Quantum Field Theory

Transcription

1 207 S-semester Quantum Field Theory Koichi Hamaguchi Last updated: July 3, 207

2 Contents 0 Introduction 0. Course objectives Quantum mechanics and quantum field theory Notation and convention Various fields Outline: what we will learn S-matrix, amplitude M = observables σ and Γ Scalar spin 0 Field 6. Lorentz transformation Lorentz transformation of coordinates Lorentz transformation of quantum fields Lagrangian and Canonical Quantization of Real Scalar Field Equation of Motion EOM Free scalar field Solution of the EOM Commutation relations a and a are the creation and annihilation operators Consistency check vacuum state One-particle state Lorentz transformation of a and a [ϕx, ϕy] for x 0 y Interacting Scalar Field What is ϕx? In/out states and the LSZ reduction formula Heisenberg field and Interaction picture field a and a again T ϕx ϕx n 0 =? Wick s theorem Summary, Feynman rules, examples Fermion spin /2 Field Representations of the Lorentz group Lorentz Transformation of coordinates again see infinitesimal Lorentz Transformation and generators of Lorentz group in the 4-vector basis A Other disconnected Lorentz transformations Lorentz transformations of fields, and representations of Lorentz group Spinor Fields i

3 2..5 Lorentz transformations of spinor bilinears Free Dirac Field Lagrangian Dirac equation and its solution Quantization of Dirac field ii

4 0 Introduction about this lecture i Language The rule of the department: For the graduate course, as far as there is one international 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 0, 7, 24, May, 8, 5, 29, no class on May 22 June 5, 2, 9, 26, July 3, 0, Exam on July 24. maybe an extra class on July 3. 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.

5 0. 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 + Higgs γ 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: q i t i =, 2, discrete QFT: ϕ x, t x continous infinite number of degrees of freedom Note: uncountably infinite, QM QFT operators q i t, p i t or q i t, q i t ϕ x, t, π x, t or ϕ x, t, ϕ x, t Heisenberg picture i =, 2, discrete x continous states [q i, p j ] = iħδ ij [ϕ x, t, π y, t] = iδ 3 x y e.g., Harmonic Oscillator 0 : ground state 0 : ground state a 0, a a 0, a p 0, a p a p 0, a written in terms of q and p a p 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

6 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. QM: ħ 0 important at small scale SR: c < important at large velocity QFT: ħ 0 and c < physics at small scale & large velocity: Particle Physics, Early Universe, Notation and convention We will use the natural units ħ = c =, where ħ kg m 2 sec, c = m sec. For instance, we write E 2 = p 2 + m 2 instead of E 2 = p 2 c 2 + m 2 c 4, and [x, p] = i instead of [x, p] = iħ. We will use the following metric. g µν =. The sign convention depends on the textbook. g µν here = g Peskin µν = g Srednicki µν x µ = x 0, x, x 2, x 3 = t, x x µ = x 0, x, x 2, x 3 = g µν x ν = t, x p µ = p 0, p, p 2, p 3 = E, p p µ = g µν p ν = E, p p x = p µ x ν = p µ x ν = p 0 x 0 p x p 2 x 2 p 3 x 3 = p 0 x 0 p x = Et p x 3

7 If p µ is the 4-momentum of a particle with mass m, 0.4 Various fields p 2 = p µ p µ = p 0 2 p 2 = E 2 p 2 = m 2. spin equation of motion for free fields scalar field ϕx 0 + m 2 ϕ = 0 Klein-Gordon eq. fermionic field ψ α x /2 iγ µ µ mψ = 0 Dirac eq. gauge field A µ x µ F µν = 0 part of Maxwell eq. F µν = µ A ν ν A µ 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, ν i... fermionic fields γ photon, W ±, Z weak bosons, g gluon... gauge fields H Higgs... scalar field 0.5 Outline: what we will learn quantization of free interacting field 2 operator path integral 2 ways 0 T [ϕx ϕx n ] 0 3 S-matrix, amplitude M 4 LSZ reduction short cut Feynman rule observables cross section σ and decay rate Γ First, we will learn with a scalar field. next, fermionic field,... A long way to go,... Today, let s discuss 4 in advance. 4

8 0.6 S-matrix, amplitude M = observables σ and Γ Let s consider the probability of the following process, P α β. α { } p 2 p p p 2 β p n p n If the initial and final states are normalized as α β = δ αβ, 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 = σ, p, σ 2, p 2, σ n, p n, σ i ; 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 = δ αβ. Instead, we normalize it as Comments σ, p σ, q = 2π 3 2E p δ 3 p qδ σσ. i What s the mass dimension of σ, p then? It s. ii Why E p δ 3 p q, not just δ 3 p q? E p δ 3 p q is Lorentz invariant. For instance, for a boost Lorentz transformation along the z direction, E γ γβ E = γβ γ p z p z one can show that check it yourself, β = v/c, γ = E δp z q z = Eδp z q z. 5 β 2,

9 S-matrix: The transition amplitude σ, p, σ m, p m ; out σ, p, σ n, p n ; in = σ, p, σ m, p m S σ, p, σ n, p n is called S-matrix. Comments i The definition of in and out-states will be given later. ii S-matrix is Lorentz invariant. iii Why matrix? β ; out α ; in = β S α = S βα is a matrix with an infinite dimension. iv In the following, we omit the label σ i and σ i. invariant matrix element, or scattering amplitude, M: As long as the total energy and momentum are conserved, p p m S p p n δ E f E i δ 3 f i f }{{}}{{} final final = δ 4 f p f i p i p f i p i We define the invariant matrix element, or scattering amplitude, M as p p m S p p n = 2π 4 δ 4 f p f i p i im p p n p p m Comments 2 i Since the S-matrix is Lorentz invariant, the amplitude M is also Lorentz invariant. ii The amplitude M can be calculated by the Feynman rule. In this lecture, we learn how the Feynman rule is derived from the Lagrangian. iii In this subsection, we derive the formula for M transition probability. on April 0, up to here. 6

10 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 Schrödinger equation of a free field is i / tψ = /2m / x 2 ψ, which corresponds to E = p 2 /2m. Promoting this relation to a relativistic relation, E 2 = p 2 + m 2, one obtains / t 2 ϕ = [ / x 2 + m 2 ]ϕ = 0, or + m 2 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 Relativity? A: Not much. If you understand for instance the notation of x µ p µ = g µν x µ p ν, and the fact that it is Lorentz invariant, then I hope the lecture is more or less understandable. If you are not sure what the statement x µ p µ is Lorentz invariant means, then you should probably review the basics of the special relativity, Lorentz transformation, Lorentz invariance, etc. Q: What is the difference between the lower and upper indices, x µ and x µ? A: You should review the basics of the special relativity and get used to these notations.... Q: Your normalization in Eq. is proportional to E p δ 3 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 E p = E q, and therefore δ 4 p q = δe p E q δ 3 p q becomes proportional to δ0, which is not very useful when normalizing fields. Q: You said that E p δ 3 p q is Lorentz invariant, after Eq.. 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

11 on April 7, from here. Outline quantization of free interacting field 0 T [ϕ ϕ] 0 LSZ S-matrix, amplitude M We are here. 0.6 observables σ and Γ Feynman rule For simplicity, we normalize the system with a box. L V = L 3, p = 2π L n x, n y, n z Then Define δ 3 p q = 2π 3 p Box = d 3 xe i p q x = 2π V δ 3 p, q. 3 }{{} discrete 2Ep V p. 4 8

12 L{ Then, Box q p Box = 2Eq V 2Ep V q p = 2Eq V 2Ep V 2π3 2E p δ 3 p q [ ] = 2π3 V δ3 p q [ E p = E q for p = q] = δ p, 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, { Probability P p p = Box p 2 p Box = δ p, p = p = p 0 p p Thus, Probability P p p n p p m = Box p p 2 m S p p n Box 2 = 2E V 2E m V p p m S p p n 2E V 4 2Em V m n n+m = p 2E i V p m S p p n 2 5 2E f= f i= But this becomes zero for V. What is the differential probability that the final state is within [ p f, p f + d p f]? dp = P p p n p p m dn }{{} number of states within [ p f, p f + d p f] p y 2π dn = dp x 2π/L dp y 2π/L dp z 2π/L = d3 p 2π 3 V p x p z dp x 9

13 For the m particle final states, p p m, m d 3 p f dn = 2π V 6 3 From 5 and 6, f= dp = P p p n p p m dn m d 3 p n f n = p 2π 3 2E f 2E i V p m S p p n 2 7 f= i= On the other hand, from the definition of M 2, p p m S p p n 2 = 2π 4 δ 4 p f p i 2π 4 δ 4 p f p i M 2 = 2π 4 δ 4 p f p i 2π 4 δ 4 0 M 2 d δ 4 4 x 0 = ei0 x = V T T : time 2π 4 2π 4 = 2π 4 δ 4 p f p i V T M 2 Substituting it in Eq.7 and dividing by T, we obtain the differential transition rate n dp T = V m n d 3 p f 2π 4 δ 4 p 2E i= i 2π 3 2E f p i M p p n p p m 2 f= f }{{} dφ m 8 Now let s discuss the cases n = and n = 2. n = q q 2 A particle decay q m From Eq.8, the probability that the particle A decays into the range of final states [ p f, p f + d p f] per unit time is dp p A q q m T = 2E A dφ m Mp A q q m 2 0

14 Integrating over the final momenta, we have Decay Rate Comments ΓA, 2, = dφ m Mp A q q m 2 2m A m d 3 q f = 2π 4 δ 4 p 2m }{{ A 2π } 3 A 2E f f= f at rest frame symmetry factor q f Mp A q q m 2 i The mass dimension of Γ is energy + time. CHECK q p = 2π 3 2E }{{} p δ 3 p q p E. }{{} E E 3 q q m S p A = 2π 4 δ 4 q pa imp }{{} A q }{{}}{{} E m E 4 E 3 m Γ = m d 3 q f 2π 4 δ 4 p 2m }{{ A 2π } A q f Mp A q q m 2 2E f }{{} f= f }{{}}{{} E 6 2m E } E 2m E {{ 4 } E + ii If there is more than one decay modes, their sum ΓA all = ΓA, 2 + ΓA, 2 + is called the total decay rate, and its inverse gives the lifetime of A. Example: muon. τ A = ΓA all Γµ all Γµ e ν e ν µ GeV. τ µ GeV sec.

15 iii If not in the rest frame, Γ = dφ m M 2 2E A }{{} Lorentz inv. In the frame which is boosted by a velocity β, the energy is E A = γm A. γ = / β 2. Γ becomes smaller by a factor of /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 and 2 are identical, A 2 and Thus, we should A 2 are indistinguishable. reduce the integration range θ = [0, π] [0, π/2], or 2 divide by a symmetry factor = 2 after integration. n = 2 p A q q 2 particle scattering p A q m From Eq.8, the probability that the final particles are in the range of [ p f, p f + d p f] per unit time is dp p A, p B q q m T = V 2E A 2E B dφ m Mp A q q m 2 9 In this case, we consider a quantity called scattering cross section or just cross section. A B 2

16 Suppose that a particle A collides with a bunch of particles B with number density n B with a relative velocity v rel. The number that the scattering A, B, 2 occurs per unit time is given by P p A, p B, 2 T Why cross section? = n B v rel σp A, p B, 2 }{{} cross section 0 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 N B = σ v rel T n B. This is consistent with 0. For small T, N B < and it gives the probability. In the situation of Eq.9, there is only one B particle, so n B = /V. Thus, the differential cross section that the final state goes within [ p f, p f + d p f] is dσp A, p B, 2 = V dp p A, p B, 2 [ 0] v rel T = dφ m Mp A q q m 2 [ 9] v rel 2E A 2E B on April 7, up to here. Questions after the lecture: only some of them Q: Is the integration only over the final state momenta q f or p f but not over the initial ones p i? A: Right. The initial momenta are specified by the collider experiment, for instance. Q: Does the amplitude M depend on the final state momenta q f? 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? 3

17 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 conservation, δ E f E i, gives for E f = E i, δ0 = T/2 T/2 dt/2πei0t = 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 of Srednicki s textbook [] 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 quantization of free interacting field 0 T [ϕ ϕ] 0 LSZ S-matrix, amplitude M We are still here. 0.6 observables σ and Γ Feynman rule 4

18 Integrating over the final momenta, Cross Section σp A, p B, 2 = dφ m Mp A q q m 2 2E A 2E B v rel m d 3 q f = 2π 4 δ 4 p 2E A 2E B v rel 2π 3 A + p B 2E f f= f Comments symmetry factor q f Mp A, p B q 2 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 v rel is given by v rel = p A p B E A E B For a head-on collision with speeds of light, v rel = 2. iv E A E B v rel = E B p A E A p B is not Lorentz inv., and therefore σ in the above formula is not Lorentz inv. either. Lorentz inv. cross section can be defined by replacing as E A E B v rel p A p B 2 m 2 A m2 B. Problems a Show that the mass dim. of σ is 2. b Show that p A p B 2 m 2 A m2 B = E AE B v rel for p A p B. 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. 5

19 Scalar spin 0 Field We consider a real scalar field ϕx. real: ϕx = ϕx Hermitian operator. scalar: Lorentz transformation of the field is given by ϕx ϕ x = ϕλ x. Now let s briefly review the Lorentz transformation before starting QFT.. Lorentz transformation.. Lorentz transformation of coordinates is a linear, homogeneous change of coordinates from x µ to x µ, x µ = Λ µ νx ν, where Λ is a 4 4 matrix satisfying g µν Λ µ ρλ ν σ = g ρσ or Λ T gλ = g in matrix notation. Comments i It preserves inner products of four vectors: x y = g µν x µ y ν x y = g µν x µ y ν = g µν Λ µ ρλ ν σx ρ y σ = g ρσ x ρ y σ = x y. This is similar to orthogonal transformation v v = R v where R is an orthogonal cos θ sin θ matrix satisfying R T R =. e.g., R = in 2-dim. sin θ cos θ Inner products are preserved: u v u v = R u R v = u T R T R v = u v. = ii The set of all Lorentz transformations LTs forms a group Lorentz group. Product of two LTs Λ and Λ 2 is defines as Λ 2 Λ µ ν = Λ 2 µ ρ Λ ρ ν. closure: if Λ T gλ = g and Λ T 2 gλ 2 = g, then Λ 2 Λ T gλ 2 Λ = g. 6

20 associativity: Λ Λ 2 Λ 3 = Λ Λ 2 Λ 3. identity: Λ µ ν = δ µ ν =. inverse: Λ µ ν = gµρ g νσ Λ σ ρ = Λ ν µ. Problems a Write the explicit form Λ for a rotation along the z-axis. Show that it satisfies Λ T gλ = g. b Write the explicit form of Λ for a boost along the z-axis. Show that it satisfies Λ T gλ = g. c Show that the above Λ satisfies Λ Λ =, i.e., Λ µ ν Λν ρ = δ ν ρ. d For an infinitesimal LT, we can write Λ µ ν = δ µ ν + δω µ ν. Show that δω µν = δω νµ and hence there are six independent δω. We will discuss more on this later in Sec Lorentz transformation of quantum fields is represented by unitary operators acting on fields: Φx Φ x = UΛΦxUΛ Φx : generic field Scalar fields are the fields which transform as Comments ϕx ϕ x = UΛϕxUΛ = ϕλ x i Note that it transforms the field ϕx at all the spacetime x. ii Substituting x = y = Λy, it means ϕ y = ϕy for all y. iii Why Φ = UΦU for fields Φ? Suppose that a state transforms as = U. Then, with operators O i, O O 2 O n UO O 2 O n = UO U UO 2 U UO n U U = O O 2 O n 7

21 .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 L = d 3 x L[ ϕ x, t, ϕ x, t] }{{} Lagrangian density In this lecture, we consider the following Lagrangian called ϕ 4 theory: L = d 3 x L[ ϕ x, t, ϕ x, t] = d 3 x 2 µϕ µ ϕ 2 m2 ϕ 2 λ 24 ϕ4 = d 3 x 2 ϕ 2 ϕ 2 ϕ 2 m2 ϕ 2 λ 24 ϕ4 where λ is real and positive constant, and µ = x µ µ ϕ µ ϕ = g µν µ ϕ ν ϕ = 2 x ϕ 0 3 i= 2 x ϕ = ϕ 2 ϕ ϕ i The λϕ 4 term represents the interaction. For λ = 0, it becomes the Lagrangian of free scalar field: L free = d 3 x 2 µϕ µ ϕ 2 m2 ϕ 2 = d 3 x 2 ϕ 2 ϕ 2 ϕ 2 m2 ϕ 2 If we regard x as just a label, ϕ ϕ x, t = {ϕ x t, ϕ x2 t, } x x 2 x QM of infinite number of degrees of freedom L = 2 ϕ x t 2 + x 2 q it 2 + i 8

22 QM QFT conjugate momentum p i = L π x, t = δl q i δ ϕ x, t = ϕ x, t functional derivative Hamiltonian H = p i q i L H = d 3 x π x, t ϕ x, t L i = d 3 x π 2 2 π2 + 2 ϕ m2 ϕ 2 + λ24 ϕ4 = d 3 x 2 π2 + 2 ϕ m2 ϕ 2 + λ 24 ϕ4 Canonical Quantization: Equal Time Commutation Relation [q i t, p j t] = iδ ij [ϕ x, t, π y, t] = iδ 3 x y [q i t, q j t] = 0 [ϕ x, t, ϕ y, t] = 0 [p i t, p j t] = 0 [π x, t, π y, t] = 0 Comments equal time i The action S = dtl = dtd 3 xl = d 4 xl is Lorentz invariant. Check: Under ϕx ϕ x = ϕλ x, [ d 4 xl[ϕx, µ ϕx] d 4 xl ϕλ x, ϕ ] x µ Λ x By a change of variable, x = Λy or y ν = Λ ν µ xµ, µ ϕx ϕ x µ Λ x = yν x µ ϕ y ν y = Λ ν µ [ νϕ]y and hence µ ϕ µ µρ ϕ ϕx g x µ Λ x ϕ x ρ Λ x = [ ν ϕ ν ϕ]y = g µρ Λ ν µ Λ σ ρ }{{} =g νσ [ ν ϕ]y[ ρ ϕ]y 9

23 Thus, S = d 4 x 2 µϕ µ ϕx + 2 m2 ϕx 2 + d 4 x 2 µϕ µ ϕy + 2 m2 ϕy 2 + = d 4 y 2 µϕ µ ϕy + 2 m2 ϕy 2 + d 4 x = x det y d4 y = det Λ d 4 y = d 4 y = S. ii Why L interaction = λ 24 ϕ4? For L interaction ϕ 3 or ϕ 3 or +ϕ 4, the corresponding Hamiltonian term becomes H interaction d 3 x ϕ 3 or ϕ 3 or ϕ 4 and hence the energy becomes unbounded below. Take, for instance ϕ x, t = const ±. Thus, L interaction ϕ 4 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 ϕ 4 interaction of the Higgs boson is assumed in the Standard Model, but it is not yet experimentally tested. There may also be interactions like ϕ 6, ϕ 8, or other forms.... Q: I am confused with the LT of field, ϕ x = ϕλ x... 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 Z 2 transformation, ϕx ϕx. The action of the ϕ 4 theory is invariant under this Z 2 transformation of the field. Similarly, the action is invariant under the LT of the field, ϕx ϕλ x. The latter seems more complicated because it is accompanied by a change of the argument, but they are both just transformations of field. 20

24 on May, from here. Outline quantization of free interacting field.2 We are here. 0 T [ϕ ϕ] 0 LSZ S-matrix, amplitude M 0.6 observables σ and Γ Feynman rule iii Schrödinger representation and Heisenberg representation: In QFT, usually the Heisenberg representation is used. state operator S-rep. Ψt S O S time-dependent time-independent H-rep. Ψ H O H t time-independent time-dependent S-rep. i d dt Ψt S = Hp, q Ψt S Ψt S = e iht t 0 Ψt 0 S Expectation value of an operator: S Ψt O S Ψt S H-rep. { Ψ H Ψt 0 S = e iht t0 Ψt S O H t e iht t0 O S e iht t 0 Expectation value: H Ψ O H t Ψ H = = S Ψt O S Ψt S i d dt O Ht = He iht t 0 O S e iht t 0 + e iht t 0 O S He iht t 0 = HO H t + O H th = [O H t, H]. Heisenberg eq. 2

25 .3 Equation of Motion EOM There are two ways to derive the EOM. i From the action principle δs = δ dtl = 0, µ δl δ µ ϕ δl δϕ = 0 Euler-Langrange eq. ii From Heisenberg eq., { i ϕ = [ϕ, H] i π = [π, H] From i, for the Lagrangian L = d 3 x 2 µϕ µ ϕ 2 m2 ϕ 2 λ 24 ϕ4, we obtain µ δl δ µ ϕ δl δϕ = µ µ ϕ + m 2 ϕ + λ 6 ϕ3 = 0 or EOM + m 2 ϕx = λ 6 ϕx3 Later we will derive the EOM with ii again see.5.3. Here, there is an important difference between λ = 0 and λ 0. For free field λ = 0, the EOM is linear in ϕx, and can be solved exactly.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 ϕx in this case? more on this later 22

26 .4 Free scalar field.4. Solution of the EOM Starting from + m 2 ϕx = 0 Klein-Gordon eq., one can show that ϕx can be expressed as d 3 p ϕx = 2π 3 a pe ip x + a pe ip x 2E p where ϕx, a p, a p are operators, p x = p µ x µ = p 0 t p x, and p 0 = E p = p 2 + m 2. Eq. is the solution of the EOM, because + m 2 e ±ip x = µ µ + m 2 e ±ip x 2 = t 2 + m 2 2 e ±ip x = Ep 2 + p 2 + m }{{} 2 e ±ip x = 0. Ep 2 proof: Now let s show that Eq. is the general solution of the EOM. i Fourier transform FT ϕx with respect to x: ϕ x, t = }{{} d 3 p C p, t e i p x }{{} 2 operator operator ii From the condition ϕ = ϕ real field, d 3 pc p, te i p x = d 3 pc p, te i p x = d 3 p C p, te i p x = d 3 pc p, te i p x p = p Using inverse FT, C p, t = C p, t 3 iii From + m 2 2 ϕ = t 2 + m 2 ϕ = 0 and 2, 2 d 3 p C p, t + C p, t p 2 + m 2 e i p x = 0 }{{} Ep 2 23

27 Using inverse FT, C p, t + EpC p, 2 t = 0 C p, t = C pe iept + C pe +iept. From 3, C p = C p, and hence C p, t = C pe iept + C pe +iept. iv Substituting it to 2 and changing p p in the 2nd term, ϕ x, t = d 3 p C pe iept e i p x + C pe +iept e i p x = d 3 p C pe ip x + C pe +ip x Finally by normalizing as a p 2π 3 2E p C p, we obtain. Note that the normalization depends on the convention textbook. ahere = apeskin = 2Ep asrednicki = 2π 3/2 aweinberg. From, we can express the operators a p and a p in terms of ϕx: a p = d 3 xe [i +ip x ϕx ] + E p ϕx 2Ep a p = d 3 xe [ i ip x ϕx ] 4 + E p ϕx 2Ep Problems a Substitute to the right-hand side RHS of 4 and show that it gives a & a. b The RHS of 4 seems to depend on x 0 = t, but the LHS does not. Show that [RHS of 4]= 0, using the EOM. Hint: integration by parts t c Substitute 4 to the RHS of and show that it gives LHS. Pay attention to which variables are just the integration variable. For instance, let s solve a: d 3 q from, ϕx = 2π 3 a qe iq x + a qe +iq x 2E q }{{} =[dq] i ϕx = [dq] E q a qe iq x E q a qe +iq x i ϕx + E p ϕx = [dq] E q + E p a qe iq x + E q + E p a qe ip x 24

28 Thus, RHS of 4 = 2Ep = 2Ep d 3 xe +ip x [dq] E q + E p a qe iq x + E q + E p a qe +iq x d 3 xe iepx0 e i p x e ieqx0 e i q x = 2π 3 δ 3 p q e iep Eqx0 2π 3 δ 3 p + q e iep+eqx0 d 3 q E q + E p a qδ 3 p q + E q + E p a qδ 3 p qe iep+eqx0 2Eq }{{} 0 = a p = LHS of Commutation relations From the commutation relation in.2, we have the following commutation relations recall πx = ϕ: [ϕ x, t, ϕ y, t] = iδ 3 x y [ϕ x, t, ϕ y, t] = 0 [ ϕ x, t, ϕ y, t] = 0 Problems a Show that RHS of 5 = LHS of 5, using. b Show that LHS of 5 = RHS of 5, using 4. [a p, a q] = 2π 3 δ 3 p q [a p, a q] = 0 [a p, a q] = a and a are the creation and annihilation operators. In this section we will see that a p annihilate a particle with energy E p, momentum p. a p create a particle with energy E p, momentum p. 5 First, we can show that [H, a p] = E p a p [H, a p] = E p a p 6 We will show it later. We can also show [ ˆ P, a p] = pa p [ ˆ P, a p] = pa p where ˆ P is the momentum operator. ˆ P = d 3 xπ ϕ. We skip the details here. Consider a state with energy E X and momentum p X ; X : { H X = EX X ˆ P X = p X X 25

29 Then, for the state a p X, H a p X = [H, a p] + a ph X = E p a p + a pe X X = E p + E X a p X, ˆ P a p X = [ ˆ P, a p] + a p ˆ P X = pa p + a p p X X = p + p X a p X. Thus, the state a p X has energy E p + E X and momentum p + p X, namely, a p adds energy E p and momentum p. creation operator Similarly, we can show H a p X = E X E p a p X, ˆ P a p X = p X 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, up to here. Questions and comments after the lecture: only some of them comment: There is a typo at the end of.2. In Schrödinger rep., the argument of the RHS of ψt S should be t 0, not t. A: Thanks! corrected. d 3 p Q: When you write, does E p 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? d 3 p A: Good question! One can indeed define a number operator N = 2π 3 a pa p, similar to the case of harmonic oscillator, N = i a i a i. Now, next week we will 26

30 see that a one-particle state is proportional to a p 0. You can check explicitly by using the commutation relation that Na p 0 = a p 0, which means the number is one. Similarly, you can check Na pa p 0 = 2a pa p 0, etc. on May 8, from here. Outline quantization of.4 free We are here. interacting field.4.3, a and a. Showing 6. 0 T [ϕ ϕ] 0 LSZ S-matrix, amplitude M 0.6 observables σ and Γ Feynman rule First, ϕx = = d 3 p 2π 3 2E p a pe ie pt+i p x + a pe iept i p x d 3 p 2π 3 2E p a pe ie pt + a pe iept e i p x p p in the 2nd term Let s define, A p, t 2π 3 2E p a pe iept and omit t for simplicity: A p = A p, t. Then ϕx = d 3 p A p + A p e i p x ϕx = d 3 p A p + A p i pe i p x ϕx = d 3 p ie p A p A p e i p x Ȧ p, t = ie pa p, t 27

31 Therefore, H = = = d 3 x 2 π2 + 2 ϕ m2 ϕ 2 d 3 x 2 ϕ ϕ m2 ϕ 2 d 3 x d 3 p d 3 q e i p x e i q x 2π 3 δ 3 p+ q [ 2 ie p ie q A p A p A q A q + 2 i pi q A p + A p A q + A q + 2 m2 A p + A p A q + A q ] = d 3 q2π 3 [ 2 E2 q A q A q A q A q + 2 q 2 + m 2 A q + A q A q + A q ] }{{} Eq 2 [ = d 3 q2π 3 Eq 2 A qa q + A qa q ] [ = d 3 q2π 3 Eq 2 A qa q + A qa q ] q q in the st term = d 3 q2π 3 Eq 2 [ a qa q + a qa q ] 2π 6 2E q Here, note that the t-dependence of A q, t cancels in H, and hence H is time independent. By using a qa q = a qa q + 2π 3 δ 3 0, we obtain H = d 3 q 2π E 3 q a qa q + 2 2π3 δ 3 0 The constant term, d 3 qe q 2 δ3 0 28

32 is the zero-point energy. This corresponds to the ħω term in the energy spectrum of 2 the harmonic oscillator, E = ħωa 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 [ H =: d 3 x 2 ϕ ϕ 2 + ] 2 m2 ϕ 2 : : aa :=: a a : normal ordering then there is no zero-point energy. In any case, we have H = Therefore, d 3 q 2π 3 E qa qa q + const. [H, a p] = d 3 q 2π E qa q [ a q, a p ] 3 }{{} 2π 3 δ 3 q p = E p a p similarly [H, a p] = E p a p.4.4 Consistency check Now that ϕx and H are expressed in terms of a and a, let s do some consistency check. i Heisenberg eq. i ϕx = [ϕx, H]. ii ϕx is a Heisenberg operator: ϕx = ϕt, x = e iht t0 ϕt 0, xe iht t0. Problems a Show i by using and 6. b Show that e iht a p e iht = a p e iept and e iht a pe iht = a p e iept by using 6. c Show ii by using the result of b and eq.. A typo here was corrected May

33 .4.5 vacuum state The operator a p decreases the energy: X a p X a qa p X energy E X E X E p E X E p E q The ground state lowest energy state 0 is a state which satisfies and Lorentz invariant.4.6 One-particle state a p 0 = 0 UΛ 0 = 0. The one-particle state in 0.6 is given by for free theory normalization p = 2E p a p 0. q p = 2E q 2Ep 0 a qa p 0 = 2E q 2Ep 0 [a q, a p] +a p a q 0 }{{}}{{} 2π 3 δ 3 p q 0 = 2π 3 2E p δ 3 p q, reproducing the normalization in 0.6. Lorentz transform: From the Lorentz transformation UΛϕxUΛ = ϕλ x, we expect UΛ p = p. where p = Λ p. This may be opposite to ordinary convention. Let s show it. LHS = 2E p UΛa puλ UΛ 0 = 2E p UΛa puλ 0 RHS = 2E p a p 0 So it is sufficient to show UΛa puλ = E p E p a p. 30

34 .4.7 Lorentz transformation of a and a Let s show. i First of all, for any f p, d 3 p f p = 2E p d 4 p δp 2 m 2 p 0 >0 f p This is because δp 2 m 2 = δp 0 2 p 2 m 2 dp 0 δp 2 m 2 p 0 >0 = = δp0 p 2 + m 2 + δp0 + p 2 + m 2 2p 0 2p 0 δfx = δx x i f x i x i ;fx i =0 dp 0 δp0 p 2 + m 2 2p 0 = 2E p ii Therefore ϕx = = d 3 p 2π 3 a pe ip x + a pe ip x 2E p d 4 pδp 2 m 2 2Ep p 0 >0 a pe ip x + a pe ip x 2π 3 and its 4-momentum FT is given by ϕk d 4 xe ik x ϕx = d 4 pδp 2 m 2 2Ep p 0 >0 a p2π 4 δ 4 p k + a p2π 4 δ 4 p + k 2π 3 = 2πδk 2 m 2 2E k θk 0 a k + θ k 0 a k. 3

35 iii and its LT is UΛ ϕkuλ = 2πδk 2 m 2 2E k θk 0 UΛa kuλ + θ k 0 UΛa kuλ. LHS = d 4 xe ik x UΛϕxUΛ = d 4 xe ik x ϕλ x = d 4 ye ik Λy ϕy x = Λy, d 4 x = det Λd 4 y = d 4 y = d 4 ye iλk Λy ϕy k = Λ k = d 4 ye ik y ϕy = ϕk = 2πδk 2 m 2 2E k θk 0 a k + θ k 0 a k. Using k 2 = k 2 and θk 0 = θk 0 and comparing it with RHS, we obtain UΛa kuλ = Ek E k a k.4.8 [ϕx, ϕy] for x 0 y 0 For x 0 = y 0 = t, we have [ϕx, ϕy] = 0. What if x 0 y 0? From Eq. and the commutation relations of a and a in.4.2, we have d 3 p [ϕx, ϕy] = e ip x y e +ip x y i x y 2π 3 2E p d 3 p x = i e ip x e +ip x 2π 3 2E p Properties of x: a + m 2 x = 0. b Lorentz invariant: Λx = x. c Local causality: x = 0 for x 2 = x 0 2 x 2 < 0 space-like. t x y x x y = 0 32

36 Among them, a is clear from the definition of x. b can be shown by using the equation in i of the previous section.4.7: d 4 p x = i 2π 4 δp2 m 2 θp 0 e ip x e ip x x d 4 p = i 2π 4 δp2 m 2 θp 0 e ip x e ip x x = Λx d 4 q = i 2π 4 δq2 m 2 θq 0 e iλq Λx e iλq Λx p = Λq, d 4 p = d 4 q d 4 q = i 2π 4 δq2 m 2 θq 0 e iq x e iq x p = Λq, d 4 p = d 4 q = x. Finally, c can be shown as follows. Fist, x 0 = 0, x = i d 3 p 2π 3 2E p e i p x e i p x = 0. On the other hand, for space-like x x 2 = x 0 2 x 2 < 0, one can always Lorentz transform it to a frame with x 0 = 0. For a Lorentz boost in the opposite direction to x, x 0 is transformed as x 0 = Taking β = x0 x x, we have 2 x 0 = 0. Note that this is impossible for a time-like x, where x 2 = x 0 2 x 2 > 0, because x 0 x x 2 > in that case. Therefore, we have x = x 0 = 0, x = 0 for x 2 < 0. on May 8, up to here. on May 5, from here. β 2 x0 β x. 33

37 .5 Interacting Scalar Field Lagrangian L = d 3 x 2 ϕ 2 2 ϕ 2 2 m2 ϕ 2 ϕ x, t π x, t = δl δ ϕ x, t = ϕ x, t } {{ } same as free theory λ 24 ϕ4 }{{} Interaction λ : real and positive constant 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 same as free theory.5. What is ϕx? In the case of free theory λ = 0, d 3 p ϕx = 2π 3 e iept ei p x a p + e iept e i p x a p 2E p We could exactly solve the t-dependence by using Fourier transform and the Klein- Gordon eq. + m 2 ϕ = 0. With the interaction, ϕ x, t =?? The EOM is see.3 + m 2 ϕx = λ 6 ϕx3. This is non-linear. Let s try Fourier transform at t = 0. d 3 p ϕt = 0, x = C pe i p x + C pe i p x 2π 3 }{{} from ϕ=ϕ Defining a p by C p = a p/ 2E p, d 3 p ϕt = 0, x = a pe i p x + a pe. i p x 2π 3 2Ep Note that, at this stage, a p and a p are just coefficients of the Fourier transformation. 34

38 In the Heisenberg picture, ϕt, x = e iht ϕ0, xe iht d 3 p = e iht a pe iht e i p x + e iht a pe iht e. i p x 2π 3 2Ep What happened in the case of free theory? free theory [H, a p] = E p a p e iht a pe iht = a pe iept This is the problem b of.4.4. An example solution is as follows: define ft = e iht a pe iht then ft = e iht i[h, a p]e iht = e iht ie p a pe iht = ie p ft Thus, ft = e iept f0 = e iept a p. Similarly, e iht a pe iht = a pe iept. Therefore, d 3 p free theory ϕt, x = a pe iept e i p x + a pe iept e, i p x 2π 3 2Ep which is a linear combination of a p and a p. However, with the interaction term, H = H 0 + H int ϕ 4 a + a 4 [H, a p] = E p a p + Oa 3, a 2 a, aa 2, a 3 e iht a pe iht includes infinitely many a and a. ϕt, x also includes infinitely many a and a. Thus, ϕx cannot be written as a linear combination of a and a. It cannot be considered as a field to create/annihilate just -particle state. It includes infinitely many particle creation/annihilation. We can t discuss scatterings just in terms of ϕx

39 Comment Here, we have used [a, a ] etc, but a and a are just Fourier coefficients in Eq.. What are these a and a? [a, a ] =? ϕt = 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 d 3 p ϕt = 0, x = a p + a p e i p x. 2π 3 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..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.4.6 p = 2E p a p 0. where see.4. a p = 2Ep = i 2Ep d 3 x e i ip x ϕx + E p ϕx d 3 x e ip x 0 ϕx. f 0 g f 0 g 0 fg, 0 = t We consider the same operator in the interacting theory. a p, t = i d 3 x e ip x 0 ϕx, 2Ep which is now time-dependent. RHS 0 for λ 0. t 36

40 And we define the in/out states by where p p n ; in = 2E p a p, 2E pn a p n, 0 q q m ; out = 2E q a q, + 2E qn a q m, + 0 a p, = a p, + = lim x 0 lim x 0 + i 2Ep i 2Ep d 3 x e ip x 0 ϕx, d 3 x e ip x 0 ϕx. Comments i One can think of operators with wave-packets: ã t = d 3 pf pa p, t with f p exp p p 2 4σ 2 and then later take σ 0. Peskin [2]. See e.g., the textbooks by Srednicki [] and/or ii Here, the vacuum state 0 is the ground state of the full Hamiltonian H = H 0 + H int. This comment is added after the lecture. See the comment at the end of this subsection. Then, one can show LSZ reduction formula where p p n ; in q q m ; out m [ = i d 4 x i e +iq i x i xi + m 2] i= n i= [ i d 4 y i e ip i y i yi + m 2] 0 T ϕx i ϕx n ϕy ϕy m 0 T : time-ordering { ϕxϕy for x 0 > y 0 T ϕxϕy = ϕyϕx for y 0 > x 0 T ϕx ϕx 2 ϕx 3 = ϕx i ϕx i2 ϕx i3 for x 0 i > x 0 i 2 > x 0 i 3 > 37

41 The LSZ reduction formula shows the relation between the S-matrix in out see 0.6 and the time-ordered correlation function 0 T ϕx 0. Let s show it. First, by the definition of in/out states, LHS of the LSZ formula = 2E p 2E q 0 a q, + a q m, + a p, a p n, 0 This is already time-ordered. Thus, we can put in the time-ordering operator T without changing anything: LHS of the LSZ formula = 2E p 2E q 0 T a q, + a q m, + a p, a p n, 0 Next, a p, + a p, = = = i 2Ep dt 0 a p, t [ i dt 0 2Ep d 3 x e ip x 0 ϕx d 4 x 0 e ip x 0 ϕx }{{} ] = e ip x E 2 pϕx = e ip x p 2 + m 2 ϕx = e ip x m 2 ϕx = e ip x m 2 ϕx d 3 xe i p x 2 f x = d 3 xe i p x 2 d 3 qe i q x d 3 y ei q y f y 2π 3 = d 3 xe i p x d 3 q q 2 e i q x d 3 y ei q y f y 2π 3 = d 3 q q 2 2π 3 δ 3 p q = p 2 d 3 ye i p y f y = d 3 y 2 e i p y f y d 3 y 2π 3 ei q y f y 38

42 and therefore a p, + a p, = i 2Ep d 4 x e ip x 2 + m 2 ϕx or a p, = a p, + + i 2Ep d 4 x e ip x 2 + m 2 ϕx Similarly, one can show a p, + = a p, + i 2Ep d 4 x e ip x 2 + m 2 ϕx Substituting these equations to, LHS of the LSZ formula = 2E p 2E q 0 T a q, + a q m, + a p, a p n, 0 a q, + i 2Eq a p, + + i 2Eq = 0 i i time-ordering a q, 0 = 0 d 4 x e iq x 2 x + m 2 ϕx i d 4 y e ip y 2 y + m 2 ϕy i time-ordering 0 a p, + = 0 d 4 x m e iqm xm 2 x m + m 2 ϕx m d 4 y n e ipn yn 2 y n + m 2 ϕy n 0 = RHS of the LSZ formula on May 5, up to here. Questions and comments after the lecture: only some of them comment: There is a typo at.4.4. The Heisenberg equation should be i ϕx = [ϕx, H] instead of i ϕx = [H, ϕx]. A: Thanks! corrected. Q: What happens to the LSZ formula, in particular the time-ordering, if you do a Lorentz transformation? 39

43 A: Good question. It is implicitly assumed that all the participating particles are causally connected, i.e., x i y j 2 > 0 time-like. In this case, the ordering x 0 i > y 0 j 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. Q: Why a p, ± = lim x 0 ± i 2Ep d 3 x e ip x 0 ϕx? A: That s the definition of the operator a p, ±. on May 29, from here. Outline quantization of.4 free interacting field.5 0 T [ϕ ϕ] 0 S-matrix, amplitude M LSZ.5.2 We are here. 0.6 observables σ and Γ Feynman rule 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 = H 0 +H int. 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, asymptotic completion, and LSZ asymptotic condition, one can show the above equation a p, ± 0 = 0. For instance, the asymptotic completeness says that, the Fock space spanned by the in -operators: { } V in = 0, a p, 0, a p, a p, 0, 40

44 and that by the out -operators: { } V out = 0, a p, + 0, a p, + a p, + 0, are the same as the Fock space spanned by the Heisenberg operator ϕx, V: V in = V out = V. For more details, see e.g., Kugo-san s textbook [5] I and Sakai-san s textbook [6]. They are in Japanese. I have checked several QFT textbooks in English, such as Peskin [2], Schrednicki [], Weinberg [3], etc, but I couldn t find the corresponding explanation. ii In general, the operator defined by a p, ± = lim x 0 ± i 2Ep d 3 xe ip x 0 ϕx does not give the correct normalization for the -particle state. 2E p a p, ± 0 p, but normalization is not correct in general. One should either define the operator by a p, ± = Z lim x 0 ± i 2Ep see e.g, Kugo-san s and Sakai-san s textbooks [5, 6], or rescale the field as d 3 xe ip x 0 ϕx ϕx = Zϕ r x ϕ r x : rescaled, or renormalized field see e.g., Srednicki s textbook [] where Z = p ϕx 0 2 represents how much the state ϕx 0 contains the one-particle state p. Note that p ϕx 0 = e ip x and Z = for the free theory. In this lecture, we do not discuss the renormalization, and take Z = as the leading order perturbation. 4

45 .5.3 Heisenberg field and Interaction picture field 0 Tϕx ϕx n 0 Now we want to calculate this..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 Tϕx ϕx n 0. Let s start from ϕ0, x and ϕ0, x at t = 0. The Equal-time commutation relation at t = 0 is [ϕ0, x, ϕ0, y] = iδ 3 x y. Define the Hamiltonian at t = 0. H = d 3 x 2 ϕ0, x ϕ0, x m2 ϕ0, x 2 }{{} H 0 + λ d 3 x 24 ϕ0, x4 }{{} H int Note that H 0 and H int are defined in terms of ϕ0, x and ϕ0, x at t = 0, and they are time-independent. For t 0, { Heisenberg field ϕt, x = e iht ϕ0, xe iht evolved by H Interaction piecture field new! ϕ I t, x = e ih 0t ϕ0, xe ih 0t evolved by H 0 Properties of ϕx and ϕ I x. 2 i From 2, Heisenberg equations are { ϕ = i[h, ϕ], ϕ = i[h, ϕ], ϕ I = i[h 0, ϕ I ], ϕi = i[h 0, ϕ I ]. 42

46 ii One can also show { ϕt, x = e iht ϕ0, xe iht ϕ I t, x = e ih 0t ϕi 0, xe ih 0t 3 define At = e iht ϕt, xe iht then At = e iht i[h, ϕ] + ϕ e iht = 0 so At = A0 = ϕ0, x. ϕ I is similar. ii and ϕ I 0, x = i[h 0, ϕ I 0, x] = i[h H int, ϕ0, x] = i[h, ϕ0, x] = ϕ0, x. Note that, ϕ I 0, x = ϕ0, x and ϕ I 0, x = ϕ0, x but ϕ I 0, x ϕ0, x. iii The equal-time commutation relations hold even at t 0, both for ϕ and ϕ I. [ϕt, x, ϕt, y] = e iht [ϕ0, x, ϕ0, y]e iht 2 3 = e iht iδ 3 x ye iht = iδ 3 x y [ϕ I t, x, ϕ I t, y] = e ih 0t [ϕ0, x, ϕ0, y]e ih 0t = iδ 3 x y iv H can be written in terms of ϕt, x, and H 0 can be written in terms of ϕ I t, x. H = e iht He iht = e iht d 3 x = 2 ϕ0, x λ 24 d 3 x 2 ϕt, x λ ϕt, x4 24 ϕ0, x4 e iht 2 3 Each time in the RHS is time-dependent, but the sum is time-independent. Similarly, H 0 = e ih0t H 0 e ih 0t = e ih 0t d 3 x 2 ϕ I 0, x = d 3 x 2 ϕ I t, x 2 + ϕi t, x ϕi 0, x m2 ϕ I t, x 2 2 m2 ϕ I 0, x 2 e ih 0t 2 3

47 The RHS is the sum is time-independent. Namely, if written in if written in terms of ϕt, x terms of ϕ I t, x H 0 wrong OK t-independent H int wrong wrong H = H 0 + H int OK t-independent wrong v Equation of motion: From i, ϕx = i[h, ϕx] = i d 3 y [ 2 ϕy From iv, we can take x 0 = y 0. Then 2 ϕy + 2 m2 ϕy 2 + λ ] 24 ϕy4, ϕx st term: [ ϕy 2, ϕx] x 0 =y 0 = 0. 2nd term: i d 3 y [ ϕy ] 2, ϕx 2 x 0 =y 0 = i d 3 y ϕy y [ϕy, 2 ϕx] + y [ϕy, ϕx] ϕy x 0 =y 0 = i d 3 y ϕy y iδ 3 x y x 0 =y 0 = i d 3 y 2 ϕy iδ 3 x y integration by parts = 2 ϕx. 3rd term: i d 3 y 2 m2 [ϕy 2, ϕx] x 0 =y 0 x 0 =y 0 = m 2 ϕx. [ϕ 2, ϕ] = ϕ[ϕ, ϕ] + [ϕ, ϕ]ϕ 4th term: = λ 6 ϕx3. Thus, ϕ = 2 m 2 ϕ λ 6 ϕ3 + m 2 ϕ = λ 6 ϕ3 Similarly, ϕ I = i[h 0, ϕ I ] = write H 0 in terms of ϕ I = 2 m 2 ϕ I. + m 2 ϕ I = 0 ϕ I is a free field! 44

48 .5.4 a and a again ϕ I satisfies + m 2 ϕ I = 0 and therefore it can be solved exactly see.4.. d 3 p ϕ I x = 2π 3 a pe ip x + a pe ip x 2E p p 0 =E p where a p and a p are the expansion coefficients of the interaction picture field ϕ I. We can also write them as a I and a I. Thus, the original ϕ0, x and ϕ0, x as well as H 0 and H int can be expanded in terms of a and a. ϕ0, x = ϕ I 0, x = ϕ0, x = ϕ I 0, x = all written in terms of a, a H 0 =. substitute H int = From the commutation relation of ϕ I, ϕ I, and H 0, one can show the following relations see.4.2 and.4.3 [a p, a q] = 2π 3 δ 3 p q [a, a] = 0 [a, a ] = 0 [H 0, a p] = E p a p [H 0, a p] = E p a p Note that the last two equations hold for H 0, not H. The state annihilated by a p: 0 I : a p 0 I = 0, H 0 0 I = 0 H 0 : normal ordered is NOT the ground state of the full Hamiltonian: H 0 I = H 0 + H int 0 I 0 0 I 0 on May 29, up to here. on June 5, from here. H int ϕ 4 I a + a 4 45

49 Outline quantization of.4 free interacting field.5.3, T [ϕ ϕ] we are here. S-matrix, amplitude M LSZ observables σ and Γ Feynman rule T ϕx ϕx n 0 =? We want to express it in terms of ϕ I a and a. Step i vii. i redefine the space-time points such that x 0 > x 0 2 > x 0 n, 0 T ϕx ϕx n 0 = 0 ϕx ϕx n 0. ii ϕx =? { ϕx = e iht ϕ0, xe iht ϕ I x = e ih 0t ϕ0, xe ih 0t ϕx = e iht e ih 0t ϕ I x e ih 0t e iht }{{} ut ϕx = u tϕ I xut. 2 iii 0 =? I 0 ut = I 0 e ih0t e iht = I 0 e iht H 0 0 I = 0 Insert an identity operator: = n n n= 46

50 where n represent the eigenstates of H with eigenvalues E n > E 0 = 0. The summation includes continuous parameter integral. Then [ ] I 0 ut = I n n n= e iht = I e iht + n= = I n= I 0 n n e iht I 0 n n e ient The 2nd term oscillates for t. Thus, for regularization, we take Therefore Similarly Thus t iϵ ϵ > 0, ϵ 0 then, e ient e ien iϵ e En ϵ 0 lim t iϵ I 0 ut = I lim t iϵ u t 0 I = I. 0 O 0 = I O I I }{{} I = lim t iϵ iv Substituting 2 3 to, v 0 ϕx ϕx n 0 = lim t iϵ I 0 utu t 0 I 0 utou t 0 I I 0 utu t 0 I 3 I 0 ut u t }{{} ϕ Ix ut u t 2 }{{} ϕ Ix 2 ut 2 }{{} u t n }{{} ϕ Ix n ut n u t }{{} 0 I 4 Ut, t 2 ut u t 2 t > t 2 = e ih 0t e iht t 2 e ih 0t 2 =? 47

51 It satisfies where Ut, t = 0 Ut, t 2 = ih I t Ut, t 2 5 t Ut, t 2 = iut, t 2 H I t 2 t 2 H I t e ih0t H int e ih 0t = e ih 0t d 3 x λ 24 ϕ0, x4 e ih 0t = d 3 x λ 24 ϕ Ix 4 check 5: d dt ut = d dt eih 0t e iht = e ih 0t ih 0 ihe iht = e ih 0t ih int e ih 0t e ih 0t e iht = ih I tut d dt u t = = iu th I t The solution of 5 is, if H I t at different t are commuting, t Ut, t 2 = exp i H I tdt, t 2 but this is wrong. The correct solution is [ t ] Ut, t 2 = T exp i H I tdt t [ 2 t n ] = T i H I tdt n! Let s check it. t Ut, t 2 = = n=0 n=0 n=0 [ n! T t n n! T k= t 2 t n ] i H I tdt t 2 t i H I tdt t 2 k ih I t }{{} 6 t n k i H I tdt t 2 48

52 Here, t of H I t is larger than other t t t t 2, and therefore H I t can be moved in front of the time-ordering: [ t n ] Ut, t 2 = ih I t t n! T n i H I tdt t 2 vi From 4, 0 ϕx ϕx n 0 = n= = ih I t Ut, t 2 lim t iϵ I 0 Ut, t ϕ I x Ut, t 2 ϕ I x 2 ϕ I x n Ut n, t 0 I I 0 Ut, t 0 I With 6, everything is written in terms of ϕ I x. Furthermore, the numerator is time-ordered t > t > t 2 > t n > t, and hence it can be written as 0 ϕx ϕx n 0 = lim t iϵ = lim t iϵ where we have used Ut, t 2 Ut 2, t 3 = Ut, t 3. I 0 T [ϕ I x ϕ I x n Ut, t Ut, t 2 Ut n, t] 0 I I 0 Ut, t 0 I I 0 T [ϕ I x ϕ I x n Ut, t] 0 I 7 I 0 Ut, t 0 I vii Substituting 6 to 7, we finally obtain [ t I 0 T ϕ I x ϕ I x n exp i H I t dt ] 0 I t 0 ϕx ϕx n 0 = lim [ t t iϵ I 0 T exp i H I t dt ] 0 I t 8 Everything is written in terms of ϕ I x and 0 I. By expanding exp i H I, we can do the perturbation expansion as O + Oλ + Oλ Wick s theorem All the terms in the numerator and the denominator of Eq. 8 has as the following form: Define φx as follows. ϕ I x = I 0 T [ϕ I x ϕ I x n ] 0 I. d 3 p 2π 3 a pe ip x 2E p }{{} φx + d 3 p 2π 3 2E p a pe ip x }{{} φ x. 49

53 Then ϕ I x = φx + φ x, φx 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 I x ϕ I x 2 ] = N [ φx + φ x φx 2 + φ x 2 ] [ ] = N φx φx 2 + φ x φx 2 + φx φ x 2 + φ x φ x 2 = φx φx 2 + φ x φx 2 + φ x 2 φx + φ x φ x 2 It can also be written as : ϕ I x ϕ I x 2 : by using :. POINT I 0 N [ϕ I x ϕ I x n ] 0 I = 0. Now we want to see the relation between the time-ordering and the normal ordering. In the following, we write T [ϕ I x ϕ I x n ]? N [ϕ I x ϕ I x n ]. ϕ I x i = ϕ i φx i = φ i for simplicity. Let s start from n = 2. n = 2 For x 0 > x 0 2, Tϕ ϕ 2 = ϕ ϕ 2 = φ + φ φ 2 + φ 2 = φ φ 2 + φ φ 2 + φ φ 2 + φ φ 2 = Nϕ ϕ 2 + [φ, φ 2] [φ, φ d 3 p 2] = 2π 3 e ip x 2E p d 3 p = e ip x x 2 2π 3 2E p For x 0 2 > x 0, we have a similar formula with x x 2. Therefore, 50 d 3 p 2 2π 3 2E p2 e ip 2 x 2 [a p, a p 2 ]