16

Path Integrals

16.1 Path Integrals and Classical Physics

Since Richard Feynman invented them over 60 years ago, path integrals havebeen used with increasing frequency in high-energy and condensed-matterphysics, in finance, and in biophysics (Kleinert, 2009). Feynman used themto express matrix elements of the time-evolution operator exp(�itH/~) interms of the classical action. Others have used them to compute matrixelements of the Boltzmann operator exp(�H/kT ) which in the limit of zerotemperature projects out the ground state |E

0

i of the system

limT!0

e�(H�E0)/kT = limT!0

1Xn=0

|Enie�(En

�E0)/kT hEn| = |E0

ihE0

| (16.1)

a trick used in lattice gauge theory.Path integrals magically express the quantum-mechanical probability am-

plitude for a process as a sum of exponentials exp(iS/~) of the classicalaction S of the various ways that process might occur.

16.2 Gaussian Integrals

The path integrals we can do are gaussian integrals of infinite order. So webegin by recalling the basic integral formula (5.166)

Z 1

�1exp

"� ia

✓x� b

2a

◆2

#dx =

r⇡

ia(16.2)

636 Path Integrals

which holds for real a and b, and also the one (5.167)Z 1

�1exp

� r

⇣x� c

2r

⌘2

�dx =

r⇡

r(16.3)

which is true for positive r and complex c. Equivalent formulas for real aand b, positive r, and complex c areZ 1

�1exp

�� iax2 + ibx

�dx =

r⇡

iaexp

✓ib2

4a

◆(16.4)Z 1

�1exp

�� rx2 + cx

�dx =

r⇡

rexp

✓c2

4r

◆. (16.5)

This last formula will be useful with x = p, r = ✏/(2m), and c = i✏qZ 1

�1exp

✓� ✏

p2

2m+ i✏ q p

◆dp =

r2⇡m

✏exp

✓� ✏

1

2mq2

◆(16.6)

as will (16.4) with x = p, a = ✏/(2m), and b = ✏qZ 1

�1exp

✓� i✏

p2

2m+ i✏ q p

◆dp =

r2⇡m

i✏exp

✓i✏

1

2mq2

◆. (16.7)

Doable path integrals are multiple gaussian integrals. One may show (ex-ercise 16.1) that for positive r

1

, . . . , rN and complex c1

, . . . , cN , the integral(16.5) leads toZ 1

�1exp

Xi

�rix2

i + cixi

!NYi=1

dxi =

NYi=1

r⇡

ri

!exp

1

4

Xi

c2iri

!. (16.8)

If R is the N ⇥N diagonal matrix with positive entries {r1

, r2

, . . . , rN}, andX and C are N -vectors with real {xi} and complex {ci} entries, then thisformula (16.8) in matrix notation isZ 1

�1exp

��XTRX + CTX

� NYi=1

dxi =

s⇡N

det(R)exp

✓1

4CTR�1C

◆. (16.9)

Now every positive symmetric matrix S is of the form S = OROT for somepositive diagonal matrix R. So inserting R = OTSO into the previousequation (16.9) and using the invariance of determinants under orthogonaltransformations, we findZ 1

�1exp

��XTOTSOX + CTX

� NYi=1

dxi =

s⇡N

det(S)exp

1

4CTOTS�1OC

�.

(16.10)

16.3 Path Integrals in Imaginary Time 637

The jacobian of the orthogonal transformations Y = OX and D = OC isunity, and soZ 1

�1exp

�� Y TSY +DTY

� NYi=1

dyi =

s⇡N

det(S)exp

✓1

4DTS�1D

◆(16.11)

in which S is a positive symmetric matrix, and D is a complex vector.The other basic gaussian integral (16.4) leads for real S and D to (exer-

cise 16.2)Z 1

�1exp

�� iY TSY + iDTY

� NYi=1

dyi =

s⇡N

det(iS)exp

✓i

4DTS�1D

◆.

(16.12)The vector Y that makes the argument � iY TSY + iDTY of the expo-

nential of this multiple gaussian integral (16.12) stationary is (exercise 16.3)

Y =1

2S�1D. (16.13)

The exponential of that integral evaluated at its stationary point Y is

exp�� iY TSY + iDTY

�= exp

✓i

4DTS�1D

◆. (16.14)

Thus, the multiple gaussian integral (16.12) is equal to its exponential eval-uated at its stationary point Y , apart from a prefactor involving the deter-minant det iS.

Similarly, the vector Y that makes the argument �Y TSY + DTY ofthe exponential of the multiple gaussian integral (16.11) stationary is Y =S�1D/2, and that exponential evaluated at Y is

exp�� Y TSY +DTY

�= exp

✓1

4DTS�1D

◆. (16.15)

Once again, a multiple gaussian integral is simply its exponential evaluatedat its stationary point Y , apart from a prefactor involving a determinant,detS.

16.3 Path Integrals in Imaginary Time

At the imaginary time t = �i�~, the time-evolution operator exp(�itH/~)is exp(��H) in which the inverse temperature � = 1/kT is the reciprocal ofBoltzmann’s constant k = 8.617⇥10�5 eV/K times the absolute temperatureT . In the low-temperature limit, exp(��H) is a projection operator (16.1)

638 Path Integrals

on the ground state of the system. These path integrals in imaginary timeare called euclidean path integrals.Let us consider a quantum-mechanical system with hamiltonian

H =p2

2m+ V (q) (16.16)

in which the commutator of the position q and momentum p operators is[q, p] = i in units in which ~ = 1. For tiny ✏, the corrections to the approxi-mation

exp

� ✏

✓p2

2m+ V (q)

◆�⇡ exp

✓� ✏

p2

2m

◆exp (� ✏V (q))+O(✏2) (16.17)

are of second order in ✏.To evaluate the matrix element hq00| exp(�✏H)|q0i, we insert the identity

operator I in the form of an integral over the momentum eigenstates

I =

Z 1

�1|p0ihp0|dp0 (16.18)

and use the inner product hq00|p0i = exp(iq00p0)/p2⇡ so as to get as ✏! 0

hq00| exp(�✏H)|q0i =Z 1

�1hq00| exp

✓�✏ p2

2m

◆|p0ihp0| exp (�✏V (q)) |q0i dp0

= e�✏V (q0)Z 1

�1exp

�✏ p02

2m+ i p0 (q00 � q0)

�dp0

2⇡. (16.19)

We now adopt the suggestive notation

q00 � q0

✏= q0 (16.20)

and use the integral formula (16.6) so as to obtain

hq00| exp(� ✏H)|q0i = 1

2⇡e�✏V (q0)

Z 1

�1exp

✓� ✏

p02

2m+ i ✏ p0 q0

◆dp0

=⇣ m

2⇡✏

⌘1/2

exp�� ✏

⇥1

2

m q 02 + V (q0)⇤

(16.21)

in which q00 enters through the notation (16.20).The next step is to link two of these matrix elements together

hq000|e�2✏H |q0i =Z 1

�1hq000|e�✏H |q00ihq00|e�✏H |q0idq00 (16.22)

=m

2⇡✏

Z 1

�1exp

�� ✏

⇥1

2

m q 002 + V (q00) + 1

2

m q 02 + V (q0)⇤

dq00.

16.3 Path Integrals in Imaginary Time 639

Linking three of these matrix elements together and using subscripts insteadof primes, we have

hq3

|e�3✏H |q0

i =ZZ 1

�1hq

3

|e�✏H |q2

ihq2

|e�✏H |q1

ihq1

|e�✏H |q0

i dq1

dq2

(16.23)

=⇣ m

2⇡✏

⌘3/2

ZZ 1

�1exp

8<:� ✏2X

j=0

⇥1

2

m q2j + V (qj)⇤9=; dq

1

dq2

.

Boldly passing from 3 to n and suppressing some integral signs, we get

hqn|e�n✏H |q0

i =ZZZ 1

�1hqn|e�✏H |qn�1

i . . . hq1

|e�✏H |q0

i dq1

. . . dqn�1

(16.24)

=⇣ m

2⇡✏

⌘n/2ZZZ 1

�1exp

8<:� ✏n�1Xj=0

⇥1

2

m q2j + V (qj)⇤9=; dq

1

. . . dqn�1

.

Writing dt for ✏ and taking the limits ✏ ! 0 and n ⌘ �/✏ ! 1, we findthat the matrix element hq� |e��H |q

0

i is a path integral of the exponentialof the average energy multiplied by ��

hq� |e��H |q0

i =Z

exp

�Z �

0

1

2

mq2(t) + V (q(t)) dt

�Dq (16.25)

in which Dq ⌘ (nm/2⇡ �)n/2 dq1

dq2

· · · dqn�1

. We sum over all paths q(t)that go from q(0) = q

0

at inverse temperature � = 0 to q(�) = q� at inversetemperature �.In the limit � ! 1, the operator exp(��H) becomes proportional to a

projection operator (16.1) on the ground state of the theory.In three dimensions with q(�) = dq(�)/d�, and ~ 6= 1, the analog of

equation (16.25) is (exercise 16.28)

hq� |e��H |q0

i =Z

exp

�Z �

0

m

2~2 q2(�0) + V (q(�0)) d�0

�Dq (16.26)

where Dq ⌘ (nm/2⇡ � ~2)3n/2 dq1

dq2

· · · dqn�1

.Path integrals in imaginary time are called euclidean mainly to distinguish

them from Minkowski path integrals, which represent matrix elements of thetime-evolution operator exp(�itH) in real time.

640 Path Integrals

16.4 Path Integrals in Real Time

Path integrals in real time represent the time-evolution operator exp(�itH).Using the integral formula (16.7), we find in the limit ✏! 0

hq00|e�i ✏H |q0i =Z 1

�1hq00| exp

� i ✏

p2

2m

�|p0ihp0| exp [� i ✏V (q)] |q0i dp0

=1

2⇡e�i ✏V (q0)

Z 1

�1exp

� i ✏

p02

2m+ i p0 (q00 � q0)

�dp0

=1

2⇡e�i ✏V (q0)

Z 1

�1exp

� i ✏

p02

2m+ i ✏ p0q0

�dp0

=⇣ m

2⇡i✏

⌘1/2

exp

i ✏

✓m q02

2� V (q0)

◆�. (16.27)

When we link together n of these matrix elements, we get the real-timeversion of (16.25)

hqn|e�in✏H |q0

i =⇣ m

2⇡i✏

⌘n/2ZZZ 1

�1exp

8<:i✏n�1Xj=0

⇥1

2

m q2j � V (qj)⇤9=; dq

1

. . . dqn�1

.

(16.28)Writing dt for ✏ and taking the limits ✏! 0 and n✏! t, we find that the

amplitude hqt|e�itH |q0

i is the path integral

hqt|e�itH |q0

i =Z

exp

i

Z t

0

1

2

m q2 � V (q) dt0�Dq (16.29)

in which Dq di↵ers from the one that appears in euclidian path integrals bythe substitution � ! it

Dq = limn!1

⇣ nm

2⇡it

⌘n/2dq

1

dq2

· · · dqn�1

. (16.30)

The integral in the exponent is the classical action

S[q] =

Z t

0

1

2

m q2 � V (q) dt0 (16.31)

for a process q(t0) that runs from q(0) = q0

to q(t) = qt. We sum over allsuch processes.In three-dimensional space

hqt|e�itH |q0

i =Z

exp

i

Z t

0

1

2

m q2 � V (q) dt0�Dq (16.32)

replaces (16.29).

16.4 Path Integrals in Real Time 641

The units of action are energy⇥ time, and the argument of the exponentialmust be dimensionless, so in ordinary units the amplitude (16.29) is

hqt|e�itH/~|q0

i =ZeiS[q]/~Dq. (16.33)

When is this amplitude big? When is it tiny? Suppose there is a processq(t) = qc(t) that goes from qc(0) = q

0

to qc(t) = qt in time t and that obeysthe classical equation of motion (15.14–15.15)

�S[qc]

�qc= mqc + V 0(qc) = 0. (16.34)

The action of such a classical process is stationary, that is, S[qc+ dq] di↵ersfrom S[qc] only by terms of second order in �q. So there are infinitely manyother processes that have the same action to within a fraction of ~. Theseprocesses add with nearly the same phase to the path integral (16.33) andso make a huge contribution to the amplitude hqt|e�itH/~|q

0

i.But if no classical process goes from q

0

to qt in time t, then the nonclassicalprocesses from q

0

to qt in time t have actions that di↵er among themselves bylarge multiples of ~. Their amplitudes cancel each other, and so the resultingamplitude that is tiny. Thus the real-time path integral explains theprinciple of stationary action (section 11.37).Does this path integral satisfy Schrodinger’s equation? To see that it does,

we’ll use (16.27) in the more explicit form

hq00|e�i✏H |q0i =⇣ m

2⇡i✏

⌘1/2

exp

im(q00 � q0)2

2✏� i ✏V (q0)

�(16.35)

to write (q00, t+ ✏) = hq00| , t+ ✏i as an integral of (q0, t) = hq0| , ti

hq00| , t+ ✏i =Z

hq00|e�i✏H |q0ihq0| , ti dq0 (16.36)

=⇣ m

2⇡i✏

⌘1/2

Zexp

im(q00 � q0)2

2✏� i ✏V (q0)

�hq0| , ti dq0.

Keeping only leading terms in ✏, we have

(q00, t+ ✏) =⇣ m

2⇡i✏

⌘1/2

e�i✏V (q00)Z 1

�1exp

im(q00 � q0)2

2✏

� (q0, t) dq0.

(16.37)Letting x = q0 � q00 and q0 = q00 + x, we find

(q00, t+ ✏) =⇣ m

2⇡i✏

⌘1/2

e�i✏V (q00)Z

exp

imx2

2✏

� (q00 + x, t) dx. (16.38)

642 Path Integrals

We now expand (q00 + x, t) as

(q00 + x, t) = (q00, t) + x 0(q00, t) +1

2x2 00(q00, t) + . . . (16.39)

and (q00, t+ ✏) as

(q00, t+ ✏) = (q00, t) + ✏ (q00, t) + . . . . (16.40)

The integral formula (16.2) impliesZ 1

�1eimx2/2✏ dx =

✓2⇡i✏

m

◆1/2

(16.41)

and its derivative with respect to im/2✏ givesZ 1

�1x2 eimx2/2✏ dx =

i✏

m

✓2⇡i✏

m

◆1/2

while

Z 1

�1x eimx2/2✏ dx = 0. (16.42)

Substituting the expansions (16.39 & 16.40) for (q00 + x, t) and (q00, t+ ✏)into the integral (16.38) and using the integral formulas (16.41 & 16.42), weget

(q00, t) + ✏ (q00, t) =⇥1� i✏V (q00)

⇤ (q00, t) +

i✏

m 00(q00, t)

�(16.43)

which is Schrodinger’s equation

i = � 1

2m 00 + V (16.44)

in natural units or i~ = � ~2 00/2m+ V in arbitrary units.

16.5 Path Integral for a Free Particle

The amplitude for a free nonrelativistic particle to go from the origin to thepoint q in time t is the path integral (16.32)

hq|e�itH |q = 0i =Z

eiS0[q]Dq =

Zexp

✓i

Z t

0

1

2

m q2(t0) dt0◆Dq. (16.45)

The classical path that goes from 0 to q in time t is qc(t0) = (t0/t) q. The

general path q(t0) over which we integrate is q(t0) = qc(t0) + �q(t0). Since

both q(t0) and qc(t0) go from 0 to q in time t, the otherwise arbitrary path

�q(t0) must be a loop that goes from �q(0) = 0 to �q(t) = 0 in time t. The

16.5 Path Integral for a Free Particle 643

velocity q = qc + �q is the sum of the constant classical velocity qc = q/tand the velocity �q. The first-order change vanishes

m

Z t

0

qc ·d�q

dtdt = m qc ·

Z t

0

d�q

dtdt = m qc · [�q(t)� �q(0)] = 0 (16.46)

and so the action S0

[q] is the classical action plus the loop action

S0

[q] = 1

2

m

Z t

0

(qc + �q)2 dt0 = S0

[qc] + S0

[�q]. (16.47)

The path integral therefore factorizes

hq|e�i tH |0i =Z

eiS0[q]Dq =

ZeiS0[q

c

+�q]D�q

=

ZeiS0[q

c

] eiS0[�q]D�q = eiS0[qc

]

ZeiS0[�q]D�q

(16.48)

into the phase of the classical action times a path integral over the loops.The loop integral L is independent of the spatial points q and 0 and so canonly depend upon the time interval, L = L(t). Thus the amplitude is theproduct

hq|e�i tH |0i = eiS0[qc

] L(t). (16.49)

Since the classical velocity is qc = q/t, the classical action is

S0

[qc] =

Z t

0

m

2q2c(t) dt =

m

2

q2

t. (16.50)

So the amplitude is

hq|e�i(t2�t1)H |0i = eimq

2/2t L(t). (16.51)

Since position eigenstates are orthogonal, this amplitude must reduce toa delta function as t ! 0

limt!0

hq|e�i tH |0i = hq|0i = �3(q). (16.52)

One of the many representations of Dirac’s delta function is

�3(q) = limt!0

⇣ m

2⇡i~t

⌘3/2

eimq

2/2~t. (16.53)

Thus N L(t) = (m/2⇡it)3/2 and

hq|e�itH/~|0i =⇣ m

2⇡i~t

⌘3/2

eimq

2/2~t (16.54)

in unnatural units. You can verify (exercise 16.6) this result by inserting

644 Path Integrals

a complete set of momentum dyadics |pihp| and doing the resulting Fouriertransform.

Example 16.1 (The Bohm-Aharonov E↵ect) From our formula (11.311)for the action of a relativistic particle of mass m and charge q, we infer(exercise 16.7) that the action a nonrelativistic particle in an electromagneticfield with no scalar potential is

S =

Zx2

x1

⇥1

2

mv + qA⇤·dx . (16.55)

Now imagine that we shoot a beam of such particles past but not through anarrow cylinder in which a magnetic field is confined. The particles can goeither way around the cylinder of area S but cannot enter the region of themagnetic field. The di↵erence in the phases of the amplitudes is the loopintegral from the source to the detector and back to the source

�S

~ =

I hmv

2+ qA

i· dx~ =

Imv · dx

2~ +q

~

ZSB · dS =

Imv · dx

2~ +q�

~(16.56)

in which � is the magnetic flux through the cylinder.

16.6 Free Particle in Imaginary Time

If we mimic the steps of the preceding section (16.5) in which the hamiltonianis H = p2/2m, set � = it/~ = 1/kT , and use Dirac’s delta function

�3(q) = limt!0

⇣ m

2⇡~t

⌘3/2

e�mq

2/2~t (16.57)

then we get

hq|e��H |0i =✓

m

2⇡~2�

◆3/2

exp

�mq2

2~2�

�=

✓mkT

2⇡~2

◆3/2

e�mkTq

2/2~2 .

(16.58)For � = t/~, this is the solution (3.200 & 13.107) of the di↵usion equationwith di↵usion constant D = ~/(2m).

16.7 Harmonic Oscillator in Real Time

Biologists have mice; physicists have harmonic oscillators with hamiltonian

H =p2

2m+

m!2q2

2. (16.59)

16.7 Harmonic Oscillator in Real Time 645

For this hamiltonian, our formula (16.29) for the coordinate matrix elementsof the time-evolution operator exp(�itH) is

hq00|e�i tH |q0i =ZeiS[q]Dq (16.60)

with action

S[q] =

Z t

0

1

2

mq2(t0)� 1

2

m!2q2(t0) dt0. (16.61)

The classical solution qc(t) = q0 cos!t + q0

sin(!t)/! in which q0 = qc(0)and q

0

= qc(0) are the initial position and velocity makes the action S[q]stationary

d

d✏S[q + ✏h]

����✏=0

= 0 (16.62)

and satisfies the classical equation of motion mqc(t) = � !2qc(t).We now apply the trick (16.46–16.48) we used for the free particle. We

write an arbitrary process q(t) is the sum of the classical process qc(t) anda loop �q(t) with �q(0) = �q(t) = 0. Since the action S[q] is quadratic inthe variables q and q, the functional Taylor series (15.31) for S[qc + �q] hasonly two terms

S[qc + �q] = S[qc] + S[�q]. (16.63)

Thus we can write the path integral (16.60) as

hq00|e�itH |q0i =Z

eiS[q]Dq =

ZeiS[qc+�q]D�q

=

ZeiS[qc]+iS[�q]D�q = eiS[qc]

ZeiS[�q]D�q.

(16.64)

The remaining path integral over the loops �q does not involve the end pointsq0 and q00 and so must be a function L(t) of the time t but not of q0 or q00

hq00|e�itH |q0i = eiS[qc] L(t). (16.65)

The action S[qc] is (exercise 16.8)

S[qc] =m!

2 sin(!t)

⇥�q02 + q002

�cos(!t)� 2q0q00

⇤. (16.66)

The action S[�q] of a loop

�q(t0) =n�1Xj=1

aj sinj⇡t0

t(16.67)

646 Path Integrals

is (exercise 16.9)

S[�q] =n�1Xj=1

mt

4a2j

(j⇡)2

t2� !2

�. (16.68)

The path integral over the loops is then, apart from a constant jacobian J ,ZeiS[�q]D�q = J

⇣ nm

2⇡it

⌘n/2 Zexp

8<:n�1Xj=1

imt

4a2j

(j⇡)2

t2� !2

�9=;n�1Yj=1

daj

= J⇣ nm

2⇡it

⌘n/2 n�1Yj=1

Z 1

�1exp

⇢imt

4a2j

(j⇡)2

t2� !2

��daj .

(16.69)

Using the gaussian integral (16.2) and the infinite product (4.140), we getZeiS[�q]D�q = Jnn/2

rm

2⇡it

n�1Yj=1

p2

j⇡

✓1� !2t2

⇡2j2

◆�1/2

=

rm!

2⇡i sin!t

0@ limn!1

Jnn/2n�1Yj=1

p2

j⇡

1A.

(16.70)

Using (16.65) and (16.66), we see that the number within the parenthesesis unity because in that case we have (Feynman and Hibbs, 1965, ch. 3)

hq00|e�itH/~|q0i =r

m!

2⇡i~ sin(!t) exp

"im!

⇥�q02 + q002

�cos(!t)� 2q0q00

⇤2~ sin(!t)

#(16.71)

which agrees with the amplitude (16.54) in the limit t ! 0 (exercise 16.10).

16.8 Harmonic Oscillator in Imaginary Time

For the harmonic oscillator with hamiltonian (16.59), our formula (16.25)for euclidean path integrals becomes

hq00|e��H |q0i =Z

exp

⇢�Z �

0

1

2

mq2(t) + 1

2

m!2q2(t) dt

�Dq. (16.72)

The euclidean action, which is a time integral of the energy of the oscillator,

Se[q] =

Z �

0

⇥1

2

mq2(t) + 1

2

m!2q2(t)⇤dt (16.73)

16.8 Harmonic Oscillator in Imaginary Time 647

is purely quadratic, and so we may play the trick (15.32) if we can find apath qe(t) that makes it stationary

�Se[qe][h] =d

d✏

Z �

0

1

2

m(qe(t) + ✏h(t))2 + 1

2

m!2(qe(t) + ✏h(t))2 dt

����✏=0

=

Z �

0

mqe(t)h(t) +m!2qe(t)h(t) dt

=

Z �

0

⇥�mqe(t) +m!2qe(t)

⇤h(t)dt = 0. (16.74)

The path qe(t) must satisfy the euclidean equation of motion

qe(t) = !2qe(t) (16.75)

whose general solution is

qe(t) = Ae!t +Be�!t. (16.76)

The path from qe(0) = q0 to qe(�) = q00 must have

A =q00e�!� � q0e�2!�

1� e�2!�and B = q0 �A. (16.77)

Its action Se[qe] is (exercise 16.12)

Se[qe] =1

2

m!hA2

⇣e2!� � 1

⌘�B2

⇣e�2!� � 1

⌘i. (16.78)

Since the action is purely quadratic, the trick (15.32) tells us that the actionSe[q] of the arbitrary path q(t) = qe(t) + �q(t) is the sum

Se[q] = Se[qe] + Se[�q] (16.79)

in which the action Se[�q] of the loop �q(t) depends upon � but not uponq� or q

0

. It follows then that for some loop function L(�) of � alone

hq00|e��H |q0i = exp (�Se[qe]) L(�) (16.80)

= L(�) expn� 1

2

m!hA2

⇣e2!� � 1

⌘�B2

⇣e�2!� � 1

⌘io.

To study the ground state of the harmonic oscillator, we let � ! 1 in thisequation. Inserting a complete set of eigenstates H|ni = En|ni, we see thatthe limit of the left-hand side is

lim�!1

hq00|e��H |q0i = lim�!1

hq00|nie��Enhn|q0i = e��E0hq00|0ih0|q0i. (16.81)

Our formulas (16.77) for A and B say that A ! q00e�!� and B ! q0 as� ! 1, and so in this limit by (16.80 & 16.81) we have

e��E0hq00|0ih0|q0i = L(�) exp⇥� 1

2

m!�q002 + q02

�⇤(16.82)

648 Path Integrals

from which may infer our earlier formula (15.44) for the wave function ofthe ground state

hq|0i =⇣m!⇡~

⌘1/4

exp

�1

2

m!q2

~

�(16.83)

in which the prefactor ensures the normalization

1 =

Z 1

�1|hq|0i|2dq. (16.84)

Euclidean path integrals help one study ground states.

16.9 Euclidean Correlation Functions

In the Heisenberg picture, the position operator q(t) is

q(t) = eitHq e�itH (16.85)

in which q = q(0) is the position operator at time t = 0 or equivalentlythe position operator in the Schrodinger picture. The analogous operator inimaginary time is the euclidean position operator qe(t) defined as

qe(t) = etHq e�tH (16.86)

obtained from q(t) by replacing t by �it.The euclidean-time-ordered product of two euclidean position opera-

tors is

T [qe(t1) qe(t2)] = ✓(t1

� t2

)qe(t1) qe(t2) + ✓(t2

� t1

)qe(t2) qe(t1) (16.87)

in which ✓(x) = (x+|x|)/2|x| is Heaviside’s function. We can use the methodof section 16.3 to compute the matrix element of the euclidean-time-orderedproduct T [q(t

1

)q(t2

)] sandwiched between two factors of exp(�tH). Fort1

� t2

, this matrix element is

hqt|e�tHqe(t1)qe(t2)e�tH |q�ti = hqt|e�(t�t1)Hqe�(t1�t2)Hqe�(t+t2)H |q�ti.

(16.88)Then instead of the path-integral formula (16.25), we get

hqt|e�tHT [qe(t1)qe(t2)] e�tH |q�ti =

Zq(t

1

)q(t2

)e�Se

[q,t,�t]Dq (16.89)

where as in (16.25) Se[q, t,�t] is the euclidean action

Se[q, t,�t] =

Z t

�t

1

2

mq2(t0) + V (q(t0)) dt0 (16.90)

16.9 Euclidean Correlation Functions 649

or the time integral of the energy. As in the path integral (16.25), theintegration is over all paths that go from q(�t) = q�t to q(t) = qt. Theanalog of (16.25) is

hqt|e�2tH |q�ti =Z

e�Se

[q,t,�t]Dq (16.91)

and the factors (nm/2⇡�)n/2 cancel in the ratio of (16.89) to (16.91)

hqt|e�tHT [qe(t1)qe(t2)] e�tH |q�tihqt|e�2tH |q�ti

=

Zq(t

1

)q(t2

)e�Se

[q,t,�t]DqZe�S

e

[q,t,�t]Dq. (16.92)

In the limit t ! 1, the operator exp(�tH) projects out the ground state|0i of the system

limt!1

e�tH |q�ti = limt!1

1Xn=0

e�tH |nihn|q�ti = limt!1

e�tE0 |0ih0|q�ti (16.93)

which we assume to be unique and normalized to unity. In the ratio (16.92),most of these factors cancel, leaving us with

h0|T [qe(t1)qe(t2)] |0i =

Zq(t

1

)q(t2

)e�Se

[q,1,�1]DqZe�S

e

[q,1,�1]Dq. (16.94)

More generally, the mean value in the ground state |0i of any euclidean-time-ordered product of position operators q(ti) is a ratio of path integrals

h0| T [q(t1

) . . . q(tn)] |0i =

Zq(t

1

) . . . q(tn) e�Se

[q]DqZe�S

e

[q]Dq(16.95)

in which Se[q] stands for Se[q,1,�1]. Why do we need the time-orderedproduct T on the LHS? Because successive factors of exp(�(tk � t`)H) leadto the path integral of exp(�Se[q]). Why don’t we need T on the RHS?Because the q(ti)’s are real numbers which commute with each other.

The result (16.95 ) is important because it can be generalized to all quan-tum theories, including field theories.

650 Path Integrals

16.10 Finite-Temperature Field Theory

Matrix elements of the operator exp(��H) where � = 1/kT tell us what asystem is like at temperature T . In the low-temperature limit, they describethe ground state of the system.Quantum mechanics imposes upon n coordinates qi and conjugate mo-

menta ⇡k the commutation relations

[qi, pk] = i �i,k and [qi, qk] = [pi, pk] = 0. (16.96)

In quantum field theory, we associate a coordinate qx

⌘ �(x) and a con-jugate momentum p

x

⌘ ⇡(x) with each point x of space and impose uponthem the very similar commutation relations

[�(x),⇡(y)] = i �(x� y) and [�(x),�(y)] = [⇡(x),⇡(y)] = 0. (16.97)

Just as in quantum mechanics the time derivative of a coordinate is itscommutator with a hamiltonian qi = i[H, qi], so too in quantum field theorythe time derivative of a field �(x, t) ⌘ �(x) is �(x) = i[H,�(x)]. A typicalhamiltonian for a single scalar field � is

H =

Z 1

2⇡2(x) +

1

2(r�(x))2 + 1

2m2�2(x) + P (�(x))

�d3x (16.98)

in which P is a quartic polynomial.Since quantum field theory is just the quantum mechanics of many vari-

ables, we can use the methods of sections 16.3 & 16.4 to write matrix ele-ments of exp(��H) as path integrals. We define a potential

V (�(x)) =1

2(r�(x))2 + 1

2m2�2(x) + P (�(x)) (16.99)

and write the hamiltonian H as

H =

Z 1

2⇡2(x) + V (�(x))

�d3x. (16.100)

Like |q0i and |p0i, the states |�0i and |⇡0i are eigenstates of the hermitianoperators �(x, 0) and ⇡(x, 0)

�(x, 0)|�0i = �0(x)|�0i and ⇡(x, 0)|⇡0i = ⇡0(x)|⇡0i. (16.101)

The analog of hq0|p0i is

h�0|⇡0i = f exp

i

Z�0(x)⇡0(x)d3x

�(16.102)

in which f is a factor which eventually will cancel.

16.10 Finite-Temperature Field Theory 651

Repeating our derivation of Eq.(16.21) with

D⇡0 ⌘Yx

d⇡0(x) (16.103)

we find in the limit ✏! 0

h�00| exp(�✏H)|�0i =Zh�00| exp

✓� ✏2

Z⇡2(x)d3x

◆|⇡0i

⇥ h⇡0| exp✓�✏

ZV (�(x)) d3x

◆|�0iD⇡0

= |f |2 exp

✓�✏

ZV (�0(x)) d3x

◆(16.104)

⇥Zexp

Z�1

2

✏⇡02(x) + i⇡0(x)[�00(x)� �0(x)]d3x

�D⇡0.

Using the abbreviation

�0(x) ⌘ �00(x)� �0(x)

✏(16.105)

and the integral formula (16.6), we get

h�00| exp(�✏H)|�0i = f 0 exp

⇢�✏

Z h1

2

�02(x) + V (�0(x))id3x

�.

Putting together n = �/✏ such terms, integrating over the intermediatestates |�000ih�000|, and absorbing the normalizing factors into D�, we have

h�� |e��H |�0

i =Z �

�

�0

exp

�Z �

0

Z1

2

�2(x) + V (�(x)) d3xdt

�D�. (16.106)

Replacing the potential V (�) with its definition (16.99), we find

h�� |e��H |�0

i =Z �

�

�0

exp

�Z �

0

Z1

2

h�2 + (r�)2 +m2�2

i+ P (�) d3xdt

�D�

(16.107)in which the limits �

0

and �� remind us that we are to integrate over allfields �(x, t) that run from �(x, 0) = �

0

(x) to �(x,�) = ��(x).In terms of the energy density

H(�) ⌘ 1

2

⇥(@a�)

2 +m2�2⇤+ P (�) (16.108)

in which a is summed from 0 to 3, the path integral (16.107) is

h�� |e��H |�0

i =Z �

�

�0

exp

�Z �

0

ZH(�) d3xdt

�D�. (16.109)

652 Path Integrals

The partition function Z(�)—defined as the trace Z(�) ⌘ Tr e��H over allstates of the system—is an integral over all loop fields (ones for which ��and �

0

coincide)

Z(�) ⌘ Tr e��H =

Z �0

�0

h�|e��H |�iD� =

Z �0

�0

exp

�Z �

0

ZH(�) d3xdt

�D�.

(16.110)Because the four space-time derivatives in H(�) occur with the same sign,finite-temperature field theory is called euclidean quantum field theory.The density operator ⇢ for the system described by the hamiltonian H inequilibrium at temperature T is ⇢ = exp(��H)/Z(�).Like the definition (16.86) of the euclidean position operator qe(t), the

euclidean field operator �e(x) is defined as

�e(x, t) = etH�(x, 0)e�tH . (16.111)

The euclidean-time-ordered product (16.87 ) of two fields is

T [�e(x1

, t1

)�e(x2

, t2

)] = ✓(t1

� t2

)et1H�e(x1

, 0)e�(t1�t2)H�e(x2

, 0)e�t2H

+ ✓(t2

� t1

)et2H�e(x2

, 0)e�(t2�t1)H�e(x1

, 0)e�t1H .

The logic of equations (16.86–16.95) leads us to write its mean value ina system described by a stationary density operator— one that commuteswith the hamiltonian—as the ratio

hT [�e(x1)�e(x2)]i = Tr {⇢ T [�e(x1)�e(x2)]}

=Tr

�e��H T [�e(x1)�e(x2)]

Tr [ e��H ]

(16.112)

=

Z �0

�0

�(x1

)�(x2

) exp

�Z �

0

ZH(�) d3x dt

�D�Z �0

�0

exp

�Z �

0

ZH(�) d3x dt

�D�

in which several factors have canceled.In the zero-temperature (� ! 1) limit, the density operator ⇢ becomes

the projection operator |0ih0| on the ground state, and mean-value formulaslike (16.112) become

h0|T [�e(x1) . . .�e(xn)] |0i =

Z�(x

1

) . . .�(xn) exp

�ZH(�) d4x

�D�Z

exp

�Z

H(�) d4x

�D�

(16.113)

16.11 Real-Time Field Theory 653

in which Hamilton’s density H(�) is integrated over all of euclidean space-time and over all fields that are periodic on the infinite time interval. Sta-tistical field theory and lattice gauge theory use formulas like (16.112 &16.113).

16.11 Real-Time Field Theory

We now follow the derivation of section 16.10 using the same notation butfor real time. In (16.104), we replace �✏H by �i✏H and follow the logic ofsections (16.4 & 16.10). We find in the limit ✏! 0 with �0 ⌘ (�00 � �0)/✏

h�00|e�i✏H |�0i =Zh�00|e�i✏

R⇡2/2 d3x|⇡0ih⇡0|e�i✏

RV (�) d3x|�0iD⇡0

= |f |2e�i✏RV (�0)d3x

Ze�i✏

R⇡02/2+i⇡0

(�00��0) d3xD⇡0

= f 0 exp

i✏

Z1

2

�02 � V (�0) d3x

�. (16.114)

Putting together 2t/✏ similar factors and integrating over all the interme-diate states |�ih�|, we arrive at the path integral

h�00|e�i2tH |�0i =Z �00

�0exp

i

Z1

2

�2(x)� V (�(x)) d4x

�D� (16.115)

in which we integrate over all fields �(x) that run from �0(x,�t) to �00(x, t).After expanding the definition (16.99) of the potential V (�), we have

h�00|e�i2tH |�0i =Z �00

�0exp

i

Z1

2

h�2 � (r�)2 �m2�2

i� P (�) d4x

�D�.

(16.116)This amplitude is a path integral

h�00|e�i2tH |�0i =Z �00

�0eiS[�]D� (16.117)

of phases exp(iS[�]) that are exponentials of the classical action

S[�] =

Z1

2

h�2 � (r�)2 �m2�2

i� P (�) d4x. (16.118)

The time dependence of the Heisenberg field operator �(x, t) is

�(x, t) = eitH�(x, 0)e�itH . (16.119)

654 Path Integrals

The time-ordered product of two fields, as in (16.87), is the sum

T [�(x1

)�(x2

)] = ✓(x01

� x02

)�(x1

)�(x2

) + ✓(x02

� x01

)�(x2

)�(x1

). (16.120)

Between two factors of exp(�itH), it is for t1

> t2

e�itHT [�(x1

)�(x2

)] e�itH = e�i(t�t1)H�(x1

, 0)e�i(t1�t2)H�(x2

, 0)e�i(t+t2)H .

So by the logic that led to the path-integral formulas (16.112) and (16.117),we can write a matrix element of the time-ordered product (16.120) as

h�00|e�itHT [�(x1

)�(x2

)] e�itH |�0i =Z �00

�0�(x

1

)�(x2

) eiS[�]D� (16.121)

in which we integrate over fields that go from �0 at time �t to �00 at time t.The time-ordered product of any combination of fields is then

h�00|e�itHT [�(x1

) . . .�(xn)] e�itH |�0i =

Z�(x

1

) . . .�(xn) eiS[�]D�.

(16.122)Like the position eigenstates |q0i of quantum mechanics, the eigenstates

|�0i are states of infinite energy that overlap most states. Yet we often areinterested in the ground state |0i or in states of a few particles. To formsuch matrix elements, we multiply both sides of equations (16.117 & 16.122)by h0|�00ih�0|0i and integrate over �0 and �00. Since the ground state is anormalized eigenstate of the hamiltonian H|0i = E

0

|0i with eigenvalue E0

,we find from (16.117)Z

h0|�00ih�00|e�i2tH |�0ih�0|0iD�00D�0 = h0|e�i2tH |0i (16.123)

= e�i2tE0 =

Zh0|�00ieiS[�]h�0|0iD�D�00D�0

and from (16.122) suppressing the di↵erentials D�00D�0

e�2itE0h0|T [�(x1

) . . .�(xn)] |0i =Zh0|�00i�(x

1

) . . .�(xn) eiS[�]h�0|0iD�.

(16.124)The mean value in the ground state of a time-ordered product of field oper-ators is then a ratio of these path integrals

h0|T [�(x1

) . . .�(xn)] |0i =

Zh0|�00i�(x

1

) . . .�(xn) eiS[�]h�0|0iD�Z

h0|�00i eiS[�]h�0|0iD�(16.125)

in which factors involving E0

have canceled and the integration is over all

16.12 Perturbation Theory 655

fields that go from �(x,�t) = �0(x) to �(x, t) = �00(x) and over �0(x) and�00(x).

16.12 Perturbation Theory

Field theories with hamiltonians that are quadratic in their fields like

H0

=

Z1

2

h⇡2(x) + (r�(x))2 +m2�2(x)

id3x (16.126)

are soluble. Their fields evolve in time as

�(x, t) = eitH0�(x, 0)e�itH0 . (16.127)

The mean value in the ground state of H0

of a time-ordered product of thesefields is by (16.125) a ratio of path integrals

h0|T [�(x1

) . . .�(xn)] |0i =

Zh0|�00i�(x

1

) . . .�(xn) eiS0[�]h�0|0iD�Z

h0|�00i eiS0[�]h�0|0iD�(16.128)

in which the action S0

[�] is quadratic in the fields

S0

[�] =

Z1

2

h�2(x)� (r�(x))2 �m2�2(x)

id4x

=

Z1

2

⇥� @a�(x)@

a�(x)�m2�2(x)⇤d4x.

(16.129)

So the path integrals in the ratio (16.128) are gaussian and doable.The Fourier transforms

�(p) =

Ze�ipx�(x) d4x and �(x) =

Zeipx�(p)

d4p

(2⇡)4(16.130)

turn the space-time derivatives in the action into a quadratic form

S0

[�] = �1

2

Z|�(p)|2 (p2 +m2)

d4p

(2⇡)4(16.131)

in which p2 = p2�p02, and �(�p) = �⇤(p) by (3.25) since the field � is real.The initial h�0|0i and final h0|�00i wave functions produce the i✏ in the

Feynman propagator (5.233). Although its exact form doesn’t matter here,the wave function h�0|0i of the ground state of H

0

is the exponential (15.53)

h�0|0i = c exp

�1

2

Z|�0(p)|2

pp2 +m2

d3p

(2⇡)3

�(16.132)

656 Path Integrals

in which �0(p) is the spatial Fourier transform

�0(p) =

Ze�ip·x �0(x) d3x (16.133)

and c is a normalization factor that will cancel in ratios of path integrals.Apart from �2i ln c which we will not keep track of, the wave functions

h�0|0i and h0|�00i add to the action S0

[�] the term

�S0

[�] =i

2

Z pp2 +m2

⇣|�(p, t)|2 + |�(p,�t)|2

⌘ d3p

(2⇡)3(16.134)

in which we envision taking the limit t ! 1 with �(x, t) = �00(x) and�(x,�t) = �0(x). The identity (Weinberg, 1995, pp. 386–388)

f(+1) + f(�1) = lim✏!0+

✏

Z 1

�1f(t) e�✏|t| dt (16.135)

allows us to write �S0

[�] as

�S0

[�] = lim✏!0+

i✏

2

Z pp2 +m2

Z 1

�1|�(p, t)|2 e�✏|t| dt d3p

(2⇡)3. (16.136)

To first order in ✏, the change in the action is (exercise 16.15)

�S0

[�] = lim✏!0+

i✏

2

Z pp2 +m2

Z 1

�1|�(p, t)|2 dt d3p

(2⇡)3

= lim✏!0+

i✏

2

Z pp2 +m2 |�(p)|2 d4p

(2⇡)4. (16.137)

The modified action is therefore

S0

[�, ✏] = S0

[�] +�S0

[�] = � 1

2

Z|�(p)|2

⇣p2 +m2 � i✏

pp2 +m2

⌘ d4p

(2⇡)4

= � 1

2

Z|�(p)|2

�p2 +m2 � i✏

� d4p

(2⇡)4(16.138)

since the square-root is positive. In terms of the modified action, our formula(16.128) for the time-ordered product is the ratio

h0|T [�(x1

) . . .�(xn)] |0i =

Z�(x

1

) . . .�(xn) eiS0[�,✏]D�Z

eiS0[�,✏]D�. (16.139)

We can use this formula (16.139) to express the mean value in the vacuum

16.12 Perturbation Theory 657

|0i of the time-ordered exponential of a space-time integral of j(x)�(x), inwhich j(x) is a classical (c-number, external) current, as the ratio

Z0

[j] ⌘ h0| T⇢exp

i

Zj(x)�(x) d4x

��|0i

=

Zexp

i

Zj(x)�(x) d4x

�eiS0[�,✏]D�Z

eiS0[�,✏]D�.

(16.140)

Since the state |0i is normalized, the mean value Z0

[0] is unity,

Z0

[0] = 1. (16.141)

If we absorb the current into the action

S0

[�, ✏, j] = S0

[�, ✏] +

Zj(x)�(x) d4x (16.142)

then in terms of the current’s Fourier transform

j(p) =

Ze�ipx j(x) d4x (16.143)

the modified action S0

[�, ✏, j] is (exercise 16.16)

S0

[�, ✏, j] = � 1

2

Z h|�(p)|2

�p2 +m2 � i✏

�� j⇤(p)�(p)� �⇤(p)j(p)

i d4p

(2⇡)4.

(16.144)Changing variables to

(p) = �(p)� j(p)/(p2 +m2 � i✏) (16.145)

we write the action S0

[�, ✏, j] as (exercise 16.17)

S0

[�, ✏, j] = � 1

2

Z | (p)|2

�p2 +m2 � i✏

�� j⇤(p)j(p)

(p2 +m2 � i✏)

�d4p

(2⇡)4

= S0

[ , ✏] + 1

2

Z j⇤(p)j(p)

(p2 +m2 � i✏)

�d4p

(2⇡)4. (16.146)

And since D� = D , our formula (16.140) gives simply (exercise 16.18)

Z0

[j] = exp

✓i

2

Z |j(p)|2p2 +m2 � i✏

d4p

(2⇡)4

◆. (16.147)

Going back to position space, one finds (exercise 16.19)

Z0

[j] = exp

i

2

Zj(x)�(x� x0) j(x0) d4x d4x0

�(16.148)

658 Path Integrals

in which �(x� x0) is Feynman’s propagator (5.233)

�(x� x0) =

Zeip(x�x0

)

p2 +m2 � i✏

d4p

(2⇡)4. (16.149)

The functional derivative (chapter 15) of Z0

[j], defined by (16.140), is

1

i

�Z0

[j]

�j(x)= h0| T

�(x) exp

✓i

Zj(x0)�(x0)d4x0

◆�|0i (16.150)

while that of equation (16.148) is

1

i

�Z0

[j]

�j(x)= Z

0

[j]

Z�(x� x0) j(x0) d4x0. (16.151)

Thus the second functional derivative of Z0

[j] evaluated at j = 0 gives

h0| T⇥�(x)�(x0)

⇤|0i = 1

i2�2Z

0

[j]

�j(x)�j(x0)

����j=0

= �i�(x� x0). (16.152)

Similarly, one may show (exercise 16.20) that

h0| T [�(x1

)�(x2

)�(x3

)�(x4

)] |0i = 1

i4�4Z

0

[j]

�j(x1

)�j(x2

)�j(x3

)�j(x4

)

����j=0

= ��(x1

� x2

)�(x3

� x4

)��(x1

� x3

)�(x2

� x4

)

��(x1

� x4

)�(x2

� x3

). (16.153)

Suppose now that we add a potential V = P (�) to the free hamiltonian(16.126). Scattering amplitudes are matrix elements of the time-ordered ex-ponential T exp

⇥�i

RP (�) d4x

⇤. Our formula (16.139) for the mean value in

the ground state |0i of the free hamiltonian H0

of any time-ordered productof fields leads us to

h0|T⇢exp

�i

ZP (�) d4x

��|0i =

Zexp

�i

ZP (�) d4x

�eiS0[�,✏]D�Z

eiS0[�,✏]D�.

(16.154)Using (16.152 & 16.153), we can cast this expression into the magical form

h0|T⇢exp

�i

ZP (�) d4x

��|0i = exp

�i

ZP

✓�

i�j(x)

◆d4x

�Z0

[j]

����j=0

.

(16.155)The generalization of the path-integral formula (16.139) to the ground

16.13 Application to Quantum Electrodynamics 659

state |⌦i of an interacting theory with action S is

h⌦|T [�(x1

) . . .�(xn)] |⌦i =

Z�(x

1

) . . .�(xn) eiS[�,✏]D�Z

eiS[�,✏]D�(16.156)

in which a term like i✏�2 is added to make the modified action S[�, ✏].These are some of the techniques one uses to make states of incoming and

outgoing particles and to compute scattering amplitudes (Weinberg, 1995,1996; Srednicki, 2007; Zee, 2010).

16.13 Application to Quantum Electrodynamics

In the Coulomb gauge r · A = 0, the QED hamiltonian is

H = Hm +

Z ⇥1

2

⇡2 + 1

2

(r ⇥ A)2 �A · j⇤d3x+ VC (16.157)

in which Hm is the matter hamiltonian, and VC is the Coulomb term

VC =1

2

Zj0(x, t) j0(y, t)

4⇡|x � y| d3x d3y. (16.158)

The operators A and ⇡ are canonically conjugate, but they satisfy theCoulomb-gauge conditions

r · A = 0 and r · ⇡ = 0. (16.159)

One may show (Weinberg, 1995, pp. 413–418) that in this theory, theanalog of equation (16.156) is

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS

C �[r · A]DAD ZeiSC �[r · A]DAD

(16.160)

in which the Coulomb-gauge action is

SC =

Z1

2

A2 � 1

2

(r ⇥ A)2 +A · j + Lm d4x �ZVC dt (16.161)

and the functional delta function

�[r · A] =Yx

�(r · A(x)) (16.162)

enforces the Coulomb-gauge condition. The term Lm is the action densityof the matter field .

660 Path Integrals

Tricks are available. We introduce a new field A0(x) and consider thefactor

F =

Zexp

i

Z1

2

�rA0+r4�1j0

�2

d4x

�DA0 (16.163)

which is just a number independent of the charge density j0 since we cancancel the j0 term by shifting A0. By 4�1, we mean � 1/4⇡|x � y|. Byintegrating by parts, we can write the number F as (exercise 16.21)

F =

Zexp

i

Z1

2

�rA0

�2 �A0j0 � 1

2

j04�1j0 d4x

�DA0

=

Zexp

i

Z1

2

�rA0

�2 �A0j0 d4x+ i

ZVC dt

�DA0.

(16.164)

So when we multiply the numerator and denominator of the amplitude(16.160) by F , the awkward Coulomb term cancels, and we get

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS0�[r · A]DAD Z

eiS0�[r · A]DAD

(16.165)

where now DA includes all four components Aµ and

S0 =

Z1

2

A2 � 1

2

(r ⇥ A)2 + 1

2

�rA0

�2

+A · j �A0j0 +Lm d4x. (16.166)

Since the delta-function �[r · A] enforces the Coulomb-gauge condition, wecan add to the action S0 the term (r · A)A0 which is � A · rA0 after weintegrate by parts and drop the surface term. This extra term makes theaction gauge invariant

S =

Z1

2

(A � rA0)2 � 1

2

(r ⇥ A)2 +A · j �A0j0 + Lm d4x

=

Z� 1

4

Fab Fab+Abjb + Lm d4x.

(16.167)

Thus at this point we have

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS �[r · A]DAD Z

eiS �[r · A]DAD (16.168)

in which S is the gauge-invariant action (16.167), and the integral is overall fields. The only relic of the Coulomb gauge is the gauge-fixing deltafunctional �[r · A].

16.13 Application to Quantum Electrodynamics 661

We now make gauge transformations

A0b(x) = Ab(x) + @b⇤(x) and 0(x) = eiq⇤(x) (x) (16.169)

and replace the fields Ab(x) and (x) everywhere in the numerator and(using a di↵erent gauge transformation ⇤0) in the denominator in the ratio(16.168) of path integrals by their gauge transforms (16.169) A0

µ(x) and 0(x). These changes of variables change nothing; they’re like replacingR1�1 f(x) dx by

R1�1 f(y) dy, and so

h⌦|T [O1

. . .On] |⌦i = h⌦|T [O1

. . .On] |⌦i0 (16.170)

in which the prime refers to the gauge transformation (16.169).We’ve seen that the action S is gauge invariant. So is the measureDAD ,

and we now restrict ourselves to operators O1

. . .On that are gauge invari-ant . So in the right-hand side of equation (16.170), the replacement of thefields by their gauge transforms a↵ects only the term �[r · A] that enforcesthe Coulomb-gauge condition

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS �[r · A+4⇤]DAD Z

eiS �[r · A+4⇤0]DAD . (16.171)

We now have two choices. If we integrate over all gauge functions ⇤(x) inboth the numerator and the denominator of this ratio (16.171), then apartfrom over-all constants that cancel, the mean value in the vacuum of thetime-ordered product is the ratio

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS DAD Z

eiS DAD (16.172)

in which we integrate over all matter fields, gauge fields, and gauges. Thatis, we do not fix the gauge.The analogous formula for the euclidean time-ordered product is

h⌦|Te [O1

. . .On] |⌦i =

ZO

1

. . .On e�S

e DAD Ze�S

e DAD (16.173)

in which the euclidean action Se is the space-time integral of the energydensity. This formula is quite general; it holds in nonabelian gauge theoriesand is important in lattice gauge theory.

662 Path Integrals

Our second choice is to multiply the numerator and the denominatorof the ratio (16.171) by the exponential exp[�i1

2

↵R(�4⇤)2 d4x] and then

integrate over ⇤(x) separately in the numerator and denominator. This op-eration just multiplies the numerator and denominator by the same constantfactor, which cancels. But if before integrating over all gauge transforma-tions ⇤(x), we shift ⇤ so that 4⇤ decreases by A0, then the exponentialfactor is exp[�i1

2

↵R(A0 � 4⇤)2 d4x]. Now when we integrate over ⇤(x),

the delta function �(r · A+4⇤) replaces 4⇤ by �r · A in the insertedexponential, converting it to exp[�i1

2

↵R(A0 + r · A)2 d4x]. The result is

to replace the gauge-invariant action (16.167) with the gauge-fixed action

S↵ =

Z� 1

4Fab F

ab � ↵

2(@bA

b)2+Abjb + Lm d4x. (16.174)

This action is Lorentz invariant and so is much easier to work with than theone (16.161) with the Coulomb term. We can use it to compute scatteringamplitudes perturbatively. The mean value of a time-ordered product ofoperators in the ground state |0i of the free theory is

h0|T [O1

. . .On] |0i =

ZO

1

. . .On eiS

↵ DAD ZeiS↵ DAD

. (16.175)

By following steps analogous to those the led to (16.149), one may show(exercise 16.22) that in Feynman’s gauge, ↵ = 1, the photon propagator is

h0|T [Aµ(x)A⌫(y)] |0i = � i4µ⌫(x� y) = � i

Z⌘µ⌫

q2 � i✏eiq·(x�y) d4q

(2⇡)4.

(16.176)

16.14 Fermionic Path Integrals

In our brief introduction (1.11–1.12) and (1.43–1.45), to Grassmann vari-ables, we learned that because

✓2 = 0 (16.177)

the most general function f(✓) of a single Grassmann variable ✓ is

f(✓) = a+ b ✓. (16.178)

So a complete integral table consists of the integral of this linear functionZf(✓) d✓ =

Za+ b ✓ d✓ = a

Zd✓ + b

Z✓ d✓. (16.179)

16.14 Fermionic Path Integrals 663

This equation has two unknowns, the integralRd✓ of unity and the integralR

✓ d✓ of ✓. We choose them so that the integral of f(✓ + ⇣)Zf(✓ + ⇣) d✓ =

Za+ b (✓ + ⇣) d✓ = (a+ b ⇣)

Zd✓ + b

Z✓ d✓ (16.180)

is the same as the integral (16.179) of f(✓). Thus the integralRd✓ of unity

must vanish, while the integralR✓ d✓ of ✓ can be any constant, which we

choose to be unity. Our complete table of integrals is thenZd✓ = 0 and

Z✓ d✓ = 1. (16.181)

The anticommutation relations for a fermionic degree of freedom are

{ , †} ⌘ † + † = 1 and { , } = { †, †} = 0. (16.182)

Because has †, it is conventional to introduce a variable ✓⇤ = ✓† thatanti-commutes with itself and with ✓

{✓⇤, ✓⇤} = {✓⇤, ✓} = {✓, ✓} = 0. (16.183)

The logic that led to (16.181) now givesZd✓⇤ = 0 and

Z✓⇤ d✓⇤ = 1. (16.184)

We define the reference state |0i as |0i ⌘ |si for a state |si that is notannihilated by . Since 2 = 0, the operator annihilates the state |0i

|0i = 2|si = 0. (16.185)

The e↵ect of the operator on the state

|✓i = exp⇣ †✓ � 1

2

✓⇤✓⌘|0i =

⇣1 + †✓ � 1

2

✓⇤✓⌘|0i (16.186)

is

|✓i = (1 + †✓ � 1

2

✓⇤✓)|0i = †✓|0i = (1� † )✓|0i = ✓|0i (16.187)

while that of ✓ on |✓i is

✓|✓i = ✓(1 + †✓ � 1

2

✓⇤✓)|0i = ✓|0i. (16.188)

The state |✓i therefore is an eigenstate of with eigenvalue ✓

|✓i = ✓|✓i. (16.189)

The bra corresponding to the ket |⇣i is

h⇣| = h0|✓1 + ⇣⇤ � 1

2⇣⇤⇣

◆(16.190)

664 Path Integrals

and the inner product h⇣|✓i is (exercise 16.23)

h⇣|✓i = h0|✓1 + ⇣⇤ � 1

2⇣⇤⇣

◆✓1 + †✓ � 1

2✓⇤✓

◆|0i

= h0|1 + ⇣⇤ †✓ � 1

2⇣⇤⇣ � 1

2✓⇤✓ +

1

4⇣⇤⇣✓⇤✓|0i

= h0|1 + ⇣⇤✓ � 1

2⇣⇤⇣ � 1

2✓⇤✓ +

1

4⇣⇤⇣✓⇤✓|0i

= exp

⇣⇤✓ � 1

2(⇣⇤⇣ + ✓⇤✓)

�. (16.191)

Example 16.2 (A Gaussian Integral) For any number c, we can computethe integral of exp(c ✓⇤✓) by expanding the exponentialZ

ec ✓⇤✓ d✓⇤d✓ =

Z(1 + c ✓⇤✓) d✓⇤d✓ =

Z(1� c ✓ ✓⇤) d✓⇤d✓ = �c (16.192)

a formula that we’ll use over and over.

The identity operator for the space of states

c|0i+ d|1i ⌘ c|0i+ d †|0i (16.193)

is (exercise 16.24) the integral

I =

Z|✓ih✓| d✓⇤d✓ = |0ih0|+ |1ih1| (16.194)

in which the di↵erentials anti-commute with each other and with otherfermionic variables: {d✓, d✓⇤} = 0, {d✓, ✓} = 0, {d✓, } = 0, and so forth.The case of several Grassmann variables ✓

1

, ✓2

, . . . , ✓n and several Fermioperators

1

, 2

, . . . , n is similar. The ✓k anticommute among themselves

{✓i, ✓j} = {✓i, ✓⇤j} = {✓⇤i , ✓⇤j} = 0 (16.195)

while the k satisfy

{ k, †`} = �k` and { k, l} = { †

k, †`} = 0. (16.196)

The reference state |0i is

|0i =

nYk=1

k

!|si (16.197)

in which |si is any state not annihilated by any k (so the resulting |0i isn’t

16.14 Fermionic Path Integrals 665

zero). The direct-product state

|✓i ⌘ exp

nX

k=1

†k✓k �

1

2✓⇤k✓k

!|0i =

"nY

k=1

✓1 + †

k✓k �1

2✓⇤k✓k

◆#|0i

(16.198)is (exercise 16.25) a simultaneous eigenstate of each k

k|✓i = ✓k|✓i. (16.199)

It follows that

` k|✓i = `✓k|✓i = �✓k `|✓i = �✓k✓`|✓i = ✓`✓k|✓i (16.200)

and so too k `|✓i = ✓k✓`|✓i. Since the ’s anticommute, their eigenvaluesmust also

✓`✓k|✓i = ` k|✓i = � k `|✓ = �✓k✓`|✓i (16.201)

which is why they must be Grassmann variables.The inner product h⇣|✓i is

h⇣|✓i = h0|"

nYk=1

(1 + ⇣⇤k k �1

2⇣⇤k⇣k)

#"nY`=1

(1 + †`✓` �

1

2✓⇤` ✓`)

#|0i

= exp

"nX

k=1

⇣⇤k✓k �1

2(⇣⇤k⇣k + ✓⇤k✓k)

#= e⇣

†✓�(⇣†⇣+✓†✓)/2. (16.202)

The identity operator is

I =

Z|✓ih✓|

nYk=1

d✓⇤kd✓k. (16.203)

Example 16.3 (Gaussian Grassmann Integral) For any 2 ⇥ 2 matrix A,we may compute the gaussian integral

g(A) =

Ze�✓

†A✓ d✓⇤1

d✓1

d✓⇤2

d✓2

(16.204)

by expanding the exponential. The only terms that survive are the onesthat have exactly one of each of the four variables ✓

1

, ✓2

, ✓⇤1

, and ✓⇤2

. Thus,the integral is the determinant of the matrix A

g(A) =

Z1

2(✓⇤kAk`✓`)

2 d✓⇤1

d✓1

d✓⇤2

d✓2

=

Z(✓⇤

1

A11

✓1

✓⇤2

A22

✓2

+ ✓⇤1

A12

✓2

✓⇤2

A21

✓1

) d✓⇤1

d✓1

d✓⇤2

d✓2

= A11

A22

�A12

A21

= detA. (16.205)

666 Path Integrals

The natural generalization to n dimensionsZe�✓

†A✓nY

k=1

d✓⇤kd✓k = detA (16.206)

is true for any n ⇥ n matrix A. If A is invertible, then the invariance ofGrassmann integrals under translations implies thatZ

e�✓†A✓+✓†⇣+⇣†✓

nYk=1

d✓⇤kd✓k =

Ze�✓

†A(✓+A�1⇣)+✓†⇣+⇣†(✓+A�1⇣)nY

k=1

d✓⇤kd✓k

=

Ze�✓

†A✓+⇣†✓+⇣†A�1⇣nY

k=1

d✓⇤kd✓k

=

Ze�(✓†+⇣†A�1

)A✓+⇣†✓+⇣†A�1⇣nY

k=1

d✓⇤kd✓k

=

Ze�✓

†A✓+⇣†A�1⇣nY

k=1

d✓⇤kd✓k

= detA e⇣†A�1⇣ . (16.207)

The values of ✓ and ✓† that make the argument �✓†A✓ + ✓†⇣ + ⇣†✓ of theexponential stationary are ✓ = A�1⇣ and ✓† = ⇣†A�1. So a gaussian Grass-mann integral is equal to its exponential evaluated at its stationary point,apart from a prefactor involving the determinant detA. This result is afermionic echo of the bosonic results (16.13–16.15).

One may further extend these definitions to a Grassmann field �m(x)and an associated Dirac field m(x). The �m(x)’s anticommute amongthemselves and with all fermionic variables at all points of space-time

{�m(x),�n(x0)} = {�⇤

m(x),�n(x0)} = {�⇤

m(x),�⇤n(x

0)} = 0 (16.208)

and the Dirac field m(x) obeys the equal-time anticommutation relations

{ m(x, t), †n(x

0, t)} = �mn �(x� x0) (n,m = 1, . . . , 4)

{ m(x, t), n(x0, t)} = { †

m(x, t), †n(x

0, t)} = 0.(16.209)

As in (16.101 & 16.197), we use eigenstates of the field at t = 0. If |0iis defined in terms of a state |si that is not annihilated by any m(x, 0) as

|0i ="Ym,x

m(x, 0)

#|si (16.210)

16.14 Fermionic Path Integrals 667

then (exercise 16.26) the state

|�i = exp

Z Xm

†m(x, 0)�m(x)� 1

2�⇤m(x)�m(x) d3x

!|0i

= exp

✓Z †�� 1

2

�†� d3x

◆|0i (16.211)

is an eigenstate of the operator m(x, 0) with eigenvalue �m(x)

m(x, 0)|�i = �m(x)|�i. (16.212)

The inner product of two such states is (exercise 16.27)

h�0|�i = exp

Z�0†�� 1

2�0†�0 � 1

2�†� d3x

�. (16.213)

The identity operator is the integral

I =

Z|�ih�|D�⇤D� (16.214)

in which

D�⇤D� ⌘Ym,x

d�⇤m(x)d�m(x). (16.215)

The hamiltonian for a free Dirac field of mass m is the spatial integral

H0

=

Z (� · r+m) d3x (16.216)

in which ⌘ i †�0 and the gamma matrices (10.287) satisfy

{�a, �b} = 2 ⌘ab (16.217)

where ⌘ is the 4⇥ 4 diagonal matrix with entries (�1, 1, 1, 1). Since |�i =�|�i and h�0| † = h�0|�0†, the quantity h�0| exp(� i✏H

0

)|�i is by (16.213)

h�0|e�i✏H0 |�i = h�0|�i exp� i✏

Z�0 (� · r+m)� d3x

�(16.218)

= exp

Z1

2

(�0† � �†)�� 1

2

�0†(�0 � �)� i✏�0(� ·r+m)�d3x

�= exp

⇢✏

Z h1

2

�†�� 1

2

�0†�� i�0 (� · r+m)�id3x

�in which �0† � �† = ✏�† and �0 � � = ✏�. Everything within the square

668 Path Integrals

brackets is multiplied by ✏, so we may replace �0† by �† and �0 by � so asto write to first order in ✏

h�0|e�i✏H0 |�i = exp

✏

Z1

2

�†�� 1

2

�†�� i� (� · r+m)� d3x

�(16.219)

in which the dependence upon �0 is through the time derivatives.Putting together n = 2t/✏ such matrix elements, integrating over all

intermediate-state dyadics |�ih�|, and using our formula (16.214), we find

h�t|e�2itH0 |��ti =Zexp

Z1

2

�†�� 1

2

�†�� i� (� ·r+m)�d4x

�D�⇤D�.

(16.220)Integrating �†� by parts and dropping the surface term, we get

h�t|e�2itH0 |��ti =Zexp

Z� �†�� i� (� ·r+m)� d4x

�D�⇤D�.

(16.221)Since � �†� = � i��0�, the argument of the exponential is

i

Z� ��0�� � (� · r+m)� d4x = i

Z� � (�µ@µ +m)� d4x. (16.222)

We then have

h�t|e�2itH0 |��ti =Zexp

✓i

ZL0

(�) d4x

◆D�⇤D� (16.223)

in which L0

(�) = � � (�µ@µ +m)� is the action density (10.289) for a freeDirac field. Thus the amplitude is a path integral with phases given by theclassical action S

0

[�]

h�t|e�2itH0 |��ti =Z

eiRL0(�) d4xD�⇤D� =

ZeiS0[�]D�⇤D� (16.224)

and the integral is over all fields that go from �(x,�t) = ��t(x) to �(x, t) =�t(x). Any normalization factor will cancel in ratios of such integrals.Since Fermi fields anticommute, their time-ordered product has an extra

minus sign

T⇥ (x

1

) (x2

)⇤= ✓(x0

1

�x02

) (x1

) (x2

)�✓(x02

�x01

) (x2

) (x1

). (16.225)

The logic behind our formulas (16.122) and (16.128) for the time-orderedproduct of bosonic fields now leads to an expression for the time-ordered

16.14 Fermionic Path Integrals 669

product of 2n Dirac fields (with D�00 and D�0 and so forth suppressed)

h0|T⇥ (x

1

) . . . (x2n)

⇤|0i =

Zh0|�00i�(x

1

) . . .�(x2n) e

iS0[�]h�0|0iD�⇤D�Zh0|�00i eiS0[�]h�0|0iD�⇤D�

.

(16.226)As in (16.139), the e↵ect of the inner products h0|�00i and h�0|0i is to insert✏-terms which modify the Dirac propagators

h0|T⇥ (x

1

) . . . (x2n)

⇤|0i =

Z�(x

1

) . . .�(x2n) e

iS0[�,✏]D�⇤D�ZeiS0[�,✏]D�⇤D�

. (16.227)

Imitating (16.140), we introduce a Grassmann external current ⇣(x) anddefine a fermionic analog of Z

0

[j]

Z0

[⇣] ⌘ h0| TheR⇣ + ⇣ d4x

i|0i =

ZeR⇣�+�⇣ d4xeiS0[�,✏]D�⇤D�Z

eiS0[�,✏]D�⇤D�. (16.228)

Example 16.4 (Feynman’s fermion propagator) Since

i (�µ@µ +m)�(x� y) ⌘ i (�µ@µ +m)

Zd4p

(2⇡)4eip(x�y)�i (�i�⌫p⌫ +m)

p2 +m2 � i✏

=

Zd4p

(2⇡)4eip(x�y) (i�µpµ +m)

(�i�⌫p⌫ +m)

p2 +m2 � i✏

=

Zd4p

(2⇡)4eip(x�y) p2 +m2

p2 +m2 � i✏= �4(x� y),

(16.229)

the function �(x� y) is the inverse of the di↵erential operator i(�µ@µ+m).Thus the Grassmann identity (16.207) implies that Z

0

[⇣] is

h0| TheR⇣ + ⇣ d4x

i|0i =

ZeR[⇣�+�⇣�i�(�µ@

µ

+m)�]d4xD�⇤D�ZeiS0[�,✏]D�⇤D�

= exp

Z⇣(x)�(x� y)⇣(y) d4xd4y

�.

(16.230)

670 Path Integrals

Di↵erentiating we get

h0|T⇥ (x) (y)

⇤|0i = �(x�y) = �i

Zd4p

(2⇡)4eip(x�y) �i�⌫p⌫ +m

p2 +m2 � i✏. (16.231)

16.15 Application to nonabelian gauge theories

The action of a generic non-abelian gauge theory is

S =

Z� 1

4

Faµ⌫Fµ⌫a � (�µDµ +m) d4x (16.232)

in which the Maxwell field is

Faµ⌫ ⌘ @µAa⌫ � @⌫Aaµ + g fabcAbµAc⌫ (16.233)

and the covariant derivative is

Dµ ⌘ @µ � ig taAaµ . (16.234)

Here g is a coupling constant, fabc is a structure constant (10.63), and ta isa generator (10.55) of the Lie algebra (section 10.15) of the gauge group.One may show (Weinberg, 1996, pp. 14–18) that the analog of equation

(16.168) for quantum electrodynamics is

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS �[Aa3]DAD Z

eiS �[Aa3]DAD (16.235)

in which the functional delta function

�[Aa3] ⌘Yx,b

�(Aa3(x)) (16.236)

enforces the axial-gauge condition, and D stands for D ⇤D .Initially, physicists had trouble computing nonabelian amplitudes beyond

the lowest order of perturbation theory. Then DeWitt showed how to com-pute to second order (DeWitt, 1967), and Faddeev and Popov, using pathintegrals, showed how to compute to all orders (Faddeev and Popov, 1967).

16.16 The Faddeev-Popov trick 671

16.16 The Faddeev-Popov trick

The path-integral tricks of Faddeev and Popov are described in (Weinberg,1996, pp. 19–27). We will use gauge-fixing functions Ga(x) to impose agauge condition on our non-abelian gauge fields Aa

µ(x). For instance, wecan use Ga(x) = A3

a(x) to impose an axial gauge or Ga(x) = i@µAµa(x) to

impose a Lorentz-invariant gauge.Under an infinitesimal gauge transformation (11.514)

A�aµ = Aaµ � @µ�a � g fabcAbµ �c (16.237)

the gauge fields change, and so the gauge-fixing functions Gb(x), whichdepend upon them, also change. The jacobian J of that change at � = 0 is

J = det

✓�G�

a(x)

��b(y)

◆�����=0

⌘ DG�

D�

�����=0

(16.238)

and it typically involves the delta function �4(x� y).Let B[G] be any functional of the gauge-fixing functions Gb(x) such as

B[G] =Yx,a

�(Ga(x)) =Yx,a

�(A3

a(x)) (16.239)

in an axial gauge or

B[G] = exp

i

2

Z(Ga(x))

2 d4x

�= exp

� i

2

Z(@µA

µa(x))

2 d4x

�(16.240)

in a Lorentz-invariant gauge.We want to understand functional integrals like (16.232)

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS B[G] J DAD Z

eiS B[G] J DAD (16.241)

in which the operators Ok, the action functional S[A], and the di↵erentialsDAD (but not the gauge-fixing functional B[G] or the Jacobian J) aregauge invariant. The axial-gauge formula (16.232) is a simple example inwhich B[G] = �[Aa3] enforces the axial-gauge condition Aa3(x) = 0 and thedeterminant J = det (�ab@3�(x� y)) is a constant that cancels.

If we translate the gauge fields by gauge transformations ⇤ and ⇤0, thenthe ratio (16.238) does not change

h⌦|T [O1

. . .On] |⌦i =

ZO⇤

1

. . .O⇤

n eiS⇤B[G⇤] J⇤DA⇤D ⇤Z

eiS⇤0B[G⇤

0] J⇤

0DA⇤

0D ⇤

0(16.242)

672 Path Integrals

any more thanRf(y) dy is di↵erent from

Rf(x) dx. Since the operators Ok,

the action functional S[A], and the di↵erentials DAD are gauge invariant,most of the ⇤-dependence goes away

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS B[G⇤] J⇤DAD Z

eiS B[G⇤

0] J⇤

0DAD

. (16.243)

Let ⇤� be a gauge transformation ⇤ followed by an infinitesimal gaugetransformation �. The jacobian J⇤ is a determinant of a product of matriceswhich is a product of their determinants

J⇤ = det

✓�G⇤�

a (x)

��b(y)

◆�����=0

= det

✓Z�G⇤�

a (x)

�⇤�c(z)

�⇤�c(z)

��b(y)d4z

◆�����=0

= det

✓�G⇤�

a (x)

�⇤�c(z)

◆�����=0

det

✓�⇤�c(z)

��b(y)

◆�����=0

= det

✓�G⇤

a (x)

�⇤c(z)

◆det

✓�⇤�c(z)

��b(y)

◆�����=0

⌘ DG⇤

D⇤

D⇤�

D�

�����=0

. (16.244)

Now we integrate over the gauge transformations ⇤ (and ⇤0) with weightfunction ⇢(⇤) = (D⇤�/D�|�=0

)�1 and find, since the ratio (16.240) is ⇤-independent

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS B[G⇤]

DG⇤

D⇤D⇤DAD Z

eiS B[G⇤

0]DG⇤

0

D⇤0 D⇤0DAD

=

ZO

1

. . .On eiS B[G⇤]DG⇤DAD Z

eiS B[G⇤]DG⇤DAD

=

ZO

1

. . .On eiS DAD Z

eiS DAD . (16.245)

Thus the mean-value in the vacuum of a time-ordered product of gauge-invariant operators is a ratio of path integrals over all gauge fields withoutany gauge fixing. No matter what gauge condition G or gauge-fixing func-tional B[G] we use, the resulting gauge-fixed ratio (16.238) is equal to theratio (16.242) of path integrals over all gauge fields without any gauge fixing.All gauge-fixed ratios (16.238) give the same time-ordered products, and so

16.17 Ghosts 673

we can use whatever gauge condition G or gauge-fixing functional B[G] ismost convenient.

The analogous formula for the euclidean time-ordered product is

h⌦|Te [O1

. . .On] |⌦i =

ZO

1

. . .On e�S

e DAD Ze�S

e DAD (16.246)

where the euclidean action Se is the space-time integral of the energy density.This formula is the basis for lattice gauge theory.

The path-integral formulas (16.172 & 16.173) derived for quantum elec-trodynamics therefore also apply to nonabelian gauge theories.

16.17 Ghosts

Faddeev and Popov showed how to do perturbative calculations in which onedoes fix the gauge. To continue our description of their tricks, we return tothe gauge-fixed expression (16.238) for the time-ordered product

h⌦|T [O1

. . .On] |⌦i =

ZO

1

. . .On eiS B[G] J DAD Z

eiS B[G] J DAD (16.247)

set Gb(x) = �i@µAµb (x) and use (16.237) as the gauge-fixing functional B[G]

B[G] = exp

i

2

Z(Ga(x))

2 d4x

�= exp

� i

2

Z(@µA

µa(x))

2 d4x

�.

(16.248)This functional adds to the action density the term �(@µA

µa)2/2 which leads

to a gauge-field propagator like the photon’s (16.176)

h0|ThAa

µ(x)Ab⌫(y)

i|0i = � i�ab4µ⌫(x� y) = � i

Z⌘µ⌫�abq2 � i✏

eiq·(x�y) d4q

(2⇡)4.

(16.249)What about the determinant J? Under an infinitesimal gauge transfor-

mation (16.234), the gauge field becomes

A�aµ = Aaµ � @µ�a � g fabcAbµ �c (16.250)

and so G�a(x) = i@µA�aµ(x) is

G�a(x) = i@µAaµ(x) + i@µ

Z[��ac@µ � g fabcAbµ(x)] �

4(x� y)�c(y) d4y.

(16.251)

674 Path Integrals

The jacobian J then is the determinant (16.235) of the matrix✓�G�

a(x)

��c(y)

◆�����=0

= �i�ac2 �4(x�y)�ig fabc

@

@xµ⇥Aµ

b (x)�4(x� y)

⇤(16.252)

that is

J = det

✓�i�ac2 �

4(x� y)� ig fabc@

@xµ⇥Aµ

b (x)�4(x� y)

⇤◆. (16.253)

But we’ve seen (16.206) that a determinant can be written as a fermionicpath integral

detA =

Ze�✓

†A ✓nY

k=1

d✓⇤kd✓k. (16.254)

So we can write the jacobian J as

J =

Zexp

Zi!⇤

a2!a + igfabc!⇤a@µ(A

µb!c) d

4x

�D!⇤D! (16.255)

which contributes the terms �@µ!⇤a@

µ!a and

�@µ!⇤a g fabcA

µb!c = @µ!

⇤a g fabcA

µc!b (16.256)

to the action density.Thus we can do perturbation theory by using the modified action density

L0 = � 1

4

Faµ⌫Fµ⌫a � 1

2

(@µAµa)

2�@µ!⇤a@

µ!a+@µ!⇤a g fabcA

µc!b� ( 6D +m)

(16.257)in which 6D ⌘ �µDµ = �µ(@µ�igtaAaµ). The ghost field ! is a mathematicaldevice, not a physical field describing real particles, which would be spinlessfermions violating the spin-statistics theorem (example 10.20).

Further Reading

Quantum Field Theory (Srednicki, 2007), The Quantum Theory of FieldsI, II, & III (Weinberg, 1995, 1996, 2005), and Quantum Field Theory in aNutshell (Zee, 2010) all provide excellent treatments of path integrals.

Exercises

16.1 Derive the multiple gaussian integral (16.8) from (5.167).

16.2 Derive the multiple gaussian integral (16.12) from (5.166).

Exercises 675

16.3 Show that the vector Y that makes the argument of the multiple gaus-sian integral (16.12) stationary is given by (16.13), and that the mul-tiple gaussian integral (16.12) is equal to its exponential evaluated atits stationary point Y apart from a prefactor involving det iS.

16.4 Repeat the previous exercise for the multiple gaussian integral (16.11).

16.5 Compute the double integral (16.23) for the case V (qj) = 0.

16.6 Insert into the LHS of (16.54) a complete set of momentum dyadics|pihp|, use the inner product hq|pi = exp(iqp~)/

p2⇡~, do the resulting

Fourier transform, and so verify the free-particle path integral (16.54).

16.7 By taking the nonrelativistic limit of the formula (11.311) for the actionof a relativistic particle of mass m and charge q, derive the expression(16.55) for the action a nonrelativistic particle in an electromagneticfield with no scalar potential.

16.8 Show that for the hamiltonian (16.59) of the simple harmonic oscillatorthe action S[qc] of the classical path is (16.66).

16.9 Show that the harmonic-oscillator action of the loop (16.67) is (16.68).

16.10 Show that the harmonic-oscillator amplitude (16.71) for q0 = 0 andq00 = q reduces as t ! 0 to the one-dimensional version of the free-particle amplitude (16.54).

16.11 Work out the path-integral formula for the amplitude for a mass minitially at rest to fall to the ground from height h in a gravitationalfield of local acceleration g to lowest order and then including loops upto an overall constant. Hint: use the technique of section 16.7.

16.12 Show that the action (16.73 ) of the stationary solution (16.76) is(16.78).

16.13 Derive formula (16.131) for the action S0

[�] from (16.129 & 16.130).

16.14 Derive identity (16.135). Split the time integral at t = 0 into twohalves, use

✏ e±✏t = ± d

dte±✏t (16.258)

and then integrate each half by parts.

16.15 Derive the third term in equation (16.137) from the second term.

16.16 Use (16.142) and the Fourier transform (16.143) of the external cur-rent j to derive the formula (16.144) for the modified action S

0

[�, ✏, j].

16.17 Derive equation (16.146) from equations (16.144) and (16.145).

16.18 Derive the formula (16.147) for Z0

[j] from the expression (16.146) forS0

[�, ✏, j].

16.19 Derive equations (16.148 & 16.149) from formula (16.147).

16.20 Derive equation (16.153) from the formula (16.148) for Z0

[j].

676 Path Integrals

16.21 Show that the time integral of the Coulomb term (16.158) is the termthat is quadratic in j0 in the number F defined by (16.163).

16.22 By following steps analogous to those the led to (16.149), derive theformula (16.176) for the photon propagator in Feynman’s gauge.

16.23 Derive expression (16.191) for the inner product h⇣|✓i.16.24 Derive the representation (16.194) of the identity operator I for a

single fermionic degree of freedom from the rules (16.181 & 16.184) forGrassmann integration and the anticommutation relations (16.177 &16.183).

16.25 Derive the eigenvalue equation (16.199) from the definition (16.197 &16.198) of the eigenstate |✓i and the anticommutation relations (16.195& 16.196).

16.26 Derive the eigenvalue relation (16.212) for the Fermi field m(x, t)from the anticommutation relations (16.208 & 16.209) and the defini-tions (16.210 & 16.211).

16.27 Derive the formula (16.213 ) for the inner product h�0|�i from thedefinition (16.211) of the ket |�i.

16.28 Without setting ~ = 1, imitate the derivation of the path-integralformula (16.25) and derive its three-dimensional version (16.26).