Quantum Field Theory Koichi Hamaguchi Last Updated: July 31, 2017 Contents

2017 S-semester Quantum Field Theory Koichi Hamaguchi Last updated: July 31, 2017 Contents

§ 0 Introduction 1 § 0.1 Course objectives ...... 2 § 0.2 Quantum mechanics and quantum ﬁeld theory ...... 2 § 0.3 Notation and convention ...... 3 § 0.4 Various ﬁelds ...... 4 § 0.5 Outline: what we will learn ...... 4 § 0.6 S-matrix, amplitude M =⇒ observables (σ and Γ) ...... 5

§ 1 Scalar (spin 0) Field 16 § 1.1 Lorentz transformation ...... 16 § 1.1.1 Lorentz transformation of coordinates ...... 16 § 1.1.2 Lorentz transformation of quantum fields ...... 17 § 1.2 Lagrangian and Canonical Quantization of Real Scalar Field ...... 18 § 1.3 Equation of Motion (EOM) ...... 22 § 1.4 Free scalar field ...... 23 § 1.4.1 Solution of the EOM ...... 23 § 1.4.2 Commutation relations ...... 25 § 1.4.3 a† and a are the creation and annihilation operators...... 25 § 1.4.4 Consistency check ...... 29 § 1.4.5 vacuum state ...... 30 § 1.4.6 One-particle state ...... 30 § 1.4.7 Lorentz transformation of a and a† ...... 31 § 1.4.8 [ϕ(x), ϕ(y)] for x0 ≠ y0 ...... 32 § 1.5 Interacting Scalar Field ...... 34 § 1.5.1 What is ϕ(x)?...... 34 § 1.5.2 In/out states and the LSZ reduction formula ...... 36 § 1.5.3 Heisenberg field and Interaction picture field ...... 42 § 1.5.4 a and a† (again) ...... 45 § 1.5.5 ⟨0|T(ϕ(x1) ··· ϕ(xn)) |0⟩ =? ...... 46 § 1.5.6 Wick’s theorem ...... 49 § 1.5.7 Summary, Feynman rules, examples ...... 56

§ 2 Fermion (spin 1/2) Field 66 § 2.1 Representations of the Lorentz group ...... 66 § 2.1.1 Lorentz Transformation of coordinates (again) (see § 1.1.1) ...... 66 § 2.1.2 inﬁnitesimal Lorentz Transformation and generators of Lorentz group (in the 4-vector basis) ...... 67 § 2.1.A Other (disconnected) Lorentz transformations ...... 69 § 2.1.3 Lorentz transformations of ﬁelds, and representations of Lorentz group. 70 § 2.1.4 Spinor Fields ...... 77

i § 2.1.5 Lorentz transformations of spinor bilinears ...... 81 § 2.2 Free Dirac Field ...... 85 § 2.2.1 Lagrangian ...... 85 § 2.2.2 Dirac equation and its solution ...... 88 § 2.2.3 Quantization of Dirac ﬁeld ...... 91

ii § 0 Introduction

about this lecture (i) Language (The rule of the department: For the graduate course, as far as there is one interna- tional student who prefers English, the lecture should be given in English. But for the undergraduate course, the lectures are usually given in Japanese. Now, this is a common lecture for both students, and there is no clear rule.. . . Speaking E J Which one do you prefer? . . . (*) We choose this option. ) Writing E (*) J - (ii) Web page Google: Koichi Hamaguchi → Lectures → Quantum Field Theory I

▼ All the announcements will also be given in this web page.

▼ The lecture note will also be updated (every week, hopefully). (iii) Schedule April 10, 17, 24, May 1, 8, 15, 29, (no class on May 22) June 5, 12, 19, 26, July 3, 10, Exam on July 24. (maybe an extra class on July 31.) (the extra class is an bonus lecture after the exam and irrelevant to the grades)

(I don’t check the attendance. You don’t have to attend the classes if you can learn by yourself, submit the homework problems, and attend the exam.) (iv) Grades based on the scores of homework problems (twice?) and the exam on July 24. In the exam, you can bring notes, textbooks, laptop, etc. (v) Textbooks This course is not based on a speciﬁc textbook, but I often refer to the following textbooks during preparing the lecture note.

▼ M. Srednicki, Quantum Field Theory.

▼ M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory.

▼ S. Weinberg, The Quantum Theory of Fields volume I.

1 § 0.1 Course objectives To learn the basics of Quantum Field Theory (QFT). One of the goals is to understand how to calculate the transition probabilities (such as the cross section and the decay rate) in QFT. (See § 0.5.) Examples

▼ at colliers e+ µ+ γ

Higgs

e− µ− γ

▼ in the early universe DM q

DM q¯

§ 0.2 Quantum mechanics and quantum ﬁeld theory Quantum Field Theory (QFT) is just Quantum Mechanics (QM) applied to ﬁelds.

▼ QM: qi(t) i = 1, 2, ··· discrete

▼ QFT: ϕ(⃗x,t) ⃗x ··· continous (inﬁnite number of degrees of freedom) (Note: uncountably inﬁnite, 非可算無限)

QM QFT ˙ operators qi(t), pi(t) or qi(t), q˙i(t) ϕ(⃗x,t), π(⃗x,t) or ϕ(⃗x,t), ϕ(⃗x,t) (Heisenberg picture) i = 1, 2, ··· discrete ⃗x ··· continous

(3) [qi, pj] = iℏδij [ϕ(⃗x,t), π(⃗y, t)] = iδ (⃗x − ⃗y) states (e.g., Harmonic Oscillator) |0⟩: ground state |0⟩: ground state † † † a† |0⟩ , a†a† |0⟩ , ··· a |0⟩ , a a |0⟩ , ··· ⃗p ⃗p p⃗′

† † a written in terms of q and p a⃗p written in terms of ϕ and π observables (expectation value) ⟨·|⃗p|·⟩, ⟨·|H|·⟩, ··· ⟨·|⃗p|·⟩, ⟨·|H|·⟩, ··· transition probability P (i → f) = |⟨f|i⟩|2 P (i → f) = |⟨f|i⟩|2

2 In this lecture, we focus on the relativistic QFT. (QFT can also be applied to non-relativistic system: condensed matter, bound state,. . . )

Relativistic QFT is based on QM and SR (special relativity). ℏ ̸ ▼ QM: = 0 (important at small scale) ∞ ▼ SR: c < (important at large velocity) ℏ ̸ ∞ ▼ QFT: = 0 and c < (physics at small scale & large velocity: Particle Physics, Early Universe,. . . )

§ 0.3 Notation and convention

▼ We will use the natural units

ℏ = c = 1 ,

where

ℏ ≃ 1.055 × 10−34 kg · m2 · sec−1 , c = 2.998 × 108 m · sec−1 .

For instance, we write – E2 = p2 + m2 instead of E2 = p2c2 + m2c4, and – [x, p] = i instead of [x, p] = iℏ.

▼ We will use the following metric .   1  −1  g =   . µν  −1  −1

Peskin − Srednicki (The sign convention depends on the textbook. gµν(here) = gµν = gµν ) xµ = (x0, x1, x2, x3) = (t, ⃗x) ν xµ = (x0, x1, x2, x3) = gµνx = (t, −⃗x) pµ = (p0, p1, p2, p3) = (E, ⃗p) ν pµ = gµνp = (E, −⃗p) ν µ p · x = pµx = p xν = p0x0 − p1x1 − p2x2 − p3x3 = p0x0 − ⃗p · ⃗x = Et − ⃗p · ⃗x

3 If pµ is the 4-momentum of a particle with mass m,

2 µ 0 2 2 2 2 p = p pµ = (p ) − |⃗p| = E − |⃗p| = m2 .

§ 0.4 Various ﬁelds

spin equation of motion for free fields scalar field ϕ(x) 0 (□ + m2)ϕ = 0 Klein-Gordon eq. µ fermionic field ψα(x) 1/2 (iγ ∂µ − m)ψ = 0 Dirac eq. µ gauge field Aµ(x) 1 ∂ Fµν = 0 (part of) Maxwell eq. (Fµν = ∂µAν − ∂νAµ)

We start from the scalar ﬁeld.

The Standard Model of Particle Physics is also written in terms of QFT:

• quarks (u, d, s, c, b, t) and leptons (e, µ, τ, νi). . . fermionic fields • γ (photon), W , Z (weak bosons), g (gluon) . . . gauge fields • H (Higgs) . . . scalar field

§ 0.5 Outline: what we will learn

free ⃝1 quantization of ﬁeld interacting

• operator ⃝2 (2 ways) • path integral

•⟨0|T [ϕ(x1) ··· ϕ(xn)]|0⟩

⃝3 • LSZ reduction short cut • S-matrix, amplitude M Feynman rule ⃝4

• observables (cross section σ and decay rate Γ)

⃝⃝⃝ → ▼ First, we will learn 1 2 3 . . . with a scalar ﬁeld. ( next, fermionic ﬁeld, . . . ) ⃝ ▼ A long way to go,. . . Today, let’s discuss 4 in advance.

4 § 0.6 S-matrix, amplitude M =⇒ observables (σ and Γ) Let’s consider the probability of the following process, P (α → β).

′ p1 p1 ′ p2 p2 α { } β

′ pn pn

⟨ | ⟩ ▼ If the initial and ﬁnal states are normalized as α β = δαβ, then P (α → β) = |⟨β, out|α, in⟩|2 (The meaning of “in” and “out” will be explained later.)

▼ We are interested in states with ﬁxed momenta.

|α⟩ = |σ1, ⃗p1, σ2, ⃗p2, ··· σn, ⃗pn⟩ ,

(σi ; spins and other quantum numbers of the particle i) Let’s consider one-particle state |σ, ⃗p⟩. Since the momentum ⃗p is continuous, we cannot normalize the states as ⟨α|β⟩ = δαβ. Instead, we normalize it as

′ 3 (3) ⟨σ, ⃗p|σ , ⃗q⟩ = (2π) 2Epδ (⃗p − ⃗q)δσσ′ . ——————————(1) Comments (i) What’s the mass dimension of |σ, ⃗p⟩ then? (It’s −1.) (3) (3) (ii) Why Epδ (⃗p − ⃗q), not just δ (⃗p − ⃗q)? (3) → Epδ (⃗p − ⃗q) is Lorentz invariant. For instance, for a boost Lorentz transformation along the z direction, ( ) ( )( ) E′ γ γβ E 1 ′ = , β = v/c, γ = √ , pz γβ γ pz 1 − β2 one can show that (check it yourself) ′ ′ − ′ − E δ(pz qz) = Eδ(pz qz).

5 ▼ ▼ Comments called is S ———o pi 0 pt ee —————— here. to up 10, April on —————— Comments amplitude, scattering or element, matrix invariant edfieteivratmti lmn,o cteigamplitude, scattering or element, matrix invariant the define We conserved, are momentum and energy total the as long As (iii) (iii) (iv) (ii) (ii) (i) (i) -matrix ⟨ σ ⟨ 1 ′ p ntefloig eoi h label the omit we following, dimension. the infinite In an with “matrix” a is “matrix”? Why later. given be will out-states S and in of definition The nti uscin edrv h oml for formula the derive Lagrangian. we the subsection, from this derived In is rule Feynman the how learn amplitude The the Since ⃗ , p ′ ⃗ mti sLrnzinvariant. Lorentz is -matrix 1 ′ 1 · · · , S ⟨ · · · p ⃗ : -matrix. ′ p ⃗ 1 ′ h rniinamplitude transition The m · · · σ m ′ | S S , p mti sLrnzivrat h amplitude the invariant, Lorentz is -matrix ⃗ | p ⃗ ′ ⃗p m ′ 1 m | · · · S out ; M | ⃗p ⃗p 1 n · · · a ecluae yteFymnrl.I hslcue we lecture, this In rule. Feynman the by calculated be can | ⟩ σ (2 = 1 ⃗p ⃗p, n ⟨ ∝ ⟩ β 1 , π → M = · · · out ; ) 4 δ δ (4) σ δ (4) ( n | ( | ∑ ⃗p, rniinprobability transition ( α ∑ f ∑ final f {z 6 f n in ; σ E in ; i p p f } ′ f ′ and f ′ ⟩ − ⟩ − − = = | ∑ ∑ ∑ ⟨ σ i final i β {z ⟨ i i ′ σ . | E S 1 p ′ p } , i i ′ | i ) ) p α ⃗ ) ′ × ⟩ 1 · , i = ——————————(2) · · · δ M (3) S M M . ( σ βα ( ⃗p ∑ m ′ 1 : f sas oet invariant. Lorentz also is , · · · p ⃗ p ⃗ ′ m ′ M ⃗p f | n S − → | as σ ∑ 1 i p ⃗ ⃗p, ′ 1 p ⃗ 1 · · · , ′ i · · · ) p ⃗ ′ σ m n ) ⃗p, n ⟩ Questions after the lecture:

Q: Why do you promote the discrete label i in QM to a continuous label ⃗x in QFT? (After all, what’s the motivation of QFT?) A: A short answer is, because it works! After all the Standard Model (including the QED) has been extremely successful in explaining a large number of phenomena in particle physics and other fields.. . . In fact, in this lecture, I skip the contents of what is called “relativistic quantum mechanics”, and directly start from the QFT. A typical introduction in the relativistic quantum mechanics is as follows. In QM, the Schrödingerequation of a free field is i(∂/∂t)ψ = −(1/2m)(∂/∂x)2ψ, which corresponds to E = p2/2m. Promoting this relation to a relativistic relation, E2 = p2 + m2, one obtains −(∂/∂t)2ϕ = −[(∂/∂x)2 + m2]ϕ = 0, or (□ + m2)ϕ = 0. This is nothing but the Klein-Gordon equation. Now, this is not yet the QFT, as far as ϕ is regarded as a wave function as in QM. There is still a logical gap from here to the QFT. (In general, there should be a logical jump when learning a new theory. For instance, one can not “derive” the QM from the classical mechanics!) Here, I do not try to fill this gap (e.g., with the arguments of negative energy etc. . . ), but just directly start from quantizing the fields. Q: What are the prerequisites for this lecture, in particular about the Special Rela- tivity? µ µ ν A: Not much. If you understand for instance the notation of x pµ = gµνx p , and the fact that it is Lorentz invariant, then (I hope) the lecture is more or less µ understandable. If you are not sure what the statement “x pµ is Lorentz invariant” means, then you should probably review the basics of the special relativity, Lorentz transformation, Lorentz invariance, etc. µ Q: What is the difference between the lower and upper indices, x and xµ? A: You should review the basics of the special relativity and get used to these nota- tions.. . . (3) Q: Your normalization in Eq.(1) is proportional to Epδ (⃗p − ⃗q). Why don’t you use a (seemingly manifestly Lorentz invariant) normalization δ(4)(⃗p − ⃗q)? A: A good question. For one particle state with a definite mass m, δ(3)(⃗p−⃗q) implies (4) (3) Ep = Eq, and therefore δ (p−q) = δ(Ep −Eq)×δ (⃗p−⃗q) becomes proportional to δ(0), which is not very useful when normalizing fields. (3) Q: You said that Epδ (⃗p − ⃗q) is Lorentz invariant, after Eq.(1). Then you also used (implicitly), in the comment after Eq.(2), the fact that δ(4)(⃗p − ⃗q) is Lorentz invariant. Are these two statements compatible? A: This is also a good question. Yes, both are correct. Check them yourself.

7 —————— on April 17, from here. —————— Outline

free quantization of ﬁeld interacting

•⟨0|T [ϕ ··· ϕ]|0⟩

• LSZ

• S-matrix, amplitude M

We are here. § 0.6 Feynman rule • observables (σ and Γ)

▼ For simplicity, we normalize the system with a box.

2π V = L3, ⃗p = (n , n , n ) L x y z L

Then ∫ 1 δ(3)(⃗p − ⃗q) = d3xei(⃗p−⃗q)·⃗x (2π)3 1 = · V · δ .——————————(3) (2π)3 |{z}⃗p,⃗q discrete Deﬁne

| ⟩ √ 1 | ⟩ ⃗p Box = ⃗p .——————————(4) 2EpV

8 Then, 1 1 Box⟨⃗q|⃗p⟩Box = √ √ ⟨⃗q|⃗p⟩ 2EqV 2EpV

1 1 3 (3) = √ √ (2π) 2Epδ (⃗p − ⃗q)[∵ (1)] 2EqV 2EpV (2π)3 = δ(3)(⃗p − ⃗q)[∵ E = E for ⃗p = ⃗q] V p q = δ⃗p,⃗q. [∵ (3)] | ⟩ Therefore, ⃗p Box is the correct normalization to give the transition probability. For instance, if there is no interaction at all, for one particle state, {

2 ⃗′ ′ ′ 1 (p = ⃗p) Probability P (⃗p → p⃗ ) = Box⟨p⃗ |⃗p⟩Box = δ⃗′ = p ,⃗p 0 (p⃗′ ≠ ⃗p)

Thus,

Probability P (⃗p ··· ⃗p → p⃗′ ··· p⃗′ ) 1 n 1 m 2 ⟨⃗′ ··· ⃗′ | | ··· ⟩ = Box p 1 p m S ⃗p1 ⃗pn Box

2 1 1 ′ ′ 1 1 = √ ··· √ ⟨p⃗ ··· p⃗ |S|⃗p1 ··· ⃗pn⟩√ ··· √ ∵ (4) 2E′ V 2E′ V 1 m 2E V 2E V ( 1 )( m ) 1 m m n ( ) ∏ 1 ∏ 1 1 n+m 2 ⟨⃗′ ··· ⃗′ | | ··· ⟩ = ′ p 1 p m S ⃗p1 ⃗pn ———————(5) 2E 2Ei V f=1 f i=1

→ ∞ ▼ But this becomes zero for V . ⃗′ ⃗′ ⃗′ What is the (diﬀerential) probability that the ﬁnal state is within [p f , p f + dp f ]?

··· → ⃗′ ··· ⃗′ × dP = P (⃗p1 ⃗pn p 1 p m) |{z}dN ⃗′ ⃗′ ⃗′ number of states within [p f , p f + dp f ]

py 2π L ←→ dp dp dp d3p dN = x · y · z = · V (2π/L) (2π/L) (2π/L) (2π)3

pz { dpx

9 ⃗′ ··· ⃗′ For the m particle ﬁnal states, p 1 p m, ( ) ∏m d3p′ dN = f · V ———————(6) (2π)3 f=1 From (5) and (6), dP = P (⃗p ··· ⃗p → p⃗′ ··· p⃗′ ) × dN ( 1 n )(1 m ) m ′ n ( ) ∏ d3p 1 ∏ 1 1 n 2 f ⟨⃗′ ··· ⃗′ | | ··· ⟩ = 3 ′ p 1 p m S ⃗p1 ⃗pn ———————(7) (2π) 2E 2Ei V f=1 f i=1

▼ On the other hand, from the deﬁnition of M (2),

2 ⟨⃗′ ··· ⃗′ | | ··· ⟩ p 1 p m S ⃗p1 ⃗pn ∑ ∑ ∑ ∑ = (2π)4δ(4)( p′ − p′ ) · (2π)4δ(4)( p′ − p′ ) · |M|2 ∑ f ∑ i f i = (2π)4δ(4)( p′ − p′ ) · (2π)4δ(4)(0) · |M|2 ( ∫ f i ) d4x V · T δ(4)(0) = ei0·x = T : time (→ ∞) (2π)4 (2π)4 ∑ ∑ 4 (4) ′ − ′ · · · |M|2 = (2π) δ ( pf pi) V T

▼ Substituting it in Eq.(7) and dividing by T , we obtain the diﬀerential transition rate ( )( ) dP ∏n 1 ∏m d3p′ 1 ∑ ∑ 1−n f 4 (4) ′ − ′ ·|M ··· → ⃗′ ··· ⃗′ |2 = V 3 ′ (2π) δ ( pf pi) (⃗p1 ⃗pn p 1 p m) T 2Ei (2π) 2Ef i=1 | f=1 {z }

≡ dΦm ———————(8)

▼ Now let’s discuss the cases n = 1 and n = 2. n = 1

q2 A particle decay

From Eq.(8), the probability that the particle A decays into the range of ﬁnal states ⃗′ ⃗′ ⃗′ [p f , p f + dp f ] per unit time is → ··· dP (pA q1 qm) 1 2 = dΦm|M(pA → q1 ··· qm)| T 2EA

10 Integrating over the ﬁnal momenta, we have Decay Rate

Γ(A → 1, 2, ··· ) ∫ 1 = dΦ |M(p → q ··· q )|2 2m m A 1 m A ∫ 1 ∏m d3q ∑ f 4 (4) − |M → ··· |2 = 3 (2π) δ (pA qf ) (pA q1 qm) 2mA (2π) 2Ef | {z } f=1 f at rest frame (× symmetry factor) Comments

(i) The mass dimension of Γ is (energy)+1 ∼ (time)−1.

⟨ | ⟩ 3 (3) − −→ | ⟩ ∼ −1 (CHECK) ⃗q ⃗p = (2π) |{z}2Ep |δ ({z⃗p ⃗q}) ⃗p E . E E−3 ∑ ⟨⃗q ··· ⃗q |S|⃗p ⟩ = (2π)4δ(4)( q − p ) × iM(p → q ··· ) | 1 {zm A} | {z A} | A {z 1 } −m−1 3−m E −4 →E ∫ E 1 ∏m d3q ∑ Γ = f (2π)4δ(4)(p − q ) |M(p → q ··· q )|2 3 A f | A {z1 m } 2mA (2π) 2Ef | {z } f=1 f | {z } | {z } E6−2m E−1 | E2m E{z−4 } E+1

(ii) If there is more than one decay modes, their sum

Γ(A → all) = Γ(A → 1, 2 ··· ) + Γ(A → 1′, 2′ ··· ) + ···

is called the total decay rate, and its inverse 1 τ = A Γ(A → all)

gives the lifetime of A. (Example: muon.

−19 Γ(µ → all) ≃ Γ(µ → eν¯eνµ) ≃ 3 × 10 GeV. 1 τ ≃ ≃ 2.2 × 10−6sec. µ 3 × 10−19GeV )

11 (iii) If not in the rest frame, ∫ 1 2 Γ = dΦm|M| 2EA | {z } Lorentz inv.

In the frame√ which is boosted by a velocity β, the energy is EA = γmA. (γ = 1/ 1 − β2.) → Γ becomes smaller by a factor of 1/γ. → The lifetime becomes longer by a factor of γ. (This is consistent with the Special Relativity!) (iv) If there are identical particles in the ﬁnal state, one should divide by a symmetry factor. (Example) If particles 1 and 2 are identical,

A 2 A 1 1 and 2 are indistinguishable. Thus, we should ⃝1 reduce the integration range (θ = [0, π] → [0, π/2]), or ⃝2 divide by a symmetry factor (= 2) after integration. n = 2

q1 pA q2 particle scattering

pA qm ⃗′ ⃗′ ⃗′ From Eq.(8), the probability that the ﬁnal particles are in the range of [p f , p f + dp f ] per unit time is → ··· dP (pA, pB q1 qm) 1 1 2 = dΦm|M(pA → q1 ··· qm)| ———————(9) T V 2EA · 2EB

▼ In this case, we consider a quantity called “scattering cross section” (or just cross section). ←− ←− ←− ←− ←− B A ←−

12 Suppose that a particle A collides with a bunch of particles B (with number density nB) with a relative velocity vrel. The number that the scattering A, B → 1, 2 ··· occurs per unit time is given by

P (p , p → 1, 2 ··· ) A B = n · v · σ(p , p → 1, 2 ··· ) ———————(10) T B rel | A B {z } cross section

Why “cross section”?

←− ←− ←− ←− ←− B ←−

A If we think a disk with an area σ, the number of B particles which goes through this disk within time T is given by

NB = σ · vrel · T · nB.

This is consistent with (10). (For small T , NB < 1 and it gives the probability.)

▼ In the situation of Eq.(9), there is only one B particle, so nB = 1/V . Thus, the ⃗′ ⃗′ ⃗′ diﬀerential cross section that the ﬁnal state goes within [p f , p f + dp f ] is

1 dP (pA, pB → 1, 2 ··· ) dσ(pA, pB → 1, 2 ··· ) = V [∵ (10)] vrel T 1 1 2 = dΦm|M(pA → q1 ··· qm)| [∵ (9)] vrel 2EA · 2EB

—————— on April 17, up to here. —————— Questions after the lecture: (only some of them)

′ Q: Is the integration only over the ﬁnal state momenta ⃗qf (or ⃗pf ) but not over the initial ones pi? A: Right. The initial momenta are speciﬁed (by the collider experiment, for instance).

Q: Does the amplitude M depend on the ﬁnal state momenta ⃗qf ? A: Yes, it does. Q: If there are identical particles in the ﬁnal state, one should divide by a symmetry factor. What about the initial state?

13 A: Even if there are identical particles in the initial state, it is not necessary to divide by a symmetry factor. In the case of final state, when integrating over the momentum phase space, you should avoid the double counting. (See the figure in the comment.) For the initial state, there is no integration, and hence there is no double counting. Q: I understand the volume factor V in δ(3)(0). What is the factor T in δ(4)(0)? And what do you mean by T → ∞? A: It’s basically the same as the volume factor. Suppose that the interaction is turned on for only∑ a time∑T . Then, the delta∑ function∑ corresponding∫ to the energy − T/2 i0t conservation, δ( Ef Ei), gives for Ef = Ei, δ(0) = −T/2 dt/(2π)e = T/(2π). To all: Here, for the derivation of the formulae for the decay rate and the cross section, I referred to Section 3.4 of Weinberg’s textbook [3] (“Rates and Cross-Sections”), with a modified normalization. There are derivations without a box normalization; see for instance Section 11 of Srednicki’s textbook [1] (“Cross sections and decay rates”) and Section 4.5 of Peskin’s textbook [2] (“Cross Sections and the S-Matrix). In general, if you look at more than one textbook for a certain topic, it can help your understanding a lot.

—————— on April 24, from here. —————— Outline

free quantization of ﬁeld interacting

•⟨0|T [ϕ ··· ϕ]|0⟩

• LSZ

• S-matrix, amplitude M

We are still here. § 0.6 Feynman rule • observables (σ and Γ)

14 Integrating over the ﬁnal momenta, Cross Section

σ(p , p → 1, 2 ··· ) A B ∫ 1 = dΦ |M(p → q ··· q )|2 2E · 2E · v m A 1 m A B rel ∫ 1 ∏m d3q ∑ f 4 (4) − |M → ··· |2 = 3 (2π) δ (pA + pB qf ) (pA, pB q1 ) 2EA · 2EB · vrel (2π) 2Ef f=1 f (× symmetry factor) Comments

(i) The mass dimension of σ is (energy)−2 ∼ (length)2 ∼ (area). (ii) If there are identical particles, divide by the symmetry factor (same as Γ).

(iii) The relative velocity vrel is given by

⃗pA ⃗pB vrel = − EA EB

(For a head-on collision with speeds of light, vrel = 2.)

(iv) EAEBvrel = |EB⃗pA − EA⃗pB| is not Lorentz inv., and therefore σ in the above formula is not Lorentz inv. either. √ → · 2 − 2 2 (Lorentz inv. cross section can be deﬁned by replacing as EAEBvrel (pA pB) mAmB.) Problems

(a) Show that the mass dim. of σ is −2. √ · 2 − 2 2 ∥ (b) Show that (pA pB) mAmB = EAEBvrel for ⃗pA ⃗pB. (Sometimes I give problems. They are mostly just for exercises. Some of those “problems” may be included in the “homework problems” which will be posted later, and which you have to submit.)

15 § 1 Scalar (spin 0) Field

We consider a real scalar ﬁeld ϕ(x).

†

▼ real: ϕ(x) = ϕ(x) (Hermitian operator).

▼ scalar: Lorentz transformation of the ﬁeld is given by

ϕ(x) → ϕ′(x) = ϕ(Λ−1x) .

Now let’s brieﬂy review the Lorentz transformation (before starting QFT).

§ 1.1 Lorentz transformation

§ 1.1.1 Lorentz transformation of coordinates — is a linear, homogeneous change of coordinates from xµ to x′µ,

′µ µ ν x = Λ νx ,

where Λ is a 4 × 4 matrix satisfying

µ ν T gµνΛ ρΛ σ = gρσ (or Λ gΛ = g in matrix notation) .

Comments

(i) It preserves inner products of four vectors:

µ ν x · y = gµνx y ′ ′ ′µ ′ν µ ν ρ σ ρ σ → x · y = gµνx y = gµνΛ ρΛ σx y = gρσx y = x · y .

→ ⃗′ ▼ This is similar to orthogonal transformation( ⃗v v = R⃗v) where R is an orthogonal cos θ sin θ matrix satisfying RT R = 1. (e.g., R = in 2-dim.) − sin θ cos θ Inner products are preserved: ⃗u · ⃗v → ⃗u′ · ⃗v′ = (R⃗u) · (R⃗v) = ⃗uT RT R⃗v = ⃗u · ⃗v.

=⇒

(ii) The set of all Lorentz transformations (LTs) forms a group (Lorentz group).

µ µ ρ ▼ Product of two LTs Λ1 and Λ2 is deﬁnes as (Λ2Λ1) ν = (Λ2) ρ(Λ1) ν. T T T ▼ closure: if Λ1 gΛ1 = g and Λ2 gΛ2 = g, then (Λ2Λ1) g(Λ2Λ1) = g.

16 § (d) (b) clrfields Scalar fields: on acting operators unitary by represented is — (a) Comments (c) (iii) 1.1.2 (ii) (i) Problems ewl ics oeo hsltri Sec. in later this on more discuss will We that Show write can we LT, infinitesimal an For Λ above the the that along Show boost a for Λ of form explicit the the Write along rotation a for Λ form explicit the Write hn ihoperators with Then, Φ Why Substituting field the transforms it that Note

oet rnfraino unu fields quantum of transformation Lorentz ▼ ▼ ▼ nes:(Λ inverse: Λ identity: (Λ associativity: ′ r h ed hc rnfr as transform which fields the are = δω U Φ( µν Φ x x U = O = ) − µ − 1 → ν 1 1 − y O ϕ ) ′ = o ed ?Spoeta state a that Suppose Φ? fields for µ δω 2 ( Φ Λ = ν x · · · δ ′ 1 νµ ) − = ( O µ Λ x 1 → ν i O y 2 = ) g , n ec hr r i independent six are there hence and aifisΛ satisfies )Λ = tmeans it , µρ n ϕ · → |·⟩ g     3 ′ U ( νσ x Λ = 1 ( = = (Λ)Φ( = ) Λ σ O 1 U 1 Λ ϕ UO ρ (Λ 1 ′ U ( ( µ − Λ = O x O 1 ν x ϕ 1 (Λ) 2 taltespacetime the all at ) 2 1 ′ = Λ ) ′ 1 Λ = U ( U · · · O y 1 ν 3 17 ϕ − (Λ) ′ µ δ ). §     2 = ) 1 ( . µ O · · · 2.1. 1 x )( . ν .. (Λ i.e., , n ′ ) − UO + U ϕ |·⟩ O 1 (Λ) ( δω n y ′ 2 z z |·⟩ ) U ai.So hti aifisΛ satisfies it that Show -axis. ai.So hti aifisΛ satisfies it that Show -axis. µ − frall (for ν − ) Φ( 1 . 1 − = ) 1 x |·⟩ · · · ) eei field generic : ) µ ϕ ν rnfrsas transforms (Λ Λ ( y UO ν − ). x ρ 1 . x = n δω ) U δ − . ν 1 ρ ) . U |·⟩ · |·⟩ → |·⟩ T T ′ g g = = Λ = Λ U |·⟩ g g . . . § 1.2 Lagrangian and Canonical Quantization of Real Scalar Field In quantum mechanics, we consider a Lagrangian ∂ L = L(q, ˙ q) ˙ = . ∂t In QFT, we also start from a Lagrangian ∫ 3 L ˙ L = d x | [ϕ(⃗x,t{z), ϕ(⃗x,t)]} Lagrangian density In this lecture, we consider the following Lagrangian (called ϕ4 theory): ∫ L = d3x L[ϕ˙(⃗x,t), ϕ(⃗x,t)] ∫ ( ) 3 1 µ 1 2 2 λ 4 = d x ∂µϕ∂ ϕ − m ϕ − ϕ ∫ (2 2 24 ) 1 1 1 λ = d3x ϕ˙2 − ∇⃗ ϕ · ∇⃗ ϕ − m2ϕ2 − ϕ4 2 2 2 24 where λ is real and positive constant, and ∂ ∂ = µ ∂xµ ( ) ( ) ∂ 2 ∑3 ∂ 2 ∂ ϕ∂µϕ = gµν∂ ϕ∂ ϕ = ϕ − ϕ = ϕ˙2 − ∇⃗ ϕ · ∇⃗ ϕ µ µ ν ∂x0 ∂xi i=1 The λϕ4 term represents the interaction. For λ = 0, it becomes the Lagrangian of free scalar field: ∫ ( ) 3 1 µ 1 2 2 Lfree = d x ∂µϕ∂ ϕ − m ϕ ∫ (2 2 ) 1 1 1 = d3x ϕ˙2 − ∇⃗ ϕ · ∇⃗ ϕ − m2ϕ2 2 2 2 If we regard ⃗x as just a label, { ···} ϕ(⃗x,t) = ϕ⃗x1 (t), ϕ⃗x2 (t), ϕ

QM of inﬁnite number of degrees of freedom ( ) ∑ 1 L = ϕ˙ (t)2 + ··· 2 ⃗x ∑⃗x 1 2 ⃗x ∼ q˙i(t) + ··· ⃗x1 ⃗x2 2 i

18 QM QFT conjugate ∂L δL ˙ momentum pi = π(⃗x,t) = = ϕ(⃗x,t) (functional derivative) ∂q˙i δϕ˙(⃗x,t) ∑ ∫ ( ) 3 ˙ Hamiltonian H = piq˙i − L H = d x π(⃗x,t)ϕ(⃗x,t) − L i ∫ ( ) 1 1 1 λ = d3x π2 − π2 + (∇⃗ ϕ)2 + m2ϕ2 + ϕ4 ∫ ( 2 2 2 )24 1 1 1 λ = d3x π2 + (∇⃗ ϕ)2 + m2ϕ2 + ϕ4 2 2 2 24 Canonical Quantization: Equal Time Commutation Relation (3) − [qi(t), pj(t)] = iδij [ϕ(⃗x,t), π(⃗y, t)] = iδ (⃗x ⃗y)

[qi(t), qj(t)] = 0 [ϕ(⃗x,t), ϕ(⃗y, t)] = 0

[pi(t), pj(t)] = 0 [π(⃗x,t), π(⃗y, t)] = 0

equal time

Comments (i) The action ∫ ∫ ∫ S = dtL = dtd3xL = d4xL

is Lorentz invariant.

Check:

Under ϕ(x) → ϕ′(x) = ϕ(Λ−1x), ∫ ∫ [ ] ∂ϕ d4xL[ϕ(x), ∂ ϕ(x)] → d4xL ϕ(Λ−1x), (Λ−1x) µ ∂xµ

ν −1 ν µ By a change of variable, x = Λy or y = (Λ ) µx , ∂ϕ ∂yν ∂ϕ ∂ ϕ(x) → (Λ−1x) = (y) = (Λ−1)ν [∂ ϕ](y) µ ∂xµ ∂xµ ∂yν µ ν and hence ∂ϕ ∂ϕ ∂ ϕ∂µϕ(x) → gµρ (Λ−1x) (Λ−1x) = gµρ(Λ−1)ν (Λ−1)σ [∂ ϕ](y)[∂ ϕ](y) µ ∂xµ ∂xρ | {zµ }ρ ν ρ =gνσ ν = [∂νϕ∂ ϕ](y)

19 Thus, ∫ ( ) 4 1 µ 1 2 2 S = d x ∂µϕ∂ ϕ(x) + m ϕ(x) + ··· ∫ (2 2 ) 4 1 µ 1 2 2 → d x ∂µϕ∂ ϕ(y) + m ϕ(y) + ··· ∫ ( 2 2 )( )

4 1 µ 1 2 2 4 ∂x 4 4 4 = d y ∂µϕ∂ ϕ(y) + m ϕ(y) + ··· d x = det d y = | det Λ|d y = d y 2 2 ∂y = S.

λ (ii) Why L = − ϕ4 ? interaction 24 L ∼ 3 − 3 4 ▼ For interaction ϕ or ϕ or +ϕ , the corresponding Hamiltonian term becomes ∫ 3 3 3 4 Hinteraction ∼ d x(−ϕ or ϕ or − ϕ )

and hence the energy becomes unbounded below. (Take, for instance ϕ(⃗x,t) = const → ∞.) L ∼ − 4 ▼ Thus, interaction ϕ is the simplest possibility.

▼ 24 = 4! is for later convenience (for Feynman rule).

—————— on April 24, up to here. —————— Questions after the lecture: (only some of them)

Q: What does ϕ represent? (a particle?) And do we start from it? Just as a toy model?

A: (sorry, I should have said in the lecture.) Yes, it represents creation and annihilation of a scalar particle. It will become clearer later. It is a good example of QFT as a simple toy model, but scalar particles do exist in nature. An example of a scalar particle is the Higgs boson (and it is the only known elementary scalar particle). The ϕ4 interaction of the Higgs boson is assumed in the Standard Model, but it is not yet experimentally tested. There may also be interactions like ϕ6, ϕ8, or other forms. . . .

Q: I am confused with the LT of ﬁeld, ϕ′(x) = ϕ(Λ−1x). . .

A: (I was also confused when I learned it!) As I said during the lecture, the LT of fields should be distinguished from the LT of coordinates. The former is one of many field transformations. A simple transformation of field is the Z2 transformation, ϕ(x) → 4 −ϕ(x). The action of the ϕ theory is invariant under this Z2 transformation of the field. Similarly, the action is invariant under the LT of the field, ϕ(x) → ϕ(Λ−1x). The latter seems more complicated because it is accompanied by a change of the argument, but they are both just transformations of field.

20 —————— on May 1, from here. —————— Outline

free quantization of ﬁeld § 1.2 interacting We are here.

•⟨0|T [ϕ ··· ϕ]|0⟩

• LSZ

• S-matrix, amplitude M

§ 0.6 Feynman rule • observables (σ and Γ)

(iii) Schr¨odingerrepresentation and Heisenberg representation: In QFT, usually the Heisenberg representation is used. state operator | ⟩ O S-rep. Ψ(t) S S time-dependent time-independent | ⟩ O H-rep. Ψ H H (t) time-independent time-dependent

S-rep. d i |Ψ(t)⟩ = H(p, q) |Ψ(t)⟩ dt S S − − | ⟩ iH(t t0) | ⟩ Ψ(t) S = e Ψ(t0) S Expectation value of an operator: S⟨Ψ(t)|OS|Ψ(t)⟩S

H-rep. { − | ⟩ ≡ | ⟩ iH(t t0) | ⟩ Ψ H Ψ(t0) S = e Ψ(t) S iH(t−t0) −iH(t−t0) OH (t) ≡ e OSe

Expectation value: H ⟨Ψ|OH (t)|Ψ⟩H = ··· = S⟨Ψ(t)|OS|Ψ(t)⟩S

d − − − − − − i O (t) = −HeiH(t t0)O e iH(t t0) + eiH(t t0)O He iH(t t0) dt H S S = −HOH (t) + OH (t)H

= [OH (t),H]. Heisenberg eq.

21 § 1.3 Equation of Motion (EOM)

▼ There are two ways to derive the EOM. ∫ (i) From the action principle δS = δ dtL = 0,

δL δL ∂µ − = 0 Euler-Langrange eq. δ(∂µϕ) δϕ

(ii) From Heisenberg eq., { iϕ˙ = [ϕ, H] iπ˙ = [π, H] ∫ ( ) 3 1 µ 1 2 2 λ 4

▼ From (i), for the Lagrangian L = d x ∂ ϕ∂ ϕ − m ϕ − ϕ , we obtain 2 µ 2 24

δL δL µ 2 λ 3 ∂µ − = ∂µ(∂ ϕ) + m ϕ + ϕ = 0 δ(∂µϕ) δϕ 6 or

( ) λ EOM □ + m2 ϕ(x) = − ϕ(x)3 ——————————(⋆) 6

§ ▼ Later we will derive the EOM (⋆) with (ii) again (see 1.5.3).

▼ Here, there is an important diﬀerence between λ = 0 and λ ≠ 0.

For free ﬁeld (λ = 0), the EOM is linear in ϕ(x), and can be solved exactly (§ 1.4). → ϕ ∼ a + a† → The relations between a, a†, and H are obtained. (creation and annihilation oprators) For λ ≠ 0, the EOM (⋆) is non-linear, and it cannot be simply solved. → what is ϕ(x) in this case? (more on this later)

22 § 1.4 Free scalar ﬁeld

§ 1.4.1 Solution of the EOM

□ 2 ▼ Starting from ( + m )ϕ(x) = 0 (Klein-Gordon eq.) , one can show that ϕ(x) can be expressed as ∫ d3p ( ) ϕ(x) = √ a(⃗p)e−ip·x + a†(⃗p)eip·x ——————————(1) 3 (2π) 2Ep √ † µ 0 0 2 2 where ϕ(x), a(⃗p), a (⃗p) are operators, p·x = pµx = p t−⃗p·⃗x, and p = Ep = ⃗p + m .

▼ Eq. (1) is the solution of the EOM, because (□ + m2)eip·x = (∂ ∂µ + m2)eip·x ( µ ) ∂2 = − ∇⃗ 2 + m2 eip·x ∂t2 − 2 2 2 ip·x = ( Ep + |⃗p +{zm})e = 0. 2 Ep

▼ proof: Now let’s show that Eq.(1) is the general solution of the EOM.

(i) Fourier transform (FT) ϕ(x) with respect to ⃗x: ∫ 3 i⃗p·⃗x |ϕ({z⃗x,t}) = d p C| ({z⃗p,t}) e ——————————(2) operator operator (ii) From the condition ϕ = ϕ† (real ﬁeld), ∫ ∫ d3pC(⃗p,t)ei⃗p·⃗x = d3pC†(⃗p,t)e−i⃗p·⃗x ∫ ′ = d3p′C†(−p⃗′, t)eip⃗ ·⃗x (p⃗′ = −⃗p) ∫ = d3pC†(−⃗p,t)ei⃗p·⃗x

Using inverse FT, C(⃗p,t) = C†(−⃗p,t) ——————————(3) ( ) ( ) ∂2 (iii) From □ + m2 ϕ = − ∇⃗ 2 + m2 ϕ = 0 and (2), ∂t2   ∫   3  ¨ 2 2  i⃗p·⃗x d p C(⃗p,t) + C(⃗p,t)(|⃗p +{zm }) e = 0 2 Ep

23 Using inverse FT, ¨ 2 C(⃗p,t) + Ep C(⃗p,t) = 0 − ′ ∴ C(⃗p,t) = C(⃗p)e iEpt + C (⃗p)e+iEpt. From (3), C′(⃗p) = C†(−⃗p), and hence − † C(⃗p,t) = C(⃗p)e iEpt + C (−⃗p)e+iEpt. (iv) Substituting it to (2) (and changing ⃗p → −⃗p in the 2nd term), ∫ ( ) − · † − · ϕ(⃗x,t) = d3p C(⃗p)e iEptei⃗p ⃗x + C (⃗p)e+iEpte i⃗p ⃗x ∫ ( ) = d3p C(⃗p)e−ip·x + C†(⃗p)e+ip·x √ 3 Finally by normalizing as a(⃗p) ≡ (2π) 2Ep · C(⃗p), we obtain (1). ■

▼ Note that the normalization depends on the convention (textbook). 1 a(here) = a(Peskin) = √ a(Srednicki) = (2π)3/2a(Weinberg). 2Ep

†

▼ From (1), we can express the operators a(⃗p) and a (⃗p) in terms of ϕ(x):  ∫ [ ]  1 3 +ip·x ˙ a(⃗p) = √ d xe iϕ(x) + Epϕ(x) 2E p ∫ [ ] ——————————(4)  † 1 3 −ip·x ˙ a (⃗p) = √ d xe −iϕ(x) + Epϕ(x) 2Ep Problems (a) Substitute (1) to the right-hand side (RHS) of (4) and show that it gives a & a†. (b) The RHS of (4) seems to depend on x0 = t, but the LHS does not. Show that ∂ [RHS of (4)]= 0, using the EOM. (Hint: integration by parts (部分積分)) ∂t (c) Substitute (4) to the RHS of (1) and show that it gives LHS. Pay attention to which variables are just the integration variable. For instance, let’s solve (a): ∫ d3q ( ) from (1), ϕ(x) = √ a(⃗q)e−iq·x + a†(⃗q)e+iq·x (2π)3 2E | {z q} ∫ =[dq] ( ) ˙ −iq·x † +iq·x iϕ(x) = [dq] Eqa(⃗q)e − Eqa (⃗q)e ∫ ( ) ˙ −iq·x † −ip·x iϕ(x) + Epϕ(x) = [dq] (Eq + Ep)a(⃗q)e + (−Eq + Ep)a (⃗q)e

24 Thus, ∫ ∫ ( ) 1 √ 3 +ip·x −iq·x − † +iq·x RHS of (4) = d xe [dq] (Eq + Ep)a(⃗q)e::::: + ( Eq + Ep)a (⃗q)e::::: 2Ep :::::::::::∫ 0 − · − 0 · d3xeiEpx e i⃗p ⃗x · e iEqx ei⃗q ⃗x

− 0 0 = (2π)3δ(3)(⃗p − ⃗q) · ei(Ep Eq)x (2π)3δ(3)(⃗p + ⃗q) · ei(Ep+Eq)x   ∫ 3 1 d q † 0 √ √  (3) − − (3) − i(Ep+Eq)x  = (Eq + Ep)a(⃗q)δ (⃗p ⃗q) + (| Eq{z+ Ep}) a (⃗q)δ (⃗p ⃗q)e 2Ep 2Eq →0 = a(⃗p) = LHS of (4) ■

§ 1.4.2 Commutation relations § ▼ From the commutation relation in 1.2, we have the following commutation relations (recall π(x) = ϕ˙): [ϕ(⃗x,t), ϕ˙(⃗y, t)] = iδ(3)(⃗x − ⃗y) [a(⃗p), a†(⃗q)] = (2π)3δ(3)(⃗p − ⃗q) [ϕ(⃗x,t), ϕ(⃗y, t)] = 0 ⇐⇒ [a(⃗p), a(⃗q)] = 0 ——– (5) [ϕ˙(⃗x,t), ϕ˙(⃗y, t)] = 0 [a†(⃗p), a†(⃗q)] = 0 Problems (a) Show that RHS of (5) =⇒ LHS of (5), using (1). (b) Show that LHS of (5) =⇒ RHS of (5), using (4).

§ 1.4.3 a† and a are the creation and annihilation operators.

▼ In this section we will see that

a(⃗p) ——– annihilate a particle with energy Ep, momentum ⃗p. † a (⃗p) ——— create a particle with energy Ep, momentum ⃗p.

† † [H, a (⃗p)] = Epa (⃗p)

▼ First, we can show that † ——— (6) (We will show it later.) [H, a(⃗p)] = −Epa (⃗p)

ˆ † † [P,⃗ a (⃗p)] = ⃗pa (⃗p) We can also show ˆ † [P,⃗ a(⃗p)] = −⃗pa (⃗p) ˆ ˆ ∫ where P⃗ is the “momentum” operator. (P⃗ = − d3xπ∇⃗ ϕ. We skip the details here.)

▼ Consider a state with energy EX and momentum ⃗pX ; { H |X⟩ = EX |X⟩ |X⟩ : ⃗ˆ P |X⟩ = ⃗pX |X⟩

† Thus, the state a (⃗p) |X⟩ has energy Ep + EX and momentum ⃗p + ⃗pX , namely, † a (⃗p) adds energy Ep and momentum ⃗p. (creation operator)

▼ Similarly, we can show

H (a(⃗p) |X⟩) = (EX − Ep)(a(⃗p) |X⟩) , ⃗ˆ P (a(⃗p) |X⟩) = (⃗pX − ⃗p)(a(⃗p) |X⟩) .

and therefore a(⃗p) is an annihilation operator.

▼ Now let’s show (6). There are two ways.

(i) Express H in terms of a and a†. (ii) Use (4) and Heisenberg eq.

Here we do (i). [Problem: Do (ii): Show (6) by using (4) and Heisenberg eq.] —————— on May 1, up to here. —————— Questions and comments after the lecture: (only some of them) comment: There is a typo at the end of §1.2. In Schrödingerrep., the argument of the RHS | ⟩ of ψ(t) S should be t0, not t. A: Thanks! corrected.∫ d3p Q: When you write √ , does Ep in the denominator depend on the integration 2Ep variable ⃗p? A: Yes. Q: Can we construct a “number operator” from the creation and annihilation operator? Does it give a finite number even though the momentum is continuous? ∫ d3p A: Good question! One can indeed define a number operator N = a†(⃗p)a(⃗p), (2π)3 ∑ † similar to the case of harmonic oscillator, N = i ai ai. Now, next week we will

26 see that a one-particle state is proportional to a†(⃗p) |0⟩. You can check explicitly (by using the commutation relation) that N(a†(⃗p) |0⟩) = (a†(⃗p) |0⟩), which means the number is one. Similarly, you can check N(a†(⃗p)a†(p⃗′) |0⟩) = 2(a†(⃗p)a†(p⃗′) |0⟩), etc.

—————— on May 8, from here. —————— Outline § free 1.4 We are here. quantization of ﬁeld † interacting § 1.4.3, a and a . Showing (6).

•⟨0|T [ϕ ··· ϕ]|0⟩

• LSZ

• S-matrix, amplitude M

§ 0.6 Feynman rule • observables (σ and Γ)

First, ∫ 3 ( ) d p − · † − · ϕ(x) = √ a(⃗p)e iEpt+i⃗p ⃗x + a (⃗p)eiEpt i⃗p ⃗x (2π)3 2E ∫ p 3 ( ) d p − † · = √ a(⃗p)e iEpt + a (−⃗p)eiEpt ei⃗p ⃗x (⃗p → −⃗p in the 2nd term) 3 (2π) 2Ep

Let’s deﬁne,

1 − A(⃗p,t) ≡ √ a(⃗p)e iEpt 3 (2π) 2Ep and omit t for simplicity: A(⃗p) = A(⃗p,t). Then ∫ ( ) ϕ(x) = d3p A(⃗p) + A†(−⃗p) ei⃗p·⃗x ∫ ( ) ∇⃗ ϕ(x) = d3p A(⃗p) + A†(−⃗p) (i⃗p)ei⃗p·⃗x ∫ ( ) ˙ 3 † i⃗p·⃗x ˙ ϕ(x) = d p(−iEp) A(⃗p) − A (−⃗p) e (∵ A(⃗p,t) = (−iEp)A(⃗p,t))

27 Therefore, ∫ ( ) 1 1 1 H = d3x π2 + (∇⃗ ϕ)2 + m2ϕ2 ∫ (2 2 2 ) 1 1 1 = d3x ϕ˙2 + (∇⃗ ϕ)2 + m2ϕ2 ∫ ∫ 2 ∫ 2 2 3 3 3 i⃗p·⃗x i⃗q·⃗x = d x d p d q e::::::::e → (2π)3δ(3)(⃗p+⃗q) :::::: [ 1 ( )( ) × (−iE )(−iE ) A(⃗p) − A†(−⃗p) A(⃗q) − A†(−⃗q) 2 p q 1 ( )( ) + (i⃗p)(i⃗q) A(⃗p) + A†(−⃗p) A(⃗q) + A†(−⃗q) 2 ] 1 ( )( ) + m2 A(⃗p) + A†(−⃗p) A(⃗q) + A†(−⃗q) ∫ 2 = d3q(2π)3 [ 1 ( )( ) × (−E2) A(−⃗q) − A†(⃗q) A(⃗q) − A†(−⃗q) 2 q ] 1 ( )( ) + (⃗q 2 + m2) A(−⃗q) + A†(⃗q) A(⃗q) + A†(−⃗q) 2 | {z } 2 ∫ Eq [ ] 3 3 2 − † − † = d q(2π) Eq A( ⃗q)A ( ⃗q) + A (⃗q)A(⃗q) ∫ [ ] 3 3 2 † † → − = d q(2π) Eq A(⃗q)A (⃗q) + A (⃗q)A(⃗q) (⃗q ⃗q in the 1st term) ∫ [ ] 3 3 2 1 † † = d q(2π) Eq 6 a(⃗q)a (⃗q) + a (⃗q)a(⃗q) (2π) 2Eq Here, note that the t-dependence of A(⃗q, t) cancels in H, and hence H is time independent. By using a(⃗q)a†(⃗q) = a†(⃗q)a(⃗q) + (2π)3δ(3)(0), we obtain ∫ ( ) 3 d q † 1 3 (3) H = 3 Eq a (⃗q)a(⃗q) + (2π) δ (0) (2π) :::::::::::::2

The constant term, ∫ 1 d3qE δ(3)(0) q 2

28 1 ℏ is the zero-point energy. (This corresponds to the 2 ω term in the energy spectrum of ℏ † 1 the harmonic oscillator, E = ω(a a + 2 ).) The zero-point energy cannot be observed (except through the gravitational force), so we neglect it in the following.

In fact, there is an ordering ambiguity to quantize the theory from a classical level. If we deﬁne the Hamiltonian by ∫ [ ] 1 1 1 H =: d3x ϕ˙2 + (∇⃗ ϕ)2 + m2ϕ2 :: aa† :=: a†a : normal ordering 2 2 2

then there is no zero-point energy. ∫ d3q In any case, we have H = E a†(⃗q)a(⃗q) (+ const.) (2π)3 q Therefore, ∫ d3q [ ] [H, a†(⃗p)] = E a†(⃗q) a(⃗q), a†(⃗p) (2π)3 q | {z } (2π)3δ(3)(⃗q−⃗p) † = Epa (⃗p) † similarly [H, a(⃗p)] = −Epa (⃗p) ■

§ 1.4.4 Consistency check Now that ϕ(x) and H are expressed in terms of a and a†, let’s do some consistency check.

(i) Heisenberg eq. iϕ˙(x) = [ϕ(x),H].1

iH(t−t0) −iH(t−t0) (ii) ϕ(x) is a Heisenberg operator: ϕ(x) = ϕ(t, ⃗x) = e ϕ(t0, ⃗x)e . Problems

(a) Show (i) by using (1) and (6).

† − † − † − (b) Show that eiHta(⃗p) e iHt = a(⃗p) eiEpt and eiHta(⃗p)e iHt = a(⃗p) e iEpt by using (6).

1(A typo here was corrected (May 27).)

29 § 1.4.5 vacuum state The operator a(⃗p) decreases the energy:

|X⟩ → a(⃗p) |X⟩ → a(⃗q)a(⃗p) |X⟩··· energy EX EX − Ep EX − Ep − Eq

The ground state (lowest energy) state |0⟩ is a state which satisﬁes

a(⃗p) |0⟩ = 0

and Lorentz invariant

U(Λ) |0⟩ = |0⟩ .

§ 1.4.6 One-particle state The one-particle state in § 0.6 is given by (for free theory) √ † |⃗p⟩ = 2Epa (⃗p) |0⟩ .

normalization √ √ † ⟨⃗q|⃗p⟩ = 2Eq 2Ep⟨0|a(⃗q)a (⃗p)|0⟩ √ √ ( ) ⟨ | † † | ⟩ = 2Eq 2Ep 0 |[a(⃗q){z, a (⃗p)]} +a (⃗p) |{z}a(⃗q) 0 (2π)3δ(3)(⃗p−⃗q) →0 3 (3) = (2π) 2Epδ (⃗p − ⃗q) ,

reproducing the normalization in § 0.6.

Lorentz transform: From the Lorentz transformation U(Λ)ϕ(x)U(Λ)−1 = ϕ(Λ−1x), we expect ⟩

U(Λ) |⃗p⟩ = p⃗′ .

where p′ = Λ−1p. (This may be opposite to ordinary convention.) Let’s show it. √ LHS = 2E U(Λ)a†(⃗p)U(Λ)−1U(Λ) |0⟩ √ p = 2E U(Λ)a†(⃗p)U(Λ)−1 |0⟩ √ p † ′ RHS = 2Ep′ a (p⃗ ) |0⟩

So it is suﬃcient to show √ E ′ U(Λ)a†(⃗p)U(Λ)−1 = p a†(p⃗′) ————(⋆). Ep

30 § 1.4.7 Lorentz transformation of a and a† Let’s show (⋆).

(i) First of all, for any f(⃗p), ∫ ∫ 1 3 4 2 − 2 d p f(⃗p) = d p δ(p m ) p0>0 f(⃗p) 2Ep

This is because

δ(p2 − m2) = δ((p0)2 − ⃗p2 − m2) √ √ δ(p0 − ⃗p2 + m2) δ(p0 + ⃗p2 + m2) = + | 0| | 0|  2p 2p ∑ −  δ(x xi) δ(f(x)) = ′ f (xi) x ;f(x )=0 ∫ ∫ i √i 0 − 2 2 0 2 2 0 δ(p ⃗p + m ) 1 ∴ − | 0 dp δ(p m ) p >0 = dp 0 = |2p | 2Ep

(ii) Therefore ∫ d3p ( ) ϕ(x) = √ a(⃗p)e−ip·x + a†(⃗p)eip·x 3 (2π) 2Ep ∫ √ ( ) 4 2 2 2Ep −ip·x † ip·x = d pδ(p − m )| 0 a(⃗p)e + a (⃗p)e p >0 (2π)3

and its 4-momentum FT is given by ∫ ϕe(k) ≡ d4xeik·xϕ(x) ∫ √ ( ) 4 2 2 2Ep 4 (4) † 4 (4) = d pδ(p − m )| 0 a(⃗p)(2π) δ (p − k) + a (⃗p)(2π) δ (p + k) p >0 (2π)3 √ ( ) 2 2 0 ⃗ 0 † ⃗ = (2π)δ(k − m ) 2Ek θ(k )a(k) + θ(−k )a (−k) .

31 (iii) and its LT is √ ( ) e −1 2 2 0 ⃗ −1 0 † ⃗ −1 U(Λ)ϕ(k)U(Λ) = (2π)δ(k − m ) 2Ek θ(k )U(Λ)a(k)U(Λ) + θ(−k )U(Λ)a (−k)U(Λ) . ∫ LHS = d4xeik·xU(Λ)ϕ(x)U(Λ)−1 ∫ = d4xeik·xϕ(Λ−1x) ∫ = d4yeik·(Λy)ϕ(y)(x = Λy, d4x = (det Λ)d4y = d4y) ∫ ′ = d4yei(Λk )·(Λy)ϕ(y)(k′ = Λ−1k) ∫ ′ = d4yeik ·yϕ(y)

= ϕe(k′) √ ( ) ′2 2 ′0 ⃗′ ′0 † ⃗′ = (2π)δ(k − m ) 2Ek′ θ(k )a(k ) + θ(−k )a (−k ) .

Using k′2 = k2 and θ(k′0) = θ(k0) and comparing it with RHS, we obtain √ E ′ U(Λ)a(⃗k)U(Λ)−1 = k a(k⃗′) ■ Ek § 1.4.8 [ϕ(x), ϕ(y)] for x0 ≠ y0 For x0 = y0 = t, we have [ϕ(x), ϕ(y)] = 0. What if x0 ≠ y0? From Eq.(1) and the commutation relations of a and a† in § 1.4.2, we have ∫ d3p ( ) −ip·(x−y) − +ip·(x−y) ≡ − [ϕ(x), ϕ(y)] = 3 e e i∆(x y) ∫ (2π) 2Ep d3p ( ) − −ip·x − +ip·x ∆(x) = ( i) 3 e e (2π) 2Ep Properties of ∆(x): (a) (□ + m2)∆(x) = 0. (b) Lorentz invariant: ∆(Λx) = ∆(x). (c) Local causality: ∆(x) = 0 for x2 = (x0)2 − ⃗x2 < 0 (space-like). t

∆(x − y) = 0 ⃗y ⃗x ⃗x

32 Among them, (a) is clear from the deﬁnition of ∆(x). (b) can be shown by using the equation in (i) of the previous section § 1.4.7: ∫ d4p ( ) − 2 − 2 0 −ip·x − ip·x ∆(x) = ( i) 4 δ(p m )θ(p ) e e ∫ (2π) 4 ( ) d p ′ ′ ′ − 2 − 2 0 −ip·x − ip·x ′ ∆(x ) = ( i) 4 δ(p m )θ(p ) e e (x = Λx) ∫ (2π) d4q ( ) − 2 − 2 0 −i(Λq)·(Λx) − i(Λq)·(Λx) 4 4 = ( i) 4 δ(q m )θ(q ) e e p = Λq, d p = d q ∫ (2π) d4q ( ) = (−i) δ(q2 − m2)θ(q0) e−iq·x − eiq·x p = Λq, d4p = d4q (2π)4 = ∆(x) .

Finally, (c) can be shown as follows. Fist, ∫ d3p ( ) 0 − i⃗p·⃗x − −i⃗p·⃗x ∆(x = 0, ⃗x) = ( i) 3 e e = 0 . (2π) 2Ep

On the other hand, for space-like x (x2 = (x0)2 − ⃗x2 < 0), one can always Lorentz transform it to a frame with x′0 = 0. 1 For a Lorentz boost in the opposite direction to ⃗x, x0 is transformed as x′0 = √ (x0 − β⃗ · ⃗x). 1 − β2 x0 Taking β⃗ = ⃗x, we have x′0 = 0. Note that this is impossible for a time-like x, ⃗x2 0 x where x2 = (x0)2 − ⃗x2 > 0, because ⃗x > 1 in that case. ⃗x2

Therefore, we have ∆(x) = ∆(x′0 = 0, x⃗′) = 0 for x2 < 0. —————— on May 8, up to here. —————— —————— on May 15, from here. ——————

33 § 1.5 Interacting Scalar Field Lagrangian ∫ ( ) 1 1 1 λ L = d3x ϕ˙2 − (∇⃗ ϕ)2 − m2ϕ2 − ϕ4 |2 2 {z 2 } | 24{z } same as free theory Interaction (λ : real and positive constant) ϕ(⃗x,t) ←→ π(⃗x,t) δL = = ϕ˙(⃗x,t) (same as free theory) δϕ˙(⃗x,t) Equal-Time Commutation Relation [ϕ(⃗x,t), π(⃗y, t)] = iδ(3)(⃗x − ⃗y) [ϕ(⃗x,t), ϕ(⃗y, t)] = 0 [π(⃗x,t), π(⃗y, t)] = 0 (same as free theory)

§ 1.5.1 What is ϕ(x)?

▼ In the case of free theory (λ = 0), ∫ 3 ( ) d p − · − · † ϕ(x) = √ e iEptei⃗p ⃗xa(⃗p) + eiEpte i⃗p ⃗xa (⃗p) 3 ::::: :::: (2π) 2Ep We could exactly solve the t-dependence by using Fourier transform and the Klein- Gordon eq. (□ + m2)ϕ = 0.

▼ With the interaction, ϕ(⃗x,t) =?? § ▼ The EOM is (see 1.3) ( ) λ □ + m2 ϕ(x) = − ϕ(x)3. 6 This is non-linear.

▼ Let’s try Fourier transform at t = 0. ∫ ( ) d3p ϕ(t = 0, ⃗x) = C(⃗p)ei⃗p·⃗x + C†(⃗p)e−i⃗p·⃗x (2π)3 | {z } from ϕ=ϕ† √ Deﬁning a(⃗p) by C(⃗p) = a(⃗p)/ 2Ep, ∫ 3 ( ) d p √1 i⃗p·⃗x † −i⃗p·⃗x ϕ(t = 0, ⃗x) = 3 a(⃗p)e + a (⃗p)e . ——————————(1) (2π) 2Ep Note that, at this stage, a(⃗p) and a(⃗p)† are just coeﬃcients of the Fourier transformation.

34 ▼ ▼ ▼ ▼ hsi h rbe b of (b) problem the is This theory? free of case the in happened What picture, Heisenberg the In oee,wt h neato term, interaction the with However, Similarly, → → → Thus, hc salna obnto of combination linear a is which ecntdsussatrnsjs ntrsof terms in just state. scatterings 1-particle discuss just creation/annihilation. can’t create/annihilate particle We to many) field (infinitely a includes as It considered be cannot It fe theory) (free ϕ ( x ϕ antb rte salna obnto of combination linear a as written be cannot ) e ( iHt ,⃗xt, a = ) † ( = → → → ⃗p ) H ∫ e e iHt − ϕ e [ define ϕ ,a H, = iHt iHt ( Thus, (2 then ( ,⃗xt, ϕ d fe hoy [ theory) (free ,⃗xt, H π 3 a (0 = p ( 0 ( ) loicue nntl many infinitely includes also ) ⃗p ⃗x, ⃗p 3 = ) + § ]= )] a ) √ f f f e ...A xml ouini sfollows: as is solution example An 1.4.4. † ) ˙ H ( − e ( ( ( 2 1 ∼ ⃗p ∫ t t t − iHt int E = ) = ) = ) ) iHt − e → ϕ = ( = p a iE (2 E 4 ( ( d ⃗p p e e e e π p e 3 ∼ ::::::::::::: nldsifiieymany infinitely includes t and ) − a iHt iHt iHt − p iHt Therefore, . ) e ( 3 iE ,a H, iE ( iHt ⃗p a √ 35 ( i a a p + ) [ − t ( ( p ,a H, + a f ⃗p 2 ⃗p ) 1 iE ( ( a f ) 0 = (0) ) E ⃗p a ⃗p e O e † ( ]= )] ) † p p − ( − t ( ) e ( ⃗p ) ( ) iHt ⃗p iHt 4 a a − ). )] a 3 ( iHt ( a , ⃗p ϕ e e e − ⃗p − ) − i⃗p ( ) e E 2 x iHt iE = e · − a ⃗x ). − p p † iHt + a a iE t a , a ( →§ ( p ⃗p ( e ⃗p :::::::::::::: ( t ⃗p ) iHt ) a e a ) e i⃗p † 1.5.2. . a − ) · a and 2 ⃗x iE and † a , + ( ( p ⃗p a and t ) a ■ † a e ) † a † − 3 ( . † ) ⃗p iHt . a ) † e . e iE − p i⃗p t e · ⃗x − ) i⃗p . · ⃗x ) , Comment Here, we have used [a, a†] etc, but a and a† are just Fourier coefficients in Eq. (1). What are these a and a†?  † • [a, a ] =?  • ϕ˙(t = 0, ⃗x) =? How is it written in terms of a and a†? • H =?  • [H, a] =?, [H, a†] =?

In fact, a(⃗p) is not uniquely determined by ∫ d3p 1 ( ) ←→ √ † − i⃗p·⃗x (1) ϕ(t = 0, ⃗x) = 3 a(⃗p) + a ( ⃗p) e . (2π) 2Ep

For any operator f(⃗p), replacing ( ) a(⃗p) → a(⃗p) + i f(⃗p) + f †(−⃗p)

does not change the above equation. We will deﬁne a(⃗p) more precisely later. (In “interaction picture”).

§ 1.5.2 In/out states and the LSZ reduction formula § ▼ We want to deﬁne the in/out states in 0.6. § ▼ In the free theory, one particle state is (see 1.4.6) √ † |p⟩ = 2Epa (⃗p) |0⟩ .

where (see § 1.4.1) ∫ ( ) † 1 3 −ip·x ˙ a (⃗p) = √ d x e −iϕ(x) + Epϕ(x) 2Ep ∫ − ←→ i 3 −ip·x = √ d x e ∂0 ϕ(x). 2E ( p ) ←→ ∂ f ∂ g ≡ f∂ g − (∂ f)g, ∂ = 0 0 0 0 ∂t

▼ We consider the same operator in the interacting theory. ∫ − ←→ † i 3 −ip·x a (⃗p,t) = √ d x e ∂0 ϕ(x), 2Ep ∂ which is now time-dependent.( (RHS)≠ 0 for λ ≠ 0.) ∂t 36 ▼ ▼ where by states in/out the deﬁne we And where show can one Then, Comments (ii) (i) S euto formula reduction LSZ Ti omn saddatrtelcue e h omn tteedo this of end the at comment the See lecture. the after subsection.) added is comment (This state vacuum the Here, [2]. take Peskin later then and wave-packets: with operators of think can One ( T time-ordering : T ( T ϕ ϕ ( ( x x ) 1 ϕ | ) ⃗q ϕ ( | 1 y ( ⃗p · · · x 1 )= )) 2 · · · ) ϕ ⃗q a a m ( † † ⃗p { x out ; ( ( n ⟨ ⃗p, ⃗p, 3 ⟨ × × in ; = ϕ ϕ ⃗p ) ( ( 1 −∞ + · · · ∏ y x i ∏ i · · · =1 0 =1 m n ) ⟩ ⟩ ∞ ) σ | ϕ ϕ ( T = ) = = [ ( [ ( lim = ) lim = ) ⃗p | → with x y i n 0 i √ √ ϕ for ) for ) ∫ in ; ⟩ ∫ ϕ ( stegon tt ftefl Hamiltonian full the of state ground the is 2 2 x x .Seeg,tetxbosb rdik 1 and/or [1] Srednicki by textbooks the e.g., See 0. x ( d 0 0 d E E x i →−∞ → | 4 ) 4 i ⃗q q p y x e 1 f a + · · · 1 1 1 i ) a † i a ( ∞ e · · · ϕ e ( ⃗p † † − t + y x 37 ( ( ϕ ( ) √ √ = ) ip x 0 iq 0 ⃗q ⃗p ( ∼ 1 ⃗q i i 1 − − i x m x > · 2 y > 2 2 , · , y x ) n + i E E out ; i i −∞ exp i ϕ ∫ ) ( ( ∞ p p ( □ ϕ 0 0 □ x d ∫ ∫ ( y ) i ( ) x 3 y 3 i ⟩ i · · · pf ) · · · 1 − + d d + ) · · · 3 3 · · · ( ( e x e x m √ √ ⃗p ⃗p m ) 2 − 4 2 2 a ϕ 2 ) − − σ ) E † E for ] ( ip ip ] ( ⃗p 2 y q ⃗p, t p · · 1 n m x x n ) ←→ ←→ a 2 ∂ ∂ a )) † ) ) 0 0 † x ( ( | ϕ ϕ i 0 0 ⃗q 1 ⃗p m ( ( n ⟩ x x x > , , ) ) + −∞ . , ∞ i 0 2 ) ) x > | | 0 0 ⟩ ⟩ i 0 3 > H · · · = H 0 + H int . ▼ ▼ ▼ hsi led ieodrd hs ecnpti h ieodrn prtrTwithout T operator time-ordering the in put can anything: we changing Thus, time-ordered. already is This states, in/out of deﬁnition the by First, it. show Let’s S-matrix the between relation the shows formula reduction LSZ The Next, n h ieodrdcreainfunction correlation time-ordered the and a ———————(1) formula) LSZ the of (LHS = † ( LSo h S formula) LSZ the of (LHS ⃗p, = √ + √ 2 E ∞ 2 p E ) 1 − · · · p 1 · · · a √ † ( ⃗p, √ 2 E −∞ 2 q E 1 ⟨ · · · q 1 = ) ⟨ · · · = = 0 | ∫ ∫ √ T 0 −∞ −∞ − | ∞ ∞ 2 ( a a E i ( t∂ dt ∂ dt ( ⃗q p 1 ⃗q                     1 , ∫ , + 0 0 + = = ∵ = = ( = d ∞ a ∞ [ 4 ∫ † ∫ ∫ ∫ ∫ :::::::::: ∂ x 38 ) √ ( − ) ⃗p, t · · · − d | ⃗p d d d d · · · 2 3 2 3 3 3 3 0 ⟨ E i = = = = xe ) ) xe q y xe 0 a a ( p ( | ∫ ( ( ( T − e e e e i⃗p ( i⃗p i⃗p e ∫ ⃗q −→ − − − − ∇ m − · ⃗q · · ⃗q d ⃗x m ⃗x ⃗x ip ip ip ip ip ϕ , 2 −→ 3 ∇ −→ 2 d ∇ ∫ , · · · · ) + ye {z ( · x x x x e 3 x + (2 x ::::::::::::::: ( ( ( ( i⃗p 2 e x 2 ∞ ←→ ∂ ∂ ∂ ∂ ∂ f i⃗p ) d ∞ · π ∫ ⃗y 0 0 0 0 0 2 2 2 2 3 · · · ( · ) ) ) ⃗y q ⃗x ϕ − ) a − − + + 3 f ( d a ) ip ( f δ † − ( ) x 3 † ( (3) E ⃗p · −→ ←− ( ⃗y ∇ ∇ x qe ( | ) ⃗p ⃗y ⃗q 2 ) 0 ←→ ⃗p p 1 ) 2 ∂ ( } 2 ) 2 2 1 ⟩ + , − ) ⃗p ) 0 , . ϕ −∞ + + i⃗q e ::::: ϕ −∞ − m ( − · ( ⃗x x m m i⃗q x 2 ⃗q ∫ ) · ) ) ) ) ⃗x 2 2 ) ϕ ] ) ) · · · ∫ ∫ · · · ϕ ϕ ( (2 x d ( ( ⟨ π x x a ) (2 3 (2 in a d y † d ) ) ) † ( π 3 3 π | 3 ( out y ⃗p y e ) ) n ⃗p 3 i⃗q 3 n , e e · , −∞ ⃗y ⟩ i⃗q i⃗q −∞ f (see · · ⃗y ( ⃗y f ⃗y f ) ) ( ( ) | ⃗y ⃗y 0 § ) ) ⟩ ) 0.6) | 0                     ⟩ and therefore ∫ −i a†(⃗p, +∞) − a†(⃗p, −∞) = √ d4x e−ip·x(□2 + m2)ϕ(x) 2Ep or ∫ i a†(⃗p, −∞) = a†(⃗p, +∞) + √ d4x e−ip·x(□2 + m2)ϕ(x) 2Ep

Similarly, one can show ∫ i a(⃗p, +∞) = a(⃗p, −∞) + √ d4x eip·x(□2 + m2)ϕ(x) 2Ep

▼ Substituting these equations to (1), (LHS of the LSZ formula) ( ) √ √ † † = 2Ep ··· 2Eq ···⟨0|T a(⃗q1, +∞) ··· a( ⃗qm, +∞)a (⃗p1, −∞) ··· a ( ⃗pn, −∞) |0⟩ 1 1 :::::::::: :::::::::::

∫ ∫ i † i a(⃗q1, −∞) + √ ··· a (⃗p1, +∞) + √ ··· 2Eq1 2Eq1

time-ordering time-ordering † a(⃗q1, −∞) |0⟩ = 0 ⟨0| a (⃗p1, +∞) = 0 ( ∫ ) ( ∫ ) · · ⟨ | 4 iq1 x1 □2 2 ··· 4 iqm xm □2 2 = 0 i d x1 e ( x1 + m )ϕ(x1) i d xm e ( xm + m )ϕ(xm) ( ∫ ) ( ∫ ) − · · × 4 ip1 y1 □2 2 ··· 4 ipn yn □2 2 | ⟩ i d y1 e ( y1 + m )ϕ(y1) i d yn e ( yn + m )ϕ(yn) 0 = (RHS of the LSZ formula) ■

—————— on May 15, up to here. —————— Questions and comments after the lecture: (only some of them) comment: There is a typo at § 1.4.4. The Heisenberg equation should be iϕ˙(x) = [ϕ(x),H] instead of iϕ˙(x) = [H, ϕ(x)].

A: Thanks! corrected.

Q: What happens to the LSZ formula, in particular the time-ordering, if you do a Lorentz transformation?

39 A: Good question. It is implicitly assumed that all the participating particles are causally − 2 0 0 connected, i.e., (xi yj) > 0 (time-like). In this case, the ordering xi > yj doesn’t change, the proof can be done in the same way, and therefore the LSZ formula in the Lorentz-boosted frame has the same form as in the original frame. ∫ − ←→ † i 3 −ip·x Q: Why a (⃗p, ∞) = lim √ d x e ∂0 ϕ(x)? 0→∞ x 2Ep

A: That’s the deﬁnition of the operator a†(⃗p, ∞).

—————— on May 29, from here. —————— Outline § 1.4 free quantization of ﬁeld interacting § 1.5

•⟨0|T [ϕ ··· ϕ]|0⟩ • LSZ § 1.5.2 We are here. • S-matrix, amplitude M

§ 0.6 Feynman rule • observables (σ and Γ)

Comments

(i) In the derivation of the LSZ formula, we have used

a(⃗p, ∞) |0⟩ = 0

where |0⟩ is the ground state (lowest energy state) of the full Hamiltonian H = H0+Hint. There is a subtlety here, but we will not discuss the details in this lecture. Under certain assumptions (axioms) on the quantum ﬁeld theory, such as “spectral conditions” (スペクトル条件), “asymptotic completion” (漸近的完全性), and “LSZ asymptotic condition”, one can show the above equation a(⃗p, ∞) |0⟩ = 0. For instance, the “asymptotic completeness” (漸近的完全性) says that, the Fock space spanned by the “in”-operators: { } Vin = |0⟩ , a†(⃗p, −∞) |0⟩ , a†(⃗p, −∞)a†(p⃗′, −∞) |0⟩ , ···

40 and that by the “out”-operators: { } Vout = |0⟩ , a†(⃗p, +∞) |0⟩ , a†(⃗p, +∞)a†(p⃗′, +∞) |0⟩ , ···

are the same as the Fock space spanned by the Heisenberg operator ϕ(x), V:

Vin = Vout = V.

(For more details, see e.g., Kugo-san’s textbook [5] 「ゲージ場の量子論 I」九後汰一郎、培風館 and Sakai-san’s textbook [6] 「場の量子論」坂井典佑、裳華房. They are in Japanese. I have checked several QFT textbooks in English, such as Peskin [2], Schrednicki [1], Weinberg [3], etc, but I couldn’t ﬁnd the corresponding explanation.)

(ii) In general, the operator deﬁned by ∫ − ←→ † i 3 −ip·x a (⃗p, ∞) = lim √ d xe ∂0 ϕ(x) 0→∞ x 2Ep

does√ not give the correct normalization for the 1-particle state. † ( 2Epa (⃗p, ∞) |0⟩ ∝ |⃗p⟩, but normalization is not correct in general.) One should either deﬁne the operator by ∫ − ←→ † 1 i 3 −ip·x a (⃗p, ∞) = √ lim √ d xe ∂0 ϕ(x) 0→∞ Z x 2Ep

(see e.g, Kugo-san’s and Sakai-san’s textbooks [5, 6]), or rescale the ﬁeld as √ ϕ(x) = Zϕr(x)(ϕr(x) : rescaled, or renormalized ﬁeld)

(see e.g., Srednicki’s textbook [1]) where

Z = |⟨p|ϕ(x)|0⟩|2

represents how much the state ϕ(x) |0⟩ contains the one-particle state |p⟩. (Note that ⟨p|ϕ(x)|0⟩ = eip·x and Z = 1 for the free theory.) In this lecture, we do not discuss the renormalization, and take Z = 1 as the leading order perturbation.

41 § 1.5.3 Heisenberg ﬁeld and Interaction picture ﬁeld

⟨0|T(ϕ(x1) ··· ϕ(xn))|0⟩ Now we want to calculate this.

§ 1.5.2, LSZ formula

⟨out|in⟩

§ 0.6

σ, Γ Idea: perturbative expansion in the coupling λ. There are two ways: • Here, we perform the perturbation in terms of the “interaction picture ﬁeld”.

• Another way is the “path integral” formalism.

The two ways give the same result for ⟨0|T(ϕ(x1) ··· ϕ(xn))|0⟩. ˙ ▼ Let’s start from ϕ(0, ⃗x) and ϕ(0, ⃗x) at t = 0. The Equal-time commutation relation at t = 0 is

[ϕ(0, ⃗x), ϕ˙(0, ⃗y)] = iδ(3)(⃗x − ⃗y). ———————⃝1

Deﬁne the Hamiltonian at t = 0. ∫ ( ( ) ) ∫ ( ) 1 1 2 1 λ H = d3x ϕ˙(0, ⃗x)2 + ∇⃗ ϕ(0, ⃗x) + m2ϕ(0, ⃗x)2 + d3x ϕ(0, ⃗x)4 | 2 2 {z 2 } | 24{z }

H0 Hint ˙ Note that H0 and Hint are deﬁned in terms of ϕ(0, ⃗x) and ϕ(0, ⃗x) at t = 0, and they are time-independent.

▼ For t ≠ 0, { Heisenberg ﬁeld ϕ(t, ⃗x) = eiHtϕ(0, ⃗x)e−iHt (evolved by H)

iH0t −iH0t Interaction piecture ﬁeld new! ϕI (t, ⃗x) = e ϕ(0, ⃗x)e (evolved by H0) ———————⃝2

▼ Properties of ϕ(x) and ϕI (x).

(i) From ⃝2 , Heisenberg equations are { ϕ˙ = i[H, ϕ], ϕ¨ = i[H, ϕ˙], ˙ ¨ ˙ ϕI = i[H0, ϕI ], ϕI = i[H0, ϕI ].

42 (ii) One can also show { ϕ˙(t, ⃗x) = eiHtϕ˙(0, ⃗x)e−iHt ———————⃝3 ˙ iH0t ˙ −iH0t ϕI (t, ⃗x) = e ϕI (0, ⃗x)e ( ∵ deﬁne A(t) = e−iHtϕ˙(t, ⃗x)eiHt ( ) then A˙(t) = e−iHt −i[H, ϕ˙] + ϕ¨ eiHt = 0 so A(t) = A(0) = ϕ˙(0, ⃗x). ■ )

ϕI is similar.

(ii)’ and ˙ ϕI (0, ⃗x) = i[H0, ϕI (0, ⃗x)]

= i[H − Hint, ϕ(0, ⃗x)] = i[H, ϕ(0, ⃗x)] = ϕ˙(0, ⃗x). ˙ ˙ ¨ ¨ Note that, ϕI (0, ⃗x) = ϕ(0, ⃗x) and ϕI (0, ⃗x) = ϕ(0, ⃗x) but ϕI (0, ⃗x) ≠ ϕ(0, ⃗x).

(iii) The equal-time commutation relations hold even at t ≠ 0, both for ϕ and ϕI . [ϕ(t, ⃗x), ϕ˙(t, ⃗y)] = eiHt[ϕ(0, ⃗x), ϕ˙(0, ⃗y)]e−iHt (∵ ⃝2 ⃝3 ) = eiHtiδ(3)(⃗x − ⃗y)e−iHt (∵ ⃝1 ) = iδ(3)(⃗x − ⃗y) − ˙ iH0t ˙ iH0t (3) [ϕI (t, ⃗x), ϕI (t, ⃗y)] = e [ϕ(0, ⃗x), ϕ(0, ⃗y)]e = iδ (⃗x − ⃗y) (iv) H can be written in terms of ϕ(t, ⃗x), and H0 can be written in terms of ϕI (t, ⃗x). H = eiHtHe−iHt ∫ ( ) 1 λ = eiHt d3x ϕ˙(0, ⃗x)2 + ··· + ϕ(0, ⃗x)4 e−iHt ∫ ( 2 24 ) 1 λ = d3x ϕ˙(t, ⃗x)2 + ··· + ϕ(t, ⃗x)4 (∵ ⃝2 ⃝3 ) 2 24 Each time in the RHS is time-dependent, but the sum is time-independent. Similarly,

iH0t −iH0t H0 = e H0e ∫ ( ( ) ) 2 1 1 1 − = eiH0t d3x ϕ˙ (0, ⃗x)2 + ∇⃗ ϕ (0, ⃗x) + m2ϕ (0, ⃗x)2 e iH0t 2 I 2 I 2 I ∫ ( ( ) ) 1 1 2 1 = d3x ϕ˙ (t, ⃗x)2 + ∇⃗ ϕ (t, ⃗x) + m2ϕ (t, ⃗x)2 (∵ ⃝2 ⃝3 ) 2 I 2 I 2 I

43 The RHS is (the sum is) time-independent. Namely, if written in if written in terms of ϕ(t, ⃗x) terms of ϕI (t, ⃗x) H0 wrong OK (t-independent) Hint wrong wrong H = H0 + Hint OK (t-independent) wrong (v) Equation of motion: From (i), ϕ¨(x) = i[H, ϕ˙(x)] ∫ [ ( ) ] 1 1 2 1 λ = i d3y ϕ˙(y)2 + ∇⃗ ϕ(y) + m2ϕ(y)2 + ϕ(y)4, ϕ˙(x) 2 2 2 24 From (iv), we can take x0 = y0. Then ˙ 2 ˙ 1st term: [ϕ(y) , ϕ(x)]x0=y0 = 0. ∫ [( ) ] 1 2 2nd term: i d3y ∇⃗ ϕ(y) , ϕ˙(x) 2 0 0 ∫ ( x =y ) 3 1 ⃗ ⃗ ˙ ⃗ ˙ ⃗ = i d y ∇ϕ(y) · ∇y[ϕ(y), ϕ(x)] + ∇y[ϕ(y), ϕ(x)] · ∇ϕ(y) 2 x0=y0 ∫

3 ⃗ ⃗ (3) = i d y ∇ϕ(y) · ∇yiδ (⃗x − ⃗y) 0 0 ∫ ( ) x =y

= i d3y −∇⃗ 2ϕ(y) iδ(3)(⃗x − ⃗y) (∵ integration by parts) x0=y0 = ∇⃗ 2ϕ(x). ∫ 3 1 2 2 3rd term: i d y m [ϕ(y) , ϕ˙(x)] 0 0 2 x =y = −m2ϕ(x). (∵ [ϕ2, ϕ˙] = ϕ[ϕ, ϕ˙] + [ϕ, ϕ˙]ϕ) λ 4th term: = − ϕ(x)3. 6 Thus, λ ϕ¨ = (∇⃗ 2 − m2)ϕ − ϕ3 6 λ (□ + m2)ϕ = − ϕ3 6 Similarly, ¨ ˙ ϕI = i[H0, ϕI ]

= ··· (write H0 in terms of ϕI ) ⃗ 2 2 = (∇ − m )ϕI . 2 ∴ (□ + m )ϕI = 0 ϕI is a free ﬁeld!

44 § 1.5.4 a and a† (again) □ 2 § ▼ ϕI satisﬁes ( + m )ϕI = 0 and therefore it can be solved exactly (see 1.4.1). ∫ 3 ( ) d p −ip·x † ip·x ϕI (x) = √ a(⃗p)e + a (⃗p)e 3 (2π) 2Ep 0 p =Ep

† where a(⃗p) and a (⃗p) are the expansion coeﬃcients of the interaction picture ﬁeld ϕI . † (We can also write them as aI and aI .) ˙ Thus, the original ϕ(0, ⃗x) and ϕ(0, ⃗x) as well as H0 and Hint can be expanded in terms of a and a†.  ϕ(0, ⃗x) = ϕ (0, ⃗x) = ···  I  ϕ˙(0, ⃗x) = ϕ˙ (0, ⃗x) = ··· I all written in terms of a, a†. H = ···  −−−−−−→ 0  substitute Hint = ···

˙ ▼ From the commutation relation of ϕI , ϕI , and H0, one can show the following relations (see § 1.4.2 and § 1.4.3)

[a(⃗p), a†(⃗q)] = (2π)3δ(3)(⃗p − ⃗q) [a, a] = 0 [a†, a†] = 0 † † [H0, a (⃗p)] = Epa (⃗p)

[H0, a(⃗p)] = −Epa(⃗p)

Note that the last two equations hold for H0, not H.

▼ The state annihilated by a(⃗p):

| ⟩ | ⟩ | ⟩ 0 I : a(⃗p) 0 I = 0,H0 0 I = 0 (H0 : normal ordered) is NOT the ground state of the full Hamiltonian:

| ⟩ | ⟩ ̸ H 0 I = (H0 + Hint) 0 I = 0 ∵ ∼ 4 ∼ † 4 Hint ϕI (a + a ) | ⟩ ̸ | ⟩ 0 I = 0

—————— on May 29, up to here. —————— —————— on June 5, from here. ——————

45 Outline § 1.4 free quantization of ﬁeld interacting § 1.5.3, § 1.5.4

•⟨0|T [ϕ ··· ϕ]|0⟩ § 1.5.5 we are here.

• LSZ § 1.5.2

• S-matrix, amplitude M

§ 0.6 Feynman rule • observables (σ and Γ)

§ 1.5.5 ⟨0|T(ϕ(x1) ··· ϕ(xn)) |0⟩ =? † We want to express it in terms of ϕI (a and a ). Step (i) ∼ (vii).

0 0 ··· 0 (i) redeﬁne the space-time points such that x1 > x2 > xn,

⟨0|T(ϕ(x1) ··· ϕ(xn)) |0⟩ = ⟨0|ϕ(x1) ··· ϕ(xn)|0⟩. —————— (1)

(ii) ϕ(x) =? { ϕ(x) = eiHtϕ(0, ⃗x)e−iHt

iH0t −iH0t ϕI (x) = e ϕ(0, ⃗x)e

iHt −iH0t iH0t −iHt → ϕ(x) = e e ϕI (x) |e {ze } ≡ u(t) † ϕ(x) = u (t)ϕI (x)u(t). —————— (2)

(iii) |0⟩ =?

iH0t −iHt I ⟨0| u(t) = I ⟨0| e e ⟨ | −iHt ∵ | ⟩ = I 0 e ( H0 0 I = 0) Insert an identity operator: ∑ 1 = |0⟩⟨0| + |n⟩⟨n| n=1

46 where |n⟩ represent the eigenstates of H with eigenvalues En > E0 = 0. (The summation includes continuous parameter (integral).) Then [ ] ∑ −iHt I ⟨0| u(t) = I ⟨0| |0⟩⟨0| + |n⟩⟨n| e n=1∑ −iHt −iHt = I ⟨0|0⟩⟨0| e + I ⟨0|n⟩⟨n| e ∑n=1 −iEnt = I ⟨0|0⟩⟨0| + I ⟨0|n⟩⟨n| e n=1 The 2nd term oscillates for t → ∞. Thus, for regularization, we take

t → ∞(1 − iϵ)(ϵ > 0, ϵ → 0) − − ∞ − − ∞· then, e iEnt → e iEn (1 iϵ) ∝ e En ϵ → 0

Therefore

lim I ⟨0| u(t) = I ⟨0|0⟩⟨0| . t→∞(1−iϵ) Similarly † − | ⟩ | ⟩⟨ | ⟩ lim u ( t) 0 I = 0 0 0 I . t→∞(1−iϵ) Thus ⟨ | ⟩⟨ |O| ⟩⟨ | ⟩ ⟨ |O| ⟩ I 0 0 0 0 0 0 I 0 0 = ⟨ | ⟩⟨ | ⟩⟨ | ⟩ I 0 0 |0{z0} 0 0 I 1 † I⟨0|u(t)Ou (−t)|0⟩I = lim † —————— (3) t→∞(1−iϵ) I⟨0|u(t)u (−t)|0⟩I

(iv) Substituting (2) (3) to (1),

⟨0|ϕ(x1) ··· ϕ(xn)|0⟩ 1 = lim † t→∞(1−iϵ) I⟨0|u(t)u (−t)|0⟩ × ⟨ | · † · † ··· ······ † · † − | ⟩ I 0 u|(t) {zu (t1}) ϕI (x1) u|(t1){zu (t2}) ϕI (x2) |u(t{z2) } | u{z(tn}) ϕI (xn) |u(tn) {zu ( t}) 0 I —————— (4)

(v)

† U(t1, t2) ≡ u(t1)u (t2)(t1 > t2) − − − = eiH0t1 e iH(t1 t2)e iH0t2 =?

47 It satisﬁes  U(t, t) = 0   ∂ U(t1, t2) = −iHI (t1)U(t1, t2) —————— (5) ∂t1  ∂  U(t1, t2) = iU(t1, t2)HI (t2) ∂t2 where

− H (t) ≡ eiH0tH e iH0t I ∫ int λ − = eiH0t d3x ϕ(0, ⃗x)4e iH0t ∫ 24 λ = d3x ϕ (x)4 24 I   d d iH t −iHt  check (5): u(t) = e 0 e   dt dt  iH0t −iHt  = e (iH0 − iH)e   − −   iH0t − iH0t iH0t iHt   = e ( iHint)e e e   −   = iHI (t)u(t)   d   u†(t) = ···  dt † = iu (t)HI (t)

The solution of (5) is, if HI (t) at diﬀerent t are commuting, ( ∫ ) t1 × U(t1, t2) = exp −i HI (t)dt , t2 but this is wrong. The correct solution is [ ( ∫ )] t1 U(t1, t2) = T exp −i HI (t)dt t [ ( 2 ∫ ) ] ∑∞ t n 1 1 = T −i H (t)dt —————— (6) n! I n=0 t2 Let’s check it. [ ( ∫ ) ] ∑∞ t n ∂ 1 ∂ 1 U(t , t ) = T −i H (t)dt ∂t 1 2 n! ∂t I 1 n=0  1 t2  ( ∫ ) ( ∫ ) ∑∞ ∑n t k−1 t n−k 1 1 1 = T  −i H (t)dt (−iH (t )) −i H (t)dt  n! I | {zI 1 } I n=0 k=1 t2 t2

48 Here, t1 of HI (t1) is larger than other t (t1 ≥ t ≥ t2), and therefore HI (t1) can be moved in front of the time-ordering: [ ( ∫ ) ] ∑∞ t n−1 ∂ 1 1 U(t , t ) = −iH (t ) T n · −i H (t)dt ∂t 1 2 I 1 n! I 1 n=1 t2

= −iHI (t1)U(t1, t2) ■

(vi) From (4),

I⟨0|U(t, t1)ϕI (x1)U(t1, t2)ϕI (x2) ··· ϕI (xn)U(tn, −t)|0⟩I ⟨0|ϕ(x1) ··· ϕ(xn)|0⟩ = lim t→∞(1−iϵ) I⟨0|U(t, −t)|0⟩I

With (6), everything is written in terms of ϕI (x). Furthermore, the numerator is time-ordered (t > t1 > t2 > ··· tn > −t), and hence it can be written as

I⟨0|T[ϕI (x1) ··· ϕI (xn)U(t, t1)U(t1, t2) ··· U(tn, −t)]|0⟩I ⟨0|ϕ(x1) ··· ϕ(xn)|0⟩ = lim t→∞(1−iϵ) I⟨0|U(t, −t)|0⟩I ⟨0|T[ϕ (x ) ··· ϕ (x )U(t, −t)]|0⟩ = lim I I 1 I n I —————— (7) t→∞(1−iϵ) I⟨0|U(t, −t)|0⟩I

where we have used U(t1, t2)U(t2, t3) = U(t1, t3). (vii) Substituting (6) to (7), we ﬁnally obtain [ ( ∫ )] t ′ ′ I⟨0|T ϕI (x1) ··· ϕI (xn) exp −i HI (t )dt |0⟩I ⟨ | ··· | ⟩ [ ( ∫ −t )] 0 ϕ(x1) ϕ(xn) 0 = lim t t→∞(1−iϵ) ′ ′ I⟨0|T exp −i HI (t )dt |0⟩I −t —————— (8) ∫ | ⟩ − Everything is written in terms of ϕI (x) and 0 I . By expanding exp( i HI ), we can do the perturbation expansion as O(1) + O(λ) + O(λ2) ··· .

§ 1.5.6 Wick’s theorem

▼ All the terms in the numerator and the denominator of Eq. (8) has as the following form:

I⟨0|T[ϕI (x1) ··· ϕI (xn)] |0⟩I . Deﬁne φ(x) as follows. ∫ ∫ 3 3 d p −ip·x d p † ip·x ϕI (x) = √ a(⃗p)e + √ a (⃗p)e . (2π)3 2E (2π)3 2E | {z p } | {z p } ≡ φ(x) ≡ φ†(x)

49 Then

† ϕI (x) = φ(x) + φ (x), | ⟩ φ(x) 0 I = 0, † I ⟨0| φ (x) = 0.

▼ Now we introduce “normal ordering”. Normal Ordering move φ† to the left, and φ to the right. [( )( )] N[ϕ (x )ϕ (x )] = N φ(x ) + φ†(x ) φ(x ) + φ†(x ) I 1 I 2 [ 1 1 2 2 ] † † † † = N φ(x1)φ(x2) + φ (x1)φ(x2) + φ(x1)φ (x2) + φ (x1)φ (x2) :::::::::::: † † † † = φ(x1)φ(x2) + φ (x1)φ(x2) + φ (x2)φ(x1) + φ (x1)φ (x2) ::::::::::::

(It can also be written as : ϕI (x1)ϕI (x2) : by using “:”.)

POINT I⟨0|N[ϕI (x1) ··· ϕI (xn)] |0⟩I = 0 .

▼ Now we want to see the relation between the time-ordering and the normal ordering.

? T[ϕI (x1) ··· ϕI (xn)] ⇐⇒ N[ϕI (x1) ··· ϕI (xn)] .

In the following, we write

ϕI (xi) = ϕi φ(xi) = φi

for simplicity. Let’s start from n = 2.

▼ n = 2

0 0 For x1 > x2, T(ϕ1ϕ2) = ϕ1ϕ2 † † = (φ1 + φ1)(φ2 + φ2) † † † † = φ1φ2 + φ1φ + φ φ2 + φ φ ::::2 1 1 2 † = N(ϕ ϕ ) + [φ , φ ] ∫ 1 2 1 2 ∫ 3 3 † d p d p 1 −ip1·x1 2 ip2·x2 † [φ1, φ ] = √ e √ e [a(⃗p1), a (⃗p2)] 2 (2π)3 2E (2π)3 2E ∫ p1 p2 d3p −ip·(x1−x2) = 3 e (2π) 2Ep

0 0 ↔ For x2 > x1, we have a similar formula with x1 x2. Therefore,

50 T(ϕ1ϕ2) = N(ϕ1ϕ2) + ϕ1ϕ2 ∫ { 3 −ip·(x1−x2) 0 0 d p e (x1 > x2) ϕ1ϕ2 = × 3 −ip·(x2−x1) 0 0 (2π) 2Ep e (x2 > x1) √ p0 = ⃗p2 + m2 not an operator, but c-number.

The symbol ϕ1ϕ2 is called “Wick contraction,” and it can also be written as ∫ 4 d p i − · − ϕ ϕ = e ip (x1 x2) 1 2 4 2 − 2 (2π) p m + iϵ (ϵ > 0, ϵ → 0). Feynman propagator √ p0 is not necessarily ⃗p2 + m2. It’s just an integration variable. Check ∫ ∫ ∫ 4 3 0 d p i − · − d p dp i − · − e ip (x1 x2) = e ip (x1 x2) 4 2 − 2 3 0 2 − 2 2 (2π) p m + iϵ (2π) 2π (p ) (|⃗p +{zm }) +iϵ 2 Ep

Here, for ϵ → 0,2 i 1 1 ∼ i · · 0 2 − 2 0 − − 0 − (p ) Ep + iϵ p (Ep iϵ) p + (Ep iϵ)

0 0 which has poles at p = Ep − iϵ and p = −Ep + iϵ. (See Fig. 1.) − 0 0− 0 0 0 ip (x1 x2) → 0 → − ∞ 0 For x1 > x2, e 0 for p i , so closing the contour at Imp < 0 (red line) ∫ I 3 0 d p dp 1 1 − · − i · · e ip (x1 x2) (2π)3 2π p0 − (E − iϵ) p0 + (E − iϵ) ∫ p p 3 d p i 1 − · − · − · ip (x1 x2) = 3 ( 2πi) e (2π) 2π Ep + Ep 0 ∫ p =Ep d3p 1 −ip·(x1−x2) = 3 e . (2π) 2E 0 p p =Ep

2 2 − ∼ − 2 Here, Ep iϵ (Ep iϵ/(2Ep)) and we renamed ϵ/(2Ep) as ϵ in the right hand side. The overall coeﬃcient of ϵ doesn’t matter as far as ϵ → 0.

51 Im p0

−Ep + iϵ × × Re

Ep − iϵ

Figure 1:

− 0 0− 0 0 0 ip (x1 x2) → 0 → ∞ 0 For x2 > x1, e 0 for p +i , so closing the contour at Imp > 0 (blue line) ∫ I 3 0 d p dp 1 1 − · − i · · e ip (x1 x2) (2π)3 2π p0 − (E − iϵ) p0 + (E − iϵ) ∫ p p 3 d p i 1 − · − = · · (+2πi)e ip (x1 x2) 3 − − (2π) 2π Ep Ep 0 − ∫ p = Ep d3p 1 −ip·(x2−x1) = 3 e , (2π) 2E 0 p p =+Ep where we have changed the integration variable ⃗p → −⃗p in the las line. Therefore ∫ ∫ { 4 3 −ip·(x1−x2) 0 0 d p i − · − d p e (x > x ) e ip (x1 x2) = × 1 2 ■ 4 2 − 2 3 −ip·(x2−x1) 0 0 (2π) p m + iϵ (2π) 2Ep e (x2 > x1)

—————— on June 5, up to here. —————— Questions and comments after the lecture: (only some of them) Q: Can you explain how the time-ordering operator acts in Eq.(6) in § 1.5.5?

A: OK. Let’s write Eq.(6) in § 1.5.5 again: [( ∫ ) ] ∑∞ t n 1 1 U(t , t ) = T −i H (t)dt 1 2 n! I n=0 t2

52 where I changed the order of the summation and the time-ordering. Now, consider the term for n = 2, for simplicity. It has the following form: [( ∫ ) ] [( ∫ )( ∫ )] t1 2 t1 t1 1 1 ′ ′ U(t1, t2)|n=2 = T −i HI (t)dt = T −i HI (t)dt −i HI (t )dt 2! t2 2! t2 t2 ∫ 3 4 ′ Note that HI (t) = d x(λ/24)ϕI (t, ⃗x) . The integration variable t and t take values ′ ′ ′ between t1 and t2. In the region where t > t , T[HI (t)HI (t )] = HI (t)HI (t ). But in ′ ′ ′ the region where t > t, T[HI (t)HI (t )] = HI (t )HI (t). (See the following ﬁgure). t′

t′ > t

t > t′

t t2 t1 Thus [ ∫ ∫ ∫ ∫ ] 1 t1 t t1 t1 U(t , t )| = T (−i)2 H (t)dt H (t′)dt′ + (−i)2 H (t)dt H (t′)dt′ 1 2 n=2 2! I I I I ∫ t2 ∫ t2 t2 t t1 t1 ( ) 1 2 ′ ′ ′ ′ ′ = (−i) dt dt θ(t − t )HI (t)HI (t ) + θ(t − t)HI (t )HI (t) 2! t2 t2 Now, the 2nd term is the same as the 1st term (t ↔ t′), and hence ∫ ∫ t1 t 2 ′ ′ U(t1, t2)|n=2 = (−i) dt dt HI (t)HI (t ) t2 t2 Similarly,

∫ ∫ ∫ ′ t1 t t 3 ′ ′′ ′ ′′ U(t1, t2)|n=3 = (−i) dt dt dt HI (t)HI (t )HI (t ) t2 t2 t2 ···

∫ ∫ ∫ ′ ∫ (n−1) t1 t t t n ′ ′′ (n) ′ ′′ (n) U(t1, t2)|n = (−i) dt dt dt ··· dt HI (t)HI (t )HI (t ) ··· HI (t ) t2 t2 t2 t2

53 Note that the factor 1/n! cancels the combinatorial factor of exchanging variables. Summing over all terms, we have the explicit form: [( ∫ ) ] ∑∞ t n ∑∞ 1 1 U(t , t ) = T −i H (t)dt = U(t , t )| . 1 2 n! I 1 2 n n=0 t2 n=0

Using the above explicit expression for U(t1, t2)|n, one can easily show that ∂ U(t1, t2)|n = −iHI (t1)U(t1, t2)|n−1 ∂t1 ∂ ∴ U(t1, t2) = −iHI (t1)U(t1, t2). ∂t1 comment: You should have explained what is ϵ when introducing the Feynman propagator. ϵ > 0 and ϵ → 0, right? (You didn’t write it explicitly).

A: You are right! Thanks! I added an explanation in the note, and I will comment on it in the next week.

· − comment: There is a typo in the formula of the Feynman propagator. e+ip (x1 x2) should be − · − e ip (x1 x2).

A: Thanks! corrected.

—————— on June 12, from here. ——————

Outline § 1.4 free quantization of ﬁeld interacting § 1.5.3 ∼ § 1.5.6 we are here.

•⟨0|T [ϕ ··· ϕ]|0⟩

• LSZ § 1.5.2

• S-matrix, amplitude M

§ 0.6 Feynman rule • observables (σ and Γ)

Comments on the lecture last week (corrected/added in the note on the web)

· − − · − (i) There was a typo in Feynman propagator. e+ip (x1 x2) → e ip (x1 x2).

54 (ii) In the formula of the Feynman propagator, the parameter ϵ is a small constant which represent the contour in the complex plane. ϵ > 0 and ϵ → 0.

0 0 0 (iii) In the case of x2 > x1, the contour should be closed at Imp > 0. ∫ (iv) The time-ordering T[exp( HI )] in Eq.(6) in § 1.5.5 can be expressed more explicitly. (See the lecture note on the web.)

▼ n = 3 0 0 0 For x3 > x1, x2,

T(ϕ1ϕ2ϕ3) = ϕ3T(ϕ1ϕ2)

= ϕ3N(ϕ1ϕ2) + ϕ3ϕ1ϕ2 † = φ N(ϕ ϕ ) + φ N(ϕ ϕ ) + ϕ ϕ ϕ {3 1 2 3 1 2 3 1 2

φ3N(ϕ1ϕ2) = N(ϕ1ϕ2φ3) + ϕ1ϕ3ϕ2 + ϕ1ϕ2ϕ3 † † φ3N(ϕ1ϕ2) = N(ϕ1ϕ2φ3)

= N(ϕ1ϕ2ϕ3) + ϕ1ϕ2ϕ3 + ϕ1ϕ2ϕ3 + ϕ1ϕ2ϕ3 . 0 0 0 0 0 0 (Similar for x1 > x2, x3 and x2 > x1, x3.)

▼ n = 4 ··· ··· T(ϕ1ϕ2ϕ3ϕ4) = N(ϕ1ϕ2ϕ3ϕ4) + |ϕ1ϕ2N(ϕ3ϕ4) + ϕ{z1ϕ3N(ϕ2ϕ4) + } + ϕ| 1ϕ2ϕ3{zϕ4 + } 6 terms 3 terms

▼ In general Wick’s theorem

T(ϕ1 ··· ϕn) = N(ϕ1 ··· ϕn) ∑ + ϕiϕj N(ϕ1··· ···ϕn) ˆi ˆj pairs ∑ + ϕiϕjϕkϕℓ N(ϕ1······ϕn) ˆi ˆj kˆ ℓˆ 2 pairs + ···   ∑  ϕ ϕ ··· ϕ ϕ (n = even)  i j p q n pairs 2 ∑ +   ϕ ϕ ··· ϕ ϕ ϕ (n = odd) .  i j p q r n−1 2 pairs (Problem. Prove it by induction.)

55 § 1.5.7 ▼ ▼ ▼ e’ aclt h rs eto o 2 for section cross the calculate Let’s Therefore, osdri h etro-asframe center-of-mass the in Consider From umr,Fymnrls examples rules, Feynman Summary, p 1 § p σ 0.6 4 ( p , I 1 ⟨ p , 0 | [ T 2 → ϕ I ϕϕ ( x 1 = ) ) · · · 2 ϕ E L I 1 θ ( = p p x · 2 1 n p 2 1 2 )] 3 E ∂ 1 | 2 µ 0 | v ϕ∂ ⟩ I 1 → − p = µ 56 2 ϕ v          cteigi the in scattering 2 − 2 | 0 n/ nlstate ﬁnal 2 1 ∫ | ∑ pairs 2 m φ {z d 2 Φ ϕ p p 2 } 2 ϕ 4 3 − i |M ϕ j 24 λ · · · ( p ϕ = E | ⃗p ⃗p 1 ⃗p 4 ϕ p , 3 1 1 1 . p √ = = | ϕ = 2 ϕ = q | → 4 − − ⃗p E | 1 theory, ⃗p ⃗p ⃗p 2 | ( ( p 4 2 2 2 n n = 3 | + p , = odd) = even) = E m 4 | ) 3 ⃗p 2 | 2 = 3 | identical particles |{z} = E × ﬁnal 4 . 2 1 | ⃗p 4 | Then

⃗p1 ⃗p2 ⃗p1 |v1 − v2| = − = 2 E E E ∫ ∫ 1 2 ∫ 1 d3p d3p ∑ |M|2 3 4 4 (4) − − |M|2 dΦ2 = 3 3 (2π) δ ( p1 + p2 p3 p4) (2π) 2E3 (2π) 2E4 = ··· ( In the center-of-mass frame, this is simpliﬁed as. . . [Problem: Show it.] ) = ··· ∫ (∫ ∫ ∫ ) 1 ⃗p dΩ 2π 1 = 3 |M|2. dΩ ≡ dφ d cos θ 8π E3 4π 0 −1

▼ Therefore, ∫ 1 1 dΩ → |M|2 σ(p1, p2 ϕϕ) = 2 ——————(1). 128π E1 4π

▼ On the other hand, M is given by

4 (4) ⟨p3, p4; out|p1, p2; in⟩ = (2π) δ (p1 + p2 − p3 − p4) · iM(p1, p2 → p3, p4) ——————(2).

and ⟨p3, p4; out|p1, p2; in⟩ is given by, from the LSZ formula § 1.5.2,

⟨p3, p4; out|p1, p2; in⟩ [ ∫ ] [ ∫ ] ∏ ( ) ∏n ( ) 4 −ipi·xi 2 4 +ipi·xi 2 = i d xie □i + m × i d xie □i + m |i=1,2 {z i=3,4 } ( We call) it “LSZ factor”

× ⟨0|T ϕ(x1)ϕ(x2)ϕ(x3)ϕ(x4) |0⟩

where, from § 1.5.5, [ ( ∫ )] λ ⟨0|T ϕ (x ) ··· ϕ (x ) exp −i ϕ (x)4 |0⟩ I I 1 I 4 24 I I ⟨0|ϕ(x ) ··· ϕ(x )|0⟩ = [ ( ∫ )] ——————(3). 1 4 λ ⟨0|T exp −i ϕ (x)4 |0⟩ I 24 I I

▼ Namely,

⟨p3, p4; out|p1, p2; in⟩ = (LSZ factor) × (3).

We can expand it with respect to λ.

57 ▼ ▼ ▼ ▼ O O ntecs of case the In general, In where o,(S factor) (LSZ Now, in theorem Wick’s from ¦ § ¦ § POINT POINT ( ( λ λ ( 0 0 emo 3’ ueao = numerator (3)’s of term ) = denominator (3)’s of term ) = 2 1 i ¥ ¤ ¥ ¤ 0 = = ∫ • • S atrcnesthe cancels factor LSZ ftoetra onsaedrcl connected, directly are points external two If ϕ LZfactor) (LSZ i = = ϕ d 1 . ∫ ϕ 4 ϕ 1 x 1 ϕ e i 2 d ϕ ∓ ∫ i ϕ 2 e 4 2 ip × 3 = − x ϕ i = d ip i D · y 4 e 4 D 2 x − F • • ∫ · x F ip i ( 4 3 2 § e x ( 2 ∓ ( (2 1.5.6 · x 1 × x □ d ip ————————————(4) 1 2 − π 4 i 2 | ( D − p · ) x − + 4 x i , F x ( | 2 p p (   m =? ) 2 □ 0! = 2 1 2 x ,both ), + + {z − 1 + i 2 I ⟨ = = ) D + − 0 m m I | i | ⟨ i F T( ∫ − m ∫ x 0 2 1 2 2 ( 58 | 2 • • iδ ) } x 2 0 + ϕ ) (2 d ) i e ⟩ ϕ x {z (4) 1 d 4 I − − D iϵ 1 1 ϕ π x 4 ϕ ip 1. = ( p 2 ) i F and e y x 2 1 e ϕ 4 − ( · ϕ atro h xenlline. external the of factor ) i − x 3 p x i ip 2 3 − ϕ ip ( 2 i ϕ · − ( 4 1 x − − x 4 · y ) x 2 p 1 | ) • • 1 − 0 2 y m r tteetra ie,so lines, external the at are 4 3 ( ⟩ x } ) + e I 2 □ 2 − ) ( . + m 1 ip {z ≡ + 1 ϕ 2 iϵ · . D x ) i 2 m x e = • F − i 2 ( ip + + ϕ ) x · I D 1 ( ( x x − x 1 • F − j 2 1 i ( )). x x x • • 2 0. = 2 1 ) ϕ ) − . 1   ϕ x 2 ϕ 2 } ) 3 ϕ 4 • • 4 3 ) ▼ ▼ ▼ O Thus, hs rm(4), from Thus, ( = + including terms = = ( λ I − emi 3’ numerator (3)’s in term ) ⟨ ϕ 0 iλ 1 4pis=15combinations) 105 = pairs (4 O | LZfactor) (LSZ ϕ T ( = ( = ) ( 2 2 1 2 1 2 1 ( λ ∫ ϕ • • • • • • 0 3 ϕ − − emi 3’ ueao 0. = numerator (3)’s in term ) ϕ y d 1 iλ iλ 4 4 • ϕ yD − 2 ) )(2 24 ϕ iλ • ∫ 3 F ϕ π ( ∫ y d 4 ) x × ( 4 4 ext. 1 − ye δ d • • • • • (4) − ( 4 i 3 4 3 4 O − ) yϕ • • ( y ip ∫ ( p 4 3 ) ext. 1 λ y 1 D · • ϕ y emi 3’ numerator) (3)’s in term ) d + e y ( F 4 − ϕ y • ( 4of 24 p ip y x 24 y 2 λ ϕ 2 2 − · y y − ϕ =0) (= e y p + y ϕ 4! 3 λ ip ) y − D 3 ϕ = · 59 y !=2 terms 24 = 4! ϕ ϕ y F p e ϕ 1 1 4 ( + 24 ϕ ϕ λ y ————————————(5) ) x ip ) 2 2 3 4 ϕ ϕ a nrdcdfrthis. for introduced was · − | y 3 3 0 ϕ ϕ ⟩ y 4 4 I ) · · D ϕ ϕ F y y ( ϕ ϕ x y y 4 ϕ ϕ − y y ϕ ϕ ϕ y y y i ) = . t,7 terms 72 etc, terms 9 etc, i oa 0 terms) 105 total (in ϕ I ( x i ) ) ϕ , y = ϕ I ( . y ) ▼ ▼ ▼ ▼ hs rm() eeetal banteapiuea h edn order, leading the at amplitude the obtain eventually we (2), from Thus, Since section cross the (1), to it substituting and of term leading the is (5) 2 xenlline external (2) with diagram (1) i 3 vertex (3) ena ue for rules Feynman M ⟨ out igas= diagrams = | ⟨ in p 3 ⟩ p , = 4 • out ; LZfactor) (LSZ σ ( | = p p 1 1 ext. i 3’ demominator (3)’s p p , − M p , p i M i iλ 2 ⟨ 2 ( in ; out p → . 1 p , ⟩ • × | • ϕϕ in ( = ext. 2 3’ numerator (3)’s ⟩ = ) → p . j − 1 = 2 = = 1. = p iλ 3 128 128 p , . . )(2 λ + 60 0 5 0. = 1 2 4 × × π π · · · π = ) E ) 10 10 · 1 4 E 1 δ 2 1 − − − (4) ∫ | 1 2 30 3 iλ . λ ( =1 {z (cont’d) λ p 2 = d 4 2 GeV 1 Ω π cm } + z O | |M| =0 ( }| 2 = p {z − λ + 2 λ ( 2 0 2 − { } + ) 2 O GeV ( E + 1 p ( GeV 1 λ 3 E O z =(5) 2 − 1 }| ) O ) ( λ 2 p ) ( { + ) . λ 4 2 + ) + ) O ( O · · · λ ( 2 λ + ) 2 ) · · · , ▼ ▼ ▼ ihrodrterms: order Higher ngnrl em ihlo igasaeotndvret n eurs“renormaliza- bubbles requires with and Terms divergent, often are diagrams loop with terms general, In ngeneral, In in.Hr,w utgv ulttv discussion. qualitative give just we Here, tion”. em ihlosa xenllines external at loops with terms + te loops other + bubbles with terms + with terms = = O ( λ I ⟨ 2 6pairs (6 0 emi 3’ numerator (3)’s in term ) | 3’ ueao = numerator (3)’s T ( ϕ 1 ϕ • → 2 ϕ ext. p 09 obntos)tr n()snumerator (3)’s in term combinations!) 10395 3 i ϕ 4 + ( r,tgte ihtelaigodrterm, order leading the with together are, − 2 i 2 1 ) ext. 2 ( p ∫ • j y • d | 4 ul connected fully y 2 1 • • 24 λ y ϕ {z • z y 4 • 2 1 + ∫ = 61 d · · · 4 3 4 4 3 z } ) • 24 λ t . etc • × • ϕ 2 1 4 3 • z 4 ( z ) z • | 0 + 1 × ⟩ etc I y • ( | etc + 1 • l ubediagrams bubble all 4 3 + etc → {z • • • 0 + ) · · · } ) On the other hand,

( [ ∫ ]) ( ) λ • (3)’s denominator = ⟨0|T exp −i ϕ4 |0⟩ = 1 + • + + ··· I 24 I I •

| {z } all bubble diagrams

Therefore, all bubble diagrams canceled out between numerator and denominator.

▼ Loops in the external line

• • + • • + • • + ··· (all these diagrams)

= •

iZ = . p2 − m2

The factor Z is absorbed by the ﬁeld renormalization. (In general, the mass ”m” here is also diﬀerent from the parameter ”m” in the Lagrangian.) We do not discuss the more details here. ▼

Feynman rules (cont’d)

(4) Ignore the bubble diagrams.

(5) We can also ignore the loops in the external lines if we take into account the renormalization.

▼ The other loops. 1 3 Example: • • y z 2 4

62 1 3 The • • term in ⟨out|in⟩ 2 4 1 3 = (LSZ factor) × the • • term in (3)’s numerator 2 4 ( ∫ ∫ ) 1 3 (−i)2 λ λ = (LSZ factor) × the • • term in ⟨0|T ϕ ϕ ϕ ϕ d4y ϕ4 d4z ϕ4 |0⟩ I 1 2 3 4 2 24 y 24 z I 2 y z 4 ( ) ∫ ∫ 1 −iλ 2 = (LSZ factor) × d4y d4z ϕ ϕ · ϕ ϕ ϕ ϕ · ϕ ϕ ϕ ϕ · ϕ ϕ 2 24 1 2 y y y y z z z z 3 4 × 12 (1, 2 ↔ y4 combinations) × 12 (3, 4 ↔ z4 combinations) × 2 (remaining y2 ↔ z2 combinations) × 2 (replacing y ↔ z) ∫ ∫ 1 = (LSZ factor) × (−iλ)2 d4y d4z D (x − y)D (x − y)D (x − z)D (x − z)D (y − z)2 2 F 1 F 2 F 3 F 4 F

—————— on June 12, up to here. —————— Questions after the lecture: (only some of them)

1 3 Q: What about a diagrams like • • ?? 2 y z 4

A: Good question! In fact, after renormalization of ﬁelds are take into account, this class of diagrams also corresponds to a diagram with ext. ext. , and its pi pj contribution to iM is zero. † † comment: There is a typo in the calculation of the normal ordering in §1.5.6. φ (x1)φ (x2) → † φ (x1)φ(x2). A: Thanks! corrected.

—————— on June 19, from here. ——————

63 ∓ipiy From (4), (LSZ factor) ×DF (xi − y) = e , and hence 1 3 The • • term in ⟨out|in⟩ 2 4 ∫ ∫ 1 2 4 4 2 = (LSZ factor) × (−iλ) d y d z DF (x1 − y)DF (x2 − y)DF (x3 − z)DF (x4 − z)DF (y − z) ∫ ∫2 1 2 4 4 −ip1·y −ip2·y +ip3·z +ip4·z 2 = (−iλ) d y d z e e e e DF (y − z) 2 ∫ ∫ 1 − · − · · · = (−iλ)2 d4y d4z e ip1 ye ip2 ye+ip3 ze+ip4 z 2 ∫ ∫ d4q i d4ℓ i × e−iq·(y−z) e−iℓ·(y−z) (2π)4 q2 − m2 + iϵ (2π)4 ℓ2 − m2 + iϵ Here ∫ p1 q p3  1 3  − ·  d4ye i(p1+p2+q+ℓ) y = (2π)4δ(4)(p + p + q + ℓ) ∫ 1 2  y • • z  4 +i(p3+p4+q+ℓ)·y 4 (4) d ze = (2π) δ (p3 + p4 + q + ℓ) 2 4 p2 ℓ p4 which represents the momentum conservation at each vertex. 1 3 Thus, the • • term in ⟨out|in⟩ 2 4 ∫ ∫ 1 d4q i d4ℓ i = (−iλ)2 2 (2π)4 q2 − m2 + iϵ (2π)4 ℓ2 − m2 + iϵ × (2π)4δ(4)(p + p + q + ℓ)(2π)4δ(4)(p + p + q + ℓ) ∫ 1 2 3 4 1 d4q i i = (−iλ)2 · (2π)4δ(4)(p + p − p − p ) 4 2 2 2 2 | 1 {z 2 3 4} 2 (2π) q − m + iϵ (−p1 − p2 − q) − m + iϵ the factor in (2) and ﬁnally, from (2), ∫ 1 3 1 d4q i i The • • term in iM = (−iλ)2 · . 2 (2π)4 q2 − m2 + iϵ (−p − p − q)2 − m2 + iϵ 2 4 1 2

p1 q p3 1 3

y • • z

2 4 p2 p4 −p1 − p2 − q

64 ▼ ▼ ▼ ieetdarm iedffrn em:frinstance, for terms: different give diagrams Different cancel: coefficients all and factor symmetry no is there loop, no is there if that, Note xml:2 Example: ena ue (cont’d) rules Feynman (9) (8) (7) (6) The (LSZ = (LSZ = ( = = utpyte“ymtyfco”(uha 1 as (such factor” “symmetry the Multiply by integrated be should momentum Loop line Internal vertex. each at conservation Momentum | · · · − × × × iλ 2 1 → ( 2 ( 4! ( 4! ) × × h orsodn emin term corresponding the 2 ) )( ( scattering. 4 y p z y 2 1 y • − 1 ↔ ( contraction) contraction) + iλ • z − z 24 ) p • ) 2 iλ 2 ∫ {z − ) d 2 p 6 5 4 3 p 4 6 ∫ i y ) 2 2 1 ∫ d emin term − 4 y d • m i × M 4 y • ∫ zD 2 = D − • z d F F 4 p i iϵ z 65 ( ( 2 M } x x ϕ − × 1 3 1 − − ϕ (2 m 6 5 4 3 i ( = 2 π 2 y z · ) ) ) + ϕ D D 4 δ y iϵ − F F ϕ (4) ∫ ( ( y iλ . / x x ( ϕ nteaoeexample). above the in 2 p 2 4 ) y (2 1 2 ϕ − − d ( + y π 4 p p y z ) · 1 p 4 ) ) ϕ + 2 D D . z − ϕ F F p z 2 ( ( p ϕ x x − 3 z 6 5 ϕ − − − p z 3 i p · ) y z 4 2 ϕ ) ) − D − 3 ϕ F p m 4 ( 5 ϕ y 2 5 − ϕ − − 6 p z iϵ 6 ) ) . . § 2 Fermion (spin 1/2) Field

Goal : To construct the Dirac Lagrangian

¯ µ L = ψ(iγ ∂µ − m)ψ,

solve the EOM (Dirac equation)

µ L = (iγ ∂µ − m)ψ = 0,

and quantize the Dirac ﬁeld.

§ 2.1 Representations of the Lorentz group

§ 2.1.1 Lorentz Transformation of coordinates (again) (see § 1.1.1)

µ ′µ µ ν ▼ x → x = Λ νx , where   1  −1  g Λµ Λν = g , g =   µν ρ σ νσ µν  −1  −1 µ ν ⇐⇒ Λ ρgµνΛ σ = gρσ ⇐⇒ ΛT gΛ = g.

▼ Exmaples

rotations around x, y, z axes       1 1 1    −    µ  1   cos θ2 sin θ2  cos θ3 sin θ3  Λ ν =   ,   ,   . cos θ1 sin θ1 1 − sin θ3 cos θ3 − sin θ1 cos θ1 sin θ2 cos θ2 1

boosts in the x, y, z directions       cosh η1 sinh η1 cosh η2 sinh η2 cosh η3 sinh η3       µ sinh η1 cosh η1   1   1  Λ ν =   ,   ,   . 1 sinh η2 cosh η2 1 1 1 sinh η3 cosh η3 eη + e−η cosh η = = γ 2 eη − e−η √ sinh η = = βγ = γ2 − 1 2 66 § 2.1.2 inﬁnitesimal Lorentz Transformation and generators of Lorentz group (in the 4-vector basis)

▼ Consider an inﬁnitesimal Lorentz Transformation:

µ µ µ µ Λ ν = δ ν + ω ν, (ω ν ≪ 1), or Λ = I + ω.

where I is the identity matrix. (I changed the notation from the blackboard.) Then, from ΛT gΛ = g,

(I + ω)T g(I + ω) = g ∴ ωT g + gω = 0 (up to O(ω2)) ∴ ρ ρ |ω µ{zgρν} + |gµρ{zω }ν = 0 ∥ ≡ ωµν ρ gνρω µ ∥ ωνµ ∴ ω = −ω anti-symmetric νµ µν  0 a b c 6 independent degrees −a 0 d e ω =   ⇕ µν −b −d 0 f 3 rotations and 3 boosts −c −e −f 0

µ µρ ▼ In fact, the matrix ω ν = g ωρν can be written as   0 η1 η2 η3  −  Note that µ η1 0 θ3 θ2 0 00 − ω ν =  −  ω ν = g ω0ν = ω0ν, ω0i = ηi = ωi0. η2 θ3 0 θ1 i ij ω ν = g ωjν = −ωiν, −ωij = ϵijkθk = ωji. η3 θ2 −θ1 0 ————————(1)

and the rotations and boosts in § 2.1.1 can be expanded as

rotation around x axis       1 1 0        1   1   0  O 2 Λ =   =   +   + (θ1) cos θ1 sin θ1 1 0 θ1 − sin θ1 cos θ1 1 −θ1 0 boost in the x direction       cosh η1 sinh η1 1 0 η1 sinh η cosh η   1  η 0  Λ =  1 1  =   +  1  + O(η2)  1   1   0  1 1 1 0

67 ▼ ▼ ▼ For where matrix The Problem representations general In generators The as written uniquely be can elements group Any generators the are matrices 6 These (            (cf. transformations.  discrete some to up ( K J 1 1 § x ( exp = Λ o xml,for example, For proof. the omit We ) ) θ ..,w ilsetesm omtto eain hold relations commutation same the see will we 2.1.3, µ µ i exp = ν ν η , = = i ≪ hwteaoecmuainrelations. commutation above the Show .             0 − ω 1, iθ 0 µ 0 i 1 ν 0 J [ [ [ J J J K − 1 n()cnas ewitnas written be also can (1) in 0 i 0 i i ) K , J , i 0 i i and K , − j θ j 0 = ] − 0 1 j θ = ] 0 = ] 1 0 ̸= K i fLrnzgroup. Lorentz of     0 = Λ iϵ i     iϵ θ 0 − , aif h olwn omtto relations commutation following the satisfy ijk θ , 1 ijk     , ( iϵ 2 J J ω I ( K ijk k 2 = 4 K = µ ) × k x ( exp = Λ ν µ J 2 4 θ ∑ n ν ) k = ∞ 3 + =0 µ = ν = i i n fteLrnzgopi h -etrbasis. 4-vector the in group Lorentz the of = 1 [     | ( θ η ! θ i i 68     0     i ( J 0 = J ( iθ 0 i − i § 0 ) − + {z i ω 0 µ i 2.1.A) J fLrnzgroup Lorentz of algebra Lie , i ν 0 η i + i 0 + K 0 − η iη i 0 } − i ) θ 0 ( 0 i 1 i + K i K     O i i θ ) ) 0 , 0 ( µ 1     θ     ( ν i J ] η , , 3 = ) ( i µ ) K o eeaosof generators for     ν 2 SO 3 = 1 ) µ     (1 ν 1 = 0 , 3) −     cos 0 i ) sin − 0 i − θ 0 θ 1 i 1 0 0 cos sin     0 , θ θ − 0 1 1 i                     . § 2.1.A Other (disconnected) Lorentz transformations

▼ The above discussion implicitly assumes that Λ is continuously connected to the identity element I by inﬁnitesimal Lorentz transformations (LTs). •

• I + ω2 •Λ = (1 + ωn) ··· (1 + ω2)(1 + ω1)I • I I + ω1 But there are also LTs which cannot be connected to I by inﬁnitesimal LTs.

µ ν ▼ From gµνΛ ρΛ σ = gνσ, (i) det g · (det Λ)2 = det g ∴ det Λ = 1. 0 0 i j (ii) g00 = g00Λ 0Λ 0 + gijΛ 0Λ 0 0 2 i 2 0 2 i 2 1 = (Λ 0) − (Λ 0) ∴ (Λ 0) = 1 + (Λ 0) ≥ 1.

▼ From (i) LTs are divided into two sets: { det Λ = +1 ; “proper” LTs det Λ = −1 ; “improper” LTs • Proper LTs form a subgroup of Lorentz group. (If det Λ1 = det Λ2 = 1, det(Λ1Λ2) = 1.) • Proper LTs and improper LTs are disconnected. (Inﬁnitesimal LTs cannot make det Λ = 1 → det Λ = −1.)

▼ From (ii) LTs are also divided as: { 0 Λ 0 ≥ 1 ; “orthochronous” LTs 0 Λ 0 ≤ −1 ; “anti-orthochronous” LTs • Orthochronous LTs form a subgroup. (Problem: Show it.) • Orthochronous LTs and anti-orthochronous LTs are disconnected. det Λ = +1   det Λ = −1   1 1  1   −1  Λ0 ≥ 1 connected to I =   connected to P =   0  1   −1  − ▼  1    1 −1 −1  −1   1  Λ0 ≤ −1 connected to PT =   connected to I =   0  −1   1  −1 1

0 ≥ ▼ In the following, we consider only proper-orthochronous (det Λ = +1 and Λ 0 1) LTs, which are connected to I. (They form a subgroup.)

69 § 2.1.3 Lorentz transformations of ﬁelds, and representations of Lorentz group.

▼ In § 1.2, we have shown the scalar action ∫ ∫ 4 S = dtL = d xL[ϕ(x), ∂µϕ(x)]

is invariant under the LTs of the scalar ﬁeld,

ϕ(x) → ϕ′(x) = ϕ(Λ−1x).

We’d like to generalize it as

→ ′ −1 ··· Φa(x) Φa(x) = Dab(Λ)Φb(Λ x) a, b = 1, N or Φ′(x′) = D(Λ)Φ(x) , x′ = Λx

× ▼ The matrix D(Λ) (N N matrix) must be a representation of the Lorentz group.

D(Λ2Λ1) = D(Λ2)D(Λ1)

(Proof:) For two successive LTs,

x −→ x′ −→ x′′ Λ1 Λ2 ′ ′ Φ (x ) = D(Λ1)Φ(x) ′′ ′′ ′ Φ (x ) = D(Λ2)Φ(x ) = D(Λ2)D(Λ1)Φ(x)

On the other hand,

′′ ′ x = Λ2x = Λ2(Λ1x) = (Λ2Λ1)x ′′ ′′ ∴Φ (x ) = D(Λ2Λ1)Φ(x)

Thus

D(Λ2Λ1) = D(Λ2)D(Λ1) ■

▼ What kind of representations does the Lorentz group have? (⇐⇒ What kind of fields (particles) are allowed in relativistic QFT?)  scalar field : D(Λ) = 1 (1 × 1 matrix) × × spinor field : D(Λ) =?? (2 2 or 4 4 matrix. (We construct it now!))  µ vector field : D(Λ) = Λ ν (4 × 4 matrix)

70 ▼ ▼ sentation. LT inﬁnitesimal an Consider ———o ue2,fo ee —————— week, here. Last from 26, June on —————— lecture: —————— the here. after to Questions up 19, June on —————— generators 6 The the where today hsw a expand can we Thus it. For parameters small 6 by parametrized Q: A: ω D is expression explicit its both Since as goes summation the Here, 1 factor the is What µν → ( I ,Λ= Λ 0, = + N ω or × = ) D                | N ( § § § § § ω N M I I .. Φ 2.1.3 2.1.A Λ group. 2.1.2 L. of Rep. 2.1.1 2.1 I {z µν N × arx[ matrix 4 + D µν × N × 4 and ab N D D ω aif h omtto eain fLrnzagba e’ show Let’s algebra. Lorentz of relations commutation the satisfy n rnfrain,and transformation), (no } ( ) ab + Λ as (Λ) I D = 4 Λ = (Λ) x / × M i ( µ I nteequation the in 2 M ( µ 4 ′ N |{z} ( ω ν µν ol oeo them) of some (only → + + x × I µν 01 = ′ N = ) ω ω r anti-symmetric, are M ] x D + ab δ = ) = ) ′ µ µ ab 01 2 i D ν r h eeaoso oet ru nti repre- this in group Lorentz of generators 6 the are Λ Λ = D ( ∑ + + µ,ν I µ I δ (Λ)Φ( ( ω = ) ν N ab I ω ω µν × = µ 71 ω + + 02 µ N ν µν ν x ≪ δ δ M ω 2 + i x ab ν µ = | M N = ) ω ) ν 02 (or 1 {z D 2 × µν i δ + µν N µ ω + } ( ν [ I ω µν I M + N ω ( + µ + N M θ O × 03 ν µν i i , N η , ω M ω ( [ µν × ] θ ω ab = ) 01 + i i M , 2 ( 03 N M ≪ J ) 2 i i + ) I matrix) 01 µ,ν µ 1). N ∑ ω ν 3 × =0 = 12 + N M µν ω ω η + µν i 10 12 ( = K 2 i M M + ω − i ) 10 µν µν ω µ M ν M . 23 t.Therefore, etc. ] νµ M µν 23 ? + ω 31 M 31 ) ▼ Now, LTs, successive three Consider oprn ohsides both Comparing Then, with 4 4 H f()= (1) of RHS H f()= (1) of LHS × × O RSo 2]= (2)] of [RHS LSo 2]=2 = (2)] of [LHS (1) O , O ( ω ( D 2 ω = = = ) , ) (Λ , O D D ( D I O O 3 + I 2 = 2 = 2 = = = = (¯ ( ( ( ( Λ ω (¯ I I ω ( − ω 2 2 ω ω i ω ω i 2 I i ω − ¯ + ¯ Λ eqs. ) eqs. ) i ω i i i µν µν (¯ − eq. ) 2 (¯ (¯ (¯ i 1 ω ω ω ωω ω ω µν µν = ) ωM ω M ¯ µν ω ) αγ αγ − − ρσ ω ω ¯ ¯ D )(      µν ω ¯ ρσ ρσ g ω − ω → → → ω D ( [ I ρσ γδ + Λ Λ Λ M γ ω I ¯ 2 ( 2 (( ω β (Λ ¯ + | − ρσ 3 2 1 ω ( − ¯ + O − trivial ocoe eain among relations closed no ω − µν 2 ¯ − i − δβ = = = g nismercunder anti-symmetric | g M ω 3 ( ) ( ω ω g µσ g M , ω σµ ) − αβ ω ω )( − µσ D I I I µσ ) ω ¯ ρσ ω ————————(2) {z ¯ M αγ ) D ω M I M − ¯ + + 2 (Λ M M ¯ + ω ρσ ¯ + ω ¯ ( ) − ( ω + ¯ νρ ρν αγ αβ I ω ω ω } ] 2 νρ ωω γ ¯ + νρ ωω 72 ω ) ¯ ω + − g β D I ρσ − ) ωω γδ − ) ) ) ω M g (Λ + − αβ g M ω ¯ νρ g ) νρ ) δβ αβ νρ M 1 ω αβ 2 ρσ i M M ————————(1) ) ) ωω ωM M ¯ ¯ M ω M αβ ω ¯ µσ µσ µν µν µσ {z ) αβ αβ + ) + M ω , + ) O = − g µν µν O νσ (¯ µ ( ,ω ω, 4 1 (¯ µ ≪ M ω ↔ ( M ) ωM ↔ µρ 2 1 2 ,ρ ν, ( ) ω , µν ν + . )( )) · ω ¯ g I ωM ¯ ↔ ) µρ + 2 M σ. 2 i − νσ ωM } ∵ ωM ¯ ) ω + µν · O ωM = ( − ω ) ) ω 2 νµ ) ) Comparing the ωµνω¯ρσ components in the both sides, we obtain

[M µν,M ρσ] = i (gµρM νσ − gνρM µσ − gµσM νρ + gνσM µρ) ————————(3) Lie algebra of Lorentz group

Deﬁning   1 D(J ) ≡ − ϵ M jk ⇐⇒ M ij = −ϵ D(J ) i 2 ijk ijk k ————————(4),  0i D(Ki) ≡ M [D(J ),D(J )] = iϵ D(J ) i j ijk k The same as (3) ⇐⇒ [D(J ),D(K )] = iϵ D(K ) ————(5) i j ijk k those in § 2.1.2 ! [D(Ki),D(Kj)] = −iϵijkD(Jk)

µν ▼ Recall that M ⇐⇒ D(Ji),D(Ki) are the generators deﬁned by

i µν D(I × + ω) = I × + ω M , 4 4 N N 2 µν

which represents inﬁnitesimal rotations and boosts. Using the notation ω0i = ηi and ωij = −ϵijkθk for the 6 small parameters (see § 2.1.2), the above eq. becomes

D(I4×4 + ω) = IN×N + i [θiD(Ji) + ηiD(Ki)]

which is the same form as the inﬁnitesimal LT of coordinates in § 2.1.2.

▼ Now, deﬁne

1 D(A ) = (D(J ) + iD(K )) i 2 i i 1 D(B ) = (D(J ) − iD(K )) i 2 i i

† † (Note that D(Bi) ≠ D(Ai) , because D(Ki) ≠ D(Ki) (see discussion later).) Then

[D(Ai),D(Aj)] = iϵijkD(Ak) (5) ⇐⇒ [D(Bi),D(Bj)] = iϵijkD(Bk) ————(6) [D(Ai),D(Bj)] = 0

This is the algebra of SU(2)×SU(2), and therefore we can classify the representations of Lorentz group by using representations of SU(2).

73 ▼ ▼ ▼ ▼ o ipiiy edenote we simplicity, For ? (6) satisfy which representation generic the What’s far: so discussion the summarize let’s ahead, going Before ersnaino “A-spin” of Representation and Define nt iesoa representation dimensional finite NOTE • • • • D ofar, So generators 6 the representing of ways equivalent Three LTs, infinitesimal For Φ fields: of LTs trigfo hs ecudso htgnrcrpeetto is representation generic that show could we this, from Starting ( eko hsfo M!! QM from this know We B i r o emta,btw a eiesmlrrsl assuming result similar derive can we but Hermitian, not are ) nQ,w aeue h atthat fact the used have we QM, In D spin- (Λ), aifig(3) satisfying M j ′ state where ( µν x ′ , = ) D A : ( m j . D M J D i [ 0 = i Λ = (Λ) D = ), = { | (Λ)Φ( µν ,m j, { ( ˆ ˆ j j D D | − A , 3 A A 2 e’ e this. see Let’s . 2 1 ( ( | j, ⇐⇒ ⇐⇒ i | 2 ⟩ ,m j, ) ,m j, A K [ , D , D x ˆ j − = 1 = i i i with ) ( ) , ), , ( j A 2 3 ˆ j ⟩ ( I ⟩ A 74 1 + A j D 4 , = 1 2 = D ] × 1 · · · j 2 + ( 4 ]= )] = {z j ( A m j , +1 J + N ( · · · A x i i iA iϵ j ), | ) (5) ′ ω ,m j, 2 2 D , × ijk 1) + Λ = ˆ iϵ j 2 j + = ) D i . ijk N ˆ j − ( ( r Hermitian, are A k ⟩ K B . x D 1 | I 3 2 matrix) ,m j, i i . , N j , ,aealgeneric all are ), ) ( } A × N k ⇐⇒ ⇐⇒ ⟩ ) + . 2 i : . ω D µν ( M A ˆ j i † i (6) ) µν = D , N ˆ j ( i B Here, . × i ) N matrices. D ( A i ) From (6), we can show

2 [A ,A3] = 0 ——— (i), 2 [A ,A] = 0 ——— (ii),

[A3,A] = A ——— (iii), 2 A = A3(A3 + 1) + A−A+

= A3(A3 − 1) + A+A− ——— (iv).

2 3 From (i), there exists a simultaneous eigenvector of A and A3;Φλ,µ,      

 2      A Φλ,µ = λ Φλ,µ              A3 Φλ,µ = µ Φλ,µ  N | {z } N×N Then, from (ii) and (iii), the vector

Φλ,µ1 ≡ AΦλ,µ satisfy { 2 A Φλ,µ1 = λΦλ,µ1

A3Φλ,µ1 = (µ 1)Φλ,µ1.

n Continuing further, the vector Φλ,µn = (A) Φλ,µ satisfy { 2 A Φλ,µn = λΦλ,µ1

A3Φλ,µn = (µ n)Φλ,µ1.

Now, assuming ﬁnite dimensional representation, there must be upper and lower bounds on A3’s eigenvalue { µmax = µ + n+,

µmin = µ + n−,

with {

A+Φλ,µmax = 0 ——— (v),

A−Φλ,µmin = 0 ——— (vi).

3 2 (At this stage, since A and A3 are not Hermitian, the eigenvalues λ and µ are not necessarily real numbers, but we will see they are real. I thank the student who pointed out this!)

75 From (iv) and (v), ( ) 2 A Φ = A (A + 1) +A− A Φ |{z} λ,µmax | 3 {z3 } |{z}+ λ,µmax →λ → →µmax(µmax+1) 0

∴ λ = µmax(µmax + 1) ——— (vii), Similarly from (iv) and (vi),

λ = µmin(µmin + 1) ——— (viii), From (vii)−(viii),

(µmax + µmin)(µmax − µmin + 1) = 0

Since µmax − µmin = n+ + n− ≡ n = non-negative integer (and therefore real), µmax − µmin + 1 > 0, and we have µmax = −µmin. Together with µmax − µmin = n, we thus have  n µ = = −µ  max 2 min  ( )  n n λ = + 1 2 2 We have obtained the irreducible representation of the “A-spin”. { 2 (A) (A) A Φa = A(A + 1) Φa (A) (A) A3Φa = a Φa where 1 3 A = 0, , 1, , ··· 2 3 − − ··· − a = | A, A + 1{z, A 1,A} (2A + 1) components

▼ Since we have two SU(2)s, D(Ai) and D(Bi), any irreducible representations of Lorentz group are parametrized by a set of two numbers: (A, B) A, B = integer or half-intetger

▼ The corresponding ﬁeld is (A,B) Φ{a,b (2A + 1)(2B + 1) components a = −A, −A + 1, ··· A − 1,A b = −B, −B + 1, ··· B − 1,B

76 transforming as  (A,B) (A,B) D(A2)Φ = A(A + 1)Φ  a,b a,b D(A )Φ(A,B) = aΦ(A,B) 3 a,b a,b ————(7)  2 (A,B) (A,B) D(B )Φa,b = B(B + 1)Φa,b  (A,B) (A,B) D(B3)Φa,b = bΦa,b ∑ ∑ 2 3 2 2 3 2 where D(A ) = i=1 D(Ai) , D(B ) = i=1 D(Bi) and we omitted the transformation of the argument x. ▼

(A, B) =(0, 0) scalar fields ( ) ( ) 1 1 (A, B) = 0, , , 0 spinor fields ( 2 ) 2 1 1 (A, B) = , vector fields 2 2

▼ Scalar ﬁelds

A = B = 0, a = b = 0 Φ(0,0) 1-components

D(Ai) = D(Bi) = 0

i µν ∴D(Λ) = I + ωµν |M{z} = I 2 =0 ∴Φ′(x′) = D(Λ)Φ(x) = Φ(x)

§ 2.1.4 Spinor Fields ( ) 1

▼ Consider ﬁelds with (A, B) = 0, . 2

(2A + 1)(2B + 1) = 1 × 2 = 2 components (0,1/2) − Φb , b = 1/2, 1/2.

µν Thus, D(Ai),D(Bi) ⇐⇒ D(Ji),D(Ki) ⇐⇒ M and D(Λ) are 2 × 2 matrices.

77 From (6) and (7),   D(Ai) = 0 (2 × 2),  1 ——— (8). D(Bi) = σi 2 ( ) ( ) ( ) 1 −i 1 σ : Pauli matrices σ = , σ = , σ = . i 1 1 2 i 3 −1  ( )  Φ1/2  ∵ For Φ = ,   (Φ−1/2 ) ( ) ( )   1/2 0 1 0 0  D(B )Φ = Φ,D(B )Φ = Φ,D(B−)Φ = Φ. 3 −1/2 + 0 0 1 0 Thus,   1 D(Ji) = D(Ai) + D(Bi) = σi (8) ⇐⇒ 2  1 D(Ki) = −iD(Ai) + iD(Bi) = i σi  2  1 M ij = − ϵ σ ⇐⇒ 2 ijk k  1 M 0i = i σ 2 i

(0,1/2) ▼ Denoting the 2-component ﬁeld Φb as ( ) 1 0, -ﬁeld: ξ (x) α = 1, 2 , 2 α its Lorentz transformation is given by ξ′ (x′) = D(Λ) βξ (x), α ( α β ) β β D(Λ)α = exp iθiD(Ji) + iηiD(Ki) ( ) α 1 1 β = exp iθi σi − ηi σi 2 2 α § ¤

▼ ¦Example¥ : • D(Λ) for a rotation around the z-axis ( ) ( ) ( ) i i θ 1 2 3 2 θ3 e D(Λ) = exp iθ3 σ3 = exp = i ——(i) − i − θ3 2 2 θ3 e 2 • D(Λ) for a boost in the z-direction ( ) ( ) ( ) 1 − 1 η 1 − 2 3 2 η3 e D(Λ) = exp −η3 σ3 = exp = 1 ——(ii) 1 η3 2 2 η3 e 2

78 Comment on the unitarity. 1 D(J ) = σ are Hermitian, D(J )† = D(J ), i 2 i i i 1 but D(K ) = i σ are anti-Hermitian, D(K )† = −D(K ). i 2 i i i Thus, the spinor representation of the Lorentz group D(Λ) is NOT unitary in general. (For instance, the rotation (i) is unitary, D(Λ)†D(Λ) = I, but the boost (ii) is not unitary, D(Λ)†D(Λ) ≠ I.) In general, there are no non-trivial ﬁnite dimensional unitary representation of the Lorentz group. Is this OK? → No problem, as far as Lorentz transformation of states are unitary: ⟨β|α⟩ → ⟨β|U(Λ)†U(Λ)|α⟩ = ⟨β|α⟩. have proper transformation.4

(Maybe more on this later, if we have time. . . )

—————— on June 26, up to here. —————— —————— on July 3, from here. —————— Last week,

• §2.1: Rep. of L. group. • §2.1.3: Φ′(x′) = D(Λ)Φ(x), (A,B) − ··· − − ··· − generic irreducible rep. Φ(x)a,b , a = A, A 1,A, b = B, B 1,B. • § (0,1/2) 2.1.4: Spinor ﬁelds, Φ(x)b (x), or ξα(x). 2-component ﬁeld. D(Λ) = exp(iθi(σi/2) − ηi(σi/2)). today → ( ) 1 (1/2,0)

▼ Similarly, for spinor ﬁelds with (A, B) = , 0 ,Φ , from (6) and (7), 2 a     1  1  1 D(J ) = σ M ij = − ϵ σ D(Ai) = σi ⇐⇒ i 2 i ⇐⇒ 2 ijk k  2  1  1 D(B ) = 0 (2 × 2) D(K ) = −i σ M 0i = −i σ i i 2 i 2 i

4(corrected after the lecture)

79 ▼ ▼ ▼ W mtteargument the notation omit the (We changed I that (sorry ( with ﬁelds spinor 2-component of (1 kinds and two are there summarize, To h nie f2cmoetsiosaeotndntdb notdaddte labels: dotted and undotted by denoted often are spinors 2-component of indices The indices spinor on Comment that Note ( ( hi nntsmltasomtosare transformations inﬁnitesimal Their npriua,tesio otato uhas such contraction spinor the particular, In oc o e sdt t,bti hslcue ed o s them. use not do we lecture, this in but it), to used get you (once tensors invariant with together ( 0 2 1 ,B A, , , 2 1 0 ) ) / ) 2 : : , ) n hi oet rnfrain r ie by given are transformations Lorentz their and 0), ψ ψ R L → → rm(9), From yusing by D D L R Thus, (Λ) (Λ)          ψ ←→ ψ L x R δψ δψ ϵψ ϵσ ϵδψ δψ δ exp = and exp = ( L R ϵψ i L ∗ L ∗ = = L = ∗ rnfrsi h aewyas way same the in transforms L = ∗ x = = ) ϵ − 1 2 ψ ψ 1 2 ′ ( αβ 2 1 ( ( ( L L ∗ Λ = σ 1 2 iθ iθ iθ ( ( iθ , i ∗ ∼ ∼ − ( ξ ψ 2 1 i ϵ i i − ϵ i L ( 2 1 iθ → α − + ˙ 2 1 x ( ( iθ iθ ) β where σ ˙ σ i α 80 n xeddPuimatrices Pauli extended and , 0 η η o simplicity.) for 1 2 i i i ψ − i , , i i − + − , ) + ) L 1 2 0 σ σ η .) ) ) η η η i i i η ψ ψ ) ψξ i i i i ) σ ) 2 1 L 2 1 R ( ϵ σ ϵσ i ∗ ψ σ σ i = ψ ≡ ( i R i i ∗ ) ϵψ L ) ∗ ψ ) iσ ψ ψ ψ α ˙ ψ L ∗ ψ L ∗ R R ∗ α 2 L — (9) ——— ) R ξ ∼ ∼ = α = = = ( ( ( ( ( 0 − 2 1 ψ I 1 0 , I , 0 1 α + 1 2 0 + ϵ ) ) ψ αβ iθ iθ . R ) ξ i i β 2 1 n(9) in 2 1 , σ r eyconvenient very are σ i i − + σ ,B A, . α η µ η β i ˙ i 2 1 1 2 sebelow). (see (0 = ) σ σ i i + + · · · · · · , 1 ) / ) 2) ψ ψ L R § 2.1.5 Lorentz transformations of spinor bilinears

▼ We have seen the Lorentz transformations of spinor ﬁelds ψL and ψR. In order to construct a Lorentz invariant Lagrangian, let’s consider Lorentz transformations of spinor bilinears, such as ( ) † ψ ψ ψ = (ψ∗ , ψ∗ ) R1 = ψ∗ ψ + ψ∗ ψ . L R L1 L2 ψ L1 R1 L2 R2 ( R2 ) † ∗ ∗ ψL1 ∗ − ∗ ψLσ3ψL = (ψL1, ψL2)σ3 = ψL1ψL1 ψL2ψL2. ψL2

In general, we can think of various combinations { } T T † † × × × { ∗ ∗ } ψL , ψR, ψL, ψR (2 2 matrix) ψL, ψR, ψL, ψR .

They can be classiﬁed according to SU(2)×SU(2). ( ) ∗ ··· 1 ψL, ψR 0, ( 2) 1 ψ , ψ∗ ··· , 0 R L 2 ( ) ( ) 1 1 § 0, ⊗ 0, 2 2 ( ) ( ) 1 1

▼ If there is only ψ ﬁeld, the possible bilinear terms transforming as 0, ⊗ 0, L 2 2 are

T · × · ψL (2 2 matrics) ψL.

T Among them, ψL ϵ ψL is Lorentz invariant.

∵ δ(ψT ϵ ψ ) = (δψT )ϵψ + ψT ϵ(δψ ) L L ( L L L )L ( ) 1 1 = (iθ − η )ψT σT ϵψ + ψT ϵ (iθ − η )σ ψ (∵ (9)) 2 k k L k L L 2 k k k L 1 = (iθ − η )ψT (σT ϵ + ϵσ )ψ 2 k k L | k {z }k L =0 = 0

† ∗ ▼ If there is only ψR ﬁeld, similarly, ψRϵ ψR is Lorentz invariant.

81 § ▼ Comments both are there If mn them, Among (iii) ( (ii) (i) 1 2 , 0 fw osdracagdfermion charged a consider we If and ψ that think might One to SU(2) of terms In Nurnsmyhv aoaams em(ab aoaafrin.Silun- Still fermion). Majorana (maybe term mass known.) Majorana have may (Neutrinos allowed. is term mass aihs oee,if ψ However, vanishes. ntefloigw osdracagdfrinadhneol h ia mass Dirac the only hence and fermion charged a consider term we following the In ∵ ) L 1 T ψ ψ ϵ δ ⊗ 2 ( ψ ψ ψ = ( L R † R † R † ψ 2 1 − and ψ ψ L ,     ψ L ψ L 0 ( = ) orsod oDrcms term mass Dirac to corresponds R † . ) 2 ψ while ψ symmetry) conserved (under ⇐⇒ charged is Φ Field ψ 0 = = ψ ψ 1 L L T L R † ( ψ ϵ ψ ϵ n hence and δψ sLrnzinvariant Lorentz is and ψ 2 1 arnini nain ne Φ under invariant is Lagrangian × L T ( L R ψ † R ∗ ϵψ − U2,teaoeterms, above the SU(2), ) R † sntivratunder invariant not is ψ em orsodt aoaams terms mass Majorana to correspond terms ψ iθ ψ L ( R L L k 0 ( = + ed ecnas think also can we field, ψ , + sivratunder invariant is 1 2 i ψ ψ ψ ψ η ) R † r nicmuig(si unie emo field), fermion quantized in (as anti-commuting are R † k L T 1 ) ψ , ( · ⊗ ϵψ ψ δψ (2 R † 2 ( L ) σ L × 0 osntvanish. not does ) ( sc seeto n oirn,ol h Dirac the only positron), and electron as (such k 82 , ) matrics) 2 − . 2 1 1 0 ψ ) 0 1 L (0 = + ( ) hsterm. this ψ ψ . ψ ψ R † | L L L T ψ ψ ( {z · → , ψ ϵ → 1 2 ψ 0) 1 2 ) } L ( e L e . iθ → = iα iα , k ψ ψ ⊕ ψ ψ e − R † L L 1 iα (0 ψ ϵ , ψ ψ , η Φ 2 , k . R 1) R ) ∗ − σ → and k ψ ψ 2 L e ψ , iα ) ψ 1 ψ R † R ψ ( . L ∵     correspond (9)) ▼ § ▼ r oet nain,crepnigto corresponding invariant, Lorentz are Similarly, h bv qaincnb rte as written be can equation above the Consider hr r needn obntos hc a etknas taken be can which combinations, independent 4 are There hytasomas transform They obnn them Combining hsi ohn u h rnfraino oet -etr See.1 of eq.(1) (See 4-vector! Lorentz of transformation Defining the but nothing is This eol osdrteDrcms term mass Dirac the consider only We ( 0 , δ 2 1 ( ) δ ψ ( R † ⊗ ψ σ R † ( i ψ ψ 2 1 R R , ( = ) ( = ) 0 = = = = δ ) σ     µ η · · · η · · · δψ δψ k i ( = ψ ψ ψ ( ( ψ ψ R R R † † † ψ ( R R † † R † R † σ σ σ R sn [ using † ,σ I, ) ) ψ ( ψ 1 2 3 ψ σ ψ ψ ψ ψ R R i R k 1 2 ψ R R R ψ i ψ + ) , = ) + R     R † 0 R δ ψ ) + σ ) ψ ( ψ = ϵ , ψ R i ( ψ R † ijk σ , ⊗ ψ L † R † ψ , ( R †     1 0 0 1 ψ R † δψ θ j ( · σ σ R 2 = ] k η η η (2 0 µ ( i 2 1 ψ , 3 2 1 R ψ ( ψ ) , δψ ) × R R † 0 − 83 ψ , R = ) † ) ψ ϵ η θ matrics) 2 R 0 ( ijk σ L θ R † T 2 1 j ) 3 (0 = ψ i ψ 0 1 1 0 ϵψ ψ σ ω R R R k hsterm. this R − ) . µ η θ . and ψ , ν 0 ) θ | 3 2 ( 1 ( ψ {z , i , R ( † 0) 1 = { − } σ L · † 0 i η θ σ 0 ψ ϵψ ν θ 1 3 i ψ 2 R σ , , −     L ∗ ⊕ R 2 . 0 j , ) i }     ) 3) (1 = ψ ψ ψ . , , ψ ( 0) R R R † † † σ R † σ σ σ 0 0 1 i ψ σ 1 3 2 ψ ψ ψ j R − R R R + 1     ) σ . j σ i 2 = § 2.1.2.) δ ij I ) ▼ ▼ Comments show can One defining Similarly, o ehv bandLrnzsaasadvcosfo pnrblnas ed to ready bilinears, spinor from vectors and scalars Lorentz obtained have we Now osrc h ia Lagrangian. Dirac the construct (iii) (ii) (i) ih( with as transform they transformation, Lorentz finite For 4-vector. Lorentz a is SU(2) of terms In etr u ed o osdrte.(hyaentivratunder invariant not are (They them. consider not do e we but vector, combinations, other The Similarly, where namely iα ψ R ′ ψ † L ( ψ , x J ′ i ) R ) σ µ µ ν → ψ n ( and R ′ e ( iα x ψ ′ ψ = ) K R × L Λ = ′ .) i † U2,teaoeagmn means argument above the SU(2), ) ( ψ µ x ν R † µ ′ )¯ ν ( ie in given σ D ψ x δ µ ( ) R † R ψ ψ ψ ( D Λ ( Λ exp = (Λ) D L L T ′ L † x 0 R µ ( ϵσ σ ¯ , ) R ν (Λ) x σ µ 2 1 (Λ) ′ σ µ ¯ x ( exp = ψ ν = ) ) § ψ ψ µ L † Λ = 2.1.2. R 84 σ R ⊗ † = ) ( = σ µ ( ψ and x µ D ( L † µ D ) ( I, 0 ω ν R ( , iθ x ψ , R iθ (Λ) µ − i ) 2 1 Λ Λ = (Λ) L ν † ψ J k σ | D ( ) ( i R 2 1 † ψ x i ψ + ) L σ σ ¯ L )¯ † = R (Λ) σ k µ σ ¯ iη ( ϵψ ν + ( x ν Λ ψ i ψ ( ) † K µ L 2 1 ∗ µ {z η L σ ¯ L ν ν lotasoma oet 4- Lorentz as transform also , , k ( i σ µ ) ¯ σ ) x 2 1 2 1 ν D ν ) σ ) . L k ∵ (Λ) ) ψ } R ′ ψ ( L x ( ′ x = ) ) D R (Λ) ψ R ψ ( L x )) → § 2.2 Free Dirac Field

§ 2.2.1 Lagrangian

▼ In § 2.1.5, we have seen

† † ψRψL, ψLψR

are Lorentz invariant. They can be the Lagrangian terms.

▼ On the other hand,

† µ † µ ψRσ ψR, ψLσ¯ ψL

are Lorentz vector. They can be combined with ∂µ to make the Lagrangian (more precisely, the action) Lorentz invariant. For instance, ∫ 4 † µ d x ψRσ ∂µψR

is Lorentz invariant, because ∫ ∫ 4 † µ → 4 ′ † µ ′ d x ψR(x)σ ∂µψR(x) d x ψR (x)σ ∂µψR(x) ∫ 4 ′ ′ † ′ µ ′ ′ ′ = d x ψR (x )σ ∂µψR(x ) ∂ (x′ = Λx, change of integration variable. ∂′ = ) ∫ µ ∂x′µ 4 ′ † † µ ′ = d x ψR (x)DR(Λ) σ ∂µDR(Λ)ψR(x) ∫ 4 ′ † µ ν ′ = d x ψR (x)Λ νσ ∂µψR(x) ( ) ρ ′ ∂ ∂x ∂ −1 ρ 4 ′ 4 ∂µ = ′µ = ′µ ρ = (Λ ) µ∂ρ, d x = d x ∫ ∂x ∂x ∂x 4 † µ ν −1 ρ = d x ψR (x)Λ νσ (Λ ) µ∂ρψR(x) ∫ 4 † ν = d x ψR (x)σ ∂νψR(x).

▼ We can also think other combinations with ∂µ to make Lorentz invariant terms, but

† µ – ∂µ(ψRσ ψR): total derivative and not a viable Lagrangian term. † µ † µ – (∂µψR)σ ψR: equivalent to ψRσ ∂µψR up to total derivative.

85 ▼ ▼ § of factor a added have we where field: Dirac free the of Lagrangian the obtain we all, them Combining Similarly, —————— week, here. Last from 10, —————— July here. on to —————— up 3, July on —————— and § Hermitian: sLrnzinvariant. Lorentz is where of terms in written be can Lagrangian above The . reDrcField Dirac Free 2.2 group. L. of Rep. 2.1 xr ls ( class extra § announcement .. Lagrangian 2.2.1 am arcs(ia matrices) (Dirac matrices gamma am matrices: gamma 補講 ia pnr Ψ Spinor: Dirac L L nJl 1 t ns h uniaino ia field) Dirac of quantization the finish (to 31. July on ) = ( = ( = iψ iψ Ψ( R † ψ σ R † R † iγ L µ σ ψ , ∂ µ µ = µ ∂ ∂ L † ψ µ µ ) Ψ( ψ R − [( Ψ γ ) R µ iγ † m ≡ ≡ + i ≡ = = ∫ σ µ ¯ i )Ψ ∂ Ψ ( µ 0 iψ iψ − ( ntedrvtv emt aeteLagrangian the make to term derivative the in d µ ∂ ψ † ψ 4 µ L σ † i ¯ γ − 86 R † ψ x R ( L σ ¯ µ 0 ∂ σ ) µ m µ iσ ( = µ ∂ ψ L † ∂ σ )Ψ µ µ 0 σ R ¯ µ µ ψ ∂ ψ ψ µ ) ) µ L † ∂ L † R ) σ µ ψ , − µ oa derivative total + ψ − ψ m R L † R ) ( ( orcmoetDrcspinor Dirac four-component ( ψ m R † I ψ L m I ) + ]( )] ( = ψ L † ψ ψ ψ ψ R R L R † ) ) ψ , . L † ) . ▼ ▼ satisfy The h oet rnfraino h -opnn ia edi ie by given is field Dirac 4-component the of transformation Lorentz The Problem: hr,from where, Comments: (iii) (ii) (i) γ o orvector four a for rep. above The ovnetidentity convenient A slash”: “Feynman notation a use sometimes We of (bases) representations ( are There γ µ .. ia rep. Dirac e.g., matrices            = S S ( hwit. Show 0 ij i σ ¯ D = = µ L/R ( ( σ − Λ in (Λ) 2 i µ σ ) 1 2 { i ϵ γ γ = ijk a µ µ − γ µ γ , = σ {( Ψ h ia arnini rte as written is Lagrangian Dirac The . 0 § 2 i k ν ′ = ..,tegenerators the 2.1.4, σ ( ( } x i I ) σ ¯ − ( 2 ′ 2 = = ) a / × µ I 1 2 a / 2 ≡ = ϵ g = = ijk σ ( ( exp − µν I µ 2 σ g γ ψ D ψ I ) × I µν k µ ) R L ′ L 4 ′ 2 L ) ( a × scle el(hrl rep. (chiral) Weyl called is ) ( ( a I (Λ) γ , µ x x 4 = 87 2 i γ , ′ ′ ) ω ) a / i ( ν µ Ψ( ) γ a µν = a = − ν ν D arcswihsatisfy which matrices i S σ ( ∂ = γ lffr ler n4d in algebra Clifford / = R µν 1 µ − bokdaoa:rdcberep.) reducible (block-diagonal: − (Λ) S a 1 2 a ) σ µ σ 2 µν { m i , I Ψ( 1 γ ( ) ) )Ψ µ r ie by given are σ x γ , , i . ) ψ )) ψ ( ν R L − } ( ( a σ x x µ 2 ) ) a ) ν σ 2 ) . , { ( γ − µ γ , σ 3 ν } σ 2 = 3 )} g µν I . They can be written as −i Sµν = [γµ, γν] 4 and satisfy [Sµν,Sρσ] = i (gµρSνσ − gνρSµσ − gµσSνρ + gνσSµρ) .

†

▼ Note that Ψ and Ψ transform as ( ) ′† ′ † i †µν Ψ (x ) = Ψ (x) exp − ωµνS . ( 2 ) ′ ′ † i †µν 0 Ψ (x ) = Ψ (x) exp − ωµνS γ (2 ) † 0 i µν †µν 0 0 µν = Ψ (x)γ exp − ωµνS (∵ S γ = γ S ) ( 2) i = Ψ(x) − ω Sµν 2 µν Thus, Ψ†Ψ is not Lorentz invariant (note that S†µν ≠ Sµν), while ΨΨ is Lorentz invariant.

§ 2.2.2 Dirac equation and its solution L − ▼ From the Lagrangian = Ψ(i∂/ m)Ψ, the EOM (Euler-Lagrange eq.) is ( ) δ L − δ L 0 = ∂µ † † δ(∂ Ψ ) δΨ [ µ a ] a − 0 / − = 0 γ (i∂ m) ab Ψb.

∴ (i∂/ − m)Ψ(x) = 0 Dirac equiation

Commnets ( ) δ δ (i) The other Euler-Lagrange eq. 0 = ∂µ L − L gives the same eq. δ(∂µΨa) δΨa (ii) In terms of 2-component spinors, it is ( )( ) µ −mI iσ ∂µ ψL µ = 0 iσ¯ ∂µ −mI ψR

(The mass term mixes left- and right-handed spinors. For massless fermion, ψL and ψR are diﬀerent particles.)

88 ▼ ▼ ▼ sin As Ψ( if all, of First it. solve Let’s q 1 aifistencsaycniin( condition necessary the satisfies (1) Eq. ia eq. Dirac eq. Klein-Gordon hssol estse o any for satisfied be should This rm( From ivreFT) (inverse ivreFT) (inverse ∫ ∴ ∴ Ψ d § Ψ e □ 3 ..,cnie ore rnfr fΨ( of transform Fourier consider 1.4.1, a p a ( ( + ⃗x, t ( ⃗p, t Ψ m e ¨ = = ) = ) a 2 ( )Ψ ∫ → ⃗p, t → ∫ u a d + ) ( a 3 x d ( p ( = 0 ( = 0 ⃗p 0, = ) 3 ( Ψ ) p e ¨ Ψ = e e u { a − a ( a ( ∫ ( iE u ( ( ( ⃗p, t i p / ⃗p, t p − / ⃗p ∂ a / p − ) ( d t − p / − e ⃗p + ) 3 ( = 0 ( ) + Ψ − ) p m − | m e ( = ( = m ip a ⃗p ( w − ) ( · m ( 2 E ) x ab ⃗x, t ) x iE t a p / ab ab Thus, . + − □ = ( E {z saslto fDrce. hni lostse the satisfies also it then eq., Dirac of solution a is ) p + ) 2 u p |{z} − ⃗p u ab Ψ p ∂ Ψ 2 e t ∂ b i / = ) ) m µ b + ( ≡ | w ∂ + v / a ∂ e ( b / ∂ m ⃗p b ( ( ⃗p + 2 µ a − ) m ( x ⃗p, t } v ) w I 0 = ) ( {z ) iE e 89 ⃗p ∫ ) − a ) ab − − 2 a 0 = ) m ( p e )Ψ ( iE 0 = ) u ⃗p t ⃗p m □ i⃗p ⃗p d . ) } ) ) b p ) ab · 3 ( 2 ⃗x a e t e + p ⃗p ) . ( + ( + + 0 = ac ) i Ψ ip e e x iE m ∂ / Ψ ( − · ihrsetto respect with ) a x p u − − 2 ip ( c t ) )Ψ a ⃗p, t ) γ · p ( x m e 0 0 ⃗p ( + a = i⃗p p ) ) ) ( ( E w , 0 e bc · p x ⃗x i⃗p p 0 − Ψ ,btntsffiin.From sufficient. not but 0, = ) − · a = ⃗x c ( p / γ ⃗p i − E -opnn spinor) 4-component : ) ( − ——(1) ———— p m ) p , i ) ) ab ⃗x − v , b m ( ⃗p ) ) ab e + v ip b ( · x − ) ⃗p p 0 ) = e E + p iE . p t i.e., u(⃗p) and v(⃗p) are eigenvectors of p/ with eigenvalues m and −m respectively.             p/ u(⃗p) = m u(⃗p) ,             p/ v(⃗p) = −m v(⃗p) . ———— (2)

▼ In fact, the eigenvalues of the matrix p/ are det(p/ − xI) = ··· = (x2 − m2)2 → x = m, m, −m, −m, corresponding to two independent u(⃗p) and two independent v(⃗p), satisfying (2).

▼ We can also think the “helicity” (= projection of the spin onto the direction of momentum): ( ) 1 ⃗p · ⃗σ h(p) = . 2|⃗p| ⃗p · ⃗σ whose eigenvalues are 1/2. Since [p,/ h(p)] = 0, simultaneous eigenvectors of p/ and h(p) can be taken:  u (p) u−(p) v (p) v−(p)  + + pu/ +(p) = mu+(p) p/ m m −m −m e.g., 1 h(p)u (p) = u (p) h(p) 1/2 −1/2 −1/2 1/2 + 2 +

▼ The explicit form of u(⃗p) and v(⃗p) can be written as (√ ) ( √ ) 0 0 ∗ p ∓ |⃗p| η p |⃗p| ϵη √ √ u (p) = 0 , v (p) = 0 ∗ , p |⃗p| η − p ∓ |⃗p| ϵη with  ( ) ( )  1 pˆ1 − ipˆ2 cos θ e−iϕ   √ 2 1 1 η+ = = pˆ = p /|⃗p| = sin θ cos ϕ 2(1 − pˆ3) 1 − pˆ3 sin θ ( ) ( 2 ) pˆ2 = p2/|⃗p| = sin θ sin ϕ  − 3 θ   1 1 pˆ sin 2 3 3 η− = √ = pˆ = p /|⃗p| = cos θ − 3 − 1 − 2 − θ +iϕ 2(1 pˆ ) pˆ ipˆ cos 2 e † · | | ′ ′ satisfying (⃗p ⃗σ)η = ⃗p η, ηsηs = δss They are normalized as  u¯s(p)us′ (p) = 2mδss′ v¯ (p)v ′ (p) = −2mδ ′ ,  s s ss u¯s(p)vs′ (p) = 0

90 and satisfy ∑ us(p)¯us(p) = p/ + m s∑= vs(p)¯vs(p) = p/ − m s=

▼ To summarize, there are four independent solutions to the Dirac equation, ∫ ( ) 3 −ip·x −ip·x +ip·x +ip·x Ψ(x) = d p u+(p)e , u−(p)e , v+(p)e , v−(p)e 0 . p =Ep

§ 2.2.3 Quantization of Dirac ﬁeld

L = Ψ(i∂/ − m)Ψ.

conjugate δL Ψ ←→ Π = = (Ψiγ0) = iΨ† (= iΨ∗) a Ψa ˙ a a a δΨa

Comments ←→ ∗ (i) Ψa ΠΨa = iΨa ∗ δL but then, Ψ ←→ ? (Π ∗ = = 0 ???) a Ψ a ˙ ∗ δΨa One should do the quantization of constrained sysytem with “Dirac bracket”. ( ) ∂2L(q, q˙ In general, if det = 0, pi and qi are not independent. ∂q˙i∂q˙j In such a case, Poisson bracket quantization

Dirac bracket

Here, we skip the details and do naive quantization with Ψa and ΠΨa.

(ii) When Ψa and ΠΨa are anti-commuting, right-derivative and left-derivative gives opposite sign. Here, ΠΨa is deﬁned with←− right-derivative.−→ ∂ ∂ (If A and B are anti-commuting, (BA) = B, while (BA) = −B. ∂A ∂A Quantization with Commutation relation vs Anti-commutation relation

←→ † Ψa ΠΨa = iΨa

91 Quantization with equal-time commutation relation

† (3) − [Ψa(x), ΠΨb(y)]x0=y0 = [Ψa(x), iΨb(y)]x0=y0 = iδ (⃗x ⃗y)δab does NOT work. Instead, quantization with equal-time anti-commutation relation

{ } { † } (3) − Ψa(x), ΠΨb(y) x0=y0 = Ψa(x), iΨb(y) x0=y0 = iδ (⃗x ⃗y)δab works. Let’s see it.

▼ First of all, expand the Ψ(x) with the solutions of Dirac eq.

−ip·x +ip·x u(p)e , v(p)e

as ∫ 3 ∑ ( ) d√p −ip·x +ip·x Ψa(x) = as(⃗p)us,a(p)e + ds(⃗p)vs,a(p)e 0 . 3 p =Ep (2π) 2Ep ∫ s= 3 ∑ ( ) † d√p † † +ip·x † † −ip·x Ψ (x) = a (⃗p)u (p)e + d (⃗p)v (p)e 0 . a 3 s s,a s s,a p =Ep (2π) 2Ep s=

Here, Ψ(x), as(⃗p), and ds(⃗p) are the quantum operators. At this moment as(⃗p) and ds(⃗p) are just expansion coeﬃcients.

▼ The following can be shown: { † (3) [Ψ (x), Ψ (y)] 0 0 = δ (⃗x − ⃗y)δ a b x =y ab ———— (1) others = 0

  † 3 (3) − [ar(⃗p), as(⃗q)] = (2π) δ (⃗p ⃗q)δrs [d (⃗p), d†(⃗q)] = (2π)3δ(3)(⃗p − ⃗q)δ ———— (1)’ (problematic)  r s rs others = 0 { † (3) {Ψ (x), Ψ (y)} 0 0 = δ (⃗x − ⃗y)δ a b x =y ab ———— (2) others = 0

 { † } 3 (3) −  ar(⃗p), as(⃗q) = (2π) δ (⃗p ⃗q)δrs {d (⃗p), d†(⃗q)} = (2π)3δ(3)(⃗p − ⃗q)δ ———— (2)’ (OK)  r s rs others = 0

92 —————— on July 10, up to here. —————— (In order to finish the quantization of Dirac field, I will have an extra class (補講) on July 31. It’s after the exam and the reports, and irrelevant to the grades. . . ) —————— on July 31, from here. —————— ⇒ ⇒ ▼ Let’s first show (1)’ = (1) and (2)’ = (2). Hereafter, we use a notation { [A, B] = AB − BA [A, B} = {A, B} = AB + BA

to discuss the two cases simultaneously. } } } } ▼ First of all, from (1)’ (2)’, [a, a = [d, d = [a, d = 0, and therefore [Ψ, Ψ = 0. Similarly, [Ψ†, Ψ†} = 0. †} ▼ The remaining is [Ψ, Ψ , and ∫ ∫ 3 3 † d p d q [Ψa(t, ⃗x), Ψ (t, ⃗y)} = √ √ b (2π)3 2E (2π)3 2E ∑ ∑ ( p q × † } † −ip·x iq·y [ar(⃗p), as(⃗q) ua,r(p)ub,s(q)e e r= s= ) † } † ip·x −iq·y +[dr(⃗p), ds(⃗q) va,r(p)vb,s(q)e e ( x0=)y0=t ∫ ∑ using (1)’, and performing d3p and ∫ r= 3 ∑ ( ) d q 1 † i⃗q·(⃗x−⃗y) † −i⃗q·(⃗x−⃗y) = 3 ua,s(q)ub,s(q)e + va,s(q)vb,s(q)e (2π) 2Eq ( s= ) ∑ † u (q)u (q) = [(q + m)γ0] § ∑s a,s b,s / ab from 2.2.2, † 0 va,s(q)v (q) = [(/q − m)γ ]ab ∫ s b,s 3 ( ) d q 1 i⃗q·(⃗x−⃗y) 0 −i⃗q·(⃗x−⃗y) 0 = e [(/q + m)γ ]ab + e [(/q − m)γ ]ab (2π)3 2E ∫ q d3q 1 = ei⃗q·(⃗x−⃗y)[(γ0q − γiq + m + γ0q − γi(−q ) − m)γ0] (2π)3 2E 0 i 0 i ab ∫ q d3q 1 i⃗q·(⃗x−⃗y) 0 · = 3 e 2q δab (2π) 2Eq (3) = δ (⃗x − ⃗y)δab ■

⇒ ⇒ ▼ Thus, (1)’ = (1) and (2)’ = (2) are shown. ⇒ ⇒ ▼ The other way round, (1) = (1)’ and (2) = (2)’ can be shown, by the following steps:

93 (i) By using the explicit forms of u(p) and v(p) in § 2.2.2, show that  † ur(⃗p)us(⃗p) = 2Epδrs  v†(⃗p)v (⃗p) = 2E δ r s p rs —————— (3) u†(⃗p)v (−⃗p) = 0  r s † − vr(⃗p)us( ⃗p) = 0

(ii) Using (3), show that  ∫  1 ∑  √ † 3 ip·x as(⃗p) = us,a(⃗p) d xe Ψa(x) 2Ep 0 a p =Ep ∫ —————— (4).  1 ∑  √ † 3 −ip·x ds(⃗p) = vs,a(⃗p) d xe Ψa(x) 2Ep 0 a p =Ep

(One can also show that the RHS is independent of x0.) (iii) Using (3) (4), show (1) =⇒ (1)’ and (2) =⇒ (2)’.

▼ So far we have shown (1) ⇐⇒ (1)’ and (2) ⇐⇒ (2)’.

▼ On the other hand, the Hamiltonian is given by ∫ ( ) 3 ˙ H = d x ΠΨΨ − L ←− ∂ (note that Π is deﬁned with right-derivative, and hence H = 0.) Ψ ˙ ∫ ( ) ∂Ψ 3 † ˙ − / − = d x iΨ Ψ Ψ(|i∂ {zm)Ψ} ∫ =0 = d3x iΨ†Ψ˙

= ··· (using (3)) ∫ ( ) d3p ∑ = E a†(⃗p)a (⃗p) − d†(⃗p)d (⃗p) —————— (5). (2π)3 p s s s s s=

Note that

(i) this is the case both for quantization with [•, •], (1)⇐⇒(1)’ and that with {•, •}, (2)⇐⇒(2)’, (ii) and there is a minus sign in front of d†d.

94 ▼ ▼ ▼ u o ecnecag h oe fceto n niiainoperator. annihilation and creation of roles the exchange can we now but o,i ewudqatz ih[ with quantize would we if Now, show: can one (1)’(2)’, and (5) From nteohrhn,i eqatz with quantize we if hand, other the On energy. negative infinitely with state a construct could one and xsthat fixes o oho h ae ih[ with cases the of both for oeta,w antcag h oe of roles the change cannot we that, Note I ewuddefine −|| would we (If d e † ( ⃗p ) | X d || ⟩ † ( 2 d ∴ steceto (annihilation): creation the is ) H ≥ ( d

,inconsistent!) 0,

( d d ⃗p 1 † † ) d ( d H d ⃗p 2 † e † ) | · · · ( ( = | ⃗p X d ) s † X − ⟩ ( • d [

⃗p , d †

⟩ ) , d • 2 ( )          { | ⃗p † ,(1) ], − X d ( e ) [ [ ( = [ [ [ [ ⃗p † d , ,d H, ,d H, ,d H, ,d H, a H, a H, ⟩ •

) )

d , = † d • | E ( ⇐⇒ ( ( ( = = ⃗p ⃗q ,(1) ], s s † s s † s s † ⃗p b b X (2 = ) ) ( ( ( ( d ( ( ( † ) ] ( ⃗p ⃗p ⃗p ⃗p ⃗p ⃗p ( ⃗p n en h aumby vacuum the deﬁne and , 95 − | E ]= )] ]= )] ⃗p 1’and (1)’ ]= )] = )] ]= )] = )] d X ) {• (2 = ) X s † ⇐⇒ E ≡ d ( ⟩ ≡ , −∞ → ⃗p

1 − •} π and

) d {z E − E − − E π 2 − d H ) † E 1’ then (1)’, (2) , 3 ) p ( ( p p (2 = E E E δ 3 d ⃗p ⃗p E d a p [ + δ (3) p p p d ) ) s ) s s † {• 2 d (3) d a ( ( ( ( † ⃗p s † ⃗p ⃗p because , (0) s † s ⇐⇒ · · · − ,d H, d π ( ( , ( ) ( ) ) s † ⃗p ⃗p ⃗p •} ⃗p ) ( ) 3 ) , ) ⃗p − δ } (2) , 2’ esilhave still we (2)’, ) s † (3) ) ( | d ⃗q ( ⃗p X ) s † (0) d )] , ( ⟩ ⇐⇒ 1 † ⃗p ) ) d ol eraeenergy. decrease would ) ⟨ 2 † | , X X | · · · (2)’. | ⟩ X ≥ ⟩ X ⟩ ) d 0 e . . | 0 ⟩ ,then 0, = because { } d(⃗p), d†(⃗q) = (2π)3δ(3)(⃗p − ⃗q) = dd† + d†d = b†b + bb† { } = b(⃗q), b†(⃗p)

which leads to d†(⃗p) |0⟩ = 0 because d†(⃗p)2 = 0. This is related to the idea of the “Dirac sea”.

▼ The Hamiltonian then becomes ∫ ( ) d3p ∑ H = E a†(⃗p)a (⃗p) −d†(⃗p)d (⃗p) 3 p s s | s {z s } (2π) s= − † = bs(⃗p)bs(⃗p) † − 3 (3) = +bs(⃗p)bs(⃗p) (2π) δ (0) ∫ ( ) ∫ d3p ∑ = E a†(⃗p)a (⃗p) + b†(⃗p)b (⃗p) − d3p E δ(3)(0) (2π)3 p s s s s p s=

We neglect the (inﬁnite) constant term, as in the scalar case.

▼ To summarize, quantization with anti-commutation works, and we have

∫ 3 ∑ ( ) d√p −ip·x † +ip·x Ψa(x) = as(⃗p)us,a(p)e + b (⃗p)vs,a(p)e 0 3 s p =Ep (2π) 2Ep s=  { { † } 3 (3) − † (3)  ar(⃗p), as(⃗q) = (2π) δ (⃗p ⃗q)δrs {Ψ (x), Ψ (y)} 0 0 = δ (⃗x − ⃗y)δ a b x =y ab ⇐⇒ {b (⃗p), b†(⃗q)} = (2π)3δ(3)(⃗p − ⃗q)δ others = 0  r s rs others = 0 ∫ ( ) d3p ∑ H = E a†(⃗p)a (⃗p) + b†(⃗p)b (⃗p) (2π)3 p s s s s s=

96 The anti-commutation relation implies Fermi-Dirac statistics; ( ) 2 † † − † † † ar(⃗p)as(⃗q) = as(⃗q)ar(⃗p), in particular ar(⃗p) = 0 Pauli blocking

▼ Particle and Anti-particle

U(1) Symmetry: Ψ → Ψeiα → Current: jµ = ΨγµΨ ∫ → Charge: Q = d3xj0

= ··· ∫ d3p ∑ ( ) = a†(⃗p)a(⃗p) − b†(⃗p)b(⃗p) (+constant) (2π)3 s s s and hence { † † [Q, as(⃗p)] = +as(⃗p) † − † [Q, bs(⃗p)] = bs(⃗p) Namely, • † as(⃗p) increases the charge by one. (Particle creation) • † bs(⃗p) decreases the charge by one. (Anti-particle creation)

▼ One particle state √ |ψ; ⃗p,r⟩ = 2E a†(⃗p) |0⟩ : particle ⟩ √ p r ¯ † | ⟩ ψ; ⃗p,r = 2Epbr(⃗p) 0 : anti-particle Normalization: √ √ ⟨ψ; ⃗p,r|ψ; ⃗q, s⟩ = 2E 2E ⟨0|a (⃗p)a†(⃗q)⟩0 √ p√ q r s ⟨ | { † } − † ⟩ = 2Ep 2Eq 0 |ar(⃗p{z)as(⃗q)} as(⃗p)ar(⃗p) 0 3 (3) (2π) δ (⃗p−⃗q)δrs 3 (3) = (2π) 2Epδ (⃗p − ⃗q)δrs. ¯ ¯ 3 (3) Similarly ⟨ψ; ⃗p,r|ψ; ⃗q, s⟩ = (2π) 2Epδ (⃗p − ⃗q)δrs. Lorentz transformation: (check it by yourself. . . )

▼ (That’s all for this semester. Thank you for your attendance!)

97 References

[1] M. Srednicki, Quantum Field Theory.

[2] M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory.

[3] S. Weinberg, The Quantum Theory of Fields volume I.

[4] M. D. Schwartz, Quantum Field Theory and the Standard Model.

[5] 「ゲージ場の量子論 I」九後汰一郎、培風館.

[6] 「場の量子論」坂井典佑、裳華房.