9.3 quantization of fields in curved space...

9.3 Quantization of Fields in Curved Space 213

which might arise from an interaction such as

L = 12m

2P AiA

i � 4p⇡Ai@

i�. (9.30)

The sign of the mass term is wrong, however.

9.3 Quantization of Fields in Curved Space

Some good references for these ideas are:

Einstein’s Gravity in a Nutshell by ZeeQuantum Fields in Curved Space by Birrell and Davies,Conformal Field Theory by Di Francesco, Mathieu, and Senechal

Let’s focus on scalar fields for simplicity. We usually expand a scalar fieldin flat space as Fourier might have (1.56)

�(x) =

Zd3pp

(2⇡)32p0

ha(p) eip·x + a†(p) e�ip·x

i. (9.31)

The field � obeys the Klein-Gordon equation

(r2 � @20 �m2)�(x) ⌘ (2�m2)�(x) = 0 (9.32)

because the flat-space modes, which have p2 = �m2,

fp(x) = eip·x (9.33)

do

(r2 � @20 �m2) fp(x) ⌘ (2�m2) fp(x) = 0. (9.34)

In terms of these functions fp(x), the field is

�(x) =

Zd3pp

(2⇡)32p0

ha(p) fp(x) + a†(p) f⇤

p (x)i. (9.35)

In terms of a discrete set of modes fn, it is

�(x) =X

n

han(p) fn(x) + a†n f

⇤n(x)

i. (9.36)

The action for a scalar field in a space described by the metric gik is

S = � 1

2

Z pg d4x

hgik�,i�,k +

�m2 + ⇠R

��2

i(9.37)

in which R is the scalar curvature, ⇠ is a constant, commas denote derivativesas in �,k = @k�, gij is the inverse of the metric gij , and g is the absolute

214 E↵ective field theories and gravity

value of the determinant of the metric gij . If the spacetime metric is gij , theninstead of (9.32), the field � obeys the covariant Klein-Gordon equation

@ihp

g gij @j �(x)i��m2 + ⇠R

�pg �(x) = 0. (9.38)

However, to simplify what follows, we will now set ⇠ = 0.To quantize in the new coordinates or in the gravitational field of the

metric gij , we need solutions f 0n(x) of the curved-space equation (9.38)

@ihp

g gij @j f0n(x)

i�m2pg f 0

n(x) = 0 (9.39)

which we label with primes to distinguish them from the flat-space solutions(9.33). We use these solutions to expand the field in terms of curved-spaceannihilation and creation operators, which we also label with primes

�(x) =X

n

ha0n(p) f

0n(x) + a0†n f 0⇤

n (x)i. (9.40)

The flat-space modes obey the orthonormality relations

(fp, fq) = i

Zd3x

hf⇤p (x) @tfq(x)�

�@tf

⇤p (x)

�fq(x)

i

= i

Zd3x

he�ipx (�iq0)eiqx � ip0e�ipx eiqx

i

= 2p0(2⇡)3�3(p� q),

(9.41)

(f⇤p , f

⇤q ) = � 2p0(2⇡)3�3(p� q) and (fp, f

⇤q ) = 0. (9.42)

In terms of discrete modes, the flat-space scalar product is

(fn, fm) = i

Zd3x

hf⇤n(x) @tfm(x)�

�@tf

⇤n(x)

�fm(x)

i= �nm (9.43)

and its orthonormality relations are

(fn, fm) = �nm, (f⇤n, f

⇤m) = � �nm and (fn, f

⇤m) = 0. (9.44)

The scalar product (9.43) is a special case of more general curved-spacescalar product

(f, g) = i

Z

S

pgS d3S va

hf⇤(x) @ag(x)�

�@af

⇤(x)�g(x)

i(9.45)

in which the integral is over a spacelike surface S with a future-pointingtimelike vector va, and gS is the absolute value of the spatial part of themetric gik. This more general scalar product (9.45) is hermitian

(f, g)⇤ = (g, f). (9.46)


It also satisfies the rule

(f, g)⇤ = � (f⇤, g⇤). (9.47)

One may use Gauss’s theorem to show (Hawking and Ellis, 1973, section2.8) that this inner product is independent of S and v. The curved-spacemodes fn(x) obey orthonormality relations

(f 0n, f

0m) = �nm, (f 0⇤

n , f 0⇤m) = � �nm, and (f 0

n, f0⇤m) = 0 (9.48)

like those (9.44) of the flat-space modes.The flat-space modes fn(x) = eipnx naturally describe particles of mo-

mentum pn in flat Minkowski space. In curved space, however, there are ingeneral no equally natural modes. We can consider other complete sets ofmodes f 00

n(x) that are solutions of the curved-space Klein-Gordon equation(9.39) and obey the orthonormality relations (9.48). We can use any of thesecomplete sets of mode functions to expand a scalar field �(x)

�(x) =X

n

anfn(x) + a†nf⇤n(x)

�(x) =X

n

a0nf0n(x) + a0†nf

0⇤n (x)

�(x) =X

n

a00nf00n(x) + a00†n f 00⇤

n (x).

(9.49)

The curved-space orthonormality relations (9.48) imply that

(f`,�) =X

n

an(f`, fn) + a†n(f`, f⇤n) = a`

(f 0`,�) =

X

n

a0n(f0`, f 0

n) + a0†n (f0`, f 0⇤

n ) = a0`

(f 00`,�) =

X

n

a00n(f00`, f 00

n) + a00†n (f 00`, f 00⇤

n ) = a00`

(9.50)

and that

(f⇤`,�) =

X

n

an(f⇤`, fn) + a†n(f

⇤`, f⇤

n) = �a†`

(f 0⇤`,�) =

X

n

a0n(f0⇤`, f 0

n) + a0†n (f0⇤`, f 0⇤

n ) = �a0†`

(f 00⇤`,�) =

X

n

a00n(f00⇤`, f 00

n) + a00†n (f 00⇤`, f 00⇤

n ) = �a00†`.

(9.51)

The completeness of the mode functions lets us expand them in terms of


each other. Suppressing the spacetime argument x, we have

f 0j =

X

i

�↵jifi + �ji f

⇤i

�. (9.52)

The curved-space orthonormality relations (9.48) let us identify these Bo-goliubov coe�cients

(f`, f0j) =

X

i

h↵ji(f`, fi) + �ji (f`, f

⇤i )i= ↵j`

(f⇤`, f 0

j) =X

i

h↵ji(f

⇤`, fi) + �ji (f

⇤`, f⇤

i )i= ��j`.

(9.53)

To find the inverse relations, we use the completeness of the mode functionsf 0ito expand the fj ’s

fj =X

i

�cjif

0i + dji f

0⇤i

�(9.54)

and then use the orthonormality relations (9.48) to form the inner products

(f 0`, fj) =

X

i

hcji(f

0`, f 0

i) + dji (f0`, f 0⇤

i )i= cj`

(f 0⇤`, fj) =

X

i

hcji(f

0⇤`, f 0

i) + dji (f0⇤`, f 0⇤

i )i= �dj`.

(9.55)

The hermiticity (9.46) of the scalar product tells us that the cj`’s are relatedto the ↵’s

cj` = (fj , f0`)⇤ = ↵⇤

`j. (9.56)

The hermiticity (9.46) of the scalar product and the rule (9.47) show that

dj` = � (f 0⇤`, fj) = � (fj , f

0⇤`)⇤ = (f⇤

j , f0`) = ��`j . (9.57)

So the inverse relation (9.54) is

fj =X

i

�↵⇤ijf

0i � �ij f

0⇤i

�. (9.58)

The formulas (9.52 & 9.58) that relate the mode functions of di↵erent metricsare known as Bogoliubov transformations.The vacuum state for a given metric is the state that is mapped to zero

by all the annihilation operators. Our formulas (9.51) for the annihilationand creation operators

a` = (f`,�) and a0`= (f 0

`,�)

a†`= � (f⇤

`,�) and a0†

`= � (f 0⇤

`,�)

(9.59)


let us express the annihilation and creation operators for one metric in termsof those for a di↵erent metric. Thus, using our formula (9.52) for the f 0’s interms of the f ’s, we find

a0j = (f 0j ,�) =

X

i

h↵ji (fi,�) + �ji (f

⇤i ,�)

i=

X

i

⇣↵ji ai � �ji a

†i

⌘. (9.60)

Our formula (9.58) for the f ’s in terms of the f 0’s gives us the inverse relation

aj = (fj ,�) =X

i

h↵⇤ij (f

0i ,�)� �ij (f

0⇤i ,�)

i=

X

i

⇣↵⇤ij a

0i + �ij a

0†i

⌘. (9.61)

The annihilation operators a0jdefine the vacuum state |0i0 by the rules

a0j |0i0 = 0 (9.62)

for all modes j. Thus our formula (9.61) for aj says that

aj |0i0 =X

i

⇣↵⇤ij a

0i + �ij a

0†i

⌘|0i0 =

X

i

�ij a0†i|0i0 (9.63)

The adjoint of this equation is

8h0|a†j= 8h0|

X

i

a0i�⇤ij . (9.64)

Thus the mean value of the number operator a†jaj in the |0i0 vacuum is

8h0|a†jaj |0i0 = 8h0|

X

i

a0i�⇤ij

X

k

�kj a0†k|0i0. (9.65)

The commutation relations

[a0i, a0†k] = �ik (9.66)

and the definition a0j|0i0 = 0 (9.62) of the vacuum |0i0 imply that the average

number (9.65) of particles of mode j in the (normalized) vacuum |0i0 is8h0|a†

jaj |0i0 = 8h0|

X

ik

�⇤ij�kj �ik|0i0 =

X

i

�⇤ij�ij . (9.67)

It follows that the vacuum of one metric contains particles of the othermetric unless the Bogoliubov matrix

�j` = � (f⇤`, f 0

j) (9.68)

vanishes. The value of �j` in the Minkowski-space scalar product (9.41) is

�j` = � (f⇤`, f 0

j) = �i

Zd3x

hf`(x) @tf

0j(x)�

�@tf`(x)

�f 0j(x)

i. (9.69)


This integral for �j` is nonzero, for example, when the functions f` and f 0j

have di↵erent frequencies but are not spatially orthogonal.An example due to Rindler. Let us consider the two metrics

ds2 = dx2 + dy2 + dz2 � dt2 = dr2 + dy2 + dz2 � r2du2 (9.70)

in which y and z are the same, r plays the role of x and u that of time. Thefirst metric has gik = ⌘ik and g = | det(⌘)| = 1. The second metric has

gik =

0

BB@

�r2 0 0 00 1 0 00 0 1 00 0 0 1

1

CCA and g = r2. (9.71)

The inverse metric is

gik =

0

BB@

�r�2 0 0 00 1 0 00 0 1 00 0 0 1

1

CCA . (9.72)

The equation of motion is

@ihp

ggik@kfi= m2pgf. (9.73)

So we must solve

@u(�r�1@uf) + @r(r@rf) + @y(r@yf) + @z(r@zf) = m2rf. (9.74)

Now u, r, y, z are independent coordinates, so this equation of motion is

� r�1@2uf + @r(r@rf) + r@2

yf + r@2zf = m2rf (9.75)

or

� r�2@2uf + r�1@rf + @2

rf + @2yf + @2

zf = m2f. (9.76)

Let’s make f a plane wave in the y and z directions

f(u, r, y, z) = f(u, r)eiypy+izPz . (9.77)

If we also set

M2 = m2 + P 2y + P 2

z , (9.78)

then we must solve

� r�2@2uf + r�1@rf + @2

rf = M2f (9.79)

9.4 Accelerated coordinate systems 219

where f = f(u, r). We now make the Daniel-Middleton transformation,separating the dependence of f upon r and u

f(u, r) = a(u)b(r). (9.80)

Our di↵erential equation (9.79) reduces to

� r�2ab+ r�1ab0 + ab00 = M2ab (9.81)

in which dots denote @u and primes @r. Dividing by a, we get

� a

a

b

r2+

b0

r+ b00 �M2b = 0. (9.82)

As a(u), we choose

a(u) = ei!u anda

a= � !2. (9.83)

Then our equation (9.82) for b(r) is

b00 +b0

r�⇣M2 � !2

r2

⌘b = 0 (9.84)

or equivalently

r2b00 + rb0 �⇣M2r2 � !2

⌘b = 0. (9.85)

Yi’s solution is a modified Bessel function I⌫(z) for imaginary ⌫.Bogoliubov’s � matrix is

�j` = � (f⇤`, f 0

j). (9.86)

The scalar product (9.45) uses the metric of one, not both, of the solu-tions. If we choose the flat-space metric, then we need to write the solutionf 0j(u, r, y, z) in terms of the flat-space coordinates t, x, y, z. The mean occu-

pation number (9.67) is the sumX

j

8h0|a†jaj |0i0 =

X

ij

|�ij |2 =X

ij

|(f⇤j , f

0i)|2. (9.87)

9.4 Accelerated coordinate systems

We recall the Lorentz transformations

t0 = �(t� vx), x0 = �(x� vt), y0 = y, and z0 = z

t = �(t0 + vx0), x = �(x0 + vt0), y = y0, and z = z0(9.88)


in which � = 1/p1� v2. The velocities (in the x direction) are

u =dx

dtand u0 =

dx0

dt0=

�(dx� vdt)

�(dt� vdx)=

(u� v)

(1� vu). (9.89)

So the accelerations (in the x direction) are

a =du

dt

a0 =du0

dt0= d

h (u� v)

(1� vu)

i��(dt� vdx)

=h du

(1� vu)+

(u� v)vdu

(1� vu)2

i��(dt� vdx)

=hdu(1� vu)

(1� vu)2+

(u� v)vdu

(1� vu)2

i��(dt� vdx)

=du(1� v2)

�(1� vu)2(dt� vdx)

=(1� v2)a

�(1� vu)2(1� vu)

=(1� v2)a

�(1� vu)3=

a

�3(1� vu)3.

(9.90)

Now we let the acceleration a0 be a constant. That is, the acceleration inthe (instantaneous) rest frame of the frame moving instantaneously at u = vin the laboratory frame is a constant, a0 = ↵. In this case, since u = v, weget an equation (Rindler, 2006, sec. 3.7)

a0 = ↵ =a

�3(1� vu)3=

a

�3(1� v2)3=

a

(1� v2)3/2

=1

(1� v2)3/2du

dt=

1

(1� u2)3/2du

dt=

d

dt

✓up

1� u2

◆ (9.91)

that we can integrateup

1� u2= ↵ (t� t0). (9.92)

Squaring and solving for u, we find

u =dx

dt=

↵(t� t0)p1 + ↵2(t� t0)2

(9.93)

which we can integrate to

x = ↵�1Z

dt↵2(t� t0)p

1 + ↵2(t� t0)2=

p1 + ↵2(t� t0)2 � 1

↵+ x0. (9.94)

9.5 Scalar field in an accelerating frame 221

The proper time of the accelerating frame is ⌧ = t0. An interval dt0 ofproper time is one at which dx0 = 0. So dt0 =

p1� v2dt or dt⌧ =

p1� u2dt

where u(t) is the velocity (9.93). Integrating the equation

dt0 = d⌧ =p

1� u2dt =

s

1� ↵2(t� t0)2

1 + ↵2(t� t0)2dt =

dtp1 + ↵2(t� t0)2

,

(9.95)we get

↵(t0 � t00) = ↵(⌧ � ⌧0) =

Z↵ dtp

1 + ↵2(t� t0)2= arcsinh(↵(t� t0)). (9.96)

So

↵(t� t0) = sinh(↵(⌧ � ⌧0)). (9.97)

The formula (9.94) for x now gives

↵(x� x0) = cosh(↵(⌧ � ⌧0)). (9.98)

9.5 Scalar field in an accelerating frame

For simplicity, we’ll work with a real scalar field (1.54)

�(x) =

Zd3pp

(2⇡)32p0


i. (9.99)

If the field is quantized in a box of volume V , then the expansion of the fieldis

�(x) =X

k

1p2!kV

ha(k) eik·x + a†(k) e�ik·x

i. (9.100)

For the scalar field (9.99), the zero-temperature correlation function is the


mean value in the vacuum

h0|�(x, t)�(x0, t0)|0i = h0|Z

d3pp(2⇡)32p0


i

⇥Z

d3p0p(2⇡)32p00

ha(p0) eip

0·x0+ a†(p0) e�ip

0·x0i|0i

= h0|Z

d3pd3p0

(2⇡)3p

2p02p00a(p)a†(p0) eip·x�ip

0·x0 |0i

=

Zd3pd3p0

(2⇡)3p2p02p00

�3(p� p0)eip·x�ip0·x0

=

Zd3p

(2⇡)32p0eip·(x�x

0). (9.101)

This integral is simpler for massless fields. Setting p = |p| and r = |x� x0|,we add a small imaginary part to the exponential and find

h0|�(x, t)�(x0, t0)|0i =Z

d3p

(2⇡)32peipr cos ✓�ip(t�t

0�i✏)

=

Zpdp dcos ✓

(2⇡)22eipr cos ✓�ip(t�t

0�i✏)

=

Zpdp

(2⇡)22

✓eipr � e�ipr

ipr

◆e�ip(t�t

0�i✏)

=

Z 1

0

dp

(2⇡)22ir

�eipr � e�ipr

�e�ip(t�t

0�i✏)

=1

(2⇡)22ir

✓� 1

i(r � (t� t0))� 1

i(r + (t� t0))

◆

=1

(2⇡)22r

✓1

r � (t� t0)+

1

r + (t� t0)

◆

=1

(2⇡)2[(x� x0)2 � (t� t0)2]. (9.102)

This is the zero-temperature correlation function.In the instantaneous rest frame of an accelerated observer moving in the

x direction with coordinates (9.97 & 9.98), this two-point function is

h0|�(x, t)�(x0, t0)|0i = ↵2/(2⇡)2⇥cosh(↵⌧)� cosh(↵⌧ 0)

⇤2 �⇥sinh(↵⌧)� sinh(↵⌧ 0)

⇤2

= � ↵2

(4⇡)2 sinh2[↵(⌧ � ⌧ 0)/2](9.103)

9.5 Scalar field in an accelerating frame 223

in which the identities

cosh(↵⌧) cosh(↵⌧ 0)� sinh(↵⌧) sinh(↵⌧ 0) = cosh(↵(⌧ � ⌧ 0)) (9.104)

and

cosh(↵(⌧ � ⌧ 0))� 1 = 2 sinh(↵(⌧ � ⌧ 0)/2) (9.105)

as well as cosh2 ↵⌧ � sinh2 ↵⌧ = 1 were used.Now let’s compute the same correlation function at a finite inverse tem-

perature � = 1/(kBT )

h�(0, ⌧)�(0, ⌧ 0)i� = Tr⇥�(0, 0)�(0, t)e��H

⇤.Tr

�e��H

�. (9.106)

The mean value of the number operator for a given momentum k is

ha†(k) a(k)i� = nk = Tr⇥a†(k) a(k)e��H

⇤.Tr

�e��H

�(9.107)

in which

H0 =X

k

!k

�a†(k) a(k) + 1

2

�. (9.108)

The trace over all momenta with k0 6= k is unity, and we are left with

ha† ai� = nk = Tr⇥a† ae��!ka

†a⇤.

Tr�e��!ka

†a�

= � 1

!kTr�e��!ka

†a� @

@�Tr

�e��!ka

†a� (9.109)

in which a†a ⌘ a†(k) a(k) and the 1/2 terms have cancelled. The trace is

Tr�e��!ka

†a�=

X

n

e��n!k =1

1� e��!k. (9.110)

So we have

ha† ai� = ��1� e��!k

�

!k

(�!ke��!k)�1� e��!k

�2 =1

e�!k � 1. (9.111)

In the trace (9.106), only terms that don’t change the number of quantain each mode contribute. So the mean value of the correlation function atinverse temperature � = 1/(kBT ) is

h�(0, ⌧)�(0, ⌧ 0)i� =X

k

1

2kV

⇢⇣e~!k/kBT � 1

⌘�1ei!k(⌧�⌧

0)

+

⇣e~!k/kBT � 1

⌘�1+ 1

�e�i!k(⌧�⌧

0)

�.

(9.112)


In the continuum limit, this ⌧, ⌧ 0 correlation function is two integrals. Formassless particles, the simpler one requires some regularization because itinvolves zero-point energies

A� =X

k

1

2kVe�ik(⌧�⌧

0) =

Zd3k

(2⇡)32ke�ik(⌧�⌧

0) =

Zk2dk

(2⇡)2ke�ik(⌧�⌧

0)

=

Z 1

0

kdk

(2⇡)2e�ik(⌧�⌧

0).

(9.113)

We send ⌧ � ⌧ 0 ! ⌧ � ⌧ 0 + i✏ and find

A� =

Z 1

0

kdk

(2⇡)2e�ik(⌧�⌧

0+i✏) = id

d⌧

Z 1

0

dk

(2⇡)2e�ik(⌧�⌧

0+i✏)

= id

d⌧

Z 1

0

dk

(2⇡)2e�ik(⌧�⌧

0+i✏) =1

(2⇡)2d

d⌧

1

(⌧ � ⌧ 0)

= � 1

(2⇡)21

(⌧ � ⌧ 0)2

(9.114)

as ✏ ! 0.The second integral is

B� =

Z 1

0

kdk

(2⇡)2

⇣e�k � 1

⌘�12 cos(k(⌧ � ⌧ 0)) (9.115)

which Mathematica says is

B� =1

(2⇡)2(⌧ � ⌧ 0)2� csch2(⇡(⌧ � ⌧ 0)/�)

4�2. (9.116)

Thus the finite-temperature correlation function is

h�(0, ⌧)�(0, ⌧ 0)i� = A� +B� = � 1

4�2 sinh2(⇡(⌧ � ⌧ 0)/�). (9.117)

Equating this formula to the zero–temperature correlation function (9.103)in the accelerating frame, we find

1

4�2 sinh2(⇡(⌧ � ⌧ 0)/�)=

↵2

(4⇡)2 sinh2[↵(⌧ � ⌧ 0)/2](9.118)

which says redundantly

kBT =↵

2⇡and ⇡kBT =

↵

2. (9.119)

So a detector accelerating uniformly with acceleration ↵ in the vacuum feels

9.6 Maximally symmetric spaces 225

a nonzero temperature

T =~↵

2⇡ckB. (9.120)

This result (Davies, 1975) is equivalent to the finding (Hawking, 1974) thata gravitational field of local acceleration g makes empty space radiate at atemperature

T =~g

2⇡ckB. (9.121)

Black holes are not black.

9.6 Maximally symmetric spaces

The spheres S2 and S3 and the hyperboloids H2 and H3 are maximally sym-metric spaces. A transformation x ! x0 is an isometry if g0

ik(x0) = gik(x0)

in which case the distances gik(x)dxidxk = g0ik(x0)dx0idx0k = gik(x0)dx0idx0k

are the same. To see what this symmetry condition means, we consider theinfinitesimal transformation x0` = x` + ✏y`(x) under which to lowest ordergik(x0) = gik(x)+ gik,`✏y` and dx0i = dxi+ ✏yi,jdx

j . The symmetry conditionrequires

gik(x)dxidxk = (gik(x) + gik,`✏y

`)(dxi + ✏yi,jdxj)(dxk + ✏yk,mdxm) (9.122)

or

0 = gik,` y` + gim ym,k + gjk y

j

,i. (9.123)

The vector field yi(x) must satisfy this condition if x0i = xi+ ✏yi(x) is to bea symmetry of the metric gik(x). Since the covariant derivative of the metrictensor vanishes, gik;` = 0, we may write the condition on the symmetryvector y`(x) as

0 = yi;k + yk;i. (9.124)

9.3 quantization of fields in curved space...

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