Quantum Field Theory II

GEORGE SIOPSIS

Department of Physics and Astronomy The University of Tennessee Knoxville, TN 37996-1200 U.S.A. e-mail: [email protected]

Spring 2018 ii Contents

1 Interactions 1 1.1 Scattering theory ...... 1 1.2 Wick’s theorem ...... 5 1.3 Feynman diagrams ...... 15 1.4 Reaction rates ...... 19 1.5 Unitarity ...... 22 1.6 Path Integrals ...... 24

iii iv CONTENTS UNIT 1

Interactions

When you include interactions, the plot thickens. Until now, we have only considered free fields, described by a Hamiltonian H = H0, which we were able to diagonalize (find all eigenvalues and corresponding eigenstates). But that is no physics! It is time to do some serious physics. When you include interactions, the Hamiltonian is modified to

H = H0 + HI (1.0.1) where HI describes interactions. It is no longer possible to solve the eigenvalue problem for H, except in very few special cases (mostly in two dimensions). The only thing we can do is perturbation theory, assuming HI is small. This does not answer deep questions, such as “what is a proton?”, but it provides a method for very accurate calculations (e.g., the magnetic moment of the electron is known to about 10 signiﬁcant digits both theoretically and experimentally, and they agree with each other!). Experimental results are obtained primarily through scattering. To compare with them, we need to develop scattering theory. It turns out that this type of processes (scattering) holds all the information of a quantum ﬁeld theory.

1.1 Scattering theory

Recall in quantum mechanics, H = H0 + V (1.1.1) where H0 is the kinetic energy and V the potential. If V has compact support, then at times t → ±∞, H = H0, because V = 0. Thus, we may deﬁne incoming and outgoing states, |ini (at t → −∞) and |outi (at t → +∞), respectively, which are eigenstates of H. We shall attempt to do the same in quantum ﬁeld theory. Consider a state |ini. It evolves in time as

e−iHt|ini (1.1.2)

As t → −∞, H → H0 (no interactions), so asymptotically, our state approaches a state in the Hilbert space constructed from H0. Call that state |in,0i. It evolves in time as

e−iH0t|in,0i (1.1.3)

Then the statement that this is asymptotic to the state |ini amounts to

e−iHt|ini −→ e−iH0t|in,0i , as t → −∞ (1.1.4)

Therefore, |ini = lim eiHte−iH0t|in,0i (1.1.5) t→−∞ 2 UNIT 1. INTERACTIONS

The operator U(t) ≡ eiH0te−iHt (1.1.6)

† (a unitary map, U U = I) maps a state in the Hilbert space of H0 (|in,0i) to a state in the Hilbert space of H (|ini) by |ini = lim U †(t)|in,0i (1.1.7) t→−∞

−iHI t Notice that if H and H0 commute ([H,H0] = 0), then we may write U(t) = e , where we used (1.0.1). But this rarely happens. Thus, in general, U(t) is a very complicated object. Similarly, in the inﬁnite future, we may map

|outi = lim U †(t)|out,0i (1.1.8) t→+∞

where |outi (|out,0i) is in the Hilbert space of H (H0). We wish to calculate the amplitude of the process

|ini −→ |outi (1.1.9)

i.e., † hout|ini = lim hout,0|U(t+)U (t−)|in,0i = hout,0|S|in,0i (1.1.10) t±→±∞ where † S ≡ lim U(t+)U (t−) (1.1.11) t±→±∞ is the S-matrix. S maps in-asymptotes to out-asymptotes (these two Hilbert spaces may, in general, be distinct, but not here). In fact, S contains all the information of the quantum ﬁeld theory. It is an important object to study. Let us bring it into a more convenient form. To this end, we need to establish some properties of U(t) ﬁrst. We have

dU(t) = iH eiH0te−iHt + eiH0t(−iH)e−iHt dt 0 iH0t −iHt = −ie (H − H0)e iH0t −iHt = −ie HI e iH0t −iH0t iH0t −iHt = −ie HI e e e

= −iHI (t)U(t) (1.1.12)

Thus U(t) is a true evolution operator with (time-dependent) Hamiltonian HI (t). The latter is obtained from HI by evolving with H0. The ﬁrst-order ODE (1.1.12) together with the initial condition U(0) = I uniquely determine U(t). To solve (1.1.12), convert it into an integral equation,

Z t 0 Z t dU(t ) 0 0 0 0 0 dt = −i dt HI (t )U(t ) 0 dt 0 Z t 0 0 0 U(t) = I − i dt HI (t )U(t ) (1.1.13) 0

and then use iteration,

Z t 0 0 0 U0(t) = I ,Un(t) = I − i dt HI (t )Un−1(t )(n ≥ 1) (1.1.14) 0 1.1 Scattering theory 3

We obtain

U0(t) = I Z t 0 0 U1(t) = I − i dt HI (t ) 0 Z t Z t Z t1 0 0 2 U2(t) = I − i dt HI (t ) + (−i) dt1 dt2HI (t1)HI (t2) (1.1.15) 0 0 0 etc.. This may be brought into a more compact form. Look at the second-order term. In it, t ≥ t1 ≥ t2 ≥ 0. The two-dimensional integral is over a triangle in the (t1, t2) plane, which is half of the square 0 ≤ t1, t2 ≤ t. The other half gives the same answer, but with t1 and t2 interchanged. It follows that the two-dimensional integral can be written in terms of a time-ordered product as

Z t Z t1 1 Z t Z t dt1 dt2HI (t1)HI (t2) = dt1 dt2T [HI (t1)HI (t2)] (1.1.16) 0 0 2 0 0 For the nth term, we similarly obtain

Z t Z tn−1 1 Z t Z t dt1 ··· dtnHI (t1) ··· HI (tn) = dt1 ··· dtnT [HI (t1) ··· HI (tn)] (1.1.17) 0 0 n! 0 0 It follows that ∞ X (−i)n Z t Z t U(t) = T dt ··· dt H (t ) ··· H (t ) n! 1 n I 1 I n n=0 0 0 h −i R t dt0H (t0)i = T e 0 I (1.1.18) which is Dyson’s formula. Next, deﬁne the generalized evolution operator

R t2 0 0 † h −i dt HI (t )i U(t1, t2) ≡ U (t1)U(t2) = T e t1 (1.1.19)

It is easily seen from its deﬁnition that it has the following properties

0 0 00 00 † 0 0 † 0 0 U(t, t )U(t , t ) = U(t, t ) ,U (t, t ) = U(t , t) ,U (t, t )U(t, t ) = I (1.1.20) The S-matrix can be written as

h R +∞ i −i dtHI (t) S = lim U(t+, t−) = T e −∞ (1.1.21) t±→±∞ Various important properties of the S-matrix follow. • The S-matrix is unitary, † S S = I (1.1.22)

This may look like a trivial statement (direct consequence of the unitarity of U(t+, t−)), but we need to take limits t± → ±∞, and that may spoil unitarity, unless the range of S is the entire Hilbert space. The latter is a consequence of the requirement that probability be conserved: everything that comes in should go out. There are cases where this is not true and particles may be trapped into bound states. The S-matrix has a way to address these issues. For the most part, we shall assume that no trapping occurs and the S-matrix is unitary.

• S commutes with the free Hamiltonian H0,

[S, H0] = 0 (1.1.23) i.e., a scattering process conserves unperturbed energy. 4 UNIT 1. INTERACTIONS

Proof.

eiH0 Se−iH0 = lim eiH0 eit−H0 e−it−H eit+H e−it+H0 e−iH0 t±→±∞ = lim eiH0(t−+)e−iH(t−+)eiH(t++)eiH0(t−−) t±→±∞ † = lim U(t− + )U (t+ + ) t±→±∞

But as t± → ±∞, adding an makes no difference, so

eiH0 Se−iH0 = S (1.1.24)

Expanding in , at ﬁrst order we obtain the desired result (1.1.23).

• S is Lorentz-invariant.

Proof. The Lagrangian density (a Lorentz-invariant quantity) can be split as

L = L0 + LI (1.1.25)

If LI contains no time derivatives, then Z 3 HI = − d xLI (1.1.26)

Therefore, using (1.1.21),

h −i R ∞ dtH (t)i h i R d4xL i S = T e −∞ I = T e I (1.1.27)

which is manifestly Lorentz-invariant. The integral is over the entire space-time. It should be noted that time-ordering does not spoil Lorentz invariance. Indeed, time-ordering is frame-dependent only for spacelike separations, and ﬁelds commute at spacelike separations (causality).

• Corollary: By Lorentz invariance, it follows from (1.1.23) that S commutes with the unperturbed four-momentum, µ [S,P0 ] = 0 (1.1.28) This implies momentum conservation in scattering. 0 0 Indeed, consider an incoming (outgoing) state of free particles with momenta p1, p2,... (p1, p2,... ), i.e., 0 0 |in,0i = |~p1, ~p2,... i , |out,0i = |~p1, ~p2,... i (1.1.29) (notation as in eqs. (1.1.7) and (1.1.8)). These are eigenstates of unperturbed momentum

µ µ µ µ 0µ 0µ P0 |in,0i = (p1 + p2 + ... )|in,0i ,P0 |out,0i = (p1 + p2 + ... )|out,0i (1.1.30)

It follows that

µ X µ X 0µ 0 = hout,0|[S,P0 ]|in,0i = pi − pi hout|ini

and so X µ X 0µ pi = pi (1.1.31) i.e., momentum is conserved. 1.2 Wick’s theorem 5

1.2 Wick’s theorem

Notice that in the expression for the S-matrix (1.1.27), time evolution is accomplished with the unperturbed Hamiltonian H0, and not with H. Thus the fields in LI are asymptotic (free) fields, and can be treated in the same way we have already discussed. Let us concentrate on the scalar field φ. The field φ in S does not satisfy the full equation of motion, which is usually non-linear. Instead, it obeys the Klein-Gordon equation. To distinguish it from the true φ, we shall call it φI (in interaction picture). Since we shall be always talking about φI , we might as well drop the subscript I. Hopefully the reader will not get too confused. Therefore, we may perform the familiar expansion

Z d3k h i φ(x) = √ a(~k)e−ik·x + a†(~k)eik·x (1.2.1) 3 (2π) 2ωk and define the one-particle state |~pi = Ca†(~k)|0i (1.2.2) where |0i is the asymptotic vacuum (not the true vacuum of the whole system). When calculating vacuum expectation values h0|φ(x)φ(y) · · · |0i, it is convenient to normal-order. On the other hand, S contains time-ordered products. We need to relate the two types of orderings. To this end, define the contraction of two fields as the difference between the two types of ordering,

φ(x)φ(y) = T (φ(x)φ(y))− : φ(x)φ(y): (1.2.3)

The contraction is a number (not an operator), because each time we commute a and a†, we get a number. To calculate it, we shall take the vacuum expectation value of both sides of (1.2.3). We obtain φ(x)φ(y) = h0|φ(x)φ(y)|0i = h0|T (φ(x)φ(y))|0i − h0| : φ(x)φ(y): |0i The time-ordered product is the Feynman propagator,

Z d4k i D (x − y) = h0|T (φ(x)φ(y))|0i = e−ik·(x−y) (1.2.4) F (2π)4 k2 − m2 + i

The normal-ordered product contains terms of the form a†a†, a†a and aa, all of which have vanishing vacuum expectation values. Therefore, the contraction is the Feynman propagator,

Z d4k i φ(x)φ(y) = D (x − y) = h0|T (φ(x)φ(y))|0i = e−ik·(x−y) (1.2.5) F (2π)4 k2 − m2 + i

This result can be generalized to a product of n ﬁelds. To simplify the notation, denote

φi = φ(xi) , i = 1, . . . , n (1.2.6)

Wick’s theorem states that

Tn ≡ T (φ1 ··· φn) = : φ1 ··· φn :

+ φ1φ2 : φ3 . . . φn : +all other terms with 1 contraction

+ φ1φ2φ3φ4 : φ5 . . . φn : +all other terms with 2 contractions + ... (1.2.7)

For n = 1, Wick’s theorem is the trivial statement

T1 = T (φ1) = φ1 =: φ1 :

For n = 2, it was proved above. 6 UNIT 1. INTERACTIONS

For n = 3, we need to show

T3 = T (φ1φ2φ3) =: φ1φ2φ3 : +φ1φ2φ3 + φ1φ3φ2 + φ2φ3φ3 (1.2.8)

0 0 0 To show this, ﬁrst arrange the indices 1, 2, 3 so that times are order as x1 ≥ x2 ≥ x3. Then

T3 = T (φ1φ2φ3) = φ1φ2φ3

First, let us try to normal-order φ3. To this end, split

φ3 = φ3+ + φ3− (1.2.9)

where φ3+ (φ3−) contains all the creation (annihilation) operators. Then

T3 = φ1φ2φ3 = φ1φ2 (φ3+ + φ3−)

= φ1φ2φ3− + φ3+φ1φ2 + [φ1, φ3+] φ2 + φ1 [φ2, φ3+]

The commutators are numbers. We have

[φ1, φ3+] = h0| [φ1, φ3+] |0i

= h0|φ1φ3+|0i (since h0|φ3+ = 0)

= h0|φ1φ3|0i (since φ3−|0i = 0)

= h0|T (φ1φ3)|0i

= φ1φ2

and similarly for [φ2, φ3+]. We deduce

T3 = φ1φ2φ3− + φ3+φ1φ2 + φ1φ3φ2 + φ2φ3φ1

Next, we normal order φ1φ2,

φ1φ2 =: φ1φ2 : +φ1φ2 and use : φ1φ2φ3 : = : φ1φ2 : φ3− + φ3+ : φ1φ2 : together with (1.2.9) to arrive at the desired result (1.2.8). 0 0 0 0 For n = 4, with x1 ≥ x2 ≥ x3 ≥ x4, we have

T4 = φ1 ··· φ4 = φ1φ2φ3φ4− + φ4+φ1φ2φ3

+ [φ1, φ4+] φ2φ3 + φ1 [φ2, φ4+] φ3 + φ1φ2 [φ3, φ4+]

= T3φ4− + φ4+T3 + φ1φ4φ2φ3 + φ2φ4φ1φ3 + φ3φ4φ1φ2

Now use our result for T3,

T4 = : φ1φ2φ3 : φ4− + φ4+ : φ1φ2φ3 :

+ φ1φ2(φ3φ4− + φ4+φ3) + ··· + ...

+ φ1φ4φ2φ3 + ··· + ...

= : φ1φ2φ3φ4 : +φ1φ2 : φ3φ4 : + ··· + ...

+ φ1φ4φ2φ3 + ··· + ...

where the dots represent terms that are obtained by permutations of the indices (123). 1.2 Wick’s theorem 7

We still need to normal order the ﬁelds in the last line. Using our result for T2, we arrive at the desired result for T4,

T4 = : φ1φ2φ3φ4 : +φ1φ2 : φ3φ4 : + ··· + ...

+ φ1φ4 : φ2φ3 : + ··· + ...

+ φ1φ4φ2φ3 + ··· + ... The above argument can be generalized to arbitrary n by induction (assume Wick’s theorem holds for all n < N; show it holds for N).

EXAMPLE 1: External source The simplest interaction we can have is of the form Z 3 HI = d xφ(x)J(x) (1.2.10) where J(x) is a given function (source) representing some ﬁxed distribution of “matter” 1 The Klein-Gordon gets modiﬁed to

µ 2 ∂µ∂ φ + m φ = −J (1.2.11) which may be solved exactly with the aid of the Feynman propagator. We obtain Z 4 φ(x) = d yDF (x − y)J(y) (1.2.12)

The S-matrix is ∞ h R 4 i X S = T e−i d xφ(x)J(x) = S(n) (1.2.13) n=0 where (−i)n Z S(n) = d4x ··· d4x J(x ) ··· J(x )T (φ(x ) ··· φ(x )) n! 1 n 1 n 1 n It can also be written as (−i)n Z S(n) = T [S ··· S ] ,S = d4x J(x )φ(x )(k = 1, . . . , n) n! 1 n k k k k

0 Of course, all Sk are the same operator, but before we integrate over time (the xks), we need to remember to time-order. Since Sk is linear in φ, Wick’s theorem holds for the Sks. This allows us to turn time-ordering to normal-ordering,

T (S1 ··· Sn) = : S1 ··· Sn :

+ S1S2 : S3 ...Sn : +permutations

+ S1S2S3S4 : S5 ...Sn : +permutations + ... (1.2.14)

Since all Sks are the same object, all permutations give the same result; we just need to count them. With one contraction, the number of permutations is n n! N = = 1 2 2!(n − 2)!

1E.g., if you think of φ as the electromagnetic potential, J is the current that “creates” it. Of course the electromagnetic potential (photon) has two polarizations, so there are complications, but our discussion is still valid, as it captures the essence of the ﬁeld. 8 UNIT 1. INTERACTIONS

With two contractions, the number of permutations is the number of ways of choosing two pairs from n objects, 1nn − 2 n! N = = 2 2 2 2 23(n − 4)! where we divided by 2 because of over-counting (identical pairs). Generalizing to p contractions, the number is

1 nn − 2 n − (2p − 2) n! N = ··· = p p! 2 2 2 p!2p(n − 2p)!

The sum of terms with p contractions in (1.2.14) is

p n−2p NpS1S2S3S4 ··· S2p−1S2p : S2p+1 ··· Sn := Np (S1S2) : S3 :

Its contribution to S(n) is (−i)n (−i)n N (S S )p : Sn−2p : = (S S )p : Sn−2p : n! p 1 2 3 p!2p(n − 2p)! 1 2 3

Now let us collect all the terms with p contractions in S. Such terms exist in all S(n) with n ≥ 2p. We obtain the following contribution to S:

∞ X (−i)n A = (S S )p : Sn−2p : p p!2p(n − 2p)! 1 2 3 n=2p ∞ n0+2p X (−i) 0 = (S S )p : Sn : p!2p(n0)! 1 2 3 n0=0 (−i)2p = (S S )p : e−iS3 : p!2p 1 2

Summing over p, we obtain a simple expression for the S-matrix,

∞ 1 R 4 X − (S1S2) −iS3 −α/2 −i d xJ(x)φ(x) S = Ap = e 2 : e : = e : e : (1.2.15) 0

where the coefﬁcient e−α/2 is a c-number, and Z 4 4 α = d xd yJ(x)DF (x − y)J(y) (1.2.16)

We can now study the physical properties of the system. Evidently, the source creates particles.2 To see this, calculate the probability amplitude that the true vacuum |0iin will evolve to an n-particle state |~p1 . . . ~pniout. We have

outh~p1 . . . ~pn|0iin = h~p1 . . . ~pn|S|0i (1.2.17)

where the states on the right-hand side are free particle states. • n = 0. We have −α/2 outh0|0iin = h0|S|0i = e so the probability that no particle will be created is

P (0 → 0) = e−β , β = <α

2This is similar to the creation of photons (bremsstrahlung radiation) by an accelerating charged particle. 1.2 Wick’s theorem 9

To show that this is an honest probability, i.e., P (0 → 0) ≤ 1, we need to show that β ≥ 0. Using (1.2.4), we obtain Z d4k i α = Je(k) Je(−k) (2π)4 k2 − m2 + i in terms of the Fourier transform of J, Z d4k J(x) = Je(k)e−ik·x (2π)4

Since J(x) is real, we have Je(−k) = Je∗(k), therefore 1 Z d4k i i β = |Je(k)|2 − 2 (2π)4 k2 − m2 + i k2 − m2 − i Using 1 1 lim − = −2πiδ(x) →0+ x + i x − i we deduce 1 Z d4k β = |Je(k)|22πδ(k2 − m2) 2 (2π)4

Integrating over k0, we obtain two identical contributions at k0 = ±ωk, and so Z 3 d k 2 β = 3 |Je(k)| ≥ 0 (1.2.18) (2π) 2ωk • n = 1. In this case only the ﬁrst-order term in the expansion of the normal-ordered exponential in (1.2.15) contributes. We obtain Z −α/2 4 outh~p1|0iin = h~p1|S|0i = e (−i) d xJ(x)h~p1|φ(x)|0i

ip1·x Using h~p1|φ(x)|0i = e , which we showed a while ago, we deduce

−α/2 outh~p1|0iin = −ie Je(p1) The probability that a single particle will be created from the vacuum is Z 3 Z 3 d p1 2 −β d p1 2 −β P (0 → 1) = 3 |outh~p1|0iin| = e 3 |Je(p1)| = βe (1.2.19) (2π) 2ωp1 (2π) 2ωp1 For a general (given) n ≥ 1, the nth-order term in the expansion of the normal-ordered exponential in (1.2.15) contributes. We obtain

Only the term with n creation operators in : S1 ··· Sn : contributes. By commuting all creation operators through to the left, it is easy to see that

Notice that we divided by n! in the expression for the probability, because these are identical particles. Thus, we have a Poisson distribution. Notice that

∞ X P (0 → n) = 1 n=0 Probability is conserved and the S-matrix is unitary. The expected number of emitted quanta (photons) is

∞ X hni = nP (0 → n) = β (1.2.21) n=0

If the source is localized (J(x) ∼ δ4(x)), then the Fourier transform is Je(k) = 1, so β → ∞. The divergence comes from large k (ultraviolet (UV) divergence). Thus a source can never be per- fectly localized, and no experiment can measure φ(x) precisely, in accordance with Heisenberg’s uncertainty principle.

EXAMPLE 2: Static external source As a special case, consider a static source. Well, it will not be completely static, because it has to be switched on at a certain time and then switched off again, after a long time. So let

J(x) = f(t)ρ(~x) (1.2.22)

where ρ(~x) represents the static (“charge”) distribution, and f(t) is a smooth function of time of compact support, so that f(t) = 0 for t > |T |, say, and f(t) = 1 for t < |T |, with a smooth transition from 1 to 0 near t = ±T . We shall assume that interactions are turned on and off adiabatically. If the cutoff time T is very large, then the Fourier transform of f(t), Z fe(ω) = dteiωtf(t)

will have narrow support (Riemann-Lebesgue Lemma). Without the smooth transition at t = ±T , we have Z T 2 sin ωT fe(ω) = dteiωt = −T ω It does not have compact support, but its width is ∆ω ∼ 1/T (i.e., ∆ω∆t ∼ 1). As T → ∞, fe(ω) → 2πδ(ω), which has compact support (the point ω = 0), and the frequency can be measured with inﬁnite accuracy. If we smoothen out the above f(t), then it will have compact support of width ∼ ∆ω. Let us arrange things (choose a large enough cutoff time T ) so that fe(ω) has support within the interval |ω| < m, i.e., fe(ω) = 0 , ω ≥ m (1.2.23) To ﬁnd the probability of particle production (radiation), notice that Z 4 ik·x ~ Je(k) = d xJ(x)e = fe(ω)ρe(k)

p~ 2 2 In the expression for β (1.2.18), ω = ωk = k + m ≥ m, therefore, ω is outside the support of fe, by our assumption (1.2.23), and fe(ωk) = 0. It follows that β = 0, and no particles can be created. This makes physical sense: no radiation is emitted by static charges. Thus the S-matrix is pretty boring, but we may still talk about the vacuum-to-vacuum transition amplitude h0|S|0i = e−α/2 1.2 Wick’s theorem 11 where α is purely imaginary (we just showed that its real part β = 0). The probability is |e−α| = 1, as expected, since this is the only possible process, but what is the physical meaning of the amplitude? To answer this question, use (1.1.6) and (1.1.11) to write the amplitude as

H|0iphys = E0|0iphys where E0 is the true ground state energy. From the adiabatic theorem in quantum mechanics, we know that if interactions are turned on very slowly, the state |0i (which is the ground state in the absence of interactions) evolves into the vacuum state |0iphys without getting excited. It follows that

−α/2 −iH(t+−t−) −2iE0T e = physh0|e |0iphys = e t±=±T

This is true for large cutoff time T , more precisely, as T → ∞. We deduce

i α E0 = − lim (1.2.24) 4 T →∞ T Thus the S-matrix contains information about the ground state energy of the system. For an explicit expression, use (1.2.16) together with (1.2.4). The integrals over times x0 and y0 are

T T Z Z 0 0 dx0 dy0e−ik0(x −y ) −T −T

Evidently, in the limit T → ∞, this expression has support consisting of the single point k0 = 0. It follows that for large T , it is approximately

T T Z Z 0 0 0 0 −ik0(x −y ) dx dy e ≈ 2πCδ(k0) −T −T

The constant C is found by integrating both sides over k0. We obtain

Z ∞ Z T Z T Z ∞ 2 dk0 0 0 dk0 2 sin k0T C = dx0 dy0e−ik0(x −y ) = = 2T −∞ 2π −T −T −∞ 2π k0

It follows that for large T , Z α ≈ 2iT d3xd3yρ(~x)V (~x − ~y)ρ(~y) and 1 Z E = d3xd3yρ(~x)V (~x − ~y)ρ(~y) (1.2.25) 0 2 which is the energy of two static (“charge”) distributions interacting via the potential

Z d3k ei~k·~x e−mr V (~x) = − = − (1.2.26) (2π)3 ~k2 + m2 4πr

This is a Yukawa potential. As m → 0, it turns into the Coulomb potential. 12 UNIT 1. INTERACTIONS

EXAMPLE 3: A non-linear quantum ﬁeld theory

Consider three distinct Klein Gordon ﬁelds φ1 (i = 1, 2, 3) with interaction Lagrangian density

LI = −λφ1(x)φ2(x)φ3(x) (1.2.27)

where λ is a coupling constant. The S-matrix can be expanded in powers of λ,

∞ h R 4 i X S = T e−i d xλφ1φ2φ3 = S(n) (1.2.28) n=0

The nth term is proportional to λn, and Z (0) (1) 4 S = 1 ,S = −iλ d xφ1φ2φ3 (no need to normal − order) ,... (1.2.29)

We shall represent S(1) by a diagram thusly:

1 @@ 3

2 r

Each line is a ﬁeld. Lines that meet at a vertex have common argument which is integrated over and multiplied by −iλ. S(1) contributes to the process

1(p1) → 2(p2) + 3(p3) (1.2.30) µ where pi is the momentum of the particle of type i (i = 1, 2, 3). The probability amplitude for this process is

(1) outh2, p2; 3, p3|1, p1iin = h2, p2; 3, p3|S|1, p1i = h2, p2; 3, p3|S |1, p1i + ... (1.2.31)

where the dots represent contributions of higher order in λ. Using h0|φ(x)|pi = e−ip·x thrice, we obtain Z (1) 4 −ip1·x ip2·x ip3·x 4 4 h2, p2; 3, p3|S |1, p1i = −iλ d xe e e = −iλ(2π) δ (p1 − p2 − p3)

Therefore momentum is conserved, as expected (see discussion following eq. (1.1.28)). For the general process |ii → |fi, we deﬁne the scattering amplitude Afi by

4 4 hf|S|ii − hf|ii = (2π) δ (pf − pi)iAfi (1.2.32)

µ µ where pi (pf ) is the total momentum of the initial (ﬁnal) state |ii (|fi). Notice that we subtracted the amplitude representing no scattering, hf|ii. Earlier, for |ii = |1, p1i, |fi = |2, p2; 3, p3i, we found

2 iAfi = −iλ + O(λ )

Also, momentum conservation implies that the decay (1.2.30) is only allowed (to all orders in λ!) if

m1 ≥ m2 + m3 (1.2.33) 1.2 Wick’s theorem 13

2 2 Proof. By momentum conservation and using pi = mi (i = 1, 2, 3),

2 2 2 m1 = m2 + m3 + 2p2 · p3 Also p2 · p3 = E2E3 − ~p2 · ~p3 ≥ E2E3 − |~p2||~p3| and q 2 2 2 E2E3 = (m2 + ~p2)(m3 + ~p3) q 2 2 2 2 2 2 2 2 = m2m3 + ~p2~p3 + m2~p3 + m3~p2 q 2 2 2 2 ≥ m2m3 + ~p2~p3 + 2m2m3|~p2||~p3|

= m2m3 + |~p2||~p3|

Therefore p2 · p3 ≥ E2E3 − |~p2||~p3| ≥ m2m3 and q q 2 2 2 2 m1 = m2 + m3 + 2p2 · p3 ≥ m2 + m3 + 2m2m3 = m2 + m3

At O(λ2), we have

(−iλ)2 Z S(2) = d4xd4yT [φ (x)φ (x)φ (x)φ (y)φ (y)φ (y)] (1.2.34) 2 1 2 3 1 2 3 We need to re-express this in terms of normal-ordered products. This introduces contractions in the above expression. Here are the various possibilities. • 0 contractions 1 1 @@ 3 × @@ 3

2 r 2 r

Same rules as before, but be sure to normal order! Topologically, this is a disconnected diagram. If we calculate a matrix element of S, we will get 2 δ-functions conserving momentum (of 6 particles).

• 1 contraction. Denote the contraction (propagator) φ(x)φ(y) by

x y r r There are 3 possible contractions, each joining the two vertices,

2 2 1 1 1 1 @@ @@ @@ 3 1 @@3 3 2 @@3 2 3 @@2 r r r r r r

An intermediate line labeled i denotes the contraction φi(x)φi(y) (i = 1, 2, 3). These are not Feynman diagrams (yet). • 2 contractions. There are 3 possibilities. 14 UNIT 1. INTERACTIONS

2 1 1 1 1 2 2 3 3 n3 n3 n2

The ﬁrst diagram is proportional to φ2(x)φ2(y) × φ3(x)φ3(y), etc. • 3 contractions. Only 1 possibility. 1 2 r r 3

SYMMETRY FACTORS In general, if a diagram has a symmetry (a permutation of its vertices that does not give rise to a new contraction), we need to divide by the corresponding symmetry factor. All of the diagrams above have a symmetry: just interchange the two vertices. This explains the 1 (2) factor of 2 in the expression (1.2.34) for S . At higher orders, this symmetry need not be present. For example, consider the contribution to S(3),

1 1 1 @@ 3 2 3 @@2 rr r The vertices are distinguishable, so there is no symmetry. The symmetry factor we ought to divide by is 1. 1 (3) To see this in a little more detail, recall that there is a factor of 3! in the deﬁnition of S . On the other hand, there are 3! ways of doing the two contractions, here is one:

(φ1φ2φ3)(φ1φ2φ3)(φ1φ2φ3)

1 Thus, we obtain a factor of 3! × 3! = 1.

EXAMPLE 3: A non-linear quantum ﬁeld theory of a single ﬁeld

Suppose the 3 ﬁelds are identical, φ1 = φ2 = φ3 = φ, and let the interaction Lagrangian density be 1 L = − λφ3 (1.2.35) I 3!

1 where we re-deﬁned the coupling constant λ in order to expose a factor of 3! for convenience. This time, the symmetry factor is the number of permutations of vertices times the number of permutations of lines.

@@ • has symmetry factor 3! (number of permutations of (123)).

@@ 2 • @@ has symmetry factor 2 (vertices) × 2 (two pairs of external lines) = 8.

@@ • 3! × 23 = 48 @ has symmetry factor (vertices) (three pairs of external lines) .

• has symmetry factor 2 (vertices) × 2 (internal lines) = 4. n 1.3 Feynman diagrams 15

A general graph is disconnected and consists of lower-order connected graphs. Let S(graph) be the contribution of an individual graph to the S-matrix. It can be written as

n n (graph) 1 (c) 1 (c) 2 S = S1 S2 ··· (1.2.36) n1!n2! ···

(c) where Si (i = 1, 2,... ) is the contribution to the S-matrix of the corresponding connected sub- graph. Notice that we had to divide by symmetry factors ni! (i = 1, 2,... ), because we have ni identical connected sub-graphs. Be reminded that sub-graphs represent operators which do not necessarily commute with each other. However the product is normal-ordered, so the ordering of the subgraphs is immaterial. Also note that the above expression includes connected graphs, if we set n1 = 1, ni = 0 (i ≥ 2). The entire S-matrix is the sum over all possibilities,

n n X (graph) X 1 (c) 1 (c) 2 S = S = S1 S2 ··· (1.2.37) n1!n2! ··· n1,n2,...

This looks hideous, but can, in fact, be neatly organized, because the sums over n1, n2,... , are independent of each other. Thus S can be written as an inﬁnite product,

∞ n ! ∞ n ! X 1 (c) 1 X 1 (c) 2 (c) (c) (c) S1 S2 S S = S1 S2 ··· = e e ··· = e (1.2.38) n1! n2! n1=0 n2=0 where S(c) is the connected part of the S-matrix,

(c) (c) (c) S = S1 + S2 + ... (1.2.39) consisting of connected graphs only. EXAMPLE: Let us re-visit the system with the external source (1.2.10) whose S-matrix we found exactly (eqs. (1.2.15) and (1.2.16)), albeit rather painfully. There are only two connected graphs:

(c) S2 one contraction. r r The connected S-matrix is 1 S(c) = S(c) + S(c) 1 2 2 where I included the requisite symmetry factor. The S-matrix is S(c) 1 S(c) S(c) S = e = e 2 2 e 1 The two factors can be separated because they commute with each other. The ﬁrst factor is a number, (c) 2 in fact S2 = (−i) α = −α. The second factor is a (normal-ordered!) operator identical to the expression (1.2.15) derived earlier.

1.3 Feynman diagrams

So far, we have been developing tools to calculate the S-matrix, which is an operator. Now, we shall apply these rules to actual scattering processes and ﬁnd the corresponding rules due to Feynman. 16 UNIT 1. INTERACTIONS

Continuing to work with the interaction Lagrangian density (1.2.35), which is physically useless, but good enough for the moment, let us consider the 2 → 2 interaction of particles with initial momenta µ µ µ µ k1 , k2 and ﬁnal momenta k3 , k4 . The amplitude is

(2) hk3, k4|S|k1, k2i = hk3, k4|S |k1, k2i + ... (1.3.1)

where the dots represent terms of higher order in λ. Thus, the lowest-order contribution is O(λ2). Recall that there are various contributions, with 0, 1, 2, or 3 contractions. Since the process of interest involves 4 particles, only the part of S(2) with 4 external legs will contribute (the one with 1 contraction, of symmetry factor 23 = 8). We need to match the momenta with these 4 external lines. There are 3 possibilities:

k1 k3 k1 k2 k1 k2 @@ @@ @@ k2 @@k4 k3 @@k4 k4 @@k3 r r r r r r There are 8 ways to assign momenta in each of the 3 cases, so the overall symmetry factor of a 1 diagram including momenta (i.e., a Feynman diagram) is 8 × 8 = 1. This symmetry factor may also be deduced by looking at the Feynman diagram itself: it is the number of permutations of internal lines and vertices (leaving external lines unchanged).

2 • EXAMPLE: The diagram contributing to hk1|S|k2i at O(λ )

k1 k2 n

has symmetry factor 2 (for the internal lines). Recall that the corresponding S-matrix diagram has symmetry factor 2 × 2 = 4, since there are 2 ways of assigning momenta k1 and k2, we 1 1 obtain 2 × 4 = 2 , which is the correct symmetry factor of the Feynman diagram. Let us now compute the ﬁrst diagram. It is Z 2 4 4 2 2 (−iλ) d xd yφ(x)φ(y)hk3, k4| : φ (x)φ (y): |k1, k2i (1.3.2)

I omitted a symmetry factor anticipating that it will eventually be 1. −ip·x Using φ−(x)|~pi = e |0i, the above expression becomes Z 2 4 4 ik3·x ik4·x −ik1·y −ik2·y (−iλ) d xd yDF (x − y)e e e e (1.3.3)

Notice that we have a factor e+ip·x (e−ip·x) for each outgoing (incoming) momentum. Introducing the Fourier transform of the Feynman propagator, we obtain

Z d4k i (−iλ)2 d4xd4y e−k·(x−y) eik3·xeik4·xe−ik1·ye−ik2·y (1.3.4) (2π)4 k2 − m2 + i

4 4 which yields 2 δ-functions, δ (k3 + k4 − k) and δ (k − k1 − k2), allowing us to draw the diagram

k1 k3 @@ k2 k @@k4 r r where the intermediate line may be thought of as representing a particle of four-momentum kµ. Momentum is conserved at each vertex, and consequently, overall,

k1 + k2 = k3 + k4 (1.3.5) 1.3 Feynman diagrams 17 as expected on general grounds. The intermediate “particle” is not real. Indeed consider its decay to the two outgoing particles. For this to happen, we need m1 ≥ m2 + m3, i.e., m ≥ 2m, which is not satisﬁed, unless m = 0. The latter is also impossible for kinematical reasons. We say that the intermediate particle is “virtual” and its existence is allowed by the Heisenberg Uncertainty Principle (k2 6= m2 is OK in quantum mechanics). After doing all the integrals, the diagram becomes

2 4 4 i (−iλ) (2π) δ (k1 + k2 − k3 − k4) 2 2 (1.3.6) (k1 + k2) − m + i The other 2 diagrams are similarly obtained. We deduce the scattering amplitude 2 i i i iA = (−iλ) 2 2 + 2 2 2 2 (1.3.7) (k1 + k2) − m + i (k1 − k3) − m + i (k1 − k4) − m + i Notice that it is a Lorentz-invariant expression, as it ought to be. Generalizing to more complex diagrams is straightforward. One obtains the following general rules.

FEYNMAN RULES (in momentum space)

• To each external line of a diagram assign an incoming or outgoing momentum.

• To each internal line assign the most general momentum consistent with momentum conservation at each vertex.

• Include a factor (−iλ) for each vertex.

i µ • Include a factor k2−m2+i for each internal line of momentum k .

R d4k 3 • Integrate with (2π)4 for each undetermined momentum. • Divide by the appropriate symmetry factor.

4 4 These rules yield the scattering amplitude iA after removing the overall factor (2π) δ (kf − ki). µ 2 EXAMPLE: Consider the propagation of a single particle of momentum k1 . At O(λ ), the diagram that contributes is k k1 k1 k −nk1

1 There is 1 arbitrary momentum, and the symmetry factor is 2 , so (−iλ)2 Z d4k i i iA = 4 2 2 2 2 (1.3.8) 2 (2π) k − m + i (k − k1) − m + i We managed to derive an expression for this amplitude without much effort, but alas, it is an inﬁnite integral. We will come back to this later. GRAPH TOPOLOGY: The topology of a graph is related to the number of undetermined momenta, so it is important when one tries to determine the complexity of the integral(s), as well as the order in the quantum mechanical expansion of matrix elements of the S-matrix. EXAMPLES

@@ • @@ has no undetermined momenta. r r 3Notice that internal (“virtual”) particles are not on the mass shell (k2 6= m2), so we integrate over all four-momenta. 18 UNIT 1. INTERACTIONS

• has one undetermined momentum. n In general, consider a connected graph with V vertices and I internal lines. • Each vertex gives a δ-function conserving momentum. One of them is the overall factor 4 4 (2π) δ (kf − ki) (for connected graphs), so we have V − 1 δ-functions constraining momenta. • Each internal line provides an arbitrary momentum. Therefore, we have L integrals to do, where L = I − (V − 1) = I − V + 1 (1.3.9)

The following theorem, relating the integrals to the topology of the graph, holds. Theorem. L is the number of loops in the graph. Proof. The proof is by induction on L. When L = 0, we have a tree graph and I = V − 1 (because each time you add a vertex, you add an internal line, so V − I =const. in a tree; if I = 0, then obviously V = 1, therefore V − I = 1). To add a loop, join two external lines. This increases I → I + 1 and V is unchanged. So if we start with I − V = L − 1, we end up with (I + 1) − V = L = (L + 1) − 1, which completes the inductive step.

PHYSICAL CONSIDERATIONS. Let us go back to the 2 → 2 scattering process. In the center of mass frame, ~ ~ ~ ~ k1 = −k2 , k3 = −k4 ,E1 + E2 = E3 + E4 therefore, ~ ~ ~ ~ E1 = E2 = E3 = E4 = E, |k1| = |k2| = |k3| = |k4| = k showing that we have an elastic collision. E is the beam energy in an accelerator. The total center- of-mass energy (a Lorentz-invariant quantity) is Ecm = 2E, because

2 2 2 2 Ecm = (k1 + k2) = (E1 + E2) = 3E It is also convenient to deﬁne the momentum transfers

~ ~ 0 ~ ~ ~q = k1 − k3 , ~q = k1 − k4

Their norms are Lorentz-invariant quantities, and found to be

2 2 2 02 2 2 ~q = (k1 − k3) = 2k (1 + cos θ) , ~q = (k1 − k4) = 2k (1 − cos θ) ~ ~ where θ is the angle between the vectors k1 and k3. The scattering amplitude (1.3.7) reads

λ2 λ2 λ2 A = − 2 2 + 2 2 + 02 2 (1.3.10) Ecm − m ~q + m ~q + m The second term can be easily understood. It is the Fourier transform of a Yukawa potential (eq. (1.2.26)). This matches the result from non-relativistic quantum mechanics (Born approximation) for scattering off of a potential, Z ~ ~ 3 i~q·~x hk1|V (~x)|k3i = d xe V (~x)

The third term has a similar interpretation. Its presence is possible quantum mechanically because we have identical particles. The ﬁrst term has poles at Ecm = ±m. It represents a relativistic effect. Note that the poles are at 2 Ecm = ±mc , so in the non-relativistic limit, c → ∞, and the poles go away to inﬁnity. 1.4 Reaction rates 19

1.4 Reaction rates

To get in touch with reality and calculate quantities that an experimentalist can measure, we will need the relativistic analog of Fermi’s Golden Rule. To this end, it is advantageous to put the world in a box, both in space and time (recall that we need to turn off interactions for |t| ≥ T , where T is large; a time box is just as good and more convenient). In one dimension, we have a basis |ki with R dk |ki hk| = 1 and hx|ki = √1 e−ikx so hx|yi = 2π R R dk −ik(x−y) dk hx |ki hk| yi = 2π e = δ(x − y). We now demand 0 ≤ x ≤ L so we get discrete 1 i 2πn x 2πn eigenstates ϕ (x) = √ e L ≡ hx|ni with P |ni hn| = 1. As L → ∞, → k, and the n L L 2π P L R difference in successive n’s, L → ∆k → dk and n → 2π dk. In three dimensions, we repeat these substitutions 3 times with k = 2π (n , n , n ) to get ϕ (x) = √1 eik·x and, as L → ∞, L 1 2 3 k V P V R 3 k → (2π)3 d k. To make everything concrete, we use a K-G ﬁeld ϕ(x) = 1 P 1 e−ik·xa(k) − eik·xa†(k). V k 2k0 z }| { 1 P 1 −ik·(x−y) R d3k −ik·(x−y) † 0 ϕ(x)ϕ(y) = e → 3 e ; a(k), a (k ) = δk,k0 and h0 |ϕ(x)| ki = V k 2k0 (2π) 2k0 √ 1 e−ik·x. We introduce a new Feynman rule: assign to each external leg a factor of √ 1 . The 2V k0 2EV transition amplitude is

4 4 Y 1 hf |S − 1| ii = iA(2π) δ (pf − pi) p j 2EjV so the transition probability is

Y 1 | hf |S − 1| ii |2 = |A|2((2π)4δ4(p − p ))2 f i 2E V j j Y 1 = |A|2VT (2π)4δ4(p − p ) f i 2E V j j where the square of the delta function is

2 Z (2π)4 δ4(k) = (2π)4 δ4(k) d4yeik·y Z = (2π)4 δ4(k) d4y

= (2π)4 δ4(k)VT because δ4(k) forces k = 0. Thus the probability per unit time (the decay rate), as V → ∞, becomes

Z 3 X 2 4 4 Y 1 Y d pjout 2 4 4 Y 1 |A| V (2π) δ (pout−pin) → 3 |A| V (2π) δ (pout−pin) 2EjV (2π) 2E 2Ejin V jout j j out jout j in

P R d3k 2 because k → V (2π)3 as VT → ∞. All the details of the theory are contained in |A| .

Example 1. One incoming particle pin and two outgoing particles p1 and p2.

The decay rate is

Z 3 3 d p1 d p2 2 4 4 1 Γ = 3 3 |A| V (2π) δ (pin − p1 − p2) (2π) 2E1 (2π) 2E2 2EinV Z 3 1 d p1 2 = 3 |A| 2πδ(Ein − E1 − E2) 2m (2π) 2E12E2 20 UNIT 1. INTERACTIONS

In the rest frame of the incoming particle pin = (m, 0) , p1 = (E1, p1) , and p2 = (E2, p2). Using 3 2 0 spherical coordinates, d p1 = p1dp1dΩ and writing f(p1) = m − E1 − E2, we have f (p1) = dE1 dE2 p1 p1 E1+E2 m − − = − − = −p1 = −p1 . Then dp1 dp1 E1 E2 E1E2 E1E2 Z 2 1 p1dp1dΩ 2 E1E2 Γ = 3 2π|A| δ (p1 − p1root) 2m (2π) 4E1E2 mp1root p Z = 1 dΩ|A|2 32π2m2 In an arbitrary frame m 1 Γarb = Γ = Γ Ein γ using E = √ m = γm. If the incoming particle has spin 0, A is independent of Ω so Γ = 1−v2 2 p1 2 8πm2 |A| . Example 2. One incoming particle and three outgoing particles.

3 d p3 Much the same reasoning applies. We have to include three more integrals for 3 and four (2π) 2E3 4 more delta functions δ (pin − p1 − p2 − p3). In the rest frame of the incoming particle Z 3 3 1 d p1d p2 2 Γ = 6 2πδ(m − E1 − E2 − E3)|A| 2m (2π) 2E12E22E3

Use spherical coordinates with z-axis parallel to p1 and let θ12 be the angle between p1 and p2; 3 2 d p2 = p2dp2dΩ12; dr12 = d(cos θ12)dϕ12. We will eventually use spherical coordinates to represent p1 so we can think of the momenta as positive magnitudes. 2 2 Now specialize to the case where all outgoing particles have mass 0. Note pi = Ei . The argument of the delta function m − E1 − E2 − E3 = m − p1 − p2 − |p1 + p2| = m − p1 − p2 − p 2 2 0 −p1p2 E1E2 p + p − 2p1p2 cos θ12 ≡ f(cos θ12); f (cos θ12) = √ = − . 1 2 2 2 E3 p1+p2−2p1p2 cos θ12 Z 3 2 1 d p1p2dp2dϕ12 E3 2 Γ = 5 |A| 2m (2π) 8E1E2E3 E1E2 Z 3 2 1 d p1p2dp2dϕ12 2 = 5 2 2 |A| 2m (2π) 8E1 E2 Z 2 1 p1dp1dΩdp2 2 = 5 2 |A| 16(2π) m E1 1 Z = dE dΩdE dϕ |A|2 16(2π)5m 1 2 12

2 2 3 2 where we have used pi = Ei and converted p1 to spherical coordinates: d p1 = p1dp1dΩ. The p 2 2 delta function imposes a constraint: m = E1 + E2 + E3 = E1 + E2 + E1 + E2 − 2E1E2 cos θ12 so m ≥ E1 + E2. 2 2 2 2 Furthermore (m − E1 − E2) = E1 + E2 − 2E1E2 cos θ12 so m + 2E1E2 − 2mE1 − 2mE2 = −2E1E2 cos θ12 which gives m(m − 2E1 − 2E2) = −2E1E2(1 + cos θ12) which implies a negative m left hand side, i.e. E1 + E2 ≥ 2 . The limits on cos θ12 also mean m(m − 2E1 − 2E2) ≥ −4E1E2 m m which rearranges to 2 − E1 2 − E2 ≥ 0. Both terms negative contradicts m ≥ E1 + E2 so m E1 and E2 are less than or equal to 2 . Thus integration is over the upper triangle in the Dalitz plot below.

In the case where all three particles have mass, the area of integration becomes a ﬁrst quadrant shape (its Dalitz plot). Example 3. Two in, two out. 1.4 Reaction rates 21

We have the formula for the transition probability. We now calculate the cross-section. Let k2 = 0 L #particles (rest frame of particle 2); The volume is area times length V = AL; v = t so flux = area·time = 1 v v At = AL = V . The cross-section is

Γ σ = flux V = Γ v Z 3 3 V d k3 d k4 2 4 4 1 1 = 3 3 |A| V (2π) δ (k1 + k2 − k3 − k4) v (2π) 2E3 (2π) 2E4 2E1V 2E2V Z 3 3 1 d k3 d k4 2 4 4 1 1 = 3 3 |A| (2π) δ (k1 + k2 − k3 − k4) vE1E2 (2π) 2E3 (2π) 2E4 2 2

The integral is clearly Lorentz invariant as is the factor in front of it since the following is Lorentz invariant q q 2 2 2 2 2 2 (k1 · k2) − m1m2 = (E1m2) − m1m2 q 2 2 = m2 E1 − m1

= m2E1v

= E1E2v where the left hand side has been evaluated in the rest frame of particle 2. The third equality comes 2 2 2 2 2 2 2 2 2 2 |k| from E = γ m so E (1 − v ) = m which gives E − m = E v or v = E . 0 We also have, in the center of mass frame, −k2 = k1 = k and −k4 = k3 = k , q q 2 2 2 2 2 2 2 (k1 · k2) − m1m2 = (E1E2 + k ) − m1m2 q 2 2 4 2 2 2 = E1 E2 + k + 2E1E2k − m1m2 q 2 2 2 2 4 2 2 2 = (k + m1)(k + m2) + k + 2E1E2k − m1m2 q 4 2 2 2 2 = 2k + (m1 + m2)k + 2E1E2k q 2 2 2 2 = (2k + m1 + m2 + 2E1E2)k q 2 2 2 = (E1 + E2 + 2E1E2)k

= (E1 + E2) |k|

= ECM |k|

(E1+E2) |k| |k| The relative velocity of the particles is v = |k| = + = |v1 − v2| so E1E2 E1 E2

Z 3 3 1 d k3 d k4 2 4 4 σ = 3 3 |A| (2π) δ (k1 + k2 − k3 − k4) 4vE1E2 (2π) 2E3 (2π) 2E4 Z 3 3 1 d k3 d k4 2 4 4 = 3 3 |A| (2π) δ (k1 + k2 − k3 − k4) 4ECM |k| (2π) 2E3 (2π) 2E4 Z 3 1 d k3 2 = 3 |A| 2πδ(E1 + E2 − E3 − E4) 4ECM |k| (2π) 2E32E4 Z 1 dΩ 2 1 2 = 2 k3 0 |A| 16π ECM |k| E3E4 |f (k3)| 22 UNIT 1. INTERACTIONS

p 2 2 p 2 2 0 k3 k3 where f(k3) = E1 + E2 − E3 − E4 = ECM − m + k − m + k so f (k3) = + = 3 3 4 3 E3 E4 E3+E4 ECM k3 = k3. Therefore E3E4 E3E4 Z k3 1 2 σ = 2 2 dΩ |A| k 64π ECM where A is theory dependent. In ϕ3 theory, we have Feynman diagrams 2 i i i so iA = (−iλ) 2 2 + 2 2 + 2 2 + ... Define the Mandelstam vari- (k1+k2) −m (k1−k3) −m (k1−k4) −m 2 2 2 ables s = (k1 + k2) , t = (k1 − k3) , u = (k1 − k4) . We refer to these diagrams by the depen- dence of the propagators: the first diagram is the s-channel, the second is the t-channel, and the third is the u-channel. Cyclic permutations of the ki take s to t to u to s. Also we have crossing symmetry iA(s, t, u) = iA(u, s, t). 2 2 2 2 2 2 s + t + u = k1 + k2 + 2k1 · k2 + k1 + k3 − 2k1 · k3 + k1 + k4 − 2k1 · k4 2 = 6m + 2k1 · (k2 − k3 − k4) 2 2 = 6m − 2k1 = 4m2 which implies i i i iA = (−iλ)2 + + s − m2 t − m2 u − m2 Poles develop as s, t, u → m2 so we expect large cross sections at those values. Now let all particles be incoming so k3 → −k3 and k4 → −k4 so their energies will be negative going back in time (antiparticles). We must change the field to a complex field: two particles (b, b†) and two antiparticles (c, c†). 1 + 2 → 3 + 4, 1 + 3 → 2 + 4, 1 + 4 → 2 + 3 (different crossing symmetries). In the center of mass frame, particles 1 - 4 have 4-momentums (E, k), (E, −k), 0 0 2 2 2 0 2 (E, k ), (E, −k ) respectively. 1 + 2 → 3 + 4: s = (2E) = ECM ≥ 4m ≥ 0; t = −(k − k ) ≤ 0; u = −(k + k0)2 ≤ 0. Permute to go to another channel 1 + 3 → 2 + 4: u ≥ 4m2, s, t ≤ 0; permute again to yet another channel 1 + 4 → 2 + 3: t ≥ 4m2, s, u ≤ 0. If we reverse everything, there is no effect on A. In fact A|ii→|fi = A|fi→|ii is a consequence of the CPT theorem. The CPT transformation UCPT is not unitary but anti-unitary because time reversal is † anti-unitary. [UCPT ,H] = [UCPT ,H0] = 0. Recall S = limt±→±∞ U(t+)U (t−) where U(t) = iH0t −iHt −1 −1 e e . Now UCPT U(t)UCPT = U(−t) so a change of variables shows UCPT SUCPT = † † 4 limt±→±∞ U(−t+)U (−t−) = S . Under the CPT transformation (2π) δ (pf − pi) iA|ii→|fi = hf |S − 1| ii ∗ hf |S − 1| ii = hUCPT f |UCPT (S − 1)| ii −1 ∗ = UCPT f UCPT (S − 1)UCPT UCPT i † ∗ = f S − 1 i = i |S − 1| f

which relates to A|fi→|ii.

1.5 Unitarity

4 4 S is unitary. Deﬁne iT = S − 1 so hf |T | ii = −i hf |S − 1| ii = (2π) δ (pf − pi) Afi. Then 1 = (1 + iT )† (1 + iT ) = 1 − iT † + iT + T †T so T − T † = iT †T .

† † X † hf |T | ii − f T i = i f T T i = i f T n hn |T | ii n 1.5 Unitarity 23 which translates to amplitudes ∗ 4 4 X 4 4 ∗ 4 4 Afi − Aif (2π) δ (pf − pi) = i (2π) δ (pf − pn) Anf (2π) δ (pn − pi) Ani n

The second delta function means pn = pi so ∗ X 4 4 ∗ Afi − Aif = i (2π) δ (pi − pn) Anf Ani n When |fi = |ii, we then obtain a form of the Optical Theorem

1 X 4 2 Im A = (2π) δ4 (p − p ) |A | ii 2 i n ni n † † † Fock space gives a complete set of states |k1k2...kni = a (k1)a (k2)...a (kn) |0i, n = 1, ...∞ ∞ 3 3 3 P R d k1 d k2 d k2 so 3 3 ... 3 |k1k2...kni hk1k2...kn| is the identity. If two identical par- (2π) 2E1 (2π) 2E2 (2π) 2E2 n=1 ticles scatter to two particles that are identical to the incoming particles, the first term of Im Aii is 3 3 3 1 R d k1 4 4 1 R d k1 d k2 4 3 (2π) δ (pi −k1) (>--<); the second term is 3 3 (2π) δ(pi −k1 −k2) 2 (2π) 2E1 2 (2π) 2E1 (2π) 2E2 (>= =<) and so on. This generalizes to the Optical Theorem ∞ Z 3 3 1 X 1 d k1 d kn 2 Im A = ... |A | δ4 (p − k − ... − k ) ii 2 n! (2π)32E (2π)32E ni i 1 n n=1 1 n Example 4. A single incoming Klein-Gordon particle of momentum p transitioning to itself. The only Lorentz invariant parameter is p2 ≡ s which should not be identified with the mass of the particle because intermediate particles may not be physical. Now A is a function of s: Im A(s) = 3 1 R d k1 4 4 2 1 δ(E−E1) 2 3 (2π) δ (p − k1) |A| +... = 2π |A| which is Lorentz Invariant because the 2 (2π) 2E1 2 2E1 2 2 2 2 2 2 2 2 1 Lorentz invariant quantity δ k − p = δ E − k − p + p = δ E − p = δ(E1 −p0). 1 1 1 0 1 0 2E1 Hence 1 δ(E − E1) 2 2 2 2 Im A(s) = 2π |A| = πδ k1 − p |A| 2 2E1 to first order (−A−). The higher order corrections come from diagrams (−A = A−) which i ∗ contributes a factor p2−m2+iε and (−A = A −) which Example 5. One particle decays into two particles of the same species 3 3 1 R d k1 d k2 4 2 The decay rate calculated earlier is Γ = 3 3 (2π) δ(p − k1 − k2) |A| . In the rest 2m (2π) 2E1 (2π) 2E2 √ 0 0 frame of the incoming particle p = ( s, 0), k1 = (E , k), and k2 = (E , −k). k Z Γ = √ dΩ |A|2 32π2 s √ 2 which is Lorentz invariant if √k is. E02 = m2 + k2 = s by conservation of energy. Solve for √s 2 k in terms of s and divide by s to get r 2k s − 4m2 √ = 2 s 4s which is Lorentz invariant. Then r 1 4m2 Z mΓ = 1 − dΩ |A|2 128π2 s If s < 4m2, decay is impossible (the initial energy is less than the sum of the masses of the decay products); Γ = 0 so Im A = 0 and A(s) = A∗(s∗).When two analytic function agree on an interval, the agree everywhere except singularities. so Im A = 0 except at singularities. Graphically there is a pole at m2 and a cut at parallel and slightly below the real axis for s > 4m2. For n = 3, the pole is the same and there is still a single cut but for s > 9m2. The physical content of the theory must approach from above. 24 UNIT 1. INTERACTIONS

1.6 Path Integrals

In one-dimensional non-relativistic quantum mechanics, we have position and momentum operators iHt −iHt iHt −iHt xb and pbwith [x,b pb] = i. In the Heisenberg picture, xb(t) = e xb(0)e and pb(t) = e pb(0)e . iHtb iHtb The eigenstates of xb(t) (pb(t)) are |x, ti = e |x, 0i (|p, ti = e |p, 0i) and both collections form a complete set of states with hx|x0i = δ(x − x0), hp|p0i = 2πδ(p − p0), and hx|pi = eixp. iH(t2−t1) To compare momentum and position at different times hp, t1|x, t2i = p, t1 e x, t1 ≈ D E

p, t1 1 + iHb (t2 − t1) x, t1 for small time intervals. We can regard H as a function of p and x and write

−ixp hp, t1 |1 + iH (t2 − t1)| x, t1i = (1 + iH(p, x)(t2 − t1)) e = e−iH(p,x)(t2−t1)e−ixp

0 0 Look at the amplitude of transition from |x0, t0i to |x , t i

0 0 0 0 A = hx , t |x0, t0i ≡ ψ(x , t )

0 but the remaining brackets are hard to evaluate since t1 is not necessarily close to either t0 or t . We 0 can iterate by choosing t2 between t1 and t and inserting the identity Z Z dp dx |x , t i hx , t | 2 |p , t i hp , t | 2 2 2 2 2 2π 2 2 2 2 into the last bracket to get Z 0 dp dx1dp1 dx2dp2 0 0 A = eix p eix1p1 hp , t |x , t i hp0, t0|x , t i hx , t |p , p i hp , t |x , t i 2π 2π 2π 1 1 0 0 2 2 2 2 2 2 2 2 1 1 Z 0 dp dx1dp1 dx2dp2 0 0 = eix p eix1p1 eix2p2 hp , t |x , t i hp , t |x , t i hp0, t0|x , t i 2π 2π 2π 1 1 0 0 2 2 1 1 2 2

0 Now divide the interval [t0, t ] into equally spaced intervals [ti, ti+1] of length ε with t0 < t1 < ... 0 < tn < t , then multiple iteration yields Z 0 dp dx1dp1 dxndpn 0 0 A = ... eix p eix1p1 ..eixnpn hp , t |x , t i ... hp , t |x , t i hp0, t0|x , t i 2π 2π 2π 1 1 0 0 n n n−1 n−1 n n Z 0 dp dx1dp1 dxndpn 0 0 0 0 0 = ... eix p eix1p1 ..eixnpn e−iH(p1,x0)(t1−t0)e−ix0p1 ...e−iH(p ,xn)(t −tn)e−ixnp 2π 2π 2π Z 0 dp dx1dp1 dxndpn 0 0 0 = ... ei(x −xn)p ei(x1−x0)p1 ...ei(xn−xn−1)pn e−iε(H(p1,x0)+...+H(pn,xn−1)+H(p ,xn)) 2π 2π 2π where we have chosen n large enough to make ε small enough to apply the approximation of the ﬁrst paragraph of this section. As ε → 0, the choices of xi = x(ti) ultimately give a path x(t); similarly 1.6 Path Integrals 25

we get a path p(t). We can write H(pi, xi−1) = H(ti) = H(t). xi − xi−1 = x(ti) − x(ti−1) = εx˙(ti). Then

Z 0 dp dx1dp1 dxndpn 0 0 0 A = ... eiεx˙ (t )p(t )eiεx˙ (t1)p(t1)...eiεx˙ (tn)p(tn)e−iε(H(t1)+...+H(tn)+H(t )) 2π 2π 2π Z 0 dp dx1dp1 dxndpn = ... eiε((px˙ −H)(t1)+...(px˙ −H)(tn)) 2π 2π 2π R → ei dt(px˙ −H) R = ei dtL = eiS

0 where S = R t dtL is the action. This a solution of the Schrodinger¨ equation. Note that, in the limit t0 ε → 0, we are actually integrating over all possible paths. p2 If H = 2m + V (x) and ε has a small imaginary part

2 ∞ ∞ pi Z dp Z dp iε pix˙ i− −V (x) i eiε(px˙ −H) = i e 2m −∞ 2π −∞ 2π 2 (p −mx˙ ) Z ∞ dp iε − i + m x˙ 2−V (x) = i e 2m 2 −∞ 2π ∞ 2 Z (p −mx˙ ) iε m x˙ 2−V (x) dpi −iε i = e ( 2 ) e 2m −∞ 2π r m = eiεL(x,x˙ ) 2πiε

Q dxi If we deﬁne [dx] = limε→0 √ , we have 2πiε/m Z 0 0 i R dtL(x,x˙ ) hx , t |x0, t0i = [dx] e

Note that x1(t1) and x2(t2) are operators on the left hand side but are just numbers in the result; numbers commute but operators may not. This is not a contradiction since we started with a time ordered product. We generalize to Z 0 0 iS hx , t |T (x(t1) ...x(t2))| x0, t0i = [dx] x(t1)...x(tn)e 26 UNIT 1. INTERACTIONS

(operators on the left, numbers on the right). This is all non-relativistic quantum mechanics, but it is easily brought into QFT. (x, t) is a 4-vector, ϕ is an operator and |ϕ(x)i denotes the eigenstate of ϕ. Z 0 iS hϕ(x ) |T (ϕ(x1)...ϕ(xn))| ϕ(x0)i = [dϕ] ϕ(x1)...ϕ(xn)e

is a non-perturbative way of calculating any Green function.

iHt0 because |ϕ(x0, t0)i = e |ϕ(x0, 0)i and Ω and E are eigenvalues of H. The factor g(E) is the degeneracy at E. We have expressed |ϕ(x0)i as a function f(t0); it may be impossible to calculate, but we only need its value as t0 → −∞. f(t0) is the Fourier transform of the function g(E) hE |ϕ(x0, 0)| Ei; the Riemann-Lesbegue lemma shows that the Fourier transform of a function approaches 0 as t → ±∞. If E is large, the phase differences are large and, as t0 → ±∞, there is full destructive interference.

Z 0 iS iE0(t0−t ) 0 0 [dϕ] ϕ(x1)...ϕ(xn)e → e hϕ(x , 0)|Ωi hΩ|ϕ(x0, 0)i ϕ(x , 0) hΩ |T (ϕ(x1)...ϕ(xn))| Ωi

The last bracket is deﬁnitely a physical object, the propagator for n = 2 or the scattering amplitude. Thus R [dϕ] ϕ(x )...ϕ(x )eiS hΩ |T (ϕ(x )...ϕ(x ))| Ωi = 1 n 1 n R [dϕ] eiS In terms of diagrams, the components of R [dϕ] eiS have no external legs (the diagram is a bubble) R iS because it is a vacuum to vacuum transition. Looking at [dϕ] ϕ(x1)ϕ(x2)e , the simplest graph is just a line segment, the K-G propagator. We have the connected graphs The disconnected graphs contribute which can be written as (1+bubbles)(all graphs with no bubbles). When divided by (1+bubbles), we have the propagator. Bubbles represent vacuum to vacuum transi- tions, but physical processes involve excitations of the vacuum. Therefore bubbles do not contribute to physical reality although they presumably make up the cosmological constant. R iS (n) [dϕ]ϕ(x1)...ϕ(xn)e Let G (x1, ..., xn) ≡ hΩ |T (ϕ(x1)...ϕ(xn))| Ωi = R [dϕ]eiS (graphs without bubbles). Add an external source J so S → S + R Jϕ. R Jϕ is shorthand for R d4xJ(x)ϕ(x). Deﬁne

R R [dϕ] ei(S+ Jϕ) Z [J] = R [dϕ] eiS

Note 2 3 R Z i2 Z i3 Z ei Jϕ = 1 + i Jϕ + Jϕ + Jϕ + ... 2! 3! 1.6 Path Integrals 27 so 2 i R [dϕ] R JϕeiS i R [dϕ] eiS R Jϕ R Jϕ Z [J] = 1 + + 2! + ... R [dϕ] eiS R [dϕ] eiS Note that δZ i R [dϕ] ϕeiS | = = G(1)(x) δJ(x) J=0 R [dϕ] eiS Similarly 2 R iS δ Z i [dϕ] ϕ(x1)ϕ(x2)e (2) |J=0 = R iS = G (x1, x2) δJ(x1)δJ(x2) [dϕ] e etc.