xµ) +�

S(x0)− S(y0)�

= δS(xµ)− ˙S(x0)ξ0 . (7.15)

That is, for the scalar perturbations, the gauge gransformation only cares about the differnece in slicing.It is the result of that the background quantities are only a function of time in the FRW universe.


Metric perturbations

For the metric tensor, the coordinate transformation is defined to be

gµν(y) =∂ xρ

∂ yµ∂ xσ

∂ yνgρσ(x)'

�

δρµ − ξρ,µ

��

δσν − ξσ,ν

�

gρσ(x) . (7.16)

Decomposing the metric asgαβ = gαβ +δgαβ , (7.17)

we find thatgµν(y) +δgµν(y) =

�

δρµ − ξρ,µ

��

δσν − ξσ,ν

�

�

gρσ(x) +δgρσ(x)�

, (7.18)

then to linear order in perturbation, we relate the perturbations as

δgµν(y) =δgµν(x) +�

gµν(x)− gνν(y)�

−�

δρµξσ,ν +δ

σνξ

ρ,µ

�

gρσ(x)

=δgµν(x)− gµν,αξα −

�

gµσ(x)ξσ,ν + gρν(x)ξ

ρ,µ

�

. (7.19)

In terms of the components, the gauge transformation becomes

δg00(y) = δg00(x)− 2 g0σξσ,0→ A(y) = A(x) + ξ0 (7.20)

for the time-time component, and

δg0i(y) = δg0i(x)−�

g0σξσ,i + giσξ

σ,0

�

= δg0i(x)−�

g00ξ0,i + gi jξ

j,0

�

= δg0i(x)−�

−ξ0,i + a2ξi

,0

�

(7.21)

for the time-space component, from which we can read the scalar part transformation:

B(y) = B(x) +1aξ0 − aξ , (7.22)

as well as the vector part transformation:

Ci(y) = Ci(x)− aζi . (7.23)

Finally, the space-space part becomes

δgi j(y) =δgi j(x)− gi j,αξα −

�

giσ(x)ξσ, j + gρ j(x)ξ

ρ,i

�

=δgi j(x)− 2a2Hδi jξ0 − a2

�

ξi, j + ξ

j,i

�

, (7.24)

from which we find that

2a2D(y)δi j +�

∂i∂ j −13∇2δi j

�

2a2E(y)

=2a2D(x)δi j +�

∂i∂ j −13∇2δi j

�

2a2E(x)− 2a2Hξ0δi j − 2a2∂i∂ jξ (7.25)

2a2∂{i F j}(y) = 2a2∂{i F j}(x)− 2a2∂{iζ j} (7.26)

Hi j(y) = Hi j(x) . (7.27)

That is, tensor perturbations are invariant under gauge transformation, and the gauge transformationfor the vector mode Fi is given as

Fi(y) = Fi(x)− ζi . (7.28)

Then, for the scalar modes, by selecting i 6= j case, we can identify

E(y) = E(x)− ξ . (7.29)

and the diagonal part gives

D(y) = D(x)−Hξ0 −13∇2ξ . (7.30)


7.1.3 How to deal with the gauge ambiguity

We have shown that besides the tensor perturbations, the metric perturbations transform under thechoice of coordinate. That is, the metric perturbation depends on the choice of gauge. Then, howshould proceed to calculate the perturbation theory analysis using the Einstein’s equation? There arethree routes.

• Gauge-ready formalism includes all perturbation variables all the way to the Einstein equation sothat one can study the effect of the choice of gauge at the final equation level.

• One can fix the gauge here by imposing specific conditions to the metric perturbations.

• One can come up with the gauge invariant variable, which is invariant under the choice of gauge,and proceed only with these variables.

Here, we discuss the second and third routes.

Gauge fixing: Newtonian (longitudinal) gauge

We have four scalar degrees of freedom for both scalar-mode and vector-mode metric perturbations, andthe gauge transformation involves the choice of two scalar two vector degree of freedom. Therefore,we are left only with two genuine degrees of freedom for scalar-type and vector-type perturbations.

We can identifying these two real degrees of freedom by imposing some conditions. One popular choiceof conditions for the scalar perturbations is to setting B = E = 0. First, let us show that these twoconditions uniquely fix the coordinate. If not, there must be non-zero coordinate shifts ξ0 and ξ betweentwo coordinate systems that satisfies B = E = 0. From Eq. (7.22) and Eq. (7.29), ξ0 and ξ must satisfy

ξ0 = a2ξ, ξ= 0, (7.31)

but the second equation implies ξ0 = 0. Therefore, B = E = 0 uniquely specify the coordinate systemfor the scalar perturbations.

We are then left with the two scalar functions A and D encoding the information of spacetime perturba-tion. This choice of coordinate is called Newtonian gauge, or longitudinal gauge, where the scalar partof metric perturbation becomes

ds2 = −(1− 2A)d t2 + a2(t)(1+ 2D)δi jd x id x j . (7.32)

With the similar logic, one can easily show that setting Fi = 0 uniquly specifies the coordinate for vectorperturbations.

But, watch out! Imposing two conditions does not necessarily fix the coordinate uniquely. For example,another popular gauge condition A = B = 0 (called synchronous gauge condition) does not fix thecoordinate uniquely. You will show that in the homework.

Gauge invariant variables

One can also construct a linear combination of variables so that the result stays invariant under thegauge transformation. For example, the combinations

Φ= −A+∂ [a(B − aE)]

∂ t, (7.33)


and

Ψ = D−13∇2E + aH[B − aE] (7.34)

stay invariant under the gauge transformation, therefore qualifies as the gauge invariant variables. Notethat Φ and Ψ coincide with, respectively, −A and −D in the Newtonian gauge.

For vector perturbations, the combinatin

Vi = Ci − aFi (7.35)

stays invariant under the gauge transformation.

Of course, one can construct infinitely(!) many gauge-invariant combinations, but the variables Φ andΨ has its historical importance, and called Bardeen’s potential and Bardeen’s curvature perturbation,respectively.

7.1.4 The curvature perturbation

One can also form the gauge invariant combination by using a combination of different physical quan-tities. One particularly important such combination is called ζ, normally referred as curvature pertur-bation:

ζ= D−13∇2E −H

δρ

ρ, (7.36)

because it corresponds to the curbature perturbation in the constant-energy-density gauge (gauge definedby δρ = 0).

The curvature perturbation is extremely useful in cosmology, because it is one quantity that connects theinflationary senario to the cosmological observables. Why? because the curvature perturbation staysconstant outside of the horizon when the background universe is barotropic ρ = ρ(P), which is certainlythe case when the universe is dominated by single component.

Let me give you a simple argument for that. For that, we need to accept the idea of separate Universewhich states that the physical quantities on scales larger than the comoving horizon evolve as if theyare the quantities in an indepdndent, separate universe with the local value of density and pressure.As we consider the constant density gauge δρ = 0, across the long scales, the density stays constant:ρ(t,x) = ρ(t). To see why it happens with the barotropic fluid, let’s keep the spatial dependence ofpressure, P(t,x). The metric of the local universe is give as

ds2 = −d t2 + a2(t) [1+ 2ζ(t,x)]δi jd x id x j , (7.37)

and the role of ζ can be absorbed to redefine the local scale factor:

a(t,x) = a(t) [1+ ζ(t,x)] . (7.38)

Then, the energy-momentum conservation law in the local universe is given as

ρ(t) = −3H(t,x) [ρ(t) + 3P(t,x)] = −3a(t,x)a(t,x)

[ρ(t) + 3P(t,x)] = −3�

a(t)a(t)

+ ζ(t,x)�

[ρ(t) + 3P(t,x)]

(7.39)Now that we require the universe being barotropic, P = P(ρ) = P(t), and the only space-varyingquantity in this equation is ζ(t,x). Of course, because this equation must hold at any place in theuniverse, this means that ζ must be independent of position (ζ(t,x) = ζ(t)), and, if so, one can always


re-scale the scale factor to set ζ= 0! Rhetorically, what is the meaning of spatially constant perturbation,other than the background?

Note that, combining with the energy-momentum conservation equation of ρ = −3H(ρ + P), one canalso write

ζ= D−13∇2E +

δρ

3(ρ + P)≡ψ+

δρ

3(ρ + P). (7.40)

Here, we use another popular notation of perturbation theory:

ψ= D−13∇2E , (7.41)

when defining the scalar mode of the spatial metric perturbation as

δgSi j = a2

�

2ψδi j + ∂i∂ j E�

. (7.42)

7.1.5 Perturbed Einstein tensor

Now that we understand the subtlety of imposing the gauge condition to the metric perturbation andthe gauge invariant variable, we shall proceed the following sections only with the gauge invariantvariables. Also, we shall ignore the vector fields in this course.

That is, we start from the metric perturbations including the Bardeen’s potential Φ and curvature per-turbations Ψ, as well as the, traceless and transverse, tensor perturbations hi j:

ds2 = −(1+ 2Φ)d t2 + a2�

(1+ 2Ψ)δi j + hi j

�

d x id x j . (7.43)

For this metric, the invverse matric is,

g00 = −(1− 2Φ), g i j =1a2

�

(1− 2Ψ)δi j − hi j�

. (7.44)

In this section, we shall compute the corresponding perturbed Einstein tensor.

Christoffel symbols

First, we need Christoffel symbols. Of course, we can calculate the Christoffel symbols from the defini-tion, but here we will use another technique using the geodesic equation.

As we have learned in the first chapter, the free-particle Lagrangian is given by

L = gµνd xµ

dλd xν

dλ= gµνPµPν (7.45)

In terms of the metric perturbation variables, they are

L = −(1+ 2Φ)E2 + a2�

(1+ 2Ψ)δi j + hi j

�

P i P j . (7.46)

Let us calculate the Euler-Lagrangian equation. First,

∂ L∂ xα

= −2Φ,αE2 +�

2aaδα0

�

(1+ 2Ψ)δi j + hi j

�

+ a2�

2Ψ,αδi j + hi j,α

�

P i P j . (7.47)


Then, from

∂ L∂ Pα

= −(1+ 2Φ)2Eδ0α + 2a2

�

(1+ 2Ψ)δi j + hi j

�

P jδiα , (7.48)

we also find

ddλ

�

∂ L∂ Pα

�

=− 2(1+ 2Φ)δ0α

dEdλ− 4Φ,β Eδ0

αPβ

+∂

∂ xβ�

2a2�

(1+ 2Ψ)δi j + hi j

��

P j Pβδiα ,+2a2

�

(1+ 2Ψ)δi j + hi j

� dP j

dλδiα . (7.49)

Gathering the terms to equate the Euler-Lagrange equation:

ddλ

�

∂ L∂ Pα

�

=∂ L∂ xα

, (7.50)

we find the geodesic equation for α= 0 case as Let’s gather the terms that we find, first, for α= 0,

dEdλ− ΦE2 + 2Φ,β EPβ +

§

aa(1− 2Φ)δi j + aa�

2Ψδi j + hi j

�

+12

a2�

2Ψδi j + hi j

�

ª

P i P j = 0 , (7.51)

and for α= `,

dP i

dλ+

1a2Φ,i E2 −

12

�

2Ψ ,iδ jk + h jk,i�

P j Pk +�

2Ψ,kδij + hi

j,k

�

P j Pk +�

2Hδij + 2Ψδi

j + hij

�

P j E = 0 .

(7.52)

By comparing the equation to the geodesic equation of the form,

dPµ

dλ+ Γµ

αβPαPβ = 0, (7.53)

we can read off the Christoffel symbols as following:

Γ 000 = Φ, Γ 0

0i = Φ,i , Γ0i j = a2

§

�

H(1− 2Φ+ 2Ψ) + Ψ�

δi j +�

Hhi j +12

hi j

�ª

, (7.54)

Γ i00 =

1a2Φ,i , Γ i

j0 =�

H + Ψ�

δij + hi

j , Γijk = Ψ,kδ

ij +Ψ, jδ

ik −Ψ

,iδ jk +12

�

hij,k + hi

k, j − h,ijk

�

. (7.55)

For the later conveneince, we shall divide the Christoffel symbol as the background:

Γ 0i j = a2Hδi j , Γ

ij0 = Hδi

j , (7.56)

and perturbations:

δΓ 000 = Φ, δΓ 0

0i = Φ,i , δΓ0i j = a2

§

�

2H(−Φ+Ψ) + Ψ�

δi j +�

Hhi j +12

hi j

�ª

(7.57)

δΓ i00 =

1a2Φ,i , δΓ i

j0 = Ψδij + hi

j , δΓijk = Ψ,kδ

ij +Ψ, jδ

ik −Ψ

,iδ jk +12

�

hij,k + hi

k, j − h,ijk

�

. (7.58)


Ricci tensor

Then, we compute the Ricci tensor

Rµν = Γαµν,α − Γ

αµα,ν + Γ

βµνΓ

αβα − Γ

βµαΓ

αβν = Rµν +δRµν . (7.59)

The background Ricci tensor is, as we have calculated earlier in the homework,

R00 = −3(H2 + H), Ri j = a2�

3H2 + H�

δi j , (7.60)

and we compute the perturbated Ricci tensor from

δRµν = δΓαµν,α −δΓ

αµα,ν + Γ

αβαδΓ

βµν + Γ

βµνδΓ

αβα − Γ

αβνδΓ

βµα − Γ

βµαδΓ

αβν . (7.61)

After some pages of calculation, we arrive at

δR00 =1a2∇2Φ− 3Ψ + 3HΦ− 6HΨ ,

δR0i =− 2�

Ψ −HΦ�

,i ,

δRi j =a2�

Ψ +H(−Φ+ 6Ψ) + 2(H + 3H2)(−Φ+Ψ)−1a2∇2Ψ

�

δi j − [Φ+Ψ],i j

+a2

2

�

hi j + 3Hhi j + 2�

H + 3H2�

hi j −1a2∇2hi j

�

. (7.62)

Ricci scalar

We can calculate the Ricci scalar as:

R= Rµνgµν = R+δR . (7.63)

Here, the background Ricci scalar is

R= Rµν gµν = 3(H2 + H) + 3(3H2 + H) = 6�

2H2 + H�

. (7.64)

and the perturbation is

δR= Rµνδgµν +δRµν gµν

=6Ψ + 6H�

4Ψ − Φ�

− 12(2H2 + H)Φ−2a2∇2 [Φ+ 2Ψ] . (7.65)

Einstein tensor

Finally, we perturb the Einstein tensor,

Gµν = Rµν −12

gµνR, (7.66)

around the background value:

G00 = 3H2, Gi j = −a2(3H2 + 2H)δi j (7.67)


which gives

δGµν = δRµν −12δgµνR−

12

gµνδR . (7.68)

Component-by-component, we find that

δG00 =6HΨ −2a2∇2Ψ (7.69)

δG0i =− 2�

Ψ −HΦ�

,i (7.70)

δGi j =a2�

−2Ψ + 2H(Φ− 3Ψ) + 2(2H + 3H2)(Φ−Ψ) +1a2∇2 (Φ+Ψ)

�


+a2

2

�

hi j + 3Hhi j − 2�

2H + 3H2�

hi j −1a2∇2hi j

�

(7.71)

We can also raise the first index to find out

δGµν = δgµσGσν + gµσδGσν, (7.72)

again, component by component, we have

δG00 =2H

�

HΦ− Ψ�

+2a2∇2Ψ (7.73)

δG0i =2

�

Ψ −HΦ�

,i (7.74)

δG i0 =

2a2

�

Ψ −HΦ�,i

(7.75)

δG ij =�

−2Ψ + 2H(Φ− 3Ψ) + 2(2H + 3H2)Φ+1a2∇2 (Φ+Ψ)

�

δij −

1a2[Φ+Ψ],i, j

+12

�

hij + 3Hhi

j − 2�

2H + 3H2�

hij −

1a2∇2hi

j

�

(7.76)

7.2 Inflationary perturbation theory

Now that we have the left-hand side of Einstein equation that is applicable for the general cases, weshall first apply the equation to the case for the single-field driven inflation models. To complete theEinstein equation,

Gµν = 8πGTµν , (7.77)

we need so supply energy momentum tensor from the scalar field. We find in the previous section thatthe scalar-field action

Sϕ =

∫

d4 xp

−g�

−12

gµν∇µϕ∇νϕ − V (ϕ)�

, (7.78)

yields the energy momentum tensor of

Tµν = ∂µϕ∂νϕ − gµν

�

12

gρσ∂ρϕ∂σϕ + V (ϕ)�

. (7.79)

In this section, we shall analyze the Einstein equation and study the generation of perturbations in theinflationary universe from quantum fluctuation of scalar field ϕ.

7.2. INFLATIONARY PERTURBATION THEORY 11

7.2.1 The energy-momentum tensor of the scalar field

In the FRW universe, background scalar field can only depend on time, ϕ(t), which dictates the energymomentum tensor in the form of

Tµν = ˙ϕ2δµ0δν0 − gµν

�

−12

˙ϕ2 + V (ϕ)�

. (7.80)

Using the background FRW metric, we find the background energy-momentum tensor:

T00 =12

˙ϕ2 + V (ϕ), Ti j = a2�

12

˙ϕ2 − V (ϕ)�

δi j . (7.81)

Including the perturbations to the scalar field,

ϕ(t,x) = ϕ(t) +δϕ(t,x), (7.82)

we perturb the energy momentum tensor to linear order as

δTµν =δϕ,µϕ,ν + ϕ,µδϕ,ν −δgµν

�

12

gρσϕ,ρϕ,σ + V (ϕ)�

− gµν

�

12δgρσϕ,ρϕ,σ +

12

gρσ�

δϕ,ρϕ,σ + ϕ,ρδϕ,σ

�

+ V ′(ϕ)δϕ�

. (7.83)

Component by component, we have explicitly calculated as

δT00 = ˙(δϕ) ˙ϕ + 2ΦV (ϕ) + V ′(ϕ)δϕ

δT0i = ˙ϕδϕ,i

δTi j = a2�

2Ψ�

12

˙ϕ2 − V (ϕ)�

+ ˙ϕ�

˙(δϕ)−Φ ˙ϕ�

− V ′(ϕ)δϕ�

δi j + a2�

12

˙ϕ2 − V (ϕ)�

hi j . (7.84)

We can also raise the first index to find out

δTµν = δgµσ Tσν + gµσδTσν, (7.85)

again, component by component, we have

δT00 =− ˙ϕ

�

˙(δϕ)−Φ ˙ϕ�

− V ′(ϕ)δϕ (7.86)

δT0i =− ˙ϕδϕ,i (7.87)

δT i0 =

1a2

˙ϕδϕ,i (7.88)

δT ij =

�

˙ϕ�

˙(δϕ)−Φ ˙ϕ�

− V ′(ϕ)δϕ�

δij (7.89)

7.2.2 Einstein equation

Background Einstein equation

Of course, the background Einstein equation

Gµν = 8πGTµν (7.90)


gives the Friedmann equations with the scalar field:

3H2 = 8πGρϕ = 8πG�

12

˙ϕ2 + V�

(7.91)

andaa= −

4πG3(ρϕ + 3Pϕ) = −

8πG3( ˙ϕ2 − V ) = H +H2 (7.92)

or H = −4πG ˙ϕ2, and 8πGV = 3H2 + H, that we have studied in the previous section.

Perturbed Einstein equation

Then, the perturbed Einstein equation

δGµν = 8πGδTµν (7.93)

become, component-by-component,

[00] 6HΨ −2a2∇2Ψ = 8πG

�

˙(δϕ) ˙ϕ + 2ΦV (ϕ) + V ′(ϕ)δϕ�

(7.94)

[0i] − 2�

Ψ −HΦ�

,i = 8πG ˙ϕδϕ,i (7.95)

[ij] a2�

−2Ψ + 2H(Φ− 3Ψ) + 2(2H + 3H2)(Φ−Ψ) +1a2∇2 (Φ+Ψ)

�


+a2

2

�

hi j + 3Hhi j − 2�

2H + 3H2�

hi j −1a2∇2hi j

�

=8πGa2�

2Ψ�

12

˙ϕ2 − V (ϕ)�

+ ˙ϕ�

˙(δϕ)−Φ ˙ϕ�

− V ′(ϕ)δϕ�

δi j + 8πGa2�

12

˙ϕ2 − V (ϕ)�

hi j . (7.96)

We then, integrating the 0i component once, obtain

−Ψ +HΦ= 4πG ˙ϕδϕ, (7.97)

and take the trace of the i j component, to get

− 2Ψ + 2H(Φ− 3Ψ) + 2(2H + 3H2)(Φ−Ψ) +23

1a2∇2(Φ+Ψ)

=8πG�

2Ψ�

12

˙ϕ2 − V (ϕ)�

+ ˙ϕ�

˙(δϕ)−Φ ˙ϕ�

− V ′(ϕ)δϕ�

. (7.98)

The left over i j equation becomes

−�

∂i∂ j −13δi j∇2

�

(Φ+Ψ) +a2

2

�

hi j + 3Hhi j − 2�

2H + 3H2�

hi j −1a2∇2hi j

�

=8πGa2�

12

˙ϕ2 − V (ϕ)�

hi j . (7.99)

Just like we decompose the scalar and tensor type in the metric, we also decompose the equation toscalr part and tensor part to find

−�

∂i∂ j −13δi j∇2

�

(Φ+Ψ) = 0 . (7.100)


for the scalar mode, and

a2

2

�

hi j + 3Hhi j − 2�

2H + 3H2�

hi j −1a2∇2hi j

�

= 8πGa2�

12

˙ϕ2 − V (ϕ)�

hi j (7.101)

for the tensor mode.

To summarize, here are five equations that we find from Einstein equation:

6HΨ −2a2∇2Ψ = 8πG

�

˙(δϕ) ˙ϕ + 2ΦV (ϕ) + V ′(ϕ)δϕ�

(7.102)

− Ψ +HΦ= 4πG ˙ϕδϕ, (7.103)

− 2Ψ + 2H(Φ− 3Ψ) + 2(2H + 3H2)(Φ−Ψ) +23

1a2∇2(Φ+Ψ)

=8πG�

2Ψ�

12

˙ϕ2 − V (ϕ)�

+ ˙ϕ�

˙(δϕ)−Φ ˙ϕ�

− V ′(ϕ)δϕ�

(7.104)

−�

∂i∂ j −13δi j∇2

�

(Φ+Ψ) = 0 (7.105)

hi j + 3Hhi j − 2�

2H + 3H2�

hi j −1a2∇2hi j = 16πG

�

12

˙ϕ2 − V (ϕ)�

hi j . (7.106)

The Eq. (7.105) dictates that in the single field inflation case, we have Φ = −Ψ, and we are left withsingle scalar degree of freedom. As we shall show later, this is in general the case when the anisotropicstress vanishes. We shall then use Φ to rewrite all equations above as

− 6HΦ+2a2∇2Φ= 8πG

�

˙(δϕ) ˙ϕ + 2ΦV (ϕ) + V ′(ϕ)δϕ�

(7.107)

Φ+HΦ= 4πG ˙ϕδϕ, (7.108)

2Φ+ 8HΦ+ 4(2H + 3H2)Φ= 8πG�

−2Φ�

˙ϕ2 − V (ϕ)�

+ ˙ϕ ˙(δϕ)− V ′(ϕ)δϕ�

(7.109)

hi j + 3Hhi j − 2�

2H + 3H2�

hi j −1a2∇2hi j = 16πG

�

12

˙ϕ2 − V (ϕ)�

hi j . (7.110)

We can further simplify the equations by using the background Friedmann equations, Eqs. (7.91)–(7.92), as

− 3HΦ+1a2∇2Φ− (3H2 + H)Φ= 4πG

�

˙(δϕ) ˙ϕ + V ′(ϕ)δϕ�

(7.111)

Φ+HΦ= 4πG ˙ϕδϕ, (7.112)

Φ+ 4HΦ+ (3H2 + H)Φ= 4πG�

˙ϕ ˙(δϕ)− V ′(ϕ)δϕ�

(7.113)

hi j + 3Hhi j −1a2∇2hi j = 0 . (7.114)

First, and foremost thing to notice is that the equation for the scalar mode and the tensor mode arecompletely decoupled. Although we don’t explicitly deal with the vector modes, the same is also truefor the vector mode. That is, in linear perturbation theory, the scalar, vector, and tensor modes evolveindepdendently from each other.


Using conformal time coordinate

It turns out that the analysis of the perturbations becomes a lot easier when we change the time variablesfrom cosmic time t to the conformal time η:

η=

∫

d ta

, (7.115)

for which we shall use ′ to indicate the derivative. By the way, that means the ϕ derivative must beindicated explicitly. From the definition, ∂t = (dη/d t)∂η = 1/a∂η, and for the linear derivatives, wesimply divide them by a. For the second derivative, we transform them as

x =1a∂

∂ η

�

1a∂ x∂ η

�

=1a

�

−a′

a2x ′ +

x ′′

a

�

=1a2

�

x ′′ −H x ′�

(7.116)

Here, we define,H = a′/a = a = aH. Note that

H =1a

H ′ =1a

�

Ha

�′=

1a2

�

H ′ −H 2�

. (7.117)

Therefore, in terms of the conformal time, the Einstein equations become

− 3H Φ′ +∇2Φ− (2H 2 +H ′)Φ= 4πG�

(δϕ)′ϕ′ + a2V,ϕ(ϕ)δϕ�

(7.118)

Φ′ +H Φ= 4πGϕ′δϕ, (7.119)

Φ′′ + 3H Φ′ + (2H 2 +H ′)Φ= 4πG�

ϕ′(δϕ)′ − a2V,ϕ(ϕ)δϕ�

(7.120)

h′′i j + 2H h′i j −∇2hi j = 0 . (7.121)

We already see one simplication that the 1/a2 factors in front of the Laplacian disappear!

With the conformal time, the background Friedmann equations become:

3H 2 = 8πG�

12ϕ′2 + a2V (ϕ)

�

(7.122)

H 2 −H ′ = 4πGϕ′2, (7.123)

and the equation of motion for the background scalar field becomes

ϕ′′ + 2H ϕ′ + a2V,ϕ(ϕ) = 0 . (7.124)

7.2.3 Scalar mode of metric perturbations

First, let us focus our attention on the scalar perturbations. From the Einstein equatino, we have

− 3H Φ′ +∇2Φ− (2H 2 +H ′)Φ= 4πG�

(δϕ)′ϕ′ + a2V,ϕ(ϕ)δϕ�

(7.125)

Φ′ +H Φ= 4πGϕ′δϕ, (7.126)

Φ′′ + 3H Φ′ + (2H 2 +H ′)Φ= 4πG�

ϕ′(δϕ)′ − a2V,ϕ(ϕ)δϕ�

(7.127)

First, from Eq. (7.126),

δϕ =Φ′ +H Φ4πGϕ′

. (7.128)


We then cancel the δϕ′ terms on the right hand side by subtracting Eq. (7.125) from Eq. (7.127):

Φ′′ + 6H Φ′ + 2(2H 2 +H ′)Φ−∇2Φ= −8πGa2V,ϕ(ϕ)δϕ = −2a2V,ϕ(ϕ)�

Φ′ +H Φϕ′

�

. (7.129)

We then use the equation of motion for the background field ϕ to simplify the equation as

Φ′′ + 6H Φ′ + 2(2H 2 +H ′)Φ−∇2Φ=�

2ϕ′′ + 4H ϕ′�

�

Φ′ +H Φϕ′

�

, (7.130)

or

Φ′′ + 2�

H −ϕ′′

ϕ′

�

Φ′ + 2�

H ′ −Hϕ′′

ϕ′

�

Φ−∇2Φ= 0= 0 . (7.131)

One can simplify the equations little more as

Φ′′ + 2�

ln�

aϕ′

��′Φ′ + 2H

�

ln�

Hϕ′

��′Φ−∇2Φ= 0 . (7.132)

To conver the equation to the conventional form, we first define the variable

u=aϕ′Φ (7.133)

in order to include the Hubble friction term into the solution (thus, cancel the linear derivative term).Then, it turns out that the left-over terms can be simplified in terms of the new variable:

θ ≡Haϕ′

(7.134)

that we study further in the hydrodynamical perturbation context in the next section. Using thesevariables, the equation of motion for the metric perturbation becomes

u′′ −∇2u−θ ′′

θu= 0 . (7.135)

Asymptotic solutions for metric perturbation

Let’s study assymptotic behavior of the metric perturbation during inflation. First, on small scales,R< η, the Laplacian term dominates over the θ ′′/θ term, and the evolution of the metric perturbationΦ follows the wave equation

u′′ −∇2u= 0 , (7.136)

whose solutions are spanned by the plane waves,

u(t,x) = ei(ωt−k·x), (7.137)

with the dispersion relation ω = |k|. On the other hand, on scales larger than the typical length scaleof the Universe, R2 > θ ′′/θ , the spatial variation subdominates and equation of motion reduces to

u′′ −θ ′′

θu= 0 . (7.138)

That is, the solutions are independent on scales, and we can find the time dependence of f = u/θ :

u′′ −θ ′′

θu= f ′′θ + 2 f ′θ ′ + f θ ′′ −

θ ′′

θf θ = f ′′θ + 2 f ′θ ′ = 0. (7.139)


Then,f ′′

f ′= −2

θ ′

θ, (7.140)

yields

f ′ = C1θ−2 , (7.141)

or

f = C2 + C1

∫

θ−2dη= C1

∫ η

η0

θ−2dη . (7.142)

Note that we take C2 into the integration bound. Let’s take a closer look at the integration. Using thebackground equation, we can transform the integration as

∫

θ−2dη=

∫

a2(ϕ′)2

H 2dη=

14πG

∫

a2(H 2 −H ′)H 2

dη . (7.143)

Mean while,�

a2

H

�′

=2a2H 2 − a2H ′

H 2, (7.144)

so using this, we simplify the solution as

f = C1

∫ η

η0

θ−2dη=C1

�

a2

H−∫ η

η0

a2dη

�

. (7.145)

Using the equation above, we can also transfer the solution for Φ. Namely, using the integral solution,we have

Φ=ϕ′

au=

ϕ′

aθ f =

ϕ′

aHaϕ′

C1

�

a2

H−∫ η

η0

a2dη

�

= C1

�

1−Ha2

∫ η

η0

a2dη

�

=C1

a

�

1a

∫

a2dη

�′

.

(7.146)Finally, we transform it back to t coordinate as

Φ=Ca

�

1a

∫

a2dη

�′

= Cdd t

�

1a

∫

ad t

�

= C

�

1−Ha

∫

ad t

�

. (7.147)

Note that, for the exact de-Sitter expansion phase, a(t) = eHt , metric perturbation vanishes: Φ = 0.During the slow-roll inflation phase, a(t) = eH(t)t , with slowly-varying H(t),

1a

∫

ad t =1a

∫

daH(a)

'1

H(a), (7.148)

and

Φ= Cdd t

�

1a

∫

ad t

�

' CdH−1

d t= −C

HH2= εC , (7.149)

proportional to the slow-roll parameter ε, which is nearly constant in time in slow-roll inflation model.


Asymptotic solution for scalar field perturbation

From Eq. (7.126), we can also calcualte the amplitude of the scalar field perturbations in terms of themetic perturbation. From Eq. (7.147),

δϕ =Φ′ +H Φ4πGϕ′

=C

4πGϕ′

�

−H −�

Ha2

�′∫

a2dη+H�

1−Ha2

∫

a2dη

��

=−C

4πGϕ′

�

�

Ha2

�′+H 2

a2

�∫

a2dη= −C

4πGϕ′

�

H ′ −H 2

a2

�∫

a2dη=Cϕ′

a2

∫

a2dη .

(7.150)

In terms of the time coordinate,

δϕ =C ˙ϕa

∫

ad t 'C ˙ϕH

. (7.151)

This result shows that δϕ stays almost constatnt on large scales as well, because in slow roll inflation,

4πG ˙ϕ2 = −H = εH2, (7.152)

which leads to

δϕ = C˙ϕH= C

s

ε

4πG(7.153)

that hardly evolves in time.

It is conventional to write the constant C in terms of the scalar perturbation:

C =H˙ϕδϕ =

Hϕ′δϕ , (7.154)

then, the metric perturbation is

Φ= εC = εH˙ϕδϕ (7.155)

7.2.4 Prirmordial curvature perturbation

Curvature perturbations

In order to connect the scalar perturbations generated during inflation to the initial seed scalar fluctu-ations for the large-scale structure, we need to device a quantity well defined outside of horizon afterthe horizon exit. One such candidate is

ζ≡ Ψ −Hδρ

ρ(7.156)

for which we have already shown that it stays constant in the barotropic Universe. This quantity is alsowell-motivated as the quantity of comparison, because we can calculate the density perturbations atany subsequant time. It is however, a little awkward to carry out the calculation with ζ during duringinflation, because the density contrast during inflation is

ρϕ =12ϕ2 + V, (7.157)

and its linear order perturbation is

δρϕ = ˙ϕδϕ + V ′(ϕ)δϕ . (7.158)


With ρ = −3H(Pϕ +ρϕ) = −3Hϕ2, we find an expression for ζ:

ζ= Ψ +H˙ϕδϕ + V ′(ϕ)δϕ

3H ˙ϕ2. (7.159)

As we have shown earlier, the scalar field perturbation δϕ stays constant on large scales, therefore,

ζ= Ψ +H˙ϕδϕ + V ′(ϕ)δϕ

3H ˙ϕ2→ Ψ +H

V ′(ϕ)δϕ3H ˙ϕ2

' Ψ −Hδϕ

˙ϕ. (7.160)

and we use the slow-roll condition V ′ ' −3H ˙ϕ in the last equality. This motivates us to define a quantity

R = Ψ −Hδϕ

˙ϕ(7.161)

which is the same as the ζ on large-scales, therefore, also conserved ontside the comoving horizon. Ofcourse, we can show that R is a gauge-invariant combination.

We call both ζ and R the curvature perturbation, because they are proportional to the spatial Ricciscalar (3)R in, respectively, constant-density gauge (δρ = 0) and comoving gauge (δϕ = 0). By theway, why call the gauge defined with δϕ = 0 comoving gauge? It is because T0i ∝ ρui ∝ δϕ,i!

Equation for Curvature perturbations

Let’s find out the equation of motion for R . First,

R = Ψ − Hδϕ

˙ϕ−H

ddt

�

δϕ

˙ϕ

�

, (7.162)

and we need to compute

Hddt

�

δϕ

˙ϕ

�

=− 4πGHH

�

δϕ ˙ϕ −δϕ ¨ϕ�

. (7.163)

From Eq. (7.111) and Eq. (7.112),

−3HΦ+1a2∇2Φ− (3H2 + H)Φ=4πG

�

˙(δϕ) ˙ϕ + V ′(ϕ)δϕ�

=4πG�

˙(δϕ) ˙ϕ − ¨ϕδϕ�

− 3H�

Φ+HΦ�

, (7.164)

and using that we find Eq. (??) can be writtedn in terms of Φ as

Hddt

�

δϕ

˙ϕ

�

= HΦ−HH

1a2∇2Φ . (7.165)

Then, we find the relation between R and Φ as

R =− Φ− Hδϕ

˙ϕ−HΦ+

HH

1a2∇2Φ

=− 4πG ˙ϕδϕ − Hδϕ

˙ϕ+

H

H

1a2∇2Φ=

H

H

1a2∇2Φ , (7.166)


and we revert the equation to write formally,

Φ=HH

a2∇−2R . (7.167)

Here, ∇2 is the inverse-Laplacian, that takes exactly what you do when finding gravitational potentialfrom the mass distribution. Finally, using Eq. (7.112) once again,

Φ+HΦ= 4πG ˙ϕδϕ

→ a2 HH∇−2

�

dd t

�

ln

�

HH

a2

��

R + R +HR�

= 4πG ˙ϕ

�

δϕ

˙ϕ

�

= −4πG ˙ϕ2

H(R +Φ) =

HH(R +Φ) .

(7.168)

Using the background equation,

dd t

�

ln

�

HH

a2

��

=HH−

HH+ 2H = 2

¨ϕ˙ϕ−

HH+ 2H , (7.169)

we further simplfy the equation as

R +�

2¨ϕ˙ϕ− 2

HH+ 3H

�

R −1a2∇2R = 0 . (7.170)

Using the conformal time coordinate, the equation becomes

R ′′ +�

2ϕ′′

ϕ′− 2H ′

H+ 2H

�

R ′ −∇2R =R ′′ + 2�

ln�

ϕ′aH

��′

R ′ −∇2R = 0 . (7.171)

Defining

z =aϕ′

H=

a ˙ϕH

, u= zR , (7.172)

the equation simplifies once again as

u′′ −∇2u−z′′

zu= 0 , (7.173)

which is known as Mukhanov-Sasaki equation.

The large-scale solution for this equation is,

R = C1 + C2

∫

z−2dη . (7.174)

Noting that,

z−2 =H2

a2 ˙ϕ2 = −4πGH2

a2H= −4πG

1εa2

, (7.175)

and the slow-roll parameter stays almost constant, the second term decays exponentially. Therefore, wecan see that the curvature perturbation R stays constant on large-scales.


7.2.5 Prirmordial tensor perturbation

The equation of motion for the tensor perturbation during inflation is given by Eq. (7.114)

hi j + 3Hhi j −1a2∇2hi j = 0 , (7.176)

or in terms of the conformal time:h′′i j + 2Hh′i j −∇

2hi j = 0 . (7.177)

Remember that we define the tensor perturbations as traceless, transverse part of the spatial metricperturbations:

hii = 0, hi j,i = 0 (7.178)

these two constraint equations leave only 6(symmetric 3× 3 tensor)− 1(traceless)− 3(transverse) = 2independent degrees of freedom for the tensor perturbations. Because of the transverse condition, it ismost convenient to write down these two degrees of freedom in the Fourier space:

hi j(t,x) =

∫

d3k(2π)3

hi j(t,k)eik·x. (7.179)

The tensor mode in Fourier space must also be symmetric hi j(t,k) = h ji(t,k), and the the traceless andtransverse conditions demand that

hii = 0, kihi j(k) = 0. (7.180)

First, from the transverse condition, we must be able to write the tensor modes in terms of two or-thonormal transverse vectors k⊥1 and k⊥2 (k⊥1 · k⊥2 = 0, |k⊥i|= 1) as

hi j(k) = αk⊥1,ik⊥1, j + βk⊥2,ik⊥2, j + γ�

k⊥1,ik⊥2, j + k⊥2,ik⊥1, j

�

, (7.181)

and the traceless condition dictates that α= −β . Therefore, in general we have two independent scalarfunctions α and β to characterize the tensor modes.

We often denote the two independent degrees of freedom as amplitude hp(t,k) of the two polarizationbasis εp

i j(k):

hi j(t,x) =∑

p=+,×

∫

d3k(2π)3

hp(t,k)εpi j(k)e

ik·x . (7.182)

Here, p = +,× are two polarization modes of the tensor modes

ε+i j(k) =k⊥1,ik⊥1, j − k⊥2,ik⊥2, j

ε×i j(k) =k⊥1,ik⊥2, j + k⊥2,ik⊥1, j (7.183)

which are normalized as∑

i j

εpi jε

p′

i j = 2δpp′ . (7.184)

The equation of motion for the tensor amplitudes are then given by

h′′p + 2Hh′p − k2hp = 0 (7.185)

Change of variable to vp = ahp/p

16πG leads to the form that we have seen before:

v′′p −a′′

avp − k2vp = 0 . (7.186)

7.3. QUANTUM ORIGIN OF THE SEED PERTURBATIONS 21

Note that this funny normalization is needed to treat the tensor amplitude as if it behaves as the canon-ical scalar field. The large-scale solution for the tensor equation is,

hp = C1 + C2

∫

a−2dη , (7.187)

whose second term decays exponentially. Therefore, we can, again, see that the tensor perturbations hpstay constant on large-scales.

7.3 Quantum origin of the seed perturbations

In the previous section, we have studied the growth and evolution of the perturbations in the inflationaryuniverse, and find that scalar modes such as field perturbation δϕ, gravitational potential perturbationΦ and curvature perturbation R (also ζ) as well as the tensor modes hp stay constant on large scales,where the conformal time variation ′′ is smaller compare to ∇2 operator. Then, what sets the constantvalue?

The answer comes naturally when we consider the quantum mechanical behavior of the perturbations.In this section, we shall study the quantum mechanics of the perturbations in an expanding background.The perturbation δϕ(t,x), for example, is defined as single dynamical degree of freedom at each space-time point; therefore, for this analysis, we need to use the quantum theory of fields.

In this section, we shall build up the quantum field theory in curved spacetime from the simplest, singledegree of freedom, quantume field theory, known as quantum mechanice. In particular, we shall reviewthe ladder operators that can raise or lower the quantum state of harmonic oscillator. We then extendthe quantum mechanics to the quantum field theory in flat Minkowski background; then to the FRWbackground.

7.3.1 Review of harmonic oscillator in quantum mechanics

Let’s recap the quantum mechanics of the one-dimensional harmonic oscillator. The dynamical variableis x(t) coordinate as a function of time t, and the Hamiltonian of the system is given as

H =p2

2m+ V ( x) =

p2

2m+

12

kx2 . (7.188)

Here, p = mx is the conjugate momentum p = ∂ L/∂ x , and k = mω2 is the spring constant. We indicatethe operator by tilde as O.

One can see the quantum mechanical effects when imposing the commutation relation:

[ x , p]≡ x p− p x = i (7.189)

Defining two operators

a =s

mω2

�

x + ip

mω

�

(7.190)

a† =s

mω2

�

x − ip

mω

�

, (7.191)


so that

aa† =mω2

�

x2 +p2

m2ω2−

imω[ x , p]

�

=1ω

H +12

(7.192)

a†a =mω2

�

x2 +p2

m2ω2+

imω[ x , p]

�

=1ω

H −12

. (7.193)

We then find an interesting commutation relation:�

a, a†�

= aa† − a†a = 1 , (7.194)

which yieldus�

H, a†�

=ω�

a, a†�

a† =ωa† (7.195)�

H, a�

=ω�

a†, a�

a = −ωa . (7.196)

The commutator relation above means that we can use the operators a† and a as, respectively, raisingand lowering operator for the energy state. We can also define the number operator N ≡ a†a. Why?Consider the quantum state |n⟩ associate with the energy level n:

N |n⟩= n|n⟩, H|n⟩= En|n⟩ . (7.197)

Because H =ω(N + 12), we have the relation

En =ω�

n+12

�

, (7.198)

which states that ∆E =ω. From the commutation relation, we calculate the energy states of a†|n⟩ anda|n⟩ as

H�

a†|n⟩�

=a†�

H|n⟩�

+�

H, a†�

|n⟩= Ena†|n⟩+ωa†|n⟩= (En +ω)�

a†|n⟩�

, (7.199)

H (a|n⟩) =a�

H|n⟩�

+�

H, a�

|n⟩= Ena|n⟩ −ωa|n⟩= (En −ω) (a|n⟩) . (7.200)

Then, using the uniqueness of eigenfunction for a given engenstate, we can write

a†|n⟩= α|n+ 1⟩, a|n⟩= β |n− 1⟩ . (7.201)

We can also determine α and β by requiring that the state is properly normalized. That is,

|α†|n⟩|2 = α2⟨n+ 1|n+ 1⟩= α2→ α2 = ⟨n|αα†|n⟩= ⟨n|�

N + 1�

|n⟩= n+ 1 (7.202)

|α|n⟩|2 = β2⟨n− 1|n− 1⟩= β2→ β2 = ⟨n|α†α|n⟩= ⟨n|N |n⟩= n (7.203)

Now what? We can successfully applying the lowering operator a to the state |n⟩ to lower the state:

(a)m|n⟩=Æ

n(n− 1)(n− 2) · · · (n−m+ 1)|n−m⟩ . (7.204)

Can we go down the state indefinitely? Nope! Because the state must be normalized,

n= ⟨n|N |n⟩= ⟨n|a†a|n⟩= |a|n⟩|2 ≥ 0 , (7.205)


which means that n cannot be negative, and the lowest energy state is |0⟩. Because there is no statelower than vacuum, we have the relation

a|0⟩= 0 , (7.206)

and we can raise the eneryg state by applying a†:

a†|0⟩= |1⟩ . (7.207)

Of course, one can rever Eqs. (7.190)–(7.191) to have

x =1

p2mω

�

a+ a†�

(7.208)

p =(−i)s

mω2

�

a− a†�

(7.209)

We then calcualte the root-mean-square value of the displacement x of the simple harmonic oscillatorin ground state as

⟨x2⟩= ⟨0|x2|0⟩=1

2mω⟨0|�

a2 + a†a+ aa† + (a†)2�

|0⟩=1

2mω. (7.210)

That’s all we need from quantum mechanics for now. Let’s move on to the quantum fields.

7.3.2 Review of quantum field: Klein Gordon field in the Minkowski spacetime

Equipped with the single degree-of-freedom quantum mechanics, we shall extend the theory to the field:ϕ(t,x) in the flat, Minkowski spacetime. Let us consider a scalar field associated with a particle withmass m. Following the standard procedure, we can promote

ηµνPµPν = −E2 + P2 = −m2 (7.211)

to the Klein-Gordon equation:�

∂ 2

∂ t2−∇2

�

ϕ = −m2ϕ . (7.212)

As we can see from the previous chapter, this equation of motion comes from the particle Lagrangiandensity:

L =12

�

ϕ2 − (∇ϕ)2�

−12

m2ϕ2 , (7.213)

and the conjugate momentum is

π(t,x) =∂L∂ ϕ

= ϕ(t,x) . (7.214)

Analogous to the quantum mechanics, we impose the commutation relation:�

ϕ(t,x), π(t,x′)�

= iδD(x− x′) , (7.215)

and obviously,�

ϕ(t,x), ϕ(t,x′)�

=�

π(t,x), π(t,x′)�

= 0 . (7.216)

Note that the commutators are defined all at equal time.


The general solution for the Klein-Gordon equation is written in terms of the plane waves:

ϕ(t,x) = e±i(ωk t−k·x), (7.217)

satisfying the on-shell condition:ωk =

p

k2 +m2 , (7.218)

with positive frequency modes. Keep in mind that the field ϕ(t,x) has a real value, we can write thesolution as

ϕ(t,x) =

∫

d3k(2π)3

1p

2ωk

�

ake−i(ωk t−k·x) + a†kei(ωk t−k·x)� , (7.219)

then, taking the time derivative yield

π(t,x) =

∫

d3k(2π)3

(−i)s

ωk

2

�

ake−i(ωk t−k·x) − a†kei(ωk t−k·x)� . (7.220)

Here, we take the specific normalization in order to have a proper commutation relation for the creation(raising) and annihilation (lowering) operators:

[ak, ak′] =�

a†k, a†

k′�

= 0 ,�

ak, a†k′�

= (2π)3δD(k− k′) , (7.221)

yielding the canonical commutation relation in Eq. (7.215).

In terms of the creation and annihilation operators, we calculate the Hamiltonian of the system as

H =12

∫

d3 x�

π2 + (∇ϕ)2 +m2φ2�

=

∫

d3k(2π)3

ωk

�

a†kak +

12

�

ak, a†k

�

�

. (7.222)

From the equation above, it is easy to check that�

H, a†k

�

=ωka†k,�

H, ak

�

= −ωkak . (7.223)

We can therefore, create or annihilate the particle of energy ωk by applying, respectively, a† or a. Ofcourse, the vacuum state is also defined as

a|0⟩= 0 . (7.224)

Finally, we calculate the vacuum expectation value of the field fluctuations at a given wavenumbe k asfollowing. Rewriting the ϕ(t,x) as

ϕ(t,x) =

∫

d3k(2π)3

1p

2ωk

�

ake−iωk t + a†−keiωk t

�

eik·k ≡∫

d3k(2π)3

ϕk(t)eik·k , (7.225)

we compute

⟨ϕkϕk′⟩= ⟨0|ϕkϕk′ |0⟩=1

2ωk⟨0|�

ake−iωk t + a†−keiωk t

� �

ak′e−iωk′ t + a†

−k′eiωk′ t

�

|0⟩

=1

2ωk⟨0|ake−iωk t a†

−k′eiωk′ t |0⟩

=1

2ωkei(ωk′−ωk)t⟨0|aka†

−k′ |0⟩

=1

2ωkei(ωk′−ωk)t⟨0|

��

ak, a†−k′�

+ a†−k′ ak

�

|0⟩

=(2π)31

2ωkδD(k+ k′) . (7.226)


7.3.3 Quantum fluctuations in the inflationary background: I. Scalar modes

We have shown that the curvature perturbation R satistying the equation of motion:

u′′ −∇2u−z′′

zu= 0, (7.227)

where

u= zR =a ˙ϕH

�

Ψ −Hδϕ

˙ϕ

�

= −a

�

δϕ +˙ϕHΨ

�

(7.228)

is the scale factor times the field fluctuations on the spatially-flat gauge (defined by Ψ = 0). Let’s denotethe general solution for the field u(η,x) as

u(η,x) =

∫

d3k(2π)3

�

akuk(η) + a†−ku∗k(η)

�

eik·x , (7.229)

where the Fourier-space mode function uk(η) satisfies following ordinary differential equation:

u′′k + k2uk −z′′

zuk = 0 . (7.230)

Note that uk does not depend on the direction in an isotropic universe. In order to see the quantummechanical effect, we promote the field to the quantum operator by adopting following canonical quan-tization conditions:

�

u(η,x), u′(η,x′)�

= iδD(x− x′) , (7.231)

with�

ak, a†k

�

= (2π)3δD(k− k′) . (7.232)

The two conditions above dictates the normalization of the mode function uk because�

u(η,x), u′(η,x′)�

=�

uk(η)u′k∗(η)− u∗k(η)u

′k(η)

�

δD(x− x′) . (7.233)

Therefore, the normalization condition for uk is

uk(η)u′k∗(η)− u∗k(η)u

′k(η) = i . (7.234)

On small scales, k→∞, limit, the equation of motion becomes that of the massless field in Minkowskispace,

u′′k + k2uk = 0 , (7.235)

where we have the normalized solution

uk(η) =e−ikη

p2k

, (7.236)

so thatak|0⟩= 0 (7.237)

corresponds to the usual Minkowski vacuum on small scales. This choice of vacuum state on smallscale is called Bunch-Davis vacuum state. Although this choice is not unique, it is reasonable, becausesmall-scale physics must indeed approach to Minkowski, in a sense that all curved spacetime is locallyMinkowskian.


As we have shown before, the large-scale solution is

uk = Ckz + Dkz

∫

z−2dη (7.238)

and the growing solution is

uk = Ckz = Cka ˙ϕH

. (7.239)

Note that the growing solution dictates that uk and u∗k have the same time dependence, and this allowsus to rewrite the large-scale Fourier modes as

u(η,x) =

∫

d3k(2π)3

�

akuk(η) + a†−ku∗k(η)

�

eik·x =

∫

d3k(2π)3

bka ˙ϕH

eik·x , (7.240)

withbk = Ck ak + C∗k a†

−k . (7.241)

Then,�

bk, b†k′�

=�

Ck ak + C∗k a†−k, C∗k′ a

†k′ + Ck′ a−k′

�

= 0! , (7.242)

and the large-scale Fourier modes are classical, Gaussian random variables with

⟨0|bkb†k|0⟩= (2π)

3|Ck|2δD(k− k′) . (7.243)

Coming back to the curvature perturbation, we find that

R(η,x) =

∫

d3k(2π)3

bkeik·x , (7.244)

is the classical random variables entirely specified by its power spectrum. Note that the reality of Rdemands that b†

k = b−k, which is indeed the case for Eq. (7.241). We then calculate

⟨R(k)R(k′)⟩= ⟨bk bk′⟩= ⟨bk b†−k′⟩= (2π)

3|Ck|2δD(k+ k′) . (7.245)

Let us find the solution applicable in the full range of scales for the slow-roll inflation. In slow-rollinflation, H and ˙ϕ varies slowly, and we can re-write

z′′

z=

H

a ˙ϕ

d2

dη2

�

a ˙ϕH

�

' 2a2H2�

1+ ε+32δ

�

(7.246)

in terms of the slow-roll parameters

ε= −HH2

, δ =¨ϕ

H ˙ϕ. (7.247)

With that, the equation of motion for uk becomes

u′′k +�

k2 − 2a2H2�

1+ ε+32δ

��

uk = 0 . (7.248)

To find the solution, let us rewrite the aH by using that during inflation H stays almost constant, andthe deviation is parameterized by ε = −H/H2 � 1. That is, δ ln H = −εδ ln a, and H ∝ a−ε, whichyields

η=

∫

d ta=

∫

daa2H

' −1

aH1

1− ε. (7.249)


Note that we define conformal time to be negative during inflation (but, note that conformal time alwaysincreases). Using this, we can write

2a2H2�

1+ ε+32δ

�

= 21

η2(1− ε)2

�

1+ ε+32δ

�

'2η2

�

1+ 3ε+32δ

�

, (7.250)

and the equation of motion becomes

u′′k +�

k2 −2η2

�

1+ 3ε+32δ

��

uk = 0= u′′k +�

k2 −1η2

�

ν2 −14

��

uk = 0 , (7.251)

with ν2 ≡ 9/4+6ε+3δ. The equation is the Bessel equaution and the general complex solution is givenas the Hankel function:

uk(η) =p

−η�

C1(k)H(1)ν (−kη) + C2(k)H

(2)ν (−kη)

�

. (7.252)

At the k→∞ limit, we want the solution to be Eq. (7.236). As the asymptotpic behavior of the Hankelfunctions are

H(1)ν (x � 1)'

√

√ 2πx

ei(x−π2 ν−π4 ), H(2)ν (x � 1)'

√

√ 2πx

e−i(x−π2 ν−π4 ) , (7.253)

so we take the Hankel function of the first kind, and set

C1(k) =pπ

2ei π2 (ν+

12 ) , (7.254)

to have

uk(η) =pπ

2ei π2 (ν+

12 )p

−ηH(1)ν (−kη) . (7.255)

Using the small-x behavior of the Hankel function,

H(1)ν (x � 1) =

√

√ 2π

e−i π2 2ν−32Γ (ν)Γ (3/2)

x−ν , (7.256)

we find the super-horizon solution for uk as

uk(τ) = ei(ν− 12)π2 2ν−

32Γ (ν)Γ (3/2)

1p

2k(−kη)

12−ν . (7.257)

Matching this equation to Eq. (7.239), we compute

Ck = ukH

a ˙ϕ= ei(ν− 1

2)π2 2ν−32Γ (ν)Γ (3/2)

�

H

a ˙ϕ

�

1p

2k(−kη)

12−ν . (7.258)

Note that, although there are time depending quantities in the expressin, Ck must be independent oftime for modes outside of horizon: k� η.

Finally, using the power spectrum of R from Eq. (7.245) we find that

PR(k) = |Ck|2 = 22ν−3�

Γ (ν)Γ (3/2)

�2�H˙ϕ

�21

2ka2

�

kaH(1+ ε)

�1−2ν

, (7.259)


and also compute the dimensionless power spectrum of R as

∆2R(k)≡

k3PR(k)2π2

=22ν−3�

Γ (ν)Γ (3/2)

�2�H˙ϕ

�2k2

4π2a2

�

kaH(1+ ε)

�1−2ν

' [1+ 2(2α− 1)ε+ 2αδ]�

H2π

�2�

H˙ϕ

�2�k

aH

�3−2ν

, (7.260)

to leading order in slow-roll parameters. Here, α≡ 2− log 2− γ' 0.729637.

One can also find the leading order expression

∆2R(k) =

�

H2π

�2�

H˙ϕ

�2�

�

�

�

�

k=aH

=�

H2π

�2�

−4πGH2

H

��

�

�

�

k=aH=

4πGε

�

H2π

�2�

�

�

�

k=aH(7.261)

by treating H and ˙ϕ as constants during the horizon crossing, and set η= −1/(aH). Then, the analysisgoes exactly parallel to the case for the gravitational waves in the next section.

Comparing the exact solution above during slow-roll inflation, we notice that we they coincide whencalculating the power spectrum ∆2

R at the time of horizon crossing k = aH.

The spectral tilt of the power spectrum is defined as

nR − 1≡d ln∆2

R

d ln k= 3− 2ν= −4ε− 2δ , (7.262)

evaluated at k = aH. Note that one can also calculate the spectral tilt from Eq. (7.261) directly.

Can we make sense of the result? Yes! Although, it’s only for the heuristic understanding, here’s thestory. The amplitude of Hawking radiatin from the de-Sitter space is

δϕ =H2π

, (7.263)

which generate the time shift of

δt =δϕ

˙ϕ(7.264)

corresponding to the fractional shift in time of

R 'δtt' Hδt =

H˙ϕ

H2π

, (7.265)

and the dimensionless power spectrum is the square of this quantity.

7.3.4 Quantum fluctuations in the inflationary background: II. Tensor modes

The analysis of tensor mode goes parallel to the scalar mode, except that it is much simplet! We startfrom promoting the tensor polarization amplitude as

vp(η,x) =

∫

d3k(2π)3

�

akvk,p(η) + a−kv∗k,p(η)�

eik·x (7.266)

where the amplitude satisfies the equation of

v′′k,p − k2vk,p −a′′

avk,p = 0 . (7.267)


Again, the small scale solution is Minkowskian:

vk,p(η) =e−ikη

p2k

. (7.268)

To find the general solution, we first calculate, to leading order in slow-roll parameter

a′′

a=

1η2(2+ 3ε) , (7.269)

then the the tensor equation of motion is

v′′k,p −�

k2 −1η2(2+ 3ε)

�

vk,p = v′′k,p −�

k2 −1η2

�

λ2 −14

��

vk,p == 0 . (7.270)

The case is exactly the same as the previous analysis except that now we have ν2 = 9/4 + 3ε. Thesuperhorizon solution for vk is therefore,

vk,p = ei(ν− 12)π2 2ν−

32Γ (ν)Γ (3/2)

1p

2k(−kη)

12−ν , (7.271)

and, therefore,

hk,p =p

16πGvk,p

a=p

16πGei(ν− 12)π2 2ν−

32Γ (ν)Γ (3/2)

1p

2ka(−kη)

12−ν . (7.272)

The tensor power spectrum is

Ph(k) = 4Ph,p(k) =4× 22ν−3�

Γ (ν)Γ (3/2)

�2 16πG2ka2

�

kaH(1+ ε)

�1−2ν

=64πG [1+ 2(α− 1)ε]H2

2k3

�

kaH

�3−2ν

, (7.273)

and its dimensionless power spectrum is

∆2h(k) =

k3Ph(k)2π2

= 64πG [1+ 2(α− 1)ε]�

H2π

�2� kaH

�3−2ν

, (7.274)

The spectral tilt of the tensor power spectrum is defined as

nT ≡d ln∆2

h

d ln k= 3− 2ν= −2ε , (7.275)

evaluated at k = aH. Evaluating at k = aH, the dimensionless tensor power spectrum is, to leadingorder,

∆2h(k) = 64πG

�

H2π

�2

(7.276)

which is indeed the amplitude of the Hawking radiatin! What is so striking for the primordial tensorperturbations is that by measuring them, we can directly infer the expansion rate of the Universe duringinflation. The expansion rate is, in turn, related to the energy scale of inflation!

Comparing to the scalar power spectrum, we find that

r =∆2

h(k)

∆2R(k)

= 16ε . (7.277)

That is, the tensor amplitude is suppressed by the slow-roll parameter. That’s why it is so hard to detectthe primordial tensor modes!

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