16 Path Integrals

16.1 Path Integrals and Classical Physics Since Richard Feynman invented them over 60 years ago, path integrals have been used with increasing frequency in high-energy and condensed-matter physics, in finance, and in biophysics (Kleinert, 2009). Feynman used them to express matrix elements of the time-evolution operator exp( itH/~)in terms of the classical action. Others have used them to compute matrix elements of the Boltzmann operator exp( H/kT ) which in the limit of zero temperature projects out the ground state E of the system | 0i

1 (H E0)/kT (En E0)/kT lim e =lim En e En = E0 E0 (16.1) T 0 T 0 | i h | | ih | ! ! n=0 X 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 classical action 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 we begin by recalling the basic integral formula (5.167)

b 2 ⇡ 1 exp ia x dx = (16.2) 2a ia Z1 " ✓ ◆ # r 16.2 Gaussian Integrals 671 which holds for real a and b, and also the one (5.168) c 2 ⇡ 1 exp r x dx = (16.3) 2r r Z1  ⇣ ⌘ r which is true for positive r and complex c. Equivalent formulas for real a and b, positive r, and complex c are ⇡ b2 1 exp iax2 + ibx dx = exp i (16.4) ia 4a Z1 r ✓ ◆ ⇡ c2 1 exp rx2 + cx dx = exp . (16.5) r 4r Z r ✓ ◆ 1 This last formula will be useful with x = p, r = ✏/(2m), and c = i✏q˙ p2 2⇡m 1 1 exp ✏ + i✏ qp˙ dp = exp ✏ mq˙2 (16.6) 2m ✏ 2 Z1 ✓ ◆ r ✓ ◆ as will (16.4)withx = p, a = ✏/(2m), and b = ✏q˙ p2 2⇡m 1 1 exp i✏ + i✏ qp˙ dp = exp i✏ mq˙2 . (16.7) 2m i✏ 2 Z1 ✓ ◆ r ✓ ◆ Doable path integrals are multiple gaussian integrals. One may show (ex- ercise 16.1) that for positive r1,...,rN and complex c1,...,cN , the integral (16.5) leads to

N N 2 1 2 ⇡ 1 ci exp rixi + cixi dxi = exp . (16.8) ! ri ! 4 ri ! Z1 Xi Yi=1 Yi=1 r Xi If R is the N N diagonal matrix with positive entries r ,r ,...,r , and ⇥ { 1 2 N } X and C are N-vectors with real x and complex c entries, then this { i} { i} formula (16.8) in matrix notation is

N N 1 T T ⇡ 1 T 1 exp X RX + C X dx = exp C R C . (16.9) i det(R) 4 Z i=1 s ✓ ◆ 1 Y Now every positive symmetric matrix S is of the form S = OROT for some positive diagonal matrix R.SoinsertingR = OTSO into the previous equa- tion (16.9) and using the invariance of determinants under orthogonal trans- formations, we find

N N 1 T T T ⇡ 1 T T 1 exp X O SOX + C X dxi = exp C O S OC . sdet(S) 4 Z1 i=1  Y (16.10) 672 Path Integrals The jacobian of the orthogonal transformations Y = OX and D = OC is unity, and so

N N 1 T T ⇡ 1 T 1 exp Y SY + D Y dy = exp D S D (16.11) i det(S) 4 Z i=1 s ✓ ◆ 1 Y 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)

N N 1 T T ⇡ i T 1 exp iY SY + iD Y dyi = exp D S D . sdet(iS) 4 Z1 i=1 ✓ ◆ Y (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)

1 1 Y = S D. (16.13) 2 The exponential of that integral evaluated at its stationary point Y is

T T i T 1 exp iY SY + iD Y =exp D S D . (16.14) 4 ✓ ◆ 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 of the exponential of the multiple gaussian integral (16.11) stationary is Y = 1 S D/2, and that exponential evaluated at Y is

T T 1 T 1 exp Y SY + D Y =exp D S D . (16.15) 4 ✓ ◆ Once again, a multiple gaussian integral is simply its exponential evaluated at its stationary point Y , apart from a prefactor involving a determinant, det S.

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 of Boltzmann’s constant k =8.617 10 5 eV/K times the absolute temperature ⇥ T . In the low-temperature limit, exp( H) is a projection operator (16.1) 16.3 Path Integrals in Imaginary Time 673 on the ground state of the system. These path integrals in imaginary time are called euclidian path integrals. Let us consider a quantum-mechanical system with hamiltonian p2 H = + V (q) (16.16) 2m 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 p2 p2 exp ✏ + V (q) exp ✏ exp ( ✏V(q)) + (✏2) (16.17) 2m ⇡ 2m O  ✓ ◆ ✓ ◆ are of second order in ✏. To evaluate the matrix element q exp( ✏H) q ,weinserttheidentity h 1| | 0i operator I in the form of an integral over the momentum eigenstates

1 I = p0 p0 dp0 (16.18) | ih | Z1 and use the inner product q p =exp(iq p )/p2⇡ so as to get as ✏ 0 h 1| 0i 1 0 ! 2 1 p q exp( ✏H) q = q exp ✏ p0 p0 exp ( ✏V(q)) q dp0 h 1| | 0i h 1| 2m | ih | | 0i Z1 ✓ ◆ 2 ✏V(q0) 1 p0 dp0 = e exp ✏ + ip0 (q q ) . (16.19) 2m 1 0 2⇡ Z1  If we adopt the suggestive notation q q 1 0 =˙q (16.20) ✏ 0 and use the integral formula (16.6), then we find

2 1 ✏V(q0) 1 p0 q exp( ✏H) q = e exp ✏ + i✏p0 q˙ dp0 h 1| | 0i 2⇡ 2m 0 Z1 ✓ ◆ m 1/2 = exp ✏ 1 m q˙2 + V (q ) (16.21) 2⇡✏ 2 0 0 ⇣ ⌘ ⇥ ⇤ in which q1 enters through the notation (16.20). The next step is to link two of these matrix elements together

2✏H 1 ✏H ✏H q e q = q e q q e q dq (16.22) h 2| | 0i h 2| | 1ih 1| | 0i 1 Z1 m = 1exp ✏ 1 m q˙2 + V (q )+ 1 m q˙2 + V (q ) dq . 2⇡✏ 2 1 1 2 0 0 1 Z 1 ⇥ ⇤ 674 Path Integrals Linking three of these matrix elements together and using subscripts instead of primes, we have

3✏H 1 ✏H ✏H ✏H q e q = q e q q e q q e q dq dq (16.23) h 3| | 0i h 3| | 2ih 2| | 1ih 1| | 0i 1 2 ZZ1 2 m 3/2 = 1exp ✏ 1 m q˙2 + V (q ) dq dq . 2⇡✏ 8 2 j j 9 2 1 ZZ j=0 ⇣ ⌘ 1 < X ⇥ ⇤= Boldly passing from 3 to n and suppressing: some integral signs,; we get

n✏H 1 ✏H ✏H qn e q0 = qn e qn 1 ... q1 e q0 dqn 1 ...dq1 (16.24) h | | i h | | i h | | i ZZZ 1 n 1 n/2 m 1 1 2 = exp ✏ m q˙j + V (qj) dqn 1 ...dq1. 2⇡✏ 8 2 9 ZZZ j=0 ⇣ ⌘ 1 < X ⇥ ⇤= Writing d for ✏ and taking the: limits ✏ 0 and n /;✏ ,wefind ! ⌘ !1 that the matrix element q e H q is a path integral of the exponential h | | 0i of the average energy multiplied by H 1 2 q e q = exp mq˙ (0)+V (q(0)) d0 Dq (16.25) h | | 0i 2 Z  Z0

2 n/2 in which Dq (n m/2⇡~ ) dqn 1 ...dq2 dq1 andq ˙ is the -derivative of ⌘ the coordinate q(). We sum over all paths q(t) that go from q(0) = q0 at inverse temperature =0toq()=q at inverse temperature . In the limit , the operator exp( H) becomes proportional to a !1 projection operator (16.1) on the ground state of the theory. In three dimensions with q˙()=dq()/d, and ~ = 1, the analog of 6 equation (16.25) is (exercise 16.28)

H m 2 q e q = exp q˙ (0)+V (q(0)) d0 Dq (16.26) h | | 0i 2 2 Z  Z0 ~

2 3n/2 where Dq (n m/2⇡~ ) dqn 1 ...dq2 dq1. ⌘ Path integrals in imaginary time are called euclidian mainly to distinguish them from Minkowski path integrals, which represent matrix elements of the time-evolution operator exp( itH) in real time. 16.4 Path Integrals in Real Time 675 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 ! 2 i✏H 1 p q e q = q exp i✏ p0 p0 exp [ i✏V(q)] q dp0 h 1| | 0i h 1| 2m | ih | | 0i Z1  2 1 i✏V(q0) 1 p0 = e exp i✏ + ip0 (q q ) dp0 2⇡ 2m 1 0 Z1  2 1 i✏V(q0) 1 p0 = e exp i✏ + i✏p0q˙ dp0 2⇡ 2m 0 Z1  m 1/2 m q˙2 = exp i✏ 0 V (q ) . (16.27) 2⇡i✏ 2 0 ⇣ ⌘  ✓ ◆ When we link together n of these matrix elements, we get the real-time version of (16.25)

n 1 n/2 in✏H m 1 1 2 qn e q0 = exp i✏ m q˙j V (qj) dqn 1 ...dq1. h | | i 2⇡i✏ 8 2 9 ⇣ ⌘ ZZZ 1 < Xj=0 = ⇥ ⇤ (16.28) Writing dt for ✏ and taking the limits: ✏ 0 and n✏ t, we; find that the ! ! amplitude q e itH q is the path integral h t| | 0i t itH/~ i 1 2 q e q = exp m q˙ V (q) dt0 Dq (16.29) h t| | 0i 2 Z ~ Z0 in which Dq di↵ers from the one that appears in euclidian path integrals by the substitution it/~ ! nm n/2 Dq =lim dq1dq2 ...dqn 1. (16.30) n 2⇡i~t !1 ⇣ ⌘ The integral in the exponent is the classical action

t 1 2 S[q]= m q˙ V (q) dt0 (16.31) 2 Z0 for a process q(t0) that runs from q(0) = q0 to q(t)=qt. We sum over all such processes. In three-dimensional space, the analog of (16.29) is

t itH 1 2 q e q = exp i m q˙ V (q) dt0 Dq. (16.32) h t| | 0i 2 Z  Z0 676 Path Integrals The units of action are energy time, and the argument of the exponential ⇥ must be dimensionless, so in arbitrary units the amplitude (16.29) is

itH/~ iS[q]/~ q e q = e Dq. (16.33) h t| | 0i Z

When is this amplitude big? When is it tiny? Suppose there is a process q(t)=qc(t) that goes from qc(0) = q0 to qc(t)=qt in time t and that obeys the classical equation of motion (15.14–15.15)

S[qc] = mq¨c + V 0(qc)=0. (16.34) qc

The action of such a classical process is stationary. A process qc(t)+dq(t) that di↵ers from a stationary process qc(t) by a small deviation dq(t) has an action S[qc + dq] that di↵ers from S[qc] only by terms of second order in q. Thus a stationary process has many neighboring processes that have the same action to within a fraction of ~. These processes add with nearly the same phase to the path integral (16.33) and make a huge contribution to the amplitude q e itH/~ q . h t| | 0i But if no classical process goes from q0 to qt in time t, then the nonclas- sical, nonstationary processes that go from q0 to qt in time t have actions that di↵er from each other by large multiples of ~. These amplitudes cancel each other, and the resulting amplitude is tiny. Thus the real-time path integral explains the principle of stationary action (section 11.41).

Example 16.1 (Schr¨odinger’s equation). The amplitude ~x , t = (~x , t ) h | i to get to ~x in time t from somewhere ~x 0 at time 0 is a sum over all processes

iS/~ (~x , t )= e (~x 0, 0). (16.35) X

So using our formulas (6.213 & 6.226) for the space-time partial derivatives 16.5 Path Integral for a Free Particle 677 S = p~ and @S/@t = H,wehave r

@ (~x , t ) @S iS/~ iS/~ i~ = e (~x 0, 0) = He (~x 0, 0) @t @t X X p~ 2 = H eiS/~ (~x , 0) = H (~x , t )= + V (~x ) (~x , t ) 0 2m ✓ ◆ p~X2 = + V (~x ) eiS/~ (~x , 0) (16.36) 2m 0 ✓ ◆ ( S)2 X = r + V (~x ) eiS/~ (~x , 0) 2m 0 ✓ ◆ 2 2 X ~ = r + V (~x ) eiS/~ (~x , 0) (16.37) 2m 0 ✓ ◆ 2 2 X ~ = r + V (~x ) (~x , 0) 2m 0 ✓ ◆ which is Schr¨odinger’s equation.

16.5 Path Integral for a Free Particle The amplitude for a free nonrelativistic particle to go from the origin to the point q in time t is the path integral (16.32)

t itH iS0[q] 1 2 q e q = 0 = e Dq = exp i m q˙ (t0) dt0 Dq. (16.38) h | | i 2 Z Z ✓ Z0 ◆

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 arbitrary detour q(t0) must be a loop that goes from q(0) = 0 to q(t)=0intimet.Thevelocity q˙ = q˙c + q˙ is the sum of the constant classical velocity q˙c = q/t and the loop’s velocity q˙. The first-order change vanishes

t dq t dq m q˙ dt = m q˙ dt = m q˙ [q(t) q(0)] = 0 (16.39) c · dt c · dt c · Z0 Z0 and so the action S0[q] is the classical action plus the loop action

t 1 2 S0[q]= 2 m (q˙c + q˙) dt0 = S0[qc]+S0[q]. (16.40) Z0 678 Path Integrals The path integral therefore factorizes

itH iS0[q] iS0[q +q] q e 0 = e Dq = e c Dq h | | i Z Z (16.41) = eiS0[qc] eiS0[q] Dq = eiS0[qc] eiS0[q] Dq Z Z into the phase of the classical action times a path integral over loops. The loop integral L is independent of the spatial points q and 0 and so can only depend upon the time interval, L = L(t). Thus the amplitude is the phase of the classical action times a function L(t) of the time

itH iS0[q ] q e 0 = e c L(t). (16.42) h | | i Since the classical velocity is q˙c = q/t, the classical action is t m m q2 S [q ]= q˙2(t) dt = . (16.43) 0 c 2 c 2 t Z0 So the amplitude is

2 i(t2 t1)H imq /2t q e 0 = e L(t). (16.44) h | | i Because position eigenstates are orthogonal, this amplitude must reduce to a delta function as t 0 ! itH 3 lim q e 0 = q 0 = (q). (16.45) t 0 h | | i h | i ! One of the many representations of Dirac’s delta function is

m 3/2 2 3(q)=lim eimq /2~t. (16.46) t 0 2⇡i~t ! ⇣ ⌘ Thus NL(t)=(m/2⇡it)3/2 and

3/2 itH/~ m imq2/2~t q e 0 = e (16.47) h | | i 2⇡i~t ⇣ ⌘ in arbitrary units. You can verify (exercise 16.6) this result by inserting a complete set of momentum dyadics p p into the left-hand side of the | ih | amplitude (16.47) and doing the resulting Fourier transform. Example 16.2 (The Bohm-Aharonov E↵ect). From our formula (11.322) for the action of a relativistic particle of mass m and charge q,weinfer (exercise 16.7) that the action a nonrelativistic particle in an electromagnetic field with no scalar potential is

x2 S = 1 mv + q A dx . (16.48) 2 · Zx1 ⇥ ⇤ 16.6 Free Particle in Imaginary Time 679 Now imagine that we shoot a beam of such particles past but not through a narrow cylinder in which a magnetic field is confined. The particles can go either way around the cylinder of area S but cannot enter the region of the magnetic field. The di↵erence in the phases of the amplitudes is the loop integral from the source to the detector and back to the source S mv dx mv dx q mv dx q = + q A = · + B dS = · + ~ 2 · ~ 2~ ~ S · 2~ ~ I h i I Z I (16.49) 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 hamiltonian is H = p2/2m,set = it/~ =1/kT , and use Dirac’s delta function 3/2 3 m mq2/2~t (q)=lim e (16.50) t 0 2⇡~t ! ⇣ ⌘ then we get

3/2 2 3/2 H m mq mkT mkT q2/2~2 q e 0 = exp = e . h | | i 2⇡~2 2~2 2⇡~2 ✓ ◆  ✓ ◆ (16.51) For = t/~, this is the solution (3.200 & 13.124) of the di↵usion equation with di↵usion constant D = ~/(2m).

16.7 Harmonic Oscillator in Real Time Biologists have mice; physicists have harmonic oscillators with hamiltonian p2 m!2q2 H = + . (16.52) 2m 2 For this hamiltonian, our formula (16.29) for the coordinate matrix elements of the time-evolution operator exp( itH)is itH iS[q] q e q = e Dq (16.53) h t| | 0i Z with action t 1 2 1 2 2 S[q]= mq˙ (t0) m! q (t0) dt0. (16.54) 2 2 Z0 The classical solution qc(t)=q0 cos !t +˙q0 sin(!t)/! in which q0 = qc(0) 680 Path Integrals andq ˙0 =˙qc(0) are the initial position and velocity satisfies the classical equation of motion mq¨ (t)= !2q (t) and makes the action S[q] stationary c c

d S[q + ✏h] =0. (16.55) d✏ ✏=0 We now apply the trick (16.39–16.41) we used for the free particle. We write an arbitrary process q(t) as the sum of the classical process qc(t) and a loop q(t)withq(0) = q(t) = 0. Since the action S[q] is quadratic in the variables q andq ˙, the functional Taylor series (15.31) for S[qc + q] has only two terms

S[qc + q]=S[qc]+S[q]. (16.56)

Thus we can write the path integral (16.53) as

itH iS[q] iS[qc+q] q e q = e Dq = e Dq h t| | 0i Z Z (16.57) = eiS[qc]+iS[q] Dq = eiS[qc] eiS[q] Dq. Z Z The remaining path integral over the loops q does not involve the end points q0 and qt and so must be a function L(t) of the time t but not of q0 or qt

itH iS[qc] q e q = e L(t). (16.58) h t| | 0i

The action S[qc] is (exercise 16.8) m! S[q ]= q2 + q2 cos(!t) 2q q . (16.59) c 2sin(!t) 0 t 0 t ⇥ ⇤ The action S[q] of a loop

n 1 j⇡t0 q(t0)= a sin (16.60) j t Xj=1 is (exercise 16.9)

n 1 mt (j⇡)2 S[q]= a2 !2 . (16.61) 4 j t2 Xj=1  16.8 Harmonic Oscillator in Imaginary Time 681 The path integral over the loops is then, apart from a constant jacobian J,

n 1 n 1 nm n/2 imt (j⇡)2 eiS[q] Dq = J exp a2 !2 da 2⇡it 8 4 j t2 9 j Z ⇣ ⌘ Z

Using the gaussian integral (16.2) and the infinite product (4.147), we get

n 1 1/2 m p2 !2t2 eiS[q] Dq = Jnn/2 1 2⇡it j⇡ ⇡2j2 Z r j=1 ✓ ◆ Y (16.63) n 1 m! p2 = lim Jnn/2 . 2⇡i sin !t 0n j⇡ 1 r !1 j=1 Y @ A From (16.58) and (16.59), we see that the number within the parentheses is unity because we then have (Feynman and Hibbs, 1965, ch. 3)

m! q2 + q2 cos(!t) 2q q itH/~ m! 0 t 0 t qt e q0 = exp i h | | i 2⇡i~ sin(!t) 2~ sin(!t) r " ⇥ ⇤# (16.64) which agrees with the amplitude (16.47) in the limit t 0 (exercise 16.10). !

16.8 Harmonic Oscillator in Imaginary Time For the harmonic oscillator with hamiltonian (16.52), our formula (16.25) for euclidian path integrals becomes

H 1 2 1 2 2 q e q = exp mq˙ (0)+ m! q (0) d0 Dq. (16.65) h | | 0i 2 2 Z ⇢ Z0 The euclidian action, which is a -integral of the energy of the oscillator,

1 2 1 2 2 Se[q]= 2 mq˙ (0)+ 2 m! q (0) d0 (16.66) Z0 ⇥ ⇤ 682 Path Integrals is purely quadratic, and so we may play a trick (15.32). We look for a path qe() that makes the euclidian action (16.66) stationary

d 1 2 1 2 2 S [q ][h]= m(˙q (0)+✏h˙ (0)) + m! (q (0)+✏h(0)) d0 e e d✏ 2 e 2 e Z0 ✏=0 2 = mq˙e(0)h˙ (0)+m! qe(0)h(0) d0 0 Z 2 = mq¨ (0)+m! q (0) h(0)d0 =0. (16.67) e e Z0 ⇥ ⇤ The path qe() must satisfy the euclidian equation of motion

2 q¨e()=! qe() (16.68) whose general solution is

! ! qe()=Ae + Be . (16.69)

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

q e ! q e 2! A = 0 and B = q A. (16.70) 1 e 2! 0 Its action Se[qe] is (exercise 16.12)

1 2 2! 2 2! S [q ]= m! A e 1 + B 1 e . (16.71) e e 2 h ⇣ ⌘ ⇣ ⌘i Since the action is purely quadratic, the trick (15.32) tells us that the action Se[q] of the arbitrary path q()=qe()+q()isthesum

Se[q]=Se[qe]+Se[q] (16.72) in which the action Se[q] of the loop q() depends upon but not upon q or q0. It follows then that for some loop function L() of alone

H q e q =exp( S [q ]) L() (16.73) h | | 0i e e 1 2 2! 2 2! = L()exp m! A e 1 + B 1 e . 2 n h ⇣ ⌘ ⇣ ⌘io To study the ground state of the harmonic oscillator, we let in this !1 equation. Inserting a complete set of eigenstates H n = E n , we see that | i n| i the limit of the left-hand side is

H En E0 lim q e q0 =limq n e n q0 = e q 0 0 q0 . (16.74) h | | i h | i h | i h | ih | i !1 !1 16.9 Euclidian Correlation Functions 683 Our formulas (16.70) for A and B say that A q e ! and B q as ! ! 0 , and so in this limit by (16.73 & 16.74) we have !1 E0 1 2 2 e q 0 0 q = L()exp m! q + q (16.75) h | ih | 0i 2 0 from which may infer our earlier formula⇥ (15.44) for the⇤ wave function of the ground state m! 1/4 1 m!q2 q 0 = exp (16.76) h | i ⇡~ 2 ~ ⇣ ⌘  in which the prefactor ensures the normalization

1= 1 q 0 2dq. (16.77) |h | i| Z1 euclidian path integrals help one study ground states.

16.9 Euclidian Correlation Functions In the Heisenberg picture, the position operator q(t)is

itH/~ itH/~ q(t)=e qe (16.78) in which q = q(0) is the position operator at time t = 0 or equivalently the position operator in the Schr¨odinger picture. The analogous operator in imaginary time t = i~ = i~/(kT) is the euclidian position operator qe() defined as H H qe()=e qe . (16.79) The euclidian-time-ordered product of two euclidian position operators is

[q ( ) q ( )] = ✓( )q ( ) q ( )+✓( )q ( ) q ( ) (16.80) T e 1 e 2 1 2 e 1 e 2 2 1 e 2 e 1 in which ✓(x)=(x+ x )/2 x is Heaviside’s function. We can use the method | | | | of section 16.3 to compute the matrix element of the euclidian-time-ordered product [q( )q( )] sandwiched between two factors of exp( H). For T 1 2 , this matrix element is 1 2 H H ( 1)H (1 2)H (+2)H q e qe(1)qe(2)e q = q e qe qe q . h | | i h | | i (16.81) Then instead of the path-integral formula (16.25), we get

H H Se[q,, ] q e [qe(1)qe(2)] e q = q(1)q(2)e Dq (16.82) h | T | i Z 684 Path Integrals where as in (16.25) S [q, , ]istheeuclidian action e 1 2 Se[q, , ]= 2 mq˙ (0)+V (q(0)) d0 (16.83) Z or the time integral of the energy. As in the path integral (16.25), the inte- gration is over all paths that go from q( )=q to q()=q. The analog of the single-particle path integral (16.25) is

2H Se[q,, ] q e q = e Dq (16.84) h | | i Z and the factors (nm/2⇡)n/2 cancel in the ratio of (16.82) to (16.84)

Se[q,, ] H H q(1)q(2)e Dq q e [qe(1)qe(2)] e q h | T 2H | i = Z . q e q Se[q,, ] h | | i e Dq Z (16.85) In the limit , the operator exp( H) projects out the ground state !1 0 of the system | i 1 H H E0 lim e q =lim e n n q =lime 0 0 q | i | ih | i | ih | i !1 !1 n=0 !1 X (16.86) which we assume to be unique and normalized to unity. In the ratio (16.85), most of these factors cancel, leaving us with

Se[q, , ] q(1)q(2)e 1 1 Dq 0 [qe(1)qe(2)] 0 = Z . (16.87) h |T | i Se[q, , ] e 1 1 Dq Z More generally, the mean value in the ground state 0 of any euclidian- | i time-ordered product of position operators q(ti) is a ratio of path integrals

Se[q] q(1) ...q(n) e Dq 0 [q(1) ...q(n)] 0 = Z (16.88) h |T | i Se[q] e Dq Z in which S [q] stands for S [q, , ]. Why do we need the time-ordered e e 1 1 product on the LHS? Because successive factors of exp( ( )H) lead T k ` to the path integral of exp( S [q]). Why don’t we need on the RHS? e T Because the q(i)’s are real numbers which commute with each other. 16.10 Finite-Temperature Field Theory 685 The result (16.88 ) is important because it can be generalized to all quan- tum theories, including field theories.

16.10 Finite-Temperature Field Theory Matrix elements of the operator exp( H)where =1/kT tell us what a system is like at temperature T . In the low-temperature limit, they describe the ground state of the system. Quantum mechanics imposes upon n coordinates qi and conjugate mo- menta ⇡k the commutation relations

[qi,pk]=ii,k and [qi,qk]=[pi,pk]=0. (16.89)

In quantum field theory, we associate a coordinate q (x) and a conjugate x ⌘ momentum p ⇡(x) with each point x of space and impose upon them x ⌘ the very similar commutation relations

[(x),⇡(y)] = i(x y) and [(x),(y)] = [⇡(x),⇡(y)] = 0. (16.90) Just as in quantum mechanics the time derivative of a coordinate is its commutator with a hamiltonianq ˙i = i[H, qi], so too in quantum field theory the time derivative of a field ˙(x,t) ˙(x)is˙(x)=i[H,(x)]. A typical ⌘ hamiltonian for a single scalar field is 1 1 1 H = ⇡2(x)+ ( (x))2 + m22(x)+P ((x)) d3x (16.91) 2 2 r 2 Z  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 1 1 V ((x)) = ( (x))2 + m22(x)+P ((x)) (16.92) 2 r 2 and write the hamiltonian H as 1 H = ⇡2(x)+V ((x)) d3x. (16.93) 2 Z  Like q and p , the states and ⇡ are eigenstates of the hermitian | 0i | 0i | 0i | 0i operators (x, 0) and ⇡(x, 0)

(x, 0) 0 = 0(x) 0 and ⇡(x, 0) ⇡0 = ⇡0(x) ⇡0 . (16.94) | i | i | i | i 686 Path Integrals The analog of q p is h 0| 0i 3 0 ⇡0 = f exp i 0(x)⇡0(x)d x (16.95) h | i  Z in which f is a factor which eventually will cancel. Repeating our derivation of Eq.(16.21) with

D⇡0 d⇡0(x) (16.96) ⌘ x Y we find in the limit ✏ 0 ! ✏ 2 3 exp( ✏H) = exp ⇡ (x)d x ⇡0 h 1| | 0i h 1| 2 | i Z ✓ Z ◆ 3 ⇡0 exp ✏ V ((x)) d x D⇡0 ⇥h | | 0i ✓ Z ◆ = f 2 exp ✏ V ( (x)) d3x (16.97) | | 0 ✓ Z ◆ 1 2 3 exp ✏⇡0 (x)+i⇡0(x)[ (x) (x)]d x D⇡0. ⇥ 2 1 0 Z Z Using the abbreviation (x) (x) ˙ (x) 1 0 (16.98) 0 ⌘ ✏ and the integral formula (16.6), we get

1 2 3 exp( ✏H) = f 0 exp ✏ ˙ (x)+V ( (x)) d x . h 1| | 0i 2 0 0 ⇢ Z h i Putting together n = /✏ such terms, integrating over the intermediate states from j =1toj = n 1 and absorbing the normalizing | jih j| factors into D, we get

H 1 2 3 e = exp ˙ (x)+V ((x)) d xd0 D. (16.99) h | | 0i 2 Z0  Z0 Z Replacing the potential V () with its definition (16.92), we have

H 1 2 2 2 2 3 e = exp ˙ +( ) + m + P () d xd0 D h | | 0i 2 r Z0  Z0Z h i (16.100) in which the limits 0 and remind us that we are to integrate over all fields (x,t) that run from (x, 0) = 0(x)to(x,)=(x). 16.10 Finite-Temperature Field Theory 687 In terms of the energy density

() 1 (@ )2 + m22 + P () (16.101) H ⌘ 2 a in which @ =(@/@, ), the path⇥ integral (16.100)⇤ is a r H 3 e = exp () d xd0 D. (16.102) h | | 0i H Z0  Z0 Z The partition function Z()—defined as the trace Z() Tr e H over all ⌘ the states of the system—is an integral over all periodic ( = 0)fields

0 0 H H 3 Z() Tr e = e D = exp () d xd0 D. ⌘ h | | i H Z0 Z0  Z0 Z (16.103) Because the squares of the four derivatives @a occur with the same sign, finite-temperature field theory is called euclidian quantum field theory. Like the definition (16.79) of the euclidian position operator qe(), the euclidian field operator e(x) is defined as

H H e(x,)=e (x, 0)e . (16.104)

The euclidian-time-ordered product (16.80 ) of two fields is

1H (1 2)H 2H [ (x , ) (x , )] = ✓( )e (x , 0)e (x , 0)e T e 1 1 e 2 2 1 2 e 1 e 2 2H (2 1)H 1H + ✓( )e (x , 0)e (x , 0)e . 2 1 e 2 e 1 The logic of equations (16.79–16.88) leads us to write its mean value in a system described by a maximum-entropy-at-inverse-temperature- density H operator ⇢ = e /Z() as the ratio

[e(x1)e(x2)] =Tr ⇢ [e(x1)e(x2)] hT i { TH } Tr e [e(x1)e(x2)] = T H (16.105) Tr [ e ] 0 3 (x )(x )exp () d xd0 D 1 2 H = Z0  Z0 Z 0 3 exp () d xd0 D H Z0  Z0 Z in which several factors have canceled. In the zero-temperature ( ) limit, the density operator ⇢ becomes !1 the projection operator 0 0 on the ground state, and mean-value formulas | ih | 688 Path Integrals like (16.105) become

(x ) ...(x )exp () d4x D 1 n H 0 [ (x ) ... (x )] 0 = Z  Z h |T e 1 e n | i exp () d4x D H Z  Z (16.106) in which Hamilton’s density () is integrated over all of euclidian spacetime H and over all fields that are periodic on the infinite time interval. Statistical field theory and lattice gauge theory use formulas like (16.105 & 16.106).

16.11 Real-Time Field Theory We can apply the derivation of section 16.10 to amplitudes in real time. In the amplitude (16.97), we replace ✏H by i✏H and follow the logic of sections (16.4 & 16.10). We find in the limit ✏ 0with˙ ( )/✏ ! 0 ⌘ 1 0 i✏H i✏ ⇡2/2 d3x i✏ V () d3x e = e ⇡0 ⇡0 e D⇡0 h 1| | 0i h 1| | ih | | 0i Z R R 3 2 3 2 i✏ V (0)d x i ✏⇡ /2+i⇡ (1 0) d x = f e e 0 0 D⇡0 | | R Z R 1 2 3 = f 0 exp i✏ ˙ V ( ) d x . (16.107) 2 0 0  Z Putting together 2t/✏ similar factors and integrating over all the interme- diate states , we arrive at the path integral | jih j| t i2tH 1 2 4 t e t = exp i ˙ (x) V ((x)) d x D (16.108) 2 h | | i t Z  Z in which we integrate over all fields (x) that run from t(x, t)tot(x,t). After expanding the definition (16.92) of the potential V (), we have

t i2tH 1 2 2 2 2 4 t e t = exp i ˙ ( ) m P () d x D. 2 h | | i t r Z  Z h i (16.109) This amplitude is a path integral

t i2tH/~ iS[]/~ t e t = e D (16.110) h | | i t Z of phases exp(iS[]/~) that are exponentials of the classical action

S[]= 1 ˙2 ( )2 m22 P () d4x. (16.111) 2 r Z h i 16.11 Real-Time Field Theory 689

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

itH/~ itH/~ (x,t)=e (x, 0)e . (16.112)

The time-ordered product of two fields, as in (16.80), is the sum

[(x )(x )] = ✓(x0 x0)(x )(x )+✓(x0 x0)(x )(x ). (16.113) T 1 2 1 2 1 2 2 1 2 1 Between two factors of exp( itH), it is for t >t 1 2

itH itH i(t t1)H i(t1 t2)H i(t+t2)H e [(x )(x )] e = e (x , 0)e (x , 0)e . T 1 2 1 2 So by the logic that led to the path-integral formulas (16.105) and (16.110), we can write a matrix element of the time-ordered product (16.113) as

t itH itH iS[] t e [(x1)(x2)] e t = (x1) (x2) e D (16.114) h | T | i t Z in which we integrate over fields that go from t at time t to t at time t. The time-ordered product of any combination of fields is then

itH itH iS[] t e [(x1) ...(xn)] e t = (x1) ...(xn) e D. h | T | i Z (16.115) Like the position eigenstates q of quantum mechanics, the eigenstates | 0i are states of infinite energy that overlap many states, whereas we often | 0i are interested in the ground state 0 or in states of a few particles. To form | i such matrix elements, we multiply both sides of equations (16.110 & 16.115) by 0 t t 0 and integrate over t and t. Since the ground state is a h | ih | i normalized eigenstate of the hamiltonian H 0 = E 0 with eigenvalue E , | i 0| i 0 we find from the path-integral formulas (16.110)

i2tH i2tH 0 t t e t t 0 DtD t = 0 e 0 (16.116) h | ih | | ih | i h | | i Z i2tE0 iS[] = e = 0 t e t 0 DDtD t h | i h | i Z and (16.115) (suppressing the di↵erentials DtD t)

2itE0 iS[] e 0 [(x1) ...(xn)] 0 = 0 t (x1) ...(xn) e t 0 D. h |T | i h | i h | i Z (16.117) 690 Path Integrals The mean value in the ground state of a time-ordered product of field oper- ators is then a ratio of these path integrals

iS[] 0 t (x1) ...(xn) e t 0 D h | i h | i 0 [(x1) ...(xn)] 0 = Z h |T | i iS[] 0 t e t 0 D h | i h | i Z (16.118) in which factors involving E0 have canceled and the integration is over all fields that go from (x, t)= t(x)to(x,t)=t(x) and over t(x) and t(x).

16.12 Perturbation Theory Field theories with hamiltonians that are quadratic in their fields like

H = 1 ⇡2(x)+( (x))2 + m22(x) d3x (16.119) 0 2 r Z h i are soluble. Their fields evolve in time as

itH0 itH0 (x,t)=e (x, 0)e . (16.120)

The mean value in the ground state of H0 of a time-ordered product of these fields is by (16.118) a ratio of path integrals

iS0[] 0 (x ) ...(x ) e 0 0 D h | ti 1 n h | i 0 [(x1) ...(xn)] 0 = Z h |T | i iS0[] 0 00 e 0 0 D h | i h | i Z (16.121) in which the action S0[] is quadratic in the fields

S []= 1 ˙2(x) ( (x))2 m22(x) d4x 0 2 r Z h i (16.122) = 1 @ (x)@a(x) m22(x) d4x. 2 a Z So the path integrals in⇥ the ratio (16.121) are gaussian⇤ and doable. The Fourier transforms 4 ipx 4 ipx d p ˜(p)= e (x) d x and (x)= e ˜(p) (16.123) (2⇡)4 Z Z turn the spacetime derivatives in the action into a quadratic form d4p S []= 1 ˜(p) 2 (p2 + m2) (16.124) 0 2 | | (2⇡)4 Z 16.12 Perturbation Theory 691 in which p2 = p2 p02, and ˜( p)=˜ (p) by (3.25) since the field is real. ⇤ The initial t 0 and final 0 t wave functions produce the i✏ in the h | i h | i Feynman propagator (5.236). Although its exact form doesn’t matter here, the wave function 0 of the ground state of H is the exponential (15.53) h | i 0 1 d3p 0 = c exp ˜(p) 2 p2 + m2 (16.125) h | i 2 | | (2⇡)3  Z p in which ˜(p) is the spatial Fourier transform of the eigenvalue (x)

ip x 3 ˜(p)= e · (x) d x (16.126) Z 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 t 0 and 0 t add to the action S0[]theterm h | i h | i i d3p S []= p2 + m2 ˜(p,t) 2 + ˜(p, t) 2 (16.127) 0 2 | | | | (2⇡)3 Z p ⇣ ⌘ in which we envision taking the limit t with (x,t)= (x) and !1 t (x, t)= t(x). The identity (Weinberg, 1995, pp. 386–388) 1 ✏ t f(+ )+f( )= lim ✏ f(t) e | | dt (16.128) 1 1 ✏ 0+ ! Z1 allows us to write S0[] as 3 i✏ 2 2 1 2 ✏ t d p S0[]= lim p + m ˜(p,t) e | | dt . (16.129) ✏ 0+ 2 | | (2⇡)3 ! Z p Z1 To first order in ✏, the change in the action is (exercise 16.15)

3 i✏ 2 2 1 2 d p S0[]= lim p + m ˜(p,t) dt ✏ 0+ 2 | | (2⇡)3 ! Z Z1 i✏ p d4p =lim p2 + m2 ˜(p) 2 . (16.130) ✏ 0+ 2 | | (2⇡)4 ! Z p Thus the modified action is 1 d4p S [,✏]=S []+S []= ˜(p) 2 p2 + m2 i✏ p2 + m2 0 0 0 2 | | (2⇡)4 Z ⇣ ⌘ 1 d4p p = ˜(p) 2 p2 + m2 i✏ (16.131) 2 | | (2⇡)4 Z since the square root is positive. In terms of the modified action, our formula 692 Path Integrals (16.121) for the time-ordered product is the ratio

iS0[,✏] (x1) ...(xn) e D 0 [(x1) ...(xn)] 0 = Z . (16.132) h |T | i eiS0[,✏] D Z We can use this formula (16.132) to express the mean value in the vacuum 0 of the time-ordered exponential of a spacetime integral of j(x)(x), in | i which j(x) is a classical (c-number, external) current, as the ratio

Z [j] 0 exp i j(x) (x) d4x 0 0 ⌘h |T | i ⇢  Z exp i j(x) (x) d4x eiS0[,✏] D (16.133) = Z  Z . eiS0[,✏] D Z Since the state 0 is normalized, the mean value Z [0] is unity, | i 0 Z0[0] = 1. (16.134) If we absorb the current into the action

4 S0[,✏, j]=S0[,✏]+ j(x) (x) d x (16.135) Z then in terms of the current’s Fourier transform

ipx 4 ˜j(p)= e j(x) d x (16.136) Z the modified action S0[,✏, j] is (exercise 16.16) 4 1 2 2 2 d p S [,✏, j]= ˜(p) p + m i✏ ˜j⇤(p)˜(p) ˜⇤(p)˜j(p) . 0 2 | | (2⇡)4 Z h i(16.137) Changing variables to

˜(p)=˜(p) ˜j(p)/(p2 + m2 i✏) (16.138) we write the action S0[,✏, j] as (exercise 16.17) ˜j (p)˜j(p) d4p S [,✏, j]= 1 ˜(p) 2 p2 + m2 i✏ ⇤ 0 2 | | (p2 + m2 i✏) (2⇡)4 Z  ˜j (p)˜j(p) d4p = S [ ,✏]+ 1 ⇤ . (16.139) 0 2 (p2 + m2 i✏) (2⇡)4 Z  16.12 Perturbation Theory 693 And since D = D , our formula (16.133) gives simply (exercise 16.18) i ˜j(p) 2 d4p Z [j]=exp | | . (16.140) 0 2 p2 + m2 i✏ (2⇡)4 ✓ Z ◆ Going back to position space, one finds (exercise 16.19)

i 4 4 Z [j]=exp j(x)(x x0) j(x0) d xd x0 (16.141) 0 2  Z in which (x x ) is Feynman’s propagator (5.236) 0 ip(x x ) 4 e 0 d p (x x0)= . (16.142) p2 + m2 i✏ (2⇡)4 Z The functional derivative (chapter 15) of Z0[j], defined by (16.133), is

1 Z0[j] 4 = 0 (x)exp i j(x0)(x0)d x0 0 (16.143) i j(x) h |T | i  ✓ Z ◆ while that of equation (16.141) is

1 Z0[j] 4 = Z [j] (x x0) j(x0) d x0. (16.144) i j(x) 0 Z Thus the second functional derivative of Z0[j] evaluated at j = 0 gives 2 1 Z0[j] 0 (x)(x0) 0 = = i (x x0). (16.145) h |T | i i2 j(x)j(x ) 0 j=0 ⇥ ⇤ Similarly, one may show (exercise 16.20) that 1 4Z [j] 0 [(x )(x )(x )(x )] 0 = 0 h |T 1 2 3 4 | i i4 j(x )j(x )j(x )j(x ) 1 2 3 4 j=0 = (x1 x2)(x3 x4) (x1 x3)(x2 x4) (x x )(x x ). (16.146) 1 4 2 3 Suppose now that we add a potential V = P () to the free hamiltonian (16.119). Scattering amplitudes are matrix elements of the time-ordered ex- ponential exp i P () d4x . Our formula (16.132) for the mean value in T the ground state 0 of the free hamiltonian H of any time-ordered product ⇥| iR ⇤ 0 of fields leads us to

exp i P () d4x eiS0[,✏] D 0 exp i P () d4x 0 = Z  Z . h |T | i ⇢  Z eiS0[,✏] D Z (16.147) 694 Path Integrals Using (16.145 & 16.146), we can cast this expression into the magical form 0 exp i P () d4x 0 =exp i P d4x Z [j] . h |T | i ij(x) 0 ⇢  Z  Z ✓ ◆ j=0 (16.148) The generalization of the path-integral formula (16.132) to the ground state ⌦ of an interacting theory with action S is | i iS[,✏] (x1) ...(xn) e D ⌦ [(x1) ...(xn)] ⌦ = Z (16.149) h |T | i eiS[,✏] D Z 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 A = 0, the QED hamiltonian is r· H = H + 1 ⇡2 + 1 ( A)2 A j d3x + V (16.150) m 2 2 r⇥ · C Z ⇥ ⇤ in which Hm is the matter hamiltonian, and VC is the Coulomb term 1 j0(x,t) j0(y,t) V = d3xd3y. (16.151) C 2 4⇡ x y Z | | The operators A and ⇡ are canonically conjugate, but they satisfy the Coulomb-gauge conditions A = 0 and ⇡ =0. (16.152) r· r· One may show (Weinberg, 1995, pp. 413–418) that in this theory, the analog of equation (16.149) is

... eiSC [ A] DA D O1 On r· ⌦ [ 1 ... n] ⌦ = Z (16.153) h |T O O | i eiSC [ A] DA D r· Z in which the Coulomb-gauge action is

2 S = 1 A˙ 1 ( A)2 + A j + d4x V dt (16.154) C 2 2 r⇥ · Lm C Z Z 16.13 Application to Quantum Electrodynamics 695 and the functional delta function

[ A]= ( A(x)) (16.155) r· r· x Y enforces the Coulomb-gauge condition. The term is the action density Lm of the matter field . Tricks are available. We introduce a new field A0(x) and consider the factor

1 0 1 0 2 4 0 F = exp i A + j d x DA (16.156) 2 r r4 Z  Z which is just a number independent of the charge density j0 since we can cancel the j0 term by shifting A0.By 1, we mean 1/4⇡ x y .By 4 | | integrating by parts, we can write the number F as (exercise 16.21)

1 0 2 0 0 1 0 1 0 4 0 F = exp i A A j j j d x DA 2 r 2 4 Z  Z (16.157) 2 = exp i 1 A0 A0j0 d4x + i V dt DA0. 2 r C Z  Z Z So when we multiply the numerator and denominator of the amplitude (16.153) by F , the awkward Coulomb term cancels, and we get

... eiS0 [ A] DA D O1 On r· ⌦ [ 1 ... n] ⌦ = Z (16.158) h |T O O | i eiS0 [ A] DA D r· Z where now DA includes all four components Aµ and

1 2 1 2 1 0 2 0 0 4 S0 = A˙ ( A) + A + A j A j + d x. (16.159) 2 2 r⇥ 2 r · Lm Z Since the delta-function [ A] enforces the Coulomb-gauge condition, we r· can add to the action S the term ( A˙) A0 which is A˙ A0 after we 0 r· ·r integrate by parts and drop the surface term. This extra term makes the action gauge invariant

S = 1 (A˙ A0)2 1 ( A)2 + A j A0j0 + d4x 2 r 2 r⇥ · Lm Z (16.160) = 1 F F ab+ Abj + d4x. 4 ab b Lm Z 696 Path Integrals Thus at this point we have

... eiS [ A] DA D O1 On r· ⌦ [ 1 ... n] ⌦ = Z (16.161) h |T O O | i eiS [ A] DA D r· Z in which S is the gauge-invariant action (16.160), and the integral is over all fields. The only relic of the Coulomb gauge is the gauge-fixing delta functional [ A]. r· We now make the gauge transformation

iq⇤(x) Ab0 (x)=Ab(x)+@b⇤(x) and 0(x)=e (x) (16.162) in the numerator and also, using a di↵erent gauge transformation ⇤0,inthe denominator of the ratio (16.161) of path integrals. Since we are integrating over all gauge fields, these gauge transformations merely change the order of integration in the numerator and denominator of that ratio. They are like replacing 1 f(x) dx by 1 f(y) dy. They change nothing, and so 1 1 R ⌦ [ ...R ] ⌦ = ⌦ [ ... ] ⌦ 0 (16.163) h |T O1 On | i h |T O1 On | i in which the prime refers to the gauge transformation (16.162). We’ve seen that the action S is gauge invariant. So is the measure DA D , and we now restrict ourselves to operators ... that are gauge invari- O1 On ant. So in the right-hand side of equation (16.163), the replacement of the fields by their gauge transforms a↵ects only the term [ A] that enforces r· the Coulomb-gauge condition

... eiS [ A + ⇤] DA D O1 On r· 4 ⌦ [ 1 ... n] ⌦ = Z . (16.164) h |T O O | i iS e [ A + ⇤0] DA D r· 4 Z We now have two choices. If we integrate over all gauge functions ⇤(x)in both the numerator and the denominator of this ratio (16.164), then apart from over-all constants that cancel, the mean value in the vacuum of the time-ordered product is the ratio

... eiS DA D O1 On ⌦ [ 1 ... n] ⌦ = Z (16.165) h |T O O | i eiS DA D Z in which we integrate over all matter fields, gauge fields, and gauges. That is, we do not fix the gauge. 16.13 Application to Quantum Electrodynamics 697 The analogous formula for the euclidian time-ordered product is

Se ... e DA D O1 On ⌦ e [ 1 ... n] ⌦ = Z (16.166) h |T O O | i Se e DA D Z in which the euclidian action Se is the spacetime integral of the energy density. This formula is quite general; it holds in nonabelian gauge theories and is important in lattice gauge theory. Our second choice is to multiply the numerator and the denominator of the ratio (16.164) by the exponential exp[ i 1 ↵ ( ⇤)2 d4x] and then integrate 2 4 over ⇤(x) separately in the numerator and denominator. This operation R just multiplies the numerator and denominator by the same constant factor, which cancels. But if before integrating over all gauge transformations, we shift ⇤so that ⇤changes to ⇤ A˙ 0, then the exponential factor is 4 4 exp[ i 1 ↵ (A˙ 0 ⇤)2 d4x]. Now when we integrate over ⇤(x), the delta 2 4 function ( A+ ⇤) replaces ⇤by A in the inserted exponential, Rr· 4 4 r· converting it to exp[ i 1 ↵ (A˙ 0 + A)2 d4x]. This term changes the gauge- 2 r· invariant action (16.160) to the gauge-fixed action R 1 ↵ S = F F ab (@ Ab)2+Abj + d4x. (16.167) ↵ 4 ab 2 b b Lm Z This Lorentz-invariant, gauge-fixed action is much easier to use than the Coulomb-gauge action (16.154) with the Coulomb potential (16.151). We can use it to compute scattering amplitudes perturbatively. The mean value of a time-ordered product of operators in the ground state 0 of the free | i theory is

... eiS↵ DA D O1 On 0 [ 1 ... n] 0 = Z . (16.168) h |T O O | i eiS↵ DA D Z By following steps analogous to those the led to (16.142), one may show (exercise 16.22) that in Feynman’s gauge, ↵ = 1, the photon propagator is

4 ⌘µ⌫ iq (x y) d q 0 [A (x)A (y)] 0 = i (x y)= i e · . h |T µ ⌫ | i 4µ⌫ q2 i✏ (2⇡)4 Z (16.169) 698 Path Integrals 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 (16.170) the most general function f(✓) of a single Grassmann variable ✓ is f(✓)=a + b✓. (16.171) So a complete integral table consists of the integral of this linear function

f(✓) d✓ = a + b✓d✓ = a d✓ + b ✓d✓. (16.172) Z Z Z Z This equation has two unknowns, the integral d✓ of unity and the integral ✓d✓ of ✓. We choose them so that the integral of f(✓ + ⇣) R R f(✓ + ⇣) d✓ = a + b (✓ + ⇣) d✓ =(a + b⇣) d✓ + b ✓d✓ (16.173) Z Z Z Z is the same as the integral (16.172) of f(✓). Thus the integral d✓ of unity must vanish, while the integral ✓d✓ of ✓ can be any constant, which we R choose to be unity. Our complete table of integrals is then R d✓ = 0 and ✓d✓=1. (16.174) Z Z The anticommutation relations for a fermionic degree of freedom are

, † † + † = 1 and , = †, † =0. (16.175) { }⌘ { } { } Because has †, it is conventional to introduce a variable ✓⇤ = ✓† that anti-commutes with itself and with ✓

✓⇤,✓⇤ = ✓⇤,✓ = ✓,✓ =0. (16.176) { } { } { } The logic that led to (16.174) now gives

d✓⇤ = 0 and ✓⇤ d✓⇤ =1. (16.177) Z Z We define the reference state 0 as 0 s for a state s that is not | i | i⌘ | i | i annihilated by .Since 2 = 0, the operator annihilates the state 0 | i 0 = 2 s =0. (16.178) | i | i The e↵ect of the operator on the state

1 1 ✓ =exp †✓ ✓⇤✓ 0 = 1+ †✓ ✓⇤✓ 0 (16.179) | i 2 | i 2 | i ⇣ ⌘ ⇣ ⌘ 16.14 Fermionic Path Integrals 699 is 1 ✓ = (1 + †✓ ✓⇤✓) 0 = †✓ 0 =(1 † )✓ 0 = ✓ 0 (16.180) | i 2 | i | i | i | i while that of ✓ on ✓ is | i 1 ✓ ✓ = ✓(1 + †✓ ✓⇤✓) 0 = ✓ 0 . (16.181) | i 2 | i | i The state ✓ therefore is an eigenstate of with eigenvalue ✓ | i ✓ = ✓ ✓ . (16.182) | i | i The bra corresponding to the ket ⇣ is | i 1 ⇣ = 0 1+⇣⇤ ⇣⇤⇣ (16.183) h | h | 2 ✓ ◆ and the inner product ⇣ ✓ is (exercise 16.23) h | i 1 1 ⇣ ✓ = 0 1+⇣⇤ ⇣⇤⇣ 1+ †✓ ✓⇤✓ 0 h | i h | 2 2 | i ✓ ◆✓ ◆ 1 1 1 = 0 1+⇣⇤ †✓ ⇣⇤⇣ ✓⇤✓ + ⇣⇤⇣✓⇤✓ 0 h | 2 2 4 | i 1 1 1 = 0 1+⇣⇤✓ ⇣⇤⇣ ✓⇤✓ + ⇣⇤⇣✓⇤✓ 0 h | 2 2 4 | i 1 =exp ⇣⇤✓ (⇣⇤⇣ + ✓⇤✓) . (16.184) 2  Example 16.3 (A gaussian integral). For any number c, we can compute the integral of exp(c✓⇤✓) by expanding the exponential

c✓ ✓ e ⇤ d✓⇤d✓ = (1 + c✓⇤✓) d✓⇤d✓ = (1 c✓✓⇤) d✓⇤d✓ = c. (16.185) Z Z Z

The identity operator for the space of states

c 0 + d 1 c 0 + d † 0 (16.186) | i | i⌘ | i | i is (exercise 16.24) the integral

I = ✓ ✓ d✓⇤d✓ = 0 0 + 1 1 (16.187) | ih | | ih | | ih | Z in which the di↵erentials anti-commute with each other and with other fermionic variables: d✓, d✓ = 0, d✓,✓ = 0, d✓, = 0, and so forth. { ⇤} { } { } The case of several Grassmann variables ✓1,✓2,...,✓n and several Fermi operators 1, 2,..., n is similar. The ✓k anticommute among themselves

✓ ,✓ = ✓ ,✓⇤ = ✓⇤,✓⇤ = 0 (16.188) { i j} { i j } { i j } 700 Path Integrals while the k satisfy

, † = and , = †, † =0. (16.189) { k ` } k` { k l} { k ` } The reference state 0 is | i n 0 = k s (16.190) | i ! | i kY=1 in which s is any state not annihilated by any (so the resulting 0 isn’t | i k | i zero). The direct-product state n n 1 1 ✓ exp †✓ ✓⇤✓ 0 = 1+ †✓ ✓⇤✓ 0 | i⌘ k k 2 k k | i k k 2 k k | i k=1 ! "k=1 ✓ ◆# X Y (16.191) is (exercise 16.25) a simultaneous eigenstate of each k ✓ = ✓ ✓ . (16.192) k| i k| i It follows that ✓ = ✓ ✓ = ✓ ✓ = ✓ ✓ ✓ = ✓ ✓ ✓ (16.193) ` k| i ` k| i k `| i k `| i ` k| i and so too ✓ = ✓ ✓ ✓ .Sincethe ’s anticommute, their eigenvalues k `| i k `| i must also ✓ ✓ ✓ = ✓ = ✓ = ✓ ✓ ✓ (16.194) ` k| i ` k| i k `| i k `| i (even if they commuted with the ’s in which case we’d have ✓ = ` k| i ✓ ✓ ✓ = ✓ = ✓ ✓ ✓ ). k `| i k `| i ` k| i The inner product ⇣ ✓ is h | i n n 1 1 ⇣ ✓ = 0 (1 + ⇣⇤ ⇣⇤⇣ ) (1 + †✓ ✓⇤✓ ) 0 h | i h | k k 2 k k ` ` 2 ` ` | i "k=1 #"`=1 # Yn Y 1 ⇣†✓ (⇣†⇣+✓†✓)/2 =exp ⇣k⇤✓k (⇣k⇤⇣k + ✓k⇤✓k) = e . (16.195) " 2 # Xk=1 The identity operator is n I = ✓ ✓ d✓⇤d✓ . (16.196) | ih | k k Z kY=1 Example 16.4 (Gaussian Grassmann Integral). For any 2 2 matrix A, ⇥ we may compute the gaussian integral

✓†A✓ g(A)= e d✓1⇤d✓1d✓2⇤d✓2 (16.197) Z 16.14 Fermionic Path Integrals 701 by expanding the exponential. The only terms that survive are the ones that have exactly one of each of the four variables ✓1, ✓2, ✓1⇤, and ✓2⇤.Thus,the integral is the determinant of the matrix A

1 2 g(A)= (✓⇤ A ✓ ) d✓⇤d✓ d✓⇤d✓ 2 k k` ` 1 1 2 2 Z = (✓1⇤A11✓1 ✓2⇤A22✓2 + ✓1⇤A12✓2 ✓2⇤A21✓1) d✓1⇤d✓1d✓2⇤d✓2 Z = A A A A =detA. (16.198) 11 22 12 21 The natural generalization to n dimensions

n ✓†A✓ e d✓k⇤d✓k =detA (16.199) Z kY=1 is true for any n n matrix A.IfA is invertible, then the invariance of ⇥ Grassmann integrals under translations implies that

n n 1 1 ✓†A✓+✓†⇣+⇣†✓ ✓†A(✓+A ⇣)+✓†⇣+⇣†(✓+A ⇣) e d✓k⇤d✓k = e d✓k⇤d✓k Z k=1 Z k=1 Y n Y 1 ✓†A✓+⇣†✓+⇣†A ⇣ = e d✓k⇤d✓k Z k=1 Y n 1 1 (✓†+⇣†A )A✓+⇣†✓+⇣†A ⇣ = e d✓k⇤d✓k Z k=1 n Y 1 ✓†A✓+⇣†A ⇣ = e d✓k⇤d✓k Z k=1 1 Y =detAe⇣†A ⇣ . (16.200)

The values of ✓ and ✓† that make the argument ✓†A✓+✓†⇣+⇣†✓ of the expo- 1 1 nential stationary are ✓ = A ⇣ and ✓† = ⇣†A . So a gaussian Grassmann integral is equal to its exponential evaluated at its stationary point, apart from a prefactor involving the determinant det A. This result is a fermionic 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 among them- selves and with all fermionic variables at all points of spacetime

(x), (x0) = ⇤ (x), (x0) = ⇤ (x),⇤ (x0) = 0 (16.201) { m n } { m n } { m n } 702 Path Integrals and the Dirac field m(x) obeys the equal-time anticommutation relations

m(x,t), n† (x0,t) = mn (x x0)(n, m =1,...,4) { } (16.202) (x,t), (x0,t) = † (x,t), † (x0,t) =0. { m n } { m n } As in (16.94 & 16.190), we use eigenstates of the field at t = 0. If 0 is | i defined in terms of a state s that is not annihilated by any (x, 0) as | i m

0 = (x, 0) s (16.203) | i m | i "m,x # Y then (exercise 16.26) the state

1 3 =exp † (x, 0) (x) ⇤ (x) (x) d x 0 | i m m 2 m m | i m ! Z X 1 3 =exp † †d x 0 (16.204) 2 | i ✓Z ◆ is an eigenstate of the operator m(x, 0) with eigenvalue m(x)

(x, 0) = (x) . (16.205) m | i m | i The inner product of two such states is (exercise 16.27)

1 1 3 0 =exp 0† 0†0 †dx . (16.206) h | i 2 2 Z The identity operator is the integral

I = D⇤D (16.207) | ih | Z in which

D⇤D d⇤ (x)d (x). (16.208) ⌘ m m m,x Y The hamiltonian for a free Dirac field of mass m is the spatial integral

H = ( + m) d3x (16.209) 0 ·r Z in which i 0 and the gamma matrices (10.304) satisfy ⌘ † a,b =2⌘ab (16.210) { } where ⌘ is the 4 4 diagonal matrix with entries ( 1, 1, 1, 1). Since = ⇥ | i 16.14 Fermionic Path Integrals 703 and = , the quantity exp( i✏H ) is by (16.206) | i h 0| † h 0| 0† h 0| 0 | i

i✏H0 3 0 e = 0 exp i✏ 0 ( + m) dx (16.211) h | | i h | i ·r  Z 1 1 3 =exp (0† †) 0†(0 ) i✏0( + m)d x 2 2 ·r Z 1 1 3 =exp ✏ ˙ † 0†˙ i0 ( + m) d x 2 2 ·r ⇢ Z h i in which = ✏˙ and = ✏˙. Everything within the square 0† † † 0 brackets is multiplied by ✏, so we may replace 0† by † and 0 by so as to write to first order in ✏

i✏H0 1 1 3 0 e =exp ✏ ˙ † †˙ i ( + m) dx (16.212) h | | i 2 2 ·r  Z in which the dependence upon 0 is through the time derivatives. Putting together n =2t/✏ such matrix elements, integrating over all intermediate-state dyadics , and using our formula (16.207), we find | ih |

2itH0 1 1 4 t e t = exp ˙ † †˙ i ( + m) d x D⇤D. h | | i 2 2 ·r Z Z (16.213) Integrating ˙ † by parts and dropping the surface term, we get

2itH0 4 t e t = exp †˙ i ( + m) d x D⇤D. h | | i ·r Z Z (16.214) Since ˙ = i0˙, the argument of the exponential is † i 0˙ ( + m) d4x = i (µ@ + m) d4x. (16.215) ·r µ Z Z We then have

2itH0 4 t e t = exp i 0() d x D⇤D (16.216) h | | i L Z ✓ Z ◆ in which ()= (µ@ + m) is the action density (10.306) for a free L0 µ Dirac field. Thus the amplitude is a path integral with phases given by the classical action S0[]

4 2itH0 i 0() d x iS0[] t e t = e L D⇤D = e D⇤D (16.217) h | | i Z R Z 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. 704 Path Integrals Since Fermi fields anticommute, their time-ordered product has an extra minus sign

(x ) (x ) = ✓(x0 x0) (x ) (x ) ✓(x0 x0) (x ) (x ). (16.218) T 1 2 1 2 1 2 2 1 2 1 ⇥ ⇤ The logic behind our formulas (16.115) and (16.121) for the time-ordered product of bosonic fields now leads to an expression for the time-ordered product of 2n Dirac fields (with D00 and D0 and so forth suppressed)

iS0[] 0 00 (x ) ...(x ) e 0 0 D⇤D h | i 1 2n h | i 0 (x1) ... (x2n) 0 = Z . h |T | i iS0[] 0 00 e 0 0 D⇤D ⇥ ⇤ h | i h | i Z (16.219) As in (16.132), the e↵ect of the inner products 0 and 0 is to insert h | 00i h 0| i ✏-terms which modify the Dirac propagators

iS0[,✏] (x1) ...(x2n) e D⇤D 0 (x1) ... (x2n) 0 = Z . (16.220) h |T | i eiS0[,✏] D D ⇥ ⇤ ⇤ Z Imitating (16.133), we introduce a Grassmann external current ⇣(x) and define a fermionic analog of Z0[j]

4 ⇣+⇣ d x iS0[,✏] e e D⇤D ⇣ + ⇣ d4x R Z0[⇣] 0 e 0 = Z . (16.221) ⌘h |T R | i iS0[,✏] h i e D⇤D Z Example 16.5 (Feynman’s fermion propagator). Since

4 ⌫ µ µ d p ip(x y) i ( i p⌫ + m) i ( @ + m)(x y) i ( @ + m) e µ ⌘ µ (2⇡)4 p2 + m2 i✏ 4 Z ⌫ d p ip(x y) µ ( i p⌫ + m) = e (i p + m) (2⇡)4 µ p2 + m2 i✏ Z 4 2 2 d p ip(x y) p + m 4 = e = (x y), (2⇡)4 p2 + m2 i✏ Z (16.222) the function (x y) is the inverse of the di↵erential operator i(µ@ + m). µ 16.15 Application to nonabelian gauge theories 705

Thus the Grassmann identity (16.200) implies that Z0[⇣]is

µ 4 [⇣+⇣ i ( @µ+m)]d x e D⇤D 4 0 e ⇣ + ⇣ d x 0 = Z R h |T R | i iS0[,✏] h i e D⇤D (16.223) Z =exp ⇣(x)(x y)⇣(y) d4xd4y . Z Di↵erentiating we get 4 ⌫ d p ip(x y) i p⌫ + m 0 (x) (y) 0 =(x y)= i e . (16.224) h |T | i (2⇡)4 p2 + m2 i✏ Z ⇥ ⇤

16.15 Application to nonabelian gauge theories The action of a generic non-abelian gauge theory is

S = 1 F F µ⌫ (µD + m) d4x (16.225) 4 aµ⌫ a µ Z in which the Maxwell field is F @ A @ A + gf A A (16.226) aµ⌫ ⌘ µ a⌫ ⌫ aµ abc bµ c⌫ and the covariant derivative is D @ ig t A . (16.227) µ ⌘ µ a aµ Here g is a coupling constant, fabc is a structure constant (10.65), and ta is a generator (10.56) of the Lie algebra (section 10.15) of the gauge group. One may show (Weinberg, 1996, pp. 14–18) that the analog of equation (16.161) for quantum electrodynamics is

... eiS [A ] DA D O1 On a3 ⌦ [ 1 ... n] ⌦ = Z (16.228) h |T O O | i iS e [Aa3] DA D Z in which the functional delta function [A ] (A (x)) (16.229) a3 ⌘ a3 Yx,b enforces the axial-gauge condition, and D stands for D ⇤D . 706 Path Integrals 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 path integrals, showed how to compute to all orders (Faddeev and Popov, 1967).

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 a gauge a condition on our non-abelian gauge fields Aµ(x). For instance, we can use 3 µ Ga(x)=Aa(x) to impose an axial gauge or Ga(x)=i@µAa (x) to impose a Lorentz-invariant gauge. Under an infinitesimal gauge transformation (11.544)

A = A @ gf A (16.230) aµ aµ µ a abc bµ c the gauge fields change, and so the gauge-fixing functions Gb(x), which de- pend upon them, also change. The jacobian J of that change at =0is

G(x) DG J =det a (16.231) (y) ⌘ D ✓ b ◆=0 =0 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

3 B[G]= (Ga(x)) = (Aa(x)) (16.232) x,a x,a Y Y in an axial gauge or i i B[G]=exp (G (x))2 d4x =exp (@ Aµ(x))2 d4x (16.233) 2 a 2 µ a  Z  Z in a Lorentz-invariant gauge. We want to understand functional integrals like (16.228)

... eiS B[G] JDAD O1 On ⌦ [ 1 ... n] ⌦ = Z (16.234) h |T O O | i eiS B[G] JDAD Z in which the operators , the action functional S[A], and the di↵erentials Ok DAD (but not the gauge-fixing functional B[G] or the Jacobian J) are gauge invariant. The axial-gauge formula (16.228) is a simple example in 16.16 The Faddeev-Popov trick 707 which B[G]=[Aa3] enforces the axial-gauge condition Aa3(x) = 0 and the determinant J =det( @ (x y)) is a constant that cancels. ab 3 If we translate the gauge fields by gauge transformations ⇤ and ⇤0,then the ratio (16.234) does not change

⇤ ⇤ ... ⇤ eiS B[G⇤] J ⇤ DA⇤ D ⇤ O1 On ⌦ [ 1 ... n] ⌦ = Z (16.235) h |T O O | i ⇤0 eiS B[G⇤0 ] J ⇤0 DA⇤0 D ⇤0 Z any more than f(y) dy is di↵erent from f(x) dx. Since the operators , Ok the action functional S[A], and the di↵erentials DAD are gauge invariant, R R most of the ⇤-dependence goes away

... eiS B[G⇤] J ⇤ DA D O1 On ⌦ [ 1 ... n] ⌦ = Z . (16.236) h |T O O | i eiS B[G⇤0 ] J ⇤0 DA D Z

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

G⇤(x) G⇤(x) ⇤ (z) J ⇤ =det a =det a c d4z (y) ⇤ (z) (y) ✓ b ◆=0 ✓Z c b ◆=0 ⇤ Ga (x) ⇤c(z) =det det ⇤ (z) (y) ✓ c ◆=0 ✓ b ◆=0 ⇤ ⇤ Ga (x) ⇤c(z) DG D⇤ =det det . (16.237) ⇤ (z) (y) ⌘ D⇤ D ✓ c ◆ ✓ b ◆=0 =0

Now we integrate over the gauge transformations ⇤ (and ⇤0) with weight function ⇢(⇤) = (D⇤/D ) 1 and find, since the ratio (16.236) is ⇤- |=0 708 Path Integrals independent

DG⇤ ... eiS B[G⇤] D⇤ DA D O1 On D⇤ ⌦ [ 1 ... n] ⌦ = Z ⇤0 h |T O O | i iS ⇤ DG e B[G 0 ] D⇤0 DA D D⇤ Z 0 ... eiS B[G⇤] DG⇤ DA D O1 On = Z eiS B[G⇤] DG⇤ DA D Z ... eiS DA D O1 On = Z . (16.238) eiS DA D Z 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 without any gauge fixing. No matter what gauge condition G or gauge-fixing func- tional B[G] we use, the resulting gauge-fixed ratio (16.234) is equal to the ratio (16.238) of path integrals over all gauge fields without any gauge fixing. All gauge-fixed ratios (16.234) give the same time-ordered products, and so we can use whatever gauge condition G or gauge-fixing functional B[G]is most convenient. The analogous formula for the euclidian time-ordered product is

Se ... e DA D O1 On ⌦ e [ 1 ... n] ⌦ = Z (16.239) h |T O O | i Se e DA D Z where the euclidian action Se is the spacetime integral of the energy density. This formula is the basis for lattice gauge theory. The path-integral formulas (16.165 & 16.166) 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 one does fix the gauge. To continue our description of their tricks, we return 16.17 Ghosts 709 to the gauge-fixed expression (16.234) for the time-ordered product

... eiS B[G] JDAD O1 On ⌦ [ 1 ... n] ⌦ = Z (16.240) h |T O O | i eiS B[G] JDAD Z set G (x)= i@ Aµ(x) and use (16.233) as the gauge-fixing functional B[G] b µ b i i B[G]=exp (G (x))2 d4x =exp (@ Aµ(x))2 d4x . 2 a 2 µ a  Z  Z (16.241) µ 2 This functional adds to the action density the term (@ Aa ) /2 which leads µ to a gauge-field propagator like the photon’s (16.169) 4 a b ⌘µ⌫ab iq (x y) d q 0 A (x)A (y) 0 = i (x y)= i e · . h |T µ ⌫ | i ab4µ⌫ q2 i✏ (2⇡)4 h i Z (16.242) What about the determinant J? Under an infinitesimal gauge transfor- mation (16.230), the gauge field becomes A = A @ gf A (16.243) aµ aµ µ a abc bµ c µ and so Ga(x)=i@ Aaµ(x)is

G(x)=i@µA (x)+i@µ [ @ gf A (x)] 4(x y) (y) d4y. a aµ ac µ abc bµ c Z (16.244) The jacobian J then is the determinant (16.231) of the matrix G(x) @ a = i 2 4(x y) ig f Aµ(x)4(x y) (16.245) (y) ac abc @xµ b ✓ c ◆=0 ⇥ ⇤ that is @ J =det i 2 4(x y) ig f Aµ(x)4(x y) . (16.246) ac abc @xµ b ✓ ◆ ⇥ ⇤ But we’ve seen (16.199) that a determinant can be written as a fermionic path integral n ✓†A✓ det A = e d✓k⇤d✓k. (16.247) Z kY=1 So we can write the jacobian J as

µ 4 J = exp i!a⇤2!a + igfabc!a⇤@µ(Ab !c) d x D!⇤D! (16.248) Z Z 710 Path Integrals which contributes the terms @ ! @µ! and µ a⇤ a µ µ @ !⇤ gf A ! = @ !⇤ gf A ! (16.249) µ a abc b c µ a abc c b to the action density. Thus we can do perturbation theory by using the modified action density

1 µ⌫ 1 µ 2 µ µ 0 = F F (@ A ) @ !⇤@ ! + @ !⇤ gf A ! (D + m) L 4 aµ⌫ a 2 µ a µ a a µ a abc c b 6 (16.250) in which D µD = µ(@ igtaA ). The ghost field ! is a mathematical 6 ⌘ µ µ aµ device, not a physical field describing real particles, which would be spinless fermions violating the spin-statistics theorem (example 10.21).

16.18 Integrating over the momenta When the hamiltonian is quadratic in the momenta like (16.16) and (16.91), one easily integrates over the momenta and converts the hamiltonian into the lagrangian. It may happen, however, that the hamiltonian is so complicated a function of the momenta that one can’t integrate over the momenta. In such cases, the partition function for a scalar field (x) is a path integral over both and ⇡ Z()= exp i˙(x)⇡(x) H(,⇡) dtd3x DD⇡. (16.251) 0 Z ⇢Z Z h i For some values of ⇡˙ , the exponential is not positive, and so is not a prob- ability distribution for and ⇡. The Monte Carlo methods of chapter 14 are designed to work with probability distributions, not with weight functions that assume values that are negative or complex. This is one aspect of the sign problem. The integral over the momenta P []= exp i˙(x)⇡(x) H(,⇡) dtd3x D⇡ (16.252) 0 Z ⇢Z Z h i is a probability distribution. So one can numerically integrate over the mo- menta, make a look-up table for P [], and then apply the usual Monte Carlo method to the probability functional P [] (Amdahl and Cahill, 2016).

Further Reading Quantum Field Theory (Srednicki, 2007), The Quantum Theory of Fields Exercises 711 I, II, & III (Weinberg, 1995, 1996, 2005), and Quantum Field Theory in a Nutshell (Zee, 2010) all provide excellent treatments of path integrals.

Exercises 16.1 Derive the multiple gaussian integral (16.8) from (5.168). 16.2 Derive the multiple gaussian integral (16.12) from (5.167). 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 at its 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 (16.47) a complete set of momentum dyadics p p ,usetheinnerproduct q p =exp(iqp~)/p2⇡~,dotheresulting | ih | h | i Fourier transform, and so verify the free-particle path integral (16.47). 16.7 By taking the nonrelativistic limit of the formula (11.322) for the action of a relativistic particle of mass m and charge q, derive the expression (16.48) for the action a nonrelativistic particle in an electromagnetic field with no scalar potential. 16.8 Show that for the hamiltonian (16.52) of the simple harmonic oscillator the action S[qc] of the classical path is (16.59). 16.9 Show that the harmonic-oscillator action of the loop (16.60) is (16.61). 16.10 Show that the harmonic-oscillator amplitude (16.64) for q0 = 0 and q = q reduces as t 0 to the one-dimensional version of the free- 00 ! particle amplitude (16.47). 16.11 Work out the path-integral formula for the amplitude for a mass m initially at rest to fall to the ground from height h in a gravitational field of local acceleration g to lowest order and then including loops up to an overall constant. Hint: use the technique of section 16.7. 16.12 Show that the action (16.66 ) of the stationary solution (16.69) is (16.71). 16.13 Derive formula (16.124) for the action S0[] from (16.122 & 16.123). 16.14 Derive identity (16.128). Split the time integral at t =0intotwo halves, use

✏t d ✏t ✏e± = e± (16.253) ± dt and then integrate each half by parts. 16.15 Derive the third term in equation (16.130) from the second term. 712 Path Integrals 16.16 Use (16.135) and the Fourier transform (16.136) of the external cur- rent j to derive the formula (16.137) for the modified action S0[,✏, j]. 16.17 Derive equation (16.139) from equations (16.137) and (16.138). 16.18 Derive the formula (16.140) for Z0[j] from the expression (16.139) for S0[,✏, j]. 16.19 Derive equations (16.141 & 16.142) from formula (16.140). 16.20 Derive equation (16.146) from the formula (16.141) for Z0[j]. 16.21 Show that the time integral of the Coulomb term (16.151) is the term that is quadratic in j0 in the number F defined by (16.156). 16.22 By following steps analogous to those the led to (16.142), derive the formula (16.169) for the photon propagator in Feynman’s gauge. 16.23 Derive expression (16.184) for the inner product ⇣ ✓ . h | i 16.24 Derive the representation (16.187) of the identity operator I for a single fermionic degree of freedom from the rules (16.174 & 16.177) for Grassmann integration and the anticommutation relations (16.170 & 16.176). 16.25 Derive the eigenvalue equation (16.192) from the definition (16.190 & 16.191) of the eigenstate ✓ and the anticommutation relations (16.188 | i & 16.189). 16.26 Derive the eigenvalue relation (16.205) for the Fermi field m(x,t) from the anticommutation relations (16.201 & 16.202) and the defini- tions (16.203 & 16.204). 16.27 Derive the formula (16.206 ) for the inner product from the h 0| i definition (16.204) of the ket . | i

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