Quantum Field Theory Michaelmas 2014, Prof

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University of Cambridge

Part III of the Mathematics Tripos

Quantum Field Theory Michaelmas 2014, Prof. Malcolm J. Perry Contents

Lecture Synopsis 2

Quantum mechanics: recap 4

1 Introduction to Quantum Field Theory 8 1.1 Overview ...... 8 1.2 Lorentz transformations and Lie algebras ...... 8 1.3 Quantising classical mechanics ...... 13

2 Quantisation of the scalar field: the Klein-Gordon equation 16 2.1 Quantisation of φ(x) and π(x)...... 16 2.2 Quantisation of the Hamiltonian ...... 17 2.3 From the Schrödingerpicture to the Heisenberg picture ...... 20 2.4 Propagators for the Klein-Gordon equation ...... 22 2.5 Applying Noether’s theorem for scalar fields ...... 24

3 Quantisation of the Dirac ﬁeld 27 3.1 Spinors and the Dirac equation ...... 27 3.2 Solving the Dirac equation: antiparticles ...... 30 3.3 Quantizing the Dirac equation ...... 33 3.4 Applying Noether’s theorem for the Dirac ﬁeld ...... 36 3.5 Propagators for the Dirac equation ...... 37

4 Quantisation of the electromagnetic field 39 4.1 Introduction to Maxwell’s field ...... 39 4.2 Maxwell’s equations ...... 39 4.3 Gauge invariance ...... 39 4.4 Quantisation of electromagnetism ...... 41 4.5 Propagators in electromagnetism ...... 43 4.6 Example: complex scalar fields ...... 44

5 Feynman path integrals 46 5.1 Derivation of the Feynman path Integral for a single particle ...... 46 5.2 Example: simple harmonic oscillator ...... 49 5.3 Gaussian integrals and free ﬁelds ...... 52 5.4 Non-Gaussian integrals and perturbation expansions ...... 54

6 Feynman path integrals for scalar ﬁelds 56 6.1 A further introduction ...... 56 6.2 Non-interacting systems ...... 57 6.3 Interacting systems ...... 59 6.4 Solving the path integral ...... 62 6.5 Scattering amplitudes ...... 64

7 Feynman path integrals for the Dirac ﬁeld 67 7.1 Grassmann calculus ...... 67 7.2 Road to solving the path integral ...... 69 7.3 Solution of the path integral ...... 70 7.4 Overview of solution rules ...... 71 7.5 Worked examples ...... 74

8 A little beyond perturbation theory 76 8.1 Real ﬁelds: the soliton ...... 76 8.2 Complex ﬁelds: the Higgs ...... 77

1 Lecture Synopsis

Lecture One Introduction to Quantum Field Theory. Review of special relativity and consideration of the Lorentz transform, focusing on the Lorentz boost. Lecture Two Lorentz transforms on scalars, vectors and covectors. Expressing Lorentz transformations as Lie algebras (infinitesimal transformations). Introduction of the momentum operator Lab. Lecture Three Quantising classical mechanics: Fourier transform of φ, derivation of the Klein-Gordon equation. Overview of functional derivatives and commutation relations. Introduction to Schrödingerand Heisenberg pictures. Lecture Four Derivation of operators φ(x) and π(x), leading to an expression for H in terms of creation and annihilation operators. We see that our Hamiltonian looks like an infinite sum of simple harmonic oscillators. Lecture Five Calculation of the commutation of φ(x) with π(x). Proof that the free scalar field is an infinite collection of SHO. Derivation of number operator leading to an understanding of Bose-Einstein statistics. Moving from the Schrödingerpicture to the Heisenberg picture. Lecture Six Heisenberg picture. Probability amplitude for a particle at y to be found at x (∆(x, y)). Analysis of probability amplitudes and introduction to the Feynman operator ∆F (x, y) Lecture Seven Integral representation of the Feynman propagator. Showing that the Feynman propagator is a Greens function for the Klein-Gordon equation. Overview of Green’s functions: retarded, advanced and anti-time. Introduction to Noether’s theorem. Lecture Eight Noether’s theorem examples. Introduction to the Dirac equation, spinors and different representations (we will only use the Weyl rep.). Lecture Nine Continuing with the Dirac Equation. Definition of spinor using Lorentz transformations. Con- struction of Φ.¯ Solving the Dirac equation. Lecture Ten Pauli-Lubanski spin vector. Finished solution of the Dirac equation resulting in the mathematical prediction of anti-particles. Construction of the action of the Dirac equation. Lecture Eleven Quantising the Dirac equation from ΦH to the Hamiltonian. Highlighting the presence of Fermi- Dirac statistics and the behaviour of the creation and annihilation operators. Lecture Twelve Noether’s theorem for the Dirac equation. Propagators for the Dirac equation. Introduction to the quantisation of Maxwell’s field. Lecture Thirteen Gauge invariance and the quantisation of electromagnetism. Lecture Fourteen Finishing the quantisation of electromagnetism. Studying commutation relations and the Hamil- tonian. Derivation of the propagator with an example applied to the complex scalar field.

2 Lecture Fifteen Derivation of the Feynman path integral. Lecture Sixteen Solution of the harmonic oscillator using the Feynman path integral. Lecture Seventeen Potentially optional: Feynman integral of a non-interacting system using perturbation expansion and corresponding Feynman diagrams introduced. Lecture Eighteen Finding the path integral in QFT, leading to the definition of a quantum field theory. Introduction of a source term and a 101 on how to deal with the horrible ∞ of QFT. Lecture Nineteen Probability amplitude for particle creation (and corresponding Feynman diagram). Wick’s theorem. Introduction of interactions into the path integral, examples of n = 1 and r = 0, 1, 2 are given. Feynman diagrams are also given representing connected and unconnected terms in the expression. Lecture Twenty Quick note on differentiation of Z[J]. Consideration of the four point function and note of unim- portance of unconnected terms. New expression is found by considering the zero point function in order to remove unconnected components. Given a recipe for solving the path integral by considering the Feynman diagrams in both coordinate and momentum space. Overview of scattering cross sections. Lecture Twenty-one Scattering cross sections. Solution of scattering amplitudes in QFT. Introduction to path integrals of the Dirac field and the concept of Grassmann numbers. Lecture Twenty-two A definition of Grassmann variables including introduction to the methods of calculus. Which leads us to a formal definition of path integrals for Grassmann variables. Lecture Twenty-three Application of Grassmann integrals to the Dirac field. An overview of all the different fields we have come across including their respective propagators and interaction terms. A generalised formulation of the rules needed to calculate the various matrix elements for all the different interactions we have covered is given. Finishing with an example of the scattering of two Fermions. Lecture Twenty-four We finish our discussion of Feynman diagrams for scattering fermions and look at our rule system for calculating the scattering amplitudes. The lecture series is concluded by looking at two non- perturbative solutions in Quantum Field theory. The Real Scalar field gives rise to the Soliton and by introducing a Complex field we arrive at the Higgs particle.

3 Quantum mechanics: recap Dirac Notation Basics

2 ~ 2 Schr¨odinger ⇒ i~Ψ(x, t) = − 2m ∂xΨ(x, t) + V (x)Ψ(x, t) If Ψ1,2 are solutions then λΨ1 + µΨ2 is a solution as well for λ, µ ∈ C States live in Hilbert space H: |Ψi ∈ H The Inner Product is deﬁned: hΨ | Φi = hΦ | Ψi∗. || |Ψi ||2 = hΨ | Ψi ≥ 0. hΨ | Ψi = 0 ⇐⇒ |Ψi = 0. Brakets are linear: hΨ| (λ |Φi + µ |χi) = λ hΨ | Φi + µ hΨ | χi

Observables

Pauli matrices and spin states The Pauli matrices are:

0 1 0 −i 1 0 σ = , σ = , σ = . (0.1) 1 1 0 2 i 0 3 0 −1 We deﬁne the spin states as: |↑i , |↓i. Combining these with the Pauli matrices we gain the following expressions: σ1 |↑i = |↓i , σ1 |↓i = |↑i , σ2 |↑i = i |↓i , σ2 |↓i = −i |↑i. Where:

1 0 |↑i = , |↓i = (0.2) 0 1 In one-dimension:

xˆ |xi = x |xi , position basis. (0.3)

Ψ(x) = hx | Ψi , Wavefunction (0.4)

Probability and measurement

• Measure Eigenvalue λi ⇒ state collapses to |Ψii

• Probability of this is (λ ) = P | hΨ | Ψ i2 |. P i j:λj =λi j • The expectation for a state |Ψi is hAi = hΨ | A | Ψi • The uncertainty is (∆A)2 = (A − hAi)2 .

Remark: |Ψi → eiφ |Ψi. Physics is invariant under phase change. More generally: |Ψi 7→ U |Ψi and A 7→ UAU † (i.e. U †U = UU † = I).

4 Simultaneous eigenstates

Observables: A, B, pick states |Ψi,ji with A |Ψiji = ai |Ψiji and B |P siiji = bi |Ψiji. We deﬁne the commutation relation as: [A, B] = AB − BA. Example: [ˆx, pˆ] = i~. We cannot have simultaneous eigenstates. Also, for i = { 1, 2, 3 } , [σi, σj] = 2iijkσk. A good example of this is if we take: 1 |Ψi = and measure using combinations of σ . 1 i 1 1 σ σ |Ψi = σ = (0.5) 3 1 3 1 −1

1 −1 σ σ |Ψi = σ = (0.6) 1 2 1 −1 1 Hence the two quantities are inequivalent.

Uncertainty principle 1 ∆A∆B ≥ | h[A, B]i | (0.7) 2 This is easily proved. Especially with the given hints of:

• What happens when hAi = 0 • Use the Cauchy-Schwartz inequality: hΨ|Ψi hΦ|Φi ≥ | hhΨ|Φii |2

Harmonic oscillator Starting with the Hamiltonian of the simple harmonic oscillator:

pˆ2 1 Hˆ = + mω2xˆ2 (0.8) 2m 2 Deﬁne:

rmω i a = xˆ + pˆ , (0.9) 2~ mω where a is the annihilation operator (a† is the creation operator). ˆ † 1 ˆ † ˆ ˆ 1 This allows us to express: H = ~ω(a a + 2 ); N = a a → H = ~ω(N + 2 ). Where the commutation relations are:

[a, a†] = 1, [N,ˆ a] = −a, [N,ˆ a†] = a†. (0.10) Suppose we have an eigenstate: Nˆ → N |Ψi = λ |Ψi: n 2 †n n 0 ≤ ||a |Ψi || = Ψ a a Ψ D E †n−1 ˆ n−1 = Ψ a Na Ψ = (λ − n − 1)||an−1 |Ψi ||2 < 0 if n + 1 > λ ⇒ a |ψi = 0 for some |ψi = |0i ⇒ N |0i = 0 → ground state Ψ is the vacuum state. ⇒ Na†n |0i → deﬁnes |ni := f(n)a†n |0i Spectrum of N: |0i , |1i , |2i,... Eigenvalues: 0, 1, 2, ... 1 ~ω En = ~ω(n + 2 ), with a zero-point energy of 2

5 Products, identical particles, bosons and fermions

Suppose we have quantum systems H1,H2 with bases |Ψii and |Φji respectively. We can build a space with the basis: Span { |Ψii ⊗ |Φji } = H1 ⊗ H2. Example: H1 = { | ↑1i, | ↓1i } ; H2 = { | ↑2i, | ↓2i } ⇒ H1⊗H2 = { | ↑1i ⊗ | ↑2i, | ↑1i ⊗ | ↓2i, | ↓1i ⊗ | ↑2i, | ↓1i ⊗ | ↓2i } Inner Product (hΨi| hΨj|)(|Ψki |Ψii) = hΨi|Ψki hΨj|Ψii Operators (A ⊗ B)(|Ψi ⊗ |Φi) = (A |Ψi) ⊗ (B |Φi) If H1 = H2 then we have a 1-particle systems. Lets look at H1 ⊗ H2, the two-particle system. Deﬁne swap operators: S : |Ψi ⊗ |Φi 7→ |Φi ⊗ |Φi ,S2 = 1. For identical particles in state |Ψi ,S |Ψi = α |Ψi , |α|2 ⇒ α ± 1 S |Ψi = ± |Ψi where we deﬁne Bosons to be (+) and Fermions to be (-). From this we get the Pauli exclusion principle for Fermions. (This is because Fermions must be in an antisymmetric state.)

Heisenberg picture In the classical world, x(t) = A cos(ωt) + B sin(ωt) 1 In the quantum world, |n(t)i = exp(−iω(n + 2 )t) |n(0)i. The Heisenberg picture tries to address the fact that these terminologies look so diﬀerent! In the Schr¨odingerpicture we have:

i~ |ΨS(t)i = H |ΨS(t)i (0.11)

−iHt ⇒ |ΨS(t)i = exp |ΨS(0)i (0.12) ~ We convert to the Heisenberg picture to obtain:

iHt |ΨH i = |ΨS(0)i = exp |ΨS(t)i (0.13) ~ iHt −iHt AH (t) = exp AS exp (0.14) ~ ~ Example one: dAH iHt i = [AH ,H] ”Heisenberg equation”. Note: [H, exp ] is the evolution equation. ~ dt ~ Example two: i~pˆ 2 ˙ pˆ ˙ 2 [ˆx, H] = m , [ˆp, H] = −i~mω xˆ ⇒ xˆ = m ; pˆ = −mω xˆ Which are Hamilton’s equations of motion.

Perturbation theory Hamiltonian for scattering particles:

2 2 pˆ1 pˆ1 H = + + Vˆ (ˆx1 − xˆ2) (0.15) 2m1 2m1 We can’t solve this directly most of the time, so we solve using perturbative methods. We define H = H0 + V where H0 is defined as a ”simple” Hamiltonian and V is very small. iH0t iH0t −iH0t Now we define an interaction picture: |ΨI (t)i = exp |ΨS(t)i; AI = exp AS exp ~ ~ ~

6 Exercise: a) Let |ai , |bi be eigenstates of H0 (i.e. free particles). Derive:

2 ωt 1 2 sin ( 2 ) 1 = | hb|V |ai | , ω = (Eb − Ea) (0.16) P ω 2 ~ 2 ~ Note: this is the expression for the transfer probability: |ai → |bi in time t. EV b) For the two-state system |1i , |2i, H = VE Show that: v2t2 P = (0.17) ~2

7 1 Introduction to Quantum Field Theory 1.1 Overview Conventions In our formulation of Quantum Field theory we will be using the following conventions:

η = diag(−1, 1, 1, 1) (1.1)

−2ηµν = { γµ, γν } (1.2)

Motivations So what do we use Quantum Field Theory for? 1. Both electromagnetism and gravity are described using ﬁelds 2. It allows us to unify special relativity with quantum mechanics.

3. Provides an explanation on how particles are able to change identity. E.g. Hexcited → Hgs + γ or e+e− → 2γ 4. It provides an explanation of puzzling quantum mechanics. E.g. The origin of spin, or Bose-Einstein statistics. 5. It explains all microscopic phenomena (with the exception of gravity).

1.1.1 Special relativity: recap We use co-ordinates xa = { t, x, y, z } = { t, xi } The proper distance between two points xa and xa + dxa is:

2 2 2 2 2 a b ds = −dt + dx + dy + dz = ηabdx dx (1.3) ab Where ηab is the Minkowski metric. It is the case that ηab = η and we can represent it as a 4x4 square matrix:

−1 0 0 0  0 1 0 0 ηab =   (1.4)  0 0 1 0 0 0 0 1 We deﬁne the light-cone as seen in ﬁgure (1).

1.2 Lorentz transformations and Lie algebras 1.2.1 Classical Lorentz transformations Lorentz transforms are associated to the symmetries of spacetime. The Lorentz group is the set of all transformations which preserve the structure of η such that:

ΛT ηΛ = η (1.5) is satisﬁed. Looking at the group theory axioms we see that:

1. Identity: Λ = δ 2. Composition Law: Λ, Λ0 → Λ00 = ΛΛ0 (successive Λ produce Λ.) 3. Matrix multiplication is associative: Λ(Λ0Λ00) = (ΛΛ0)Λ00

8 Future time-like seperated: ds2 < 0

Exterior: space-like Point P in Minkowski space seperated: ds2 > 0

Past time-like seperated: ds2 < 0

Figure 1: Inside the light-cone we talk about negative “Time-like” separation. Conversely, in the exterior we have a positive “Space-like” separation. Note: this sign comes from the deﬁnition of our metric and is just convention.

4. Inverse (Λ−1): det(ΛT ηΛ) = (detΛ)2(detη) = detη ⇒ detΛ = ±1.

We say that ΛT Λ = I deﬁnes an orthogonal group. ΛT IΛ = I ⇐⇒ I is a d × d identity and hence: Λ ∈ O(d). We classify the Lorenz group as a O(3, 1) with three space-like and 1 time-like dimensions. We can express the Lorentz transforms decomposed into their components:

c d Λ aΛ bηcd = ηab (1.6) and components transform as:

a a0 a b x → x = Λ bx . (1.7) Example: Boost in the x-direction. t − vx x − vt t → t0 = √ x → x0 = √ y = y0 z = z0 (1.8) 1 − v2 1 − v2  γ −γv 0 0 a −γv γ 0 0 1 Λ =   γ = √ (1.9) b  0 0 1 0 1 − v2 0 0 0 1 We set c = 1 and −1 < v < 1 equating: v = tanh φ: 1 1 γ = √ = p = cosh φ → γv = sinh φ (1.10) 1 − v2 1 − tanh2 φ allowing us to express Λ:

 cosh φ − sinh φ 0 0 a − sinh φ cosh 0 0 Λ =   (1.11) b  0 0 1 0 0 0 0 1 where φ is the rapidity. This makes our boost look like a rotation! Sometimes this will be referred to as a “Lorentz rotation” (unsurprisingly. . . ) However there are other kinds of Lorentz transforms: if we don’t disturb the time we get:

9 a 1 0 Λ b = i , (1.12) 0 R j i where R j is the ordinary rotation matrix, R ∈ SO(3) → 3 different rotations. The general Lorentz boost is given as: ! γ −γvj a 2 Λ b = vivj γ , (1.13) −γvi δij + 1+γ but don’t worry about that. We don’t need to learn it and even Perry had to look the ugly thing up while in the lecture. These are the three different Lorentz boosts which are the proper orthochronous Lorentz transformations ↑ 0 2 O+. The term “Orthochronous” here is used to differentiate between the two roots of (Λ 0) ≥ 1 such that we name:

0 orthochronous: Λ 0 ≥ 1 (1.14) 0 non-orthochronous: Λ 0 ≤ −1 “Proper orthochronous” just means that the elements are composed of only rotations and boosts. In the group theory world these are the elements of O(3, 1) which are continuously connected to the identity. Conversely, non-orthochronous transforms are not continuously connected, two given examples are:

−1 0 0 0 a  0 1 0 0 i i Time reversal: Λ =   t = −t x = x (1.15) b  0 0 1 0 0 0 0 1 1 0 0 0  a 0 −1 0 0  i i Parity transform Λ =   t = t x = −x (1.16) b 0 0 −1 0  0 0 0 −1 To see the physics of these two transformations we turn to the light-cone where we can see that the lightcone is only preserved for orthochronous transformations which maintain causality.

1.2.2 Covariant formulation Contravariant vectors are deﬁned by: a a b V = Λ bV (1.17) We obtain covariant vectors by lowering with the metric:

b Va = ηabV (1.18) We deﬁne a scalar S and use the inner product to see how a covariant vectors transforms:

a 0 0 0a S = WaV ,S = WaV = S 0 a b 0 b (1.19) S = WaΛ bV → Wa = WbΛ a

0 0b Or equivalently, we use: Wa = ηabW , which becomes: 0 b c c c Wa = ηabΛ cW = ΛacW = Λa Wc (1.20) which is the most convenient form. This generalises to the Lorentz transformation of an arbitrary tensor: T a1...ap → T 0a1...ap b1...bq b1...bq d a ...a (1.21) = Λa1 Λa2 ...Λap Λ d1 Λ d2 ...Λ q T 1 p c1 c2 cp b1 b2 bq b1...bq

10 We can ﬁnd an expression for the inverse Lorentz transformation by using the identity:

a b Λ cΛ dηbd = ηcd (1.22) −1c When we multiply on both sides by Λ e, we obtain: −1c a b −1c Λ eΛ cΛ dηab = Λ eηcd a b −1 (1.23) δe Λ dηab = Λ de b −1 Λ dηeb = Λ de −1 T −1 So we see that Λ de = Λed, or in general: Λ = Λ To summerize, our transformations are:

V a = Λa V b b (1.24) −1 b Wa = (Λ ) aWb

1.2.3 Quantum mechanical Lorentz transformations Quantum mechanically, Lorentz transformations are eﬀected by a unitary operator: i U (Λ) = exp ω Lab (1.25) 2 ab

ab Where L are a set of Hermitian operators which transform as a tensor and ωab a set of parameters describing the Lorentz transformation. We can describe inﬁnitesimal groups transformations by a Lie algebra which satisﬁes:

U(Λ0)U(Λ) = U(Λ0Λ) (1.26)

When we consider an inﬁnitesimal Lorentz transformation:

a a a 2 Λ b = (δb + ω b + O(ω )) (1.27) When we require this to satisfy 1.22, we obtain:

a a 2 b b 2 (δc + ω c + O(ω )(δd + ω d + O(ω )ηab = ηcd b a ηcd + ωbcδd + ωadδc + ... = ηcd (1.28)

ωcd + ωdc = 0 From which we conclude that the ω-matrices are antisymmetric. Since a 4x4 antisymmetric matrix has 1 6 ( 2 n(n − 1)) independent components, we can distinguish 6 diﬀerent Lorentz transformations. Since ωab is antisymmetric, Lab cannot have symmetric parts, so Lab is also antisymmetric.

1.2.4 Representations of the Lorentz group Now look at the following identity: U(Λ)−1U(Λ0)U(Λ) = U(Λ−1)U(Λ0)U(Λ) (1.29) = U(Λ−1Λ0Λ)

Suppose that Λ0 is inﬁnitesimal and Taylor expand it: i i U(Λ0) = exp ω0 Lab = 1 + ω0 Lab + ... (1.30) 2 ab 2 ab So: i i U(Λ−1) 1 + ω0 Lab U(Λ) = U Λ−1 1 + ω0 Lab Λ 2 ab 2 ab (1.31) i = I + Λ−1ωΛ Lab + ... 2 ab

11 Dropping the identity matrices on both sides, and equating terms of order ω, we obtain:

−1 ab a b cd U(Λ )L U(Λ) = Λc Λ dL (1.32)

Note: this was really unclear in the lectures, I used Weinberg for another definition, but do have a look if it’s correct. When we use the definition of the infinitesimal Lorentz transformation and its representation, it follows that: i i 1 − ω Lef Lab 1 + ω Lgh = (δa + ω a) δb + ωb Lcd (1.33) 2 ef 2 gh c c d d We collect the terms linear in ω, on the left hand side we get a commutator, while the right hand side becomes a combination of L’s and metrics: i − ω Lcd,Lab = ω Ladηbc + Lcbηad (1.34) 2 cd cd which we antisymmetrise over c and d to obtain the commutation rules:

Lab,Lcd = i(−Ladηbc + Lacηbd − Lcbηad + Ldbηac) (1.35)

This relationship deﬁnes L to be the generator of the Lorentz group.

My understanding of equation 1.31 (or 1.32) is this: we use the fact that U is a representation of the Lorentz group to conclude 1.28. The composite Lorentz transformation Λ−1Λ0Λ has components

−1 µ ρ 0ρ σ µ −1 µ 0ρ σ (Λ ) ρ(δσ + ω σ)Λ ν = δν + (Λ ) ρω σΛ ν

Then using the relation between a Lorentz transformation and its inverse:

−1 d d (Λ ) e = Λe

We can show that this composite transformation is

µ 0 ρµ σ µ µ δν + ωρσΛ Λ ν ≡ δν + Ω ν

Here Ω defines the composite transformation. Lowering the µ index on Ω and then plugging into 1.24, expanding the exponential to first order, and comparing with the left hand side of 1.30 gives the result. So in fact equations 1.31 and 1.32 are not quite correct, because of the ordering of indices on the first LT — they should both be ‘forwards’ Lorentz transformations (not one forwards and one inverse) as would be expected of an object with both indices raised. Apologies if all or most of this was obvious to you! –G. Johnson

The diﬀerential operator is deﬁned as:

∂ ∂ ∂ ∂ ∂ = , , , (1.36) a ∂t ∂x ∂y ∂z

When we act with ∂a on a scalar S, ∂aS transforms like a covariant vector. We also have: b b ∂ax = δa, ∂axb = ηab (1.37)

The coordinate representation of Lab is:

Lab = −i(xa∂b − xb∂a) = xapb − xbpa (1.38)

Where the relativistic momentum operator is deﬁned as:

pˆi = (−i∂0 − i∂i) = −i∂a (1.39)

So Lab is a generalisation of the angular momentum operator.

12 1.3 Quantising classical mechanics

In the previous lecture, we saw that Lab = xapb − xbpa, obeys the commutation rules of the Lorentz algebra. The spatial components of this operator correspond to angular momentum. For example:

L12 = xpy − ypx (1.40)

is the z-component of angular momentum. a The relativistic version of the momentum operator is deﬁned as pa = −i∂a. Where p = (E, ~p) and pa = (−E, ~p). So as an operator we can write:

∂ p = −i , −i∇~ (1.41) a ∂t

∂ From the Schr¨odingerequation, we know that i ∂t is equal to the Hamiltonian, which is equal to the total energy for many systems. Furthermore, a plane wave can be written as:

a ~ exp (ikax ) ∼ exp −iEt + ik · ~x (1.42) which is a plane wave with wavevector ~k and energy E. We can write:

a −i∂a(plane wave) = ka exp (ikax ) (1.43)

1.3.1 Fourier transforms This convention ﬁxes the Fourier transforms in space or spacetime. The 4-d Fourier transform of a function φ(x, t) is deﬁned as: Z a φˆ(p) = d4xe−ipax φ(x) (1.44) where the argument x in φ(x) represents both space and time. If applied to the plave wave state:

a φ(x) = eikax (1.45) then ˆ 4 4 φ(p) = (2π) δ (ka − pa) (1.46) This deﬁnition picks out p as begin the 4-vector of energy and momentum. Since we are working with the (-+++)-signature for the metric, the same formula holds for the 3-d transform: Z φˆ(~p) = d3xe−i~p·~xφ(~x) (1.47)

Finally, the 4-d inverse Fourier transform is:

1 Z a φ(x) = d4p φˆ(p)eipax (1.48) (2π)4

13 1.3.2 Classical ﬁeld theory

We know the following things from classical mechanics: • The Lagrangian is a function of the generalised coordinates and velocities:

L = L(qi, q˙i) (1.49)

• The action is deﬁned as: Z I = dtL(qi, q˙i) (1.50)

• The momentum conjugate to qi is deﬁned as: ∂L pi = (1.51) ∂q˙i

• The Hamiltonian is given by: X H = piq˙i − L (1.52) i • Time derivatives are expressed by Hamilton’s equations, using Poisson brackets: ∂H ∂H p˙i = {pi,H} = − q˙i = {qi,H} = (1.53) ∂qi ∂pi

The total energy of a field is defined by the sum of its kinetic energy T and its potential energy V . So we can informally write: (Total Energy − T − V ) φ = 0 (1.54) If we consider a free field (V =0), and use the special relativistic expression for the energy of a particle: p E = ~p2 + m2 (1.55) we can write the following: E2 − ~p2 − m2 φ = 0 ∂2 − + ∇~ 2 − m2 φ = 0 (1.56) ∂t2 ab 2 2 η ∂a∂b − m φ = − m φ = 0 which is an intuitive way to arrive at the Klein-Gordon equation. (Note that is the operator for the wave equation.) We now construct an action I and a Lagrangian density L: Z Z 1 1 I = d4xL = d4x − ∂ φ∂ φηab − m2φ2 2 a b 2 Z 1 1 2 1 (1.57) = dt d3x φ˙2 − ∇~ φ − m2φ2 2 2 2 | {z } The analog of the classical Lagrangian L ˙ The field theoretic analog of qi is φ(x). The momentum conjugate to φ(x) is π(x) = φ(x). With these definitions we can construct a Hamiltonian: Z H = d3x π(x)φ˙(x) − L (1.58) Z 1 1 2 1 = d3x π2 + ∇~ φ + m2φ2 2 2 2

14 Note that we converted the sum into an integral, since φ(x) is now a continuous function instead of the discrete set of coordinates qi. A better deﬁnition of the canonical momentum is: δI π(x) = (x) (1.59) δφ˙ For the ﬁeld theoretical Poisson brackets, we write:

φ˙(x) = {φ(x),H} = π(x) (1.60) δH π˙ (x) = {π(x),H} = − (1.61) δφ(x)

Where for the second Poisson bracket, we compute δH as φ(x) → φ(x) + δφ(x). Z δH = d3x ∇~ δφ · ∇~ φ + m2φδφ (1.62) Z = d3x −∇~ 2φ + m2φ δφ where in the last line we integrated by parts and discarded the boundary term. Taking the functional derivative gives: δH = −∇~ 2 + m2 φ(x) (1.63) δφ(x) So ﬁnally, we have: π˙ (x) = ∇~ 2 − m2 φ(x) (1.64)

So we see that φ˙ andπ ˙ reproduce the Klein-Gordon equation.

1.3.3 Transition to quantum mechanics When we make the transition to quantum mechanics, we make the following changes:

{X,Y } → −i [X,Y ] (1.65)

{qi, pj} = δij → [ˆqi, pˆj] = −iδij (1.66) h i x˙ = {x, H} → x˙ = −i x,ˆ Hˆ (1.67)

There is a problem: φ is not a wavefunction, but an operator, the same goes for π. The conventional wavefunction in quantum mechanics, has been turned into an operator in quantum ﬁeld theory.

1.3.4 Heisenberg & Schrödingerpicture Quantum mechanics can be looked at from two different viewpoints: 1. Schrödingerpicture: time dependent states, static operators 2. Heisenberg picture: time dependent operators, static states. We can define a map between the two pictures:

hi(t)|O |j(t)i = hi(0)|eiHtO e−iHt|j(0)i S S S S (1.68) =H hi|OH (t)|jiH where: iHt −iHt OH (t) = e Ose , (1.69) where we have “given” the time dependence to the operators. Yet there is still an explicit diﬀerence between space and time, this theory is non-relativistic.

15 2 Quantisation of the scalar ﬁeld: the Klein-Gordon equation 2.1 Quantisation of φ(x) and π(x) When quantising position and momentum in the Schr¨odingerpicture, the Poisson bracket turns into a commutator:

{pi, qj} = δij → [pi, qj] = −iδij (2.1) In ﬁeld theory, we regard φ and π as operators, deﬁned at every point in space, quantizing their Poisson bracket gives:

{π(x), φ(x0)} = δ(3)(x − x0) → [π(x), φ(x0)] = −iδ(3)(x − x0) (2.2) To ﬁnd the ﬁeld operator φ(x), we pick a basis of 3-momentum eigenstates. We write:

X i~p·~x † −i~p·~x φ(x) = ape + ape (2.3) momenta † where we added the Hermitian conjugate to make the expression for φ real. Classically, ap and ap would be numbers, but they become operators in the quantum mechanical picture. Note that the exponentials form a complete basis of functions in space, comparable to how every function can be decomposed into a Fourier series.

2.1.1 Lorentz invariant measure We can write the sum over all momentum states as:

X Z d3~p 1 ≡ 3 (2.4) (2π) 2Ep all states

2 21/2 where Ep = + p + m . This type of integral is ’secretly’ Lorentz invariant. To see that this is the case, consider the following integral, which is manifestly Lorentz invariant: Z 4 4 a 0 d pδ (pap )θ(p )g(...) (2.5) where g(...) is some Lorentz invariant function and θ(x) is the step function: ( 1 if x ≥ q0 θ(x) = (2.6) 0 if x < 0 Furthermore, we need the following identity for the delta-function: Z ∞ X g(x0) δ(f(x))g(x)dx = (2.7) |f 0(x )| −∞ −1 0 x0=f (0) We rewrite (2.5) as an integral over 3-momentum and p0:

Z 2 d3pdp0δ − p0 + ~p2 θ(p0)g(...) Z (2.8) = d3pdp0δ −p0 + |~p| p0 + |~p| θ(p0)g(...)

The integrand has zeroes at p0± = |~p|. The zero at p0 = −|~p| is wiped out by the θ-function, so only the positive zero is kept.

16 Z g(...)| 0 d3p p =|~p| | − 2p0| (2.9) Z g(...)| 0 = d3p p =|~p| 2Ep This means that the expression we chose for the sum over all states is equal to (2.5), so the expression is indeed Lorentz invariant.

2.1.2 Field operator φ For the ﬁeld operator φ(x) we now have:

Z 3 d p 1 i~p·~x † −i~p·~x φ(x) = 3 ape + ape (2.10) (2π) 2Ep ˙ Now π = φ. Taking the time derivative will bring a factor of −iEp down:

Z d3p −i π(x) = a ei~p·~x − a† e−i~p·~x (2.11) (2π)3 2 p p † This allows us to quantize the Hamiltonian and look at the commutation rules for ap and ap.

2.2 Quantisation of the Hamiltonian Classically, our Hamiltonian was equal to:

Z 1 1 2 1 H = d3x π2 + ∇~φ + m2φ2 (2.12) 2 2 2 We will compute these terms separately. In doing so, we will frequently use the identity: Z d3xeikx = (2π)3δ(3)(k) (2.13)

In the derivation of quantum mechanical Hamiltonian, all p’s and q’s are 3-vectors, even though the vector arrows are omitted. For the third term, we use (2.10) to obtain:

Z Z 3 3 3 2 3 d pd q 1 ipx † −ipx iqx † −iqx d xφ = d x 6 ape + ape aqe + aqe (2π) 4EqEp Z 3 3 d pd q 1 † † † † = 3 apaqδ(p + q) + apaqδ(q − p) + apaqδ(p − q) + apaqδ(q + p) (2.14) (2π) 4EqEp Z 3 d p 1 † † † † = 3 2 apa−p + apap + apap + apa−p (2π) 4Ep

For the second term in (2.12), we have:

Z 2 Z 3 3 3 ~ 3 d pd q 1 i(p+q)x † i(−p+q)x d x ∇φ = d x 6 apaq(−p · q)e + apaq(p · q)e (2π) 4EqEp † i(p−q)x † † −i(p+q)x +apaq(p · q)e + apaq(−p · q)e Z 3 3 d pd q p · q † † † † = 3 −apaqδ(p + q) + apaqδ(p − q) + apaqδ(p − q) − apaqδ(p + q) (2π) 4EqEp Z 3 2 d p p † † † † = 3 2 apa−p + apap + apap + apa−p (2π) 4Ep (2.15) To ﬁnish this orgy of algebra, we compute the ﬁrst term, using (2.11):

17 Z Z d3xd3pd3q −1 d3xπ2 = a eipx − a† e−ipx a eiqx − a†e−iqx (2π)6 4 p p q q (2.16) Z d3p −1 = a a − a† a − a a† + a† a† (2π)3 4 p −p p p p p p −p Collecting all the terms, we have:

Z d3p 1 1 1 1 H = a a A − + a a† A + + a† a A + + a† a† A − (2.17) (2π)3 p −p 8 p p 8 p p 8 p −p 8 where:

m2 p2 1 A = 2 + 2 = (2.18) 8Ep 8Ep 8 2 2 2 where we used that p + m = Ep . Inserting this into the Hamiltonian, the ﬁrst and last term cancel and we obtain:

Z d3p 1 1 H = a a† + a† a · 2E (2π)3 2E 4 p p p p p p (2.19) Z 3 d p 1 † 1 † 3 apap + ap, ap Ep (2π) 2Ep 2

where in the ﬁrst line we added a factor 2Ep to make the Lorentz invariant interval manifest.. It is immediately obvious that quantizing the Klein-Gordon equation looks a lot like a Simple Harmonic † Oscillator, where ap is an annihilation operator for something with momentum p and ap is a creation operator for the same object. To make the comparison with the Harmonic Oscillator more apparent, we † need to compute ap, ap .

2.2.1 Commutation rules We know the commutation rules for π(x) and φ(x0) from our canonical quantisation procedure:

0 (3) 0 [πS(x), φS(x )] = −iδ (x − x ) (2.20) where the S-subscript emphasises that we are in the Schr¨odingerpicture. Using (2.10) and (2.11) we ﬁnd an explicit expression for the commutator:

Z 3 3 0 d pd q 1 −i ipx iqx h † † i [πS(x), φS(x )] = 6 e e ap − a−p, aq + a−q (2.21) (2π) 2Eq 2

0 To ﬁnd the commutation rules, we multiply the whole expression by e−ikx and e−ilx and integrate over all x and x0. For the LHS of (2.20), we obtain:

Z 3 3 3 3 0 d pd q 1 −i (p−k)x (q−l)x0 h † † i d xd x 6 e e ap − a−p, aq + a−q (2π) 2Eq 2 Z 3 3 1 −i h † † i = d pd q δ(p − k)δ(q − l) ap − a−p, aq + a−q (2.22) 2Eq 2 −i h † † i = ak − a−k, al + a−l 4Ep While for the RHS of (2.20), we have: Z − i d3xd3x0δ(3)(x − x0)e−ikxe−ilx Z (2.23) = − i d3xe−i(k+l)x = −i(2π)3δ(3)(k + l)

Equation these two expressions, we arrive at:

18 1 h i h i h i 2E (2π)3δ(k + l) = [a , a ] − a† , a , + a , a† − a† , a† (2.24) l 2 k l −k l k −l −k −l Since the LHS of the above equation is obviously real, taking the Hermitian conjugate will not change it. We use:

[A, B]† = − A†,B† (2.25) to ﬁnd the Hermitian conjugate of the RHS. Since δ(k + l) is even, it is allowed to let k → −k and l → −l in either (2.24) or its Hermitian conjugate. Adding and subtracting the two expression makes it apparent that:

h † †i h †i 3 [ak, al] = ak, al = 0, ak, al = (2π) 2Ekδ(k − l) (2.26) This means that our Hamiltonian is very similar to the Hamiltonian for the SHO;

X 1 H ∼ E N + (2.27) p p 2 p We infer that a free scalar field is just an infinite collection of independent Simple Harmonic Oscillators if let the ground states be |0i, which we will call the vacuum state. We impose that the vacuum state is † annihilated by ap: ap |0i = 0, and that ap |0i corresponds to an excitation corresponding to energy Ep. To make idea more clear, we take a closer look at the number operator of our free scalar field.

2.2.2 Number operator From the Hamiltonian, we can deduce that the number operator is:

Z 3 d p 1 † N = 3 apap (2.28) (2π) 2Ep

† We consider the number operator on the state aq |0i:

Note that we used that ap annihilates the vacuum (ap |0i = 0)to arrive at the third line. We observe † † that the eigenvalue of N acting on the state aq |0i is 1, so ap |0i is a single excitation of momentum ~p. This prompts the following relations:

† † paˆ p |0i = pap |0i (2.30) † † Hap |0i = Epap |0i (2.31)

So a state with Nk particles with momentum k is given by:

N † ak |0i (2.32) although this state is not normalized. We represent many particles with many diﬀerent momenta ki in a similar way:

N1 Ni a† ... a† ... |0i k1 ki (2.33) Y Ni = a† |0i ki i

19 Note that the order of the a† -operators is completely irrelevant. Since identical bosons have wave- ki functions (states) that are symmetric under the interchange of any pair of particles, Bose-Einstein statistics is now automatic. Let us ﬁnally compute the energy of the vacuum state:

Z d3p 1 1 H |0i = a† a + a , a† E |0i (2π)3 2E p p 2 p p p p (2.34) Z d3p 1 Z = · (2π)3 · 2E δ(0) |0i = d3pE δ(0) |0i (2π)3 2 p p

Where again, we used that ap annihilates the vacuum. This is a problem, since this expression is badly infinite. Not only is the δ(0) an infinite spike, so is the integral over all space of d3p. Then again, it turns out that H |statei − H |0i is in fact finite so we won’t worry too much about this infinity right now.

2.3 From the Schr¨odingerpicture to the Heisenberg picture We already established the map from operators in the Sch¨odinger picture to operators in the Heisenberg picture:

iHt −iHt φH = e φSe (2.35) When we use the commutator of the operators in the Schr¨odingerpicture:

0 [πS(x), φS(x )] (2.36) we can compute the commutator in the Heisenberg picture:

0 0 iHt) −iHt iHt0 −iHt0 iHt0 −iHt0 iHt −iHt [πH (x, t), φH (x , t )] = e πSe e φSe − e φSe e πSe (2.37) Which is a mess, except for when t = t0. For this reason, we deﬁne the equal time commutator:

0 0 iHt 0 −iHt [πH (x, t), φH (x , t )] = e [πS(x), φS(x )] e = eiHt · −iδ(x − x0)e−iHt (2.38) = −iδ(x − x0)

† To ﬁnd φH (x) and πH (x): conjugate ap and ap in the expressions for φS (2.10) and πS (2.11) with the Hamiltonian. To ﬁnd the expression for ap conjugated with the Hamiltonian:

iHt −iHt e ape (2.39) we deﬁne a function of an arbitrary parameter:

λH −λH f(λ) = e ape (2.40) df = eλH [H, λ] e−λH (2.41) dλ

This prompts us to compute the commutator of the Hamiltonian and ap:

Z 3 d q 1 † 1 † [H, ap] = 3 Eq aqaq + aq, aq , ap (2π) 2Eq 2 Z d3q 1 (2.42) = a†, a a (2π)3 2 q p q

= −Epap We put this into (2.41) and obtain:

20 df = −eλH E a e−λH = −E f(λ) (2.43) dλ p p p

Using the boundary condition that f(0) = ap, we can solve this simple diﬀerential equation with:

λEp f(λ) = e ap (2.44) We ﬁll in λ = −it to obtain the creation and annihilation operators in the Heisenberg picture:

iHt −iHt −iEpt e ape = e ap (2.45)

iHt † −iHt iEpt † e ape = e ap (2.46)

We can plug these into (2.10) to obtain the ﬁeld operator in the Heisenberg picture:

Z d3p 1 −iEpt+~p·~x † iEpt−~p·~x φH (x, t) = 3 ape + ape (2π) 2Ep 3 (2.47) Z d p 1 a a ipax † −ipax = 3 ape + ape (2π) 2Ep Note that the exponentials are the plane wave solutions of the Klein-Gordon equation. This expression is manifestly covariant! For this reason, we will from now on work mainly in the Heisenberg picture. For the ﬁeld momentum we recall (2.11):

Z d3p −i π (x) = a ei~p·~x − a† e−i~p·~x (2.48) S (2π)3 2 p p † To obtain πH (x, t), we again conjugate a and a with the Hamiltonian:

3 Z d p −i a a π (x, t) = π(x) = a eipax − a† e−ipax = φ˙ (x, t) (2.49) H (2π)3 2 p p H

We are now in a position to ask what φH (x, t) |0i physically is. To answer that question, we ﬁrst note that:

ap |0i = 0 ∀ap (2.50) And we use the normalization h0|0i = 1. We then look at the explicit expression for φH (x, t) |0i:

3 Z d p 1 a a ipax † −ipax φH (x, t) |0i = 3 ape + ape |0i (2.51) (2π) 2Ep

Note that the ﬁrst term is equal to zero because, again, ap annihilates the vacuum. For the second term, † we note that a single ap acting on |0i is a one-particle state. This is not an eigenstate of momentum, as † it involves a superposition of many ap, because we are integrating over all momenta. We could check whether it is an eigenstate of position. The position operator in the momentum representation is: ∂ xˆa = i (2.52) ∂pa

We let this operator act on φH (x, t) |0i:

3 ∂ Z d p 1 a a † −ipax xˆ φH (x, t) |0i = i 3 ape |0i ∂pa (2π) 2Ep (2.53) a a = −i · ix (φH |0i) = x (φH |0i)

We conclude that φH |0i is an eigenstate of position in spacetime.

21 2.4 Propagators for the Klein-Gordon equation We can now look for the amplitude for a particle at y to be found at x, this is deﬁned as the propagator:

h0| φ(x)φ(y) |0i ∆(x, y) = (2.54) h0|0i When we plug in the explicit expressions for the ﬁelds, we obtain:

3 3 Z d pd q 1 a a a a ipax † −ipax iqay † −iqay ∆(x, y) = 6 h0| ape + ape aqe + aqe |0i (2π) 4EqEp a a Z 3 3 ipax −iqay d pd q e † = 6 h0| ap, aq |0i (2π) 4EqEp a a (2.55) Z 3 3 ipax −iqay d pd q e 3 (3) = 6 · (2π) · 2Epδ (~p − ~q) h0|0i (2π) 4EpEq 3 Z d p 1 a a ipa(x −y ) = 3 e (2π) 2Ep † In the second line, we substituted the commutator of ap and aq, since the reverse term annihilates the vacuum anyway. In the fourth line, we used that the delta function ﬁxes P~ to be equal to ~q, since p0 = p~p2 + m2 and q0 = p~q2 + m2, this means that q0 = p0 (the masses are obviously the same, since p and q are excitations of the same ﬁeld. We now assume that x and y are spacelike separated. This means we can always perform a Lorentz Transformation to obtain the coordinates: ya = (0, ~y), xa = (0, ~x). Using this, we obtain for the propagator:

Z 3 d p 1 i~p·(~x−~y) ∆(x, y) = 3 e (2.56) (2π) 2Ep To evaluate this expression, we pick polar coordinates for ~p and take |~x| = r:

Z p2dp sin θdθdφ eipr cos θ ∆(x, y) = (2π)3 2pp2 + m2 1 Z 1 eipr cos θ π = p2dp 2 p 4π 2 p2 + m2 −ipr 0 1 Z ∞ 1 e−ipr − eipr = pdp 2 p 2 2 4π 0 2 p + m −ir (2.57) 1 Z ∞ 1 = pdp · 2 sin pr 2 p 2 2 8π r −∞ 2 p + m 1 Z ∞ pdp = · sin pr 2 p 2 2 8π r −∞ p + m m ∆(x, y) = K (mr) 4πr 1 R ∞ 1 R ∞ Where in the fourth line, we used that the integrand is even to replace 0 by 2 −∞. K1(mr) is a modiﬁed Bessel function of order 1. When we look up this function, we see that it only depends on the distance between two points, and it is non-zero:

1 ∆ ∼ for small r (2.58) 4πr2 r m ∆ ∼ e−mr for large r and m 6== 0 (2.59) 32π3r3

This is a rather strange fact at ﬁrst sight: the vectors ~x and ~y are spacelike separated, yet the amplitude ∆ is non-zero (although it dies oﬀ exponentially). This would violate causality, since the points are

22 outside each other’s lightcones. We can look at this in another way: let us compute the commutator of the field operators at the two different points. If two operators commute, they don’t affect each other so only if [φH (x, t), φH (y, t)] is non-zero, then the field at x influences the field at y (note that the propagator ∆ resembles the commutator, but it is not the same, it only involves one of the combinations). The commutator of the two fields is:

[φH (x, t), φH (y, t)] = ∆(x, y) − ∆(y − x) = 0 (2.60) since ∆ is an even function if x and y are spacelike separated. We conclude that operators always commute at spacelike separation and causality is preserved.

2.4.1 Feynman propagator We now introduce the Feynman propagator, which is deﬁned as:

∆F = h0| T φ(x)φ(y) |0i (2.61) = h0| φ(x)φ(y)θ(x0 − y0) + φ(y)φ(x)θ(y0 − x0) |0i where T is a time ordering operator. Note that the Feynman propagator diﬀers from the general propagator in that it only considers future-directed amplitudes. We claim that the Feynman propagator can be written as:

Z d4p eip(x−y) ∆ = −i (2.62) F (2π)4 p2 + m2 − i 2 where is a small, positive, real number. Observe that we can write p2 + m2 as − p0 + ~p2 + m2 so, q the integral has poles at p0 = ∓ p~2 + m2. By writing d4p as d3pdp0, we can perform a contour integral over p0. This is the reason for adding the −i-term: in this way, the poles are oﬀ the real axis and we can ﬁnd a suitable contour around them. p 2 2 We use that Ep = p + m , Taylor expand the square root and absorb the factor of 2 into to discover 0 0 that the poles of the integral are located at p = Ep − i and p = −Ep + i. The exponential in the integrand can be written as:

0 eip(x−y) = e−ip (x0−y0)+i~p·(~x−~y) (2.63)

0 so if x0 − y0 > 0, the integrand blows up as Im p gets large and positive. Hence, for x0 − y0 positive, ∆F (x, y) comes from a residu at Ep − i.

Z 3 d p 1 −1 i~p·(~x−~y) ∆F (x, y) = −i 3 · −2πi e (2π) 2π 2Ep | {z } Residue at Ep − i (2.64) Z 3 d p 1 i~p·(~x−~y) = 3 e (2π) 2Ep

If x0 − y0 < 0 (so if x is to the future of y), ∆F comes from a residue at the other pole, so we need to close the integral in the top half-plane. The other contribution is:

Z 3 d p 1 −i~p·(~x−~y) 3 e (2.65) (2π) 2Ep

We see that these two expression correspond precisely to our deﬁnition of the general propagator (2.56). Adding the two together thus gives the Feynman propagor (2.61). We conclude that the relativistic expression (2.62) is indeed equivalent to (2.61).

23 2.4.2 Green’s functions If we consider the Klein-Gordon operator acting on the Feynman propagator, we can see another interpretation:

Z d4p eip(x−y) − + m2 ∆(x, y) = −i − + m2 x x (2π)4 p2 + m2 − i Z d4p eip(x−y) = · (p2 + m2) (2.66) (2π)4 p2 + m2 − i Z d4p = eip(x−y) = δ(4)(x − y) (2π)4

We see that ∆F is a Green’s function for the Klein-Gordon operator. a We can compare this with electrodynamics, where in the gauge ∂aA = 0, Maxwell’s equations are: a a A = −µ0j . We then use the same Green’s function to obtain: µ Z ja(y) Aa = 0 d4y (2.67) 4π (x − y)2 The Feynman propagator is not the only Green’s function for the Klein-Gordon operator. Since we can choose the location of the poles arbitrarily, there are other possibilities. If we take the same expression as in (2.62), but instead take both poles below the real axis, we obtain the retarded Green’s function ∆R, which has the property: ( ∆(x − y) − ∆(y − x) x0 > y0 ∆R(x, y) = (2.68) 0 x0 < y0

where ∆ is the general propagator (2.56). The retarded Green’s function is useful in electrodynamics, since it explicitly excludes eﬀects from the future. Finally, we can deﬁne the Advanced Greens’s function ∆A, which is the time reverse of ∆R.

2.5 Applying Noether’s theorem for scalar fields We finish the chapter on scalar fields by considering Noether’s theorem. We know from classical mechanics that Noether’s theorem states that a global symmetry of the action leads to a conserved quantity (although this might not always be true quantum mechanically, occasions where this happens are called ’quantum anomalies’). Let θ be a constant in spacetime and suppose that the field φ depends on θ continuously:

φ → φ(θ) φ(0) = φ (2.69)

As an example, we can consider a complex scalar ﬁeld φ, for which the action is given by: Z 4 ∗ ab ∗ I = d x −∂aφ ∂bφη − mφ φ (2.70)

We can also view this conﬁguration as two real free scalar ﬁelds by taking: φ + φ φ = 1√ 2 (2.71) 2

† The action for φ is then the sum of the actions for φ1 and φ2, resulting in φ and φ being described by two diﬀerent degrees of freedom. The symmetry of this system is the rotation of φ1 and φ2 into each other: φ(θ) = eiθφ(0) (2.72)

We immediately see that the action is invariant under this rotation: δI = 0 (2.73) δθ

24 In general, say we have a set of ﬁelds φα, for which the action is given by: Z 4 I = d xL(φα, ∂aφα) (2.74)

Then the variation in the action is given by: Z 4 δL δL δI = d x δφα + δ(∂aφα) (2.75) δφα δ(∂aφα)

Since φα varies as θ varies, we have: ∂φ ∂(∂ φ ) δφ = α δθ δ(∂ φ ) = a α δθ (2.76) α ∂θ a α ∂θ Plugging this into the action, we obtain: Z ∂L ∂φ ∂L ∂(∂ φ ) δI = d4x α δθ + a α δθ ∂φα ∂θ ∂(∂aφα) ∂θ Z 4 ∂L ∂L ∂φα ∂L ∂φα (2.77) = d x − ∂a δθ + ∂a δθ ∂φα ∂(∂aφα) ∂θ ∂(∂aφα) ∂θ | {z } =0 by virtue of the EoM where we integrated by parts to from the ﬁrst to the second line. Thus, if δI = 0, we see that: ∂L ∂φα ∂a δθ = 0 (2.78) ∂(∂aφα) ∂θ

a a a We can thus deﬁne a Noether current J which satisﬁes ∂aJ = 0, where J is given by:

∂L ∂φα Ja = (2.79) ∂(∂aφα) ∂θ J a is conserved, as long as the equations of motion hold. We can write: Z Z tf Z spatial ∞ 4 a 3 0 0 = d x∂aJ = d x J + dt Ji (2.80) V ti −spatial ∞ The second term is zero if the fields fall off fast enough. The first term is the so-called Noether charge. Noether’s theorem states that there is always a conserved charge, if there is a symmetry of the action. As an example of a global symmetry, we can take baryon number. This in contrast to local symmetries due to gauge invariance, e.g. electromagnetism and electric charge.

2.5.1 Example: Complex scalars For a complex scalar ﬁeld, we have:

∗ ab 2 ∗ L = −∂aφ ∂bφη − m φ φ (2.81) The Lagrangian (and hence the action) is invariant under the U(1) gauge transformation:

φ → eiθφ (2.82)

We regard φ and φ∗ as independent quantities, so obtain two Equations of Motion:

2 ∗ φ : − + m φ = 0 (2.83) ∗ 2 φ : − + m φ = 0 (2.84) We want to construct the Noether current for this transformation J a so we compute:

∂L ab ∂L ∗ ab ∗ = −∂aφη = −∂bφ η (2.85) ∂(∂aφ ) ∂(∂aφ)

25 Furthermore, we know how φ depends on θ:

φ(θ) = eiθφ(0) φ∗(θ) = e−iθφ∗(0) (2.86)

So for the other term in the Noether current, we have: ∂φ ∂φ∗ = iφ = −iφ∗ (2.87) ∂θ ∂θ Plugging this all in (2.79), we get: J a = iφ∗∂aφ − iφ∂aφ∗ (2.88) as the Noether current for this symmetry. We can check whether this is conserved:

a ∗ a ∗ a∗ ∗ ∂aJ = i∂aφ ∂ φ + iφ φ −i∂aφ∂ φ − iφφ (2.89) This is not trivially zero, unless we use the equations of motion. Doing that, we obtain:

a ∗ 2 2 ∗ ∂aJ = iφ m φ − iφm φ = 0 (2.90)

a It is a universal fact that ∂aJ = 0 only after using the Equations of Motion

26 3 Quantisation of the Dirac field 3.1 Spinors and the Dirac equation 3.1.1 The Dirac equation We start with the Schrödingerequation: ∂ Hψ = i ψ (3.1) ∂t We want the Schrödingerequation to be consistent with Special Relativity. Dirac noticed that the equation is first order in time derivatives, so it must also be first order in space derivatives. He came up with the following equation:

a (−iγ ∂a + m) ψ = 0 (3.2) To see what the γa represent, we use the following:

b a a b 2 (iγ ∂b + m)(−iγ ∂a + m)ψ = γ γ ∂a∂b + m ψ = 0 (3.3) Since the partial derivatives commute, the antisymmetric part must vanish under a ⇐⇒ b, hence we can write the ﬁrst term as: 1 γaγb∂ ∂ = γa, γb ∂ ∂ , (3.4) a b 2 a b where the anticommutator is { A, B } = AB + BA. We want (3.3) to be similar to the Klein-Gordon equation, to be consistent with Special Relativity. We conclude that: 1 γa, γb = γaγb + γbγa = −2ηab1 (3.5) 2 Since if we plug this into (3.3), we have:

ab 2 −η ∂a∂b + m ψ = 0 (3.6) It turns out that (3.5) can only be solved by matrices with matrices of at least 4 dimensions. 2 2 Dirac noticed that γ0 = I and γi = −I. We see that γ0 is Hermitian and γi is anti-Hermitian. The combinations of such quantities consist of a Cliﬀord algebra, this has many representations. Examples are the Weyl representation:

0 I 0 σi γ0 = γi = (3.7) I 0 −σi 0 The Dirac representation:

I 0 0 σi γ0 = γi = (3.8) 0 −I −σi 0 And the Majorana representation:

0 σ2 iσ3 0 0 −σ2 −iσ1 0 γ0 = , γ1 = , γ2 , γ3 = (3.9) σ2 0 0 iσ3 σ2 0 0 iσ1 which is used a lot in supersymmetry, where it is useful because it is purely imaginary. We will only use the Weyl representation.

27 3.1.2 Details of the γ-matrices For the Hermitian conjugate, we have:

(γa)† = γ0γaγ0 (3.10) Using the anticommutation relation, it can easily be shown that:

(γ0)† = γ0γ0γ0 = γ0 (3.11a) (γi)† = γ0γiγ0 = −γ0γ0γi = −γi (3.11b)

Which reinforces that γ0 is Hermitian and γi is anti-Hermitian. We also deﬁne the charge conjugation matrix C to look at the transpose of the γ-matrices:

(γa)T = −CγaC−1,C = −CT = −C† = −C−1 (3.12) Note: For some reason, Professor Perry uses a slightly different definition for C, which in for example Srednicki is defined with the relation:

(γa)T = −C−1γaC (3.13) it does not really matter: since C−1 = −C, the two deﬁnitions are equivalent. In the Weyl representation, C is represented by:

0 0 1 C = , = (3.14) 0 − −1 0 Another useful and important γ-matrix is:

γ5 = −iγ0γ1γ2γ3 (3.15) In the Weyl representation, we have: I 0 γ5 = (3.16) 0 −I and it can be shown that:

γ5, γa = 0 (3.17) by anticommuting γa through the expression until it can be multiplied by the other γa. So C, γ5 and γa are part of a basis of 4x4 matrices, other basis elements are:

1 γab = γ[aγb] = γa, γb (3.18) 2 γabc = γ[aγbγc] (3.19) γabcd = γ[aγbγcγd] (3.20)

So we can look at how the basis is decomposed into its separate elements:

Matrix C γa γab γabc γabcd Number of basis 1 4 6 4 1 elements

Notice that the basis is independent of γ5 as it is a composite, formed with the four basis elements of γa and the Levi-Civita tensor: i γ5 = − γaγbγcγd (3.21) 24 abcd

28 3.1.3 Spinors We recall the Dirac equation:

a (−iγ ∂a + m)ψ = 0 (3.22) We know now that to solve this equation, the γ are 4x4 matrices, so ψ should be a column vector of dimension 4. ψ is called a spinor. An index can be attached to ψ denoting the value of the spin, this is for obvious reasons called a spinor index or Dirac index. With the index attached, we write ψ as ψα. It α follows that a γ-matrix with spinor indices is given by: γβ . Recall that the charge conjugation matrix was deﬁned by:

(γa)T = −CγaC−1 (3.23) We can put back the suppressed indices on the γ’s and see that C must have two lower indices:

a T γ a α −1 βγ (γ ) δ = − (C)δα (γ ) β(C ) (3.24) β We see that C really turns a spinor into its conjugate by left multiplication: ψα = Cαβψ . It follows −1 γ −1γδ −1 that we can raise indices by left multiplication with C : ψ = C ψδ. C and C thus behave a bit like the metric in that it can be used to raise and lower spinor indices. However, it is unlike the metric in that it is antisymmetric. That is why it is always associated with left multiplication.

3.1.4 Spinor representations of the Lorentz group

We saw in 1.2 that Lab were the generators of the Lorentz group. Recall that we had:

[Lab,Lcd] = −i (Labηbc − Lacηbd − Lbdηac + Lbcηad) (3.25)

There exists a representation of Lab in terms of gamma matrices: i 1 L = − γ γ = (γ γ − γ γ ) (3.26) ab 2 ab ab 2 a b b a

This does not depend on any particular representation, it just requires that {γa, γb} = −2ηab.

We know that Lorentz transformations are given by:

i U = exp ω Lab = exp − ω γab (3.27) ab 2 ab This representation acts on spinors, with indices we can write: α α i ab U β = exp − ωabγ (3.28) 2 β This deﬁnes a Lorentz transformation on a spinorial object:

U(Λ)ψ(x) = ψ(Λx) (3.29) This really deﬁnes the idea of a spinor.

We will now try to construct a scalar from spinorial objects. To do this, we need to find a suitable object in the dual space of ψ. As a first guess, we look at the Hermitian conjugate ψ†. As our candidate for a scalar we define:

A = ψ†ψ (3.30) We know that a scalar is invariant under Lorentz transformations, so we consider an inﬁnitesimal transformation of A:

δA = δψ†ψ + ψ†δψ (3.31)

29 If we use the explicit expression for a Lorentz transformation and only use the ﬁrst order term, we see that δψ is linear in ψ: i δψ = ω γabψ (3.32) 2 ab For δψ†, we have:

i † δψ† = ψ†ω γab (3.33) 2 ab ab a b a b† b† a† Since ωab is antisymmetric, γ can be replaced by γ γ . We then use that γ γ = γ γ and (γa)† = γ0γaγ0: i δψ† = ψ†γb†γa†ω 2 ab i = ψ†γ0γbγ0γ0γaγ0ω (3.34) 2 ab i = ψ†γ0γbγaγ0ω 2 ab We can then write for the variation of our scalar δA: i ω ψ†γ0γbγaγ0ψ + ψ†γaγbψ 6= 0 (3.35) 2 ab This is not a scalar, because of all the factors of γ0. We conclude that our guess that ψ†ψ is a scalar was incorrect. Instead, we deﬁne another object in the dual space:

¯ † 0 ψD = ψ γ (3.36) This object is called the Dirac conjugate, we can compute its variation:

δψ¯ = δψ†γ0 i = γ0γbγaγ0ω γ0 2 ab i (3.37) = ψγ¯ bγaγ0ω γ0 2 ab i = ψγ¯ bγaω 2 ab We now deﬁne a new scalar S = ψψ¯ .

δS = δψψ¯ + ψδψ¯ i (3.38) = ω ψγ¯ aγbψ + ψγ¯ aγbψ = 0 2 ab a b Since ωab is completely antisymmetric, and we have a manifestly symmetric combination of γ and γ , δS is equal to zero. S is therefore a scalar under Lorentz transformations. Similarly, ψχ¯ is a scalar under Lorentz transformations for any ψ and χ, ψγ¯ aχ is a vector and ψγ¯ abχ, ψγ¯ abcχ and ψγ¯ abcdχ are tensors.

3.2 Solving the Dirac equation: antiparticles We again take a look at the Dirac equation:

a (−iγ ∂a + m) ψ = 0 (3.39) We assume plane wave solutions of the form:

ψ = u(p)eipx (3.40)

30 where u(p) is a spinor. When we substitute this into the Dirac equation:

a (γ pa + m)u(p) = 0 (3.41) We assume that m 6= 0. As long as pa is future-directed and timelike, we can perform a Lorentz a transformation such that p = (m, 0, 0, 0) and pa = (−m, 0, 0, 0). The Dirac equation then becomes:

(−mγ0 + m)u(p) = 0 (3.42) We can write this out explicitly:

 m 0 −m 0   0 m 0 −m   u(p) = 0 (3.43) −m 0 m 0  0 −m 0 m

We see that this matrix is linearly dependent and has rank 2 instead of rank√ 4. This gives rise to two solutions of the Dirac equation, which we presciently normalize by a factor of m.

1 0 √ 0 √ 1 u+ = m   u− = m   (3.44) 1 0 0 1

u± describe two spin states of the system.

3.2.1 Pauli-Lubanski spin-vector We can deﬁne the Pauli-Lubanski spin vector as follows: 1 p Sa = − abcdJ d (3.45) 2 bc m

where Jbc is the total angular momentum (orbital+spin) and pd is the 4-momentum of the particle. We see that this object is only deﬁned for m 6= 0. In the rest frame, the particle only has spin: the orbital angular momentum is zero. We recall that Lab = −i(xa∂b − xb∂a) are the generators of the Lorentz algebra for scalar objects. If we i only have spin, we need the spinorial generators, deﬁned by Sab = − 2 γaγb. Looking at the z-component of the Pauli-Lubanski vector in the rest frame, we have: 1 p 1 i Sz = − zxytS t · 2 = − · 1 · γ γ · 2 · (−1) = − γ γ (3.46) 2 xy m 2 x y 2 x y

The factor 2 in the third equality arises from the two spin states of the system: Sxy = −Syx. By writing out the γ-matrices explicitly, it is straightforward to show that: 1 S u = ± u (3.47) z ± 2 ± 1 The eigenvalue of Sz for our spinorial objects is ± 2 which shows that we are dealing with spin-1/2 objects. To ﬁnd a general expression for u(p) we act with a Lorentz boost. Suppose that the object has a a momentum in the z-direction: p = (E, 0, 0, pz). Then the equation for u is:   m 0 −E + pz 0  0 m 0 −E − pz   u(p) = 0 (3.48) −E − pz 0 m 0  0 −E + pz 0 m This gives rise to the solutions:

31 √    E − pz √ 0  0   E − pz u+ = √  and u− =   (3.49)  E + pz √ 0  0 E + pz 2 2 2 we see that this gives the correct expression as p → 0, E = pz + m . Using our deﬁnition for conjugate spinors, we have:

† 0 u¯± = u±γ (3.50) Hence √ u¯+ = m 1 0 1 0 , u− = 0 1 0 1 (3.51) The conjugate spinors look very similar to the spinors itself but this similarity is illusory, it only arises because we are working with the Weyl representation. An interesting question to ask is why the solutions of the Dirac equation have only two independent components, while the system contains 4 degrees of freedom. We can ﬁnd the other solutions by using plane wave solutions which travel the other way, deﬁning another ψ(x) as:

a −ipax ψ(x) = vs(p)e (3.52) Where the s-subscript denotes the spin state. If we plug this ψ into the Dirac equation, we obtain:

a (−γ pa + m)vs(p) = 0 (3.53) Viewing this in the rest frame again, we have:

m 0 m 0   0 m 0 m   v(p) = 0 (3.54) m 0 m 0  0 m 0 m which has solutions: √ √ v+ = m 0 1 0 −1 v− = m −1 0 1 0 (3.55) For the v-spinors, we also have that: 1 S v = ± v (3.56) z ± 2 ± If the energy of u is positive, this automatically means that the energy of v is negative. The conjugate spinorsv ¯± are constructed int he same way: √ √ v¯+ = m 0 −1 0 1 v− = m 1 0 −1 0 (3.57) Products of u and v spinors where one is a spinor and one is a conjugate spinor form Lorentz invariant scalars:

u¯s(p)us0 (p) = 2mδss0 (3.58a)

u¯s(p)vs0 (p) = 0 (3.58b)

us(p)¯vs0 (p) = 0 (3.58c)

vs(p)¯vs0 (p) = −2mδss0 (3.58d)

32 Similarly, combinations of spinors u, v and conjugate spinorsu, ¯ v¯ with a γa in between (e.g.uγ ¯ au) are spacetime vectors. For the γ0-case, we have:

0 u¯sγ us0 = 2mδss0 (3.59) 0 u¯sγ vs0 = 0 (3.60) 0 v¯sγ vs0 = −2mδss0 (3.61)

It works similarly for spatial γ-matrices, e.g.:

a a a u¯sγ us0 = 2p δss0 = −v¯sγ vs0 (3.62) If we take s = s0 and sum over s, we see that the expression:

X α us (p)¯us0β(p) (3.63) s has the same structure as a γ-matrix. Note that this is the outer product of a spinor and a conjugate spinor. We have:

1 0 1 0 0 0 0 0 1 0 1 0 X α 0 0 0 0 0 1 0 1 0 1 0 1 u (p)¯u 0 (p) = m + m = m (3.64) s s β 1 0 1 0 0 0 0 0 1 0 1 0 s       0 0 0 0 0 1 0 1 0 1 0 1 Where the ﬁrst matrix corresponds to s = s0 = + and the second to s = s0 = −. Remember that δ, γa, γab, γabc and γabcd form a basis for 4x4 matrices, they are linearly independent. We can thus write:

X 0 1 0 us(p)¯us0 (p) = m(1 + γ ) = m − p0γ (3.65) s This is true only in the rest frame, in general we have:

X a us(p)¯us0 (p) = m − γ pa = m − p¡ (3.66) s where we have introduced the Feynman slash notation:

a A = γ Aa (3.67) For the other outer products, we have:

X us(p)¯vs0 (p) = 0 (3.68) s X vs(p)¯vs0 (p) = −m − p¡ (3.69) s

3.3 Quantizing the Dirac equation 3.3.1 The action for the Dirac equation The action for the Dirac equation can be written as: Z 4 ¯ a I = d xψ (iγ ∂a − m) ψ (3.70)

If we vary this action with respect to ψ¯, we obtain the at this point familiar Dirac equation:

a (−iγ ∂a + m)ψ = 0 (3.71) We can also integrate the action by parts to obtain:

33 Z 4 ¯ a ¯ I = d x(−i∂aψγ − mψ)ψ (3.72)

Varying the action with respect to ψ now gives:

¯ a ¯ (−i∂aψγ − mψ) = 0 (3.73) Right multiplication with γ0 and taking the Hermitian conjugate will give back the Dirac equation. Note that this is the case because the ﬁrst term will pick up a minus sign to anticommute the γ0 through.

3.3.2 Dirac ﬁeld expressions In the Heisenberg picture, we can write the Dirac ﬁeld as: X ψH (x, t) = (Operator) (Solution of Dirac equation) (3.74) all solutions of the Dirac equation where we know that the solutions of the Dirac equation depend on 4-momentum and spin. We have already found solutions of the Dirac equation in (3.44) and (3.55). We add operators b and d, which we will investigate later and use our Lorentz invariant measure:

X Z d3p 1 ψ (x, t) = b (p)u (p)eipx + d†(p)v (p)e−ipx (3.75) H (2π)3 2E s s s s s p Note that classically, b and d are just numbers, but quantum mechanically, they become operators. We can compute the momentum conjugate to ψ, by observing that iψγ¯ 0ψ˙ is the only term with a time derivative in the Lagrangian:

∂L ¯ 0 πψ = = iψγ (3.76) ∂ψ˙ We see that there is no momentum conjugate to ψ¯. To quantize the Klein-Gordon equation, we replaced the Poisson brackets by commutators. It turns out that this leads to garbage for theories that are linear in the ﬁrst derivative. The correct way to quantise these theories is to replace the Poisson brackets by anticommutators. Analogous to the case for scalar ﬁelds, we can write the canonical equal time anticommutator rules:

α 0 (3) 0 α {ψ (x, t), πβ(x , t)} = −iδ (x − x )δβ (3.77) where α and β are spinor indices. Equivalent but instructive ways to write these relations are:

α ¯ 0 0 α 0 ψ (x, t), i(ψγ )β(x , t) = −iδβ δ(x − x ) (3.78a) n α † 0 o α 0 ψ (x, t), ψβ(x , t) = −δβ δ(x − x ) (3.78b) α ¯ 0 0 α 0 ψ (x, t), ψβ(x , t) = −(γ ) βδ(x − x ) (3.78c)

Finally, we have:

{ψ(x, t), ψ(x0, t)} = 0 (3.79)

34 3.3.3 Creation and annihilation operators To see what this means for b and d, we compute the quantity: Z 3 −ipx 0 d xe u¯r(p)γ ψ(x, t) (3.80)

Plugging in the expression for the Dirac ﬁeld, we obtain:

X Z Z d3q 1 d3xe−ipxu¯ (p)γ0 b (q)u (q)eiqx + d†(q)v (q)e−iqx r (2π)3 2E s s s s s q X Z d3q h i = u¯ (p)γ0 b (q)u (q)δ(p − q)ei(Ep−Eq t) + d (q)v (q)δ(p + q)e−i(Ep+Eq )t 2E r s s s s s q ( (3.81) 1 ((( X 0 0 (( 2iEpt = u¯r(p)γ bs(p)us(p) + u¯r(p)γ(ds((−(p)vs(−p)e 2E ((( s p X 1 = b (p)2E δ = b (p) 2E s p rs r s p where we used the spinor outer products (3.58) to go from the third to the fourth line. We conclude that: Z 3 −ipx 0 br(p) = d xe u¯r(p)γ ψ(x, t) (3.82)

Similarly, we have:

Z † 3 ipx ¯ 0 br(p) = d xe ψ(x, t)γ ur(p) (3.83) Z 3 −ipx 0 dr(p) = d xe v¯r(p)γ ψ(x, t) (3.84) Z † 3 ipx ¯ 0 dr(p) = d xe ψ(x, t)γ vr(p) (3.85)

Note that b and d are independent of time. Using the canonical anticommutation rules (3.77) we can compute the anticommutators for the b and d operators:

n † o 3 n † o bs(p), bs0 (q) = (2π) · 2Epδss0 δ(p − q) = ds(p), ds0 (q) (3.86) Everything else has vanishing anticommutator. This looks a lot like an inﬁnite set of Dirac creation and annihilation operators. We recall the non- relativistic version of those operators. Deﬁning a ground state |0i, we have:

† 2 † 2 b |0i = 0 bs, bs = 1 b = (b ) = 0 (3.87) If we let the creation operator b† act on the ground state, we have:

b† |0i = |1i (3.88) Note that is impossible to get more than one excitation of the same object in this formalism, conform the Pauli exclusion principle, since (b†)2 = 0. Finally, we can compute the normalisation of the excited state:

h1|1i = h0| bb† |0i = h0| 1 − b†b |0i = h0|0i = 1 (3.89)

(b†)2 |0i = b† |1i = 0 (3.90) and ﬁnally b |1i = |0i. We can now look at the possible states for our relativistic creation and annihilation operators, we have:

35 bs(p) |0i = ds(p) |0i = 0 (3.91) † bs |0i = State with particle of momentum p and spin s (3.92) † ds |0i = State with antiparticle of momentum p and spin s (3.93) We also conclude that no states exist with two particles or antiparticles with the same spin and momentum. We can write an arbitrary state with particles of momenta pi and spins si and antiparticles of momenta qi and spins ri as:

m n Y Y |p s , . . . , p s ; q r , . . . , q r i = b† (p ) d† (q ) |0i (3.94) 1 1 m m 1 1 n n si i rj j i=1 j=1 Because of the vanishing anticommutators, this is antisymmetric under interchange of adjacent pairs of operators: it follows that Fermi-Dirac statistics have to be applied to fermions.

3.3.4 The Dirac ﬁeld Hamiltonian Classically, the Hamiltonian is given by: Z H = d3x π(x)ψ˙(x) − L (3.95)

For the Dirac ﬁeld, this translates to: Z 3 ¯ i ¯ H = d x −iψγ ∂iψ + mψψ (3.96)

Note that the πψ˙-term cancels with the time derivative in the Lagrangian. We can plug in the expressions for ψ and ψ¯ to get:

Z 3 d p 1 X † † H = 3 · Ep bs(p)bs(p) − ds(p)ds(p) (2π) 2Ep s (3.97) Z d3p 1 X = · E b†(p)b (p) + d†(p)d (P ) − (2π)3 · 2E δ(0) (2π)3 2E p s s s s p p s where we used the canonical anticommutation relations to switch the d-operators. We recognise the number operator in both the b†b and the d†d-term and see that this Hamiltonian has the form:

H = #particles + #antiparticles − inﬁnite constant (3.98)

3.4 Applying Noether’s theorem for the Dirac ﬁeld We recall the action for the Dirac ﬁeld: Z 4 ¯ a I = d xψ(−iγ ∂a + m)ψ (3.99)

It is obvious from the action that the Dirac action has a global U(1)-symmetry where:

ψ → ψ0 = eiθψ ψ¯ → ψ¯0 = e−iθψ¯ (3.100) The associated Noether current is:

J a = ψγ¯ aψ (3.101) Note that is does not matter whether this is computed through varying the action with respect to ψ or with respect to ψ¯. The Noether current is conserved:

36 a ¯ a ¯ a ∂aJ = ∂aψγ ψ + ψγ ∂aψ = 0 (3.102) Which becomes apparent after using the Dirac equation. This current gives rise to a conserved Noether charge:

Z X Z d3p 1 Q = d3xJ 0 = b†(p)b (p) − d†(p)d (p) + (constant) (3.103) (2π)3 2E r r r r r p Q is the number of b-excitations minus the number of d-excitations, which is the same as the number of particles minus the number of antiparticles.

3.5 Propagators for the Dirac equation In close analogy with the scalar field propagator, we can define a Feynman propagator for Dirac fields: α α ¯ ∆F (x, y) = h0| T ψ (x)ψβ(y) |0i β (3.104) α ¯ 0 0 0 0 ¯ α = h0| ψ (x)ψβ(y) |0i θ(x − y ) − θ(y − x ) h0| ψβ(y)ψ (x) |0i Where the minus between the two terms arises because the ψ-fields anticommute. Computing the first term with the explicit expressions for the fields gives us:

X Z d3p 1 Z d3q 1 θ(x0 − y0) h0| b (q)u (q)eiqxb†(p)¯u (p)e−ipx |0i (2π)3 2E (2π)3 2E s s r r r,s p q X Z d3pd3q 1 =θ(x0 − y0) u (q)¯u (p)ei(qx−py) h0| −b(p)†b (q) + (2π)3 · 2E δ(p − q)δ |0i (2π)6 4E E s r r s p rs r,s p q X Z d3p 1 =θ(x0 − y0) u (p)¯u (p) eip(x−y) (2π)3 r r 2E r,s p X Z d3p 1 =θ(x0 − y0) eip(x−y)(m − p) (2π)3 2E ¡ r,s p (3.105) Note that in the ﬁrst line, we only kept one term from multiplying the ﬁelds, since all other terms annihilate the vacuum. We used the spinor outer products to go from the third to the fourth line. For the other term, the method is completely the same and we obtain:

Z 3 0 0 d p 1 −ip(x−y) −θ(y − x ) 3 e (m + p¡) (3.106) (2π) 2Ep Just as for the Klein-Gordon equation, one can turn this into a 4-dimensional momentum integral using the same i prescription:

Z d4p m − p ∆ (x, y) = −i eip(x−y) ¡ (3.107) F (2pi)4 p2 + m2 − i Using contour integration, this can be shown to be equal to our initial expression (3.104). Finally, like in the Klein-Gordon example, ∆F is a Green’s function for the Dirac operator:

Z d4p iγbip − m (m − γap ) (iγa∂ − m)∆ (x, y) = −i eip(x−y) b a a F (2π)4 p2 + m2 − i Z d4p −m2 + γaγbp p = −i eip(x−y) a b (3.108) (2π)4 p2 + m2 − i Z d4p −m2 − p2(+i) = −i eip(x−y) (2π)4 p2 + m2 − i where in the last line an extra i has been added to show how the fraction goes to −1 in a well-behaved manner. So the Dirac operator acting on ∆F gives:

37 Z d4p (iγa∂ + m)∆ (x, y) = i eip(x−y) = iδ(4)(x − y) (3.109) a F (2π)4 Finally, a close look at the Feynman propagator tells us that the term with θ(x0 − y0) propagates b- excitations from y to x, while the term with θ(y0 − x0) propagates d excitations from x to y. Thus the interpretation of antiparticles is that they are particles of opposites Noether charge travelling back in time, since they are created by d† at y and annihilated by d at x.

38 4 Quantisation of the electromagnetic ﬁeld 4.1 Introduction to Maxwell’s ﬁeld

We introduce a vector potential Aa which gives rise to the ﬁeld strength tensor:

Fab = ∂aAb − ∂bAa. (4.1)

This is used together with the electrical current (Ja) to form a Lagrangian for the system and hence the action: Z 1 I = d4xL, L = − F F ab + AaJ . (4.2) 4 ab a We can make contact with non-relativistic physics by noticing the following equivalences: 1 E = F = −F ,B = F ,J = { ρ, j } (4.3) i i0 0i i 2 ijk jk a i

Where Ei (Bi) are components of the Electric (Magnetic) ﬁeld, ρ is the charge density and ji is the standard electric current. We can express this in more familiar terms by removing the components and deﬁning our Vector potential: Aa = { φ, A~ }

∂A~ E~ = ∇~ φ + , B~ = ∇~ × A~ (4.4) ∂t 4.2 Maxwell’s equations We can now move from the Lagrangian into Maxwell’s equations by considering two diﬀerent equations. First we notice that due to the symmetries of partial derivatives and the antisymmetry of Fab:

∂[aFbc] = 0. (4.5) Which is the Biancchi Identity for the ﬁeld tensor. We can re-express this to give two of Maxwell’s four equations, getting us halfway back to school.

∂B~ ∇~ · B~ = 0 ∇~ × E~ = − (4.6) ∂t Secondly, we can obtain the equations of motion from the Lagrangian to give:

∂L ∂L ab − = −∂aF − Ja = 0 ∂(∂aAb) ∂Ab (4.7) b b a b = A − ∂a∂ A + J = 0 From which we obtain the last two of Maxwell’s famous equations expressed in a form that it old and familiar:

∂E~ ∇~ · E~ = ρ ∇~ × B~ = ~j + (4.8) ∂t

Note: here we are using unrationalised Gaussian units in which factors of 0 and µ0 have been set to one and the 4π has also been somewhat awkwardly set to one.

4.3 Gauge invariance

Given a field strength tensor Fab we can see that our choice of Aa is not unique. This is easily demon- 0 0 strated by transforming Aa → Aa = Aa + ∂aΛ, with Λ, an arbitrary scalar field. Subbing in Aa into 0 Fab → Fab = Fab (our field strength tensor is invariant). Turning our attention back to the Action: Z Z Z Z 4 a 4 a 0 4 a 4 a a I = d xAaJ → d xJ Aa = d xJ (Aa + ∂aΛ) = d xJ Aa − Λ∂aJ (4.9)

39 a Thus if the current which is coupled to our vector ﬁeld is conserved we have ∂aJ = 0 and have an action which is completely gauge invariant despite the explicit presence of Aa in the Lagrangian. Due to this ambiguity in our choice of Aa it is often helpful to take a particular gauge in order to help make our life easier while working with the algebra. We will use the Lorentz gauge:

a ∂aA = 0, (4.10) however this only partially ﬁxes the gauge due to a freedom in solutions of the scalar ﬁeld Λ:

a ∂aA = f(x) (4.11)

a0 a a ∂aA = ∂aA + ∂a∂ Λ = f(x) + Λ = 0. (4.12) It is always possible to solve the equation:

Λ = −f(x) (4.13) using a Green’s function for the d’Alembertian (the box operator):

(4) G = δ (x − y) (4.14)

Z Λ(x) = G(x, y)f(y)d4y (4.15)

From this we can see that the solution of Λ is not unique as any solution for Λ = 0 can be added to the functions Λ and f while maintaining an invariant theory. To properly ﬁx our vector potential for a ﬁxed Fab we must impose one last condition:

0 a i A = 0, ∂aA = 0 ⇒ ∂iA = 0. (4.16) Now, from the Lorentz gauge:

A0 → A00 = A0 + Λ˙ Λ˙ = −A0 and set A00 = 0 (4.17) We can determine A¨ by the old A0 which means that the only gauge transforms left for us which respect the conditions:

¨ 2 Λ = 0 Λ = ∇ Λ, (4.18) where we are “given” Λ¨ since the solutions of Poisson’s equations are unique and hence Λ is unique. a a0 a Thus, given any A for a fixed Fab, one can find a unique A related by a gauge transformation to A 00 i0 such that A = 0, ∂iA = 0. 0 i i In practice this means we can always set A = 0 and then find our A such that ∂iA = 0, restricting us to a much smaller class of vector potentials that must be considered.

4.3.1 Coulomb gauge Before we go any further lets just look a little deeper into the Coulomb Gauge. We won’t use it explicitly here but it is very helpful when quantising the ﬁeld. We can make use of the residual gauge transformations in Lorentz gauge to pick a diﬀerent gauge called the Coulomb gauge:

∇~ · A~ = 0 (4.19) The Coulomb gauge breaks Lorentz invariance, so may not be ideal for some purposes. However, it is very useful to exhibit the physical degrees of freedom: the 3 components of A~ satisfy the single constraint of the gauge itself, leaving behind just 2 degrees of freedom. These will be identiﬁed with the two polarisation states of the photon. The Coulomb gauge is sometimes called radiation gauge.

40 4.3.2 The projection operator Now we invent a projection operator that takes the vector potential Ai and projects into the subspace i of ∂iA = 0: ∂ ∂ P A = A0 P = δ − i j (4.20) ij j i ij ij ∇2

Projections obey PijPjk = Pik: ∂ ∂ ∂ ∂ P P = δ − i j δ − j k ij jk ij ∇2 jk ∇2

2∂ ∂ ∂ ∂ ∂ ∂ = δ − i k + i j j k ik ∇2 (∇2)2 (4.21)

2∂ ∂ ∂ ∇2∂ = δ − i k + i k ik ∇2 (∇2)2

= Pik

Acting with Pij makes Aj divergence free: ∂ ∂ ∂ (P A ) = ∂ δ − i j A i ij j i ij ∇2 j (4.22) ∇2∂ = ∂ − j = 0 j ∇2 But what about ∇−2, how do we interpret this?

∇2G(x, y) = δ(3)(x − y) ∇2f = g Z 3 −2 f(x) = d yG(x, y)g(y) f = ∇ g (4.23) Z ∇−2 = d3yG(x, y) as an operator expression.

which allows us to see that ∇−2 is just short hand for manipulations involving Green’s functions.

4.4 Quantisation of electromagnetism We can explicitly expand our action: Z 1 I = d4x − F F ab 4 ab Z 1 = d4x (∂ A − ∂ A )(∂ A − ∂ A ) (4.24) 2 0 i i 0 0 i i 0 1 − (∂ A − ∂ A )(∂ A − ∂ A ) 4 i j j i i j j i

We now follow the usual path. First we ﬁnd the momentum conjugate for Ai: ∂L πi = , ∂A˙ i i i i π = A˙ − ∂ A0, (4.25) ∂L π0 = = 0. ∂A˙ 0

41 The vanishing of this momenta reﬂects the fact that not all degrees of freedom of Aa are physical. Some are just gauge degrees of freedom which do nothing useful. From this we can construct the Hamiltonian: Z Z i ˙ 0 ˙ 0 3 1 2 2 0 ~ ~ H = π Ai + π A −L = d x (E + B ) − A (∇ · E) (4.26) | {z } 2 =0 Using Hamilton’s equations:x ˙ = { x, H } we found: ∂H π0 = 0, π˙ 0 = 0 = { π0, H} = − = ∇~ · E~ (4.27) ∂A0 where we recognise ∇~ · E~ as the Gauss Law Constraint. In the Coulomb gauge the equations of motion are:

0 i ~ ~ A = 0, A = 0, ∇ · E = 0. (4.28) We know that the Heisenberg ﬁeld will be in the form of:

i X AH (x, t) ∼ (Classical coeﬀs → operators)(sol) + (Complex conjugate) (4.29) Over all solutions For a general solution we look towards plane wave-like solutions:

Ai ∼ (polarisation vector)i eip·x,A0 = 0 (4.30)

i Since ∇~ · E~ = 0 it follows that pi = 0 (as electromagnetic waves are transverse.)

i a A = 0 ⇒ pap = 0 (4.31) where pa is a null vector, we choose i to be unit vectors. i Now suppose that p = (0, 0, pz) (i.e., lies in the x-y plane), we can deﬁne a linear operation:

x y i = (1, 0, 0) i = (0, 1, 0) (4.32) where x and y denote different polarisation states. This results in an electric field in the x, y directions. For circular polarisation we express as: 1 1 + = √ (1, −i, 0) − = √ (1, i, 0) (4.33) i 2 i 2 Polarisation vectors rotates anti-clockwise as we see in the direction of travel of the wave. Now construct the field operator:

X Z d3p 1 Ai = a (p)∗r(p)eip·x + a†(p)r(p)e−ip·x (4.34) H (2π)3 2E r i r i r p where r is the polarisation state. Now we can impose ∇~ · A~ = 0, since Pij projects in to the subspace ∇~ · A~ = 0 from all possible A we have:

Z 3 ∇i∇j X d p P A = δ − ∗r(p)a (p)eip·x + r(p)a†(p)e−ip·x ij j ij ∇2 (2π)3 j r j r r (4.35) Z 3 X d p pipj pipj = ∗r(p) δ − a (p)eip·x + r(p) δ − a†(p)e−ip·x (2π)3 j ij p2 r j ij p2 r r

∇i produces factors of ±i~p, ~ · ~p = 0 because in this gauge we are working with a transverse wave. Thus we can see PijAj = Ai. We ﬁnd the momentum operator as before by taking the time derivative (note A0 = 0 so we only work with πi):

X Z d3p −i π = A˙ = a (p)∗r(p)eip·x − a†(p)r(p)e−ip·x (4.36) (2π)3 2 r i r i r

42 Now we look to see whether the canonical momentum obeys the Gauss Law Constraint.

∇~ · ~π = 0, ∇~ = ±ip, ~p · ~ = 0 (4.37)

or equivalently we can ask if it is the case that:

Pijπj = πi (4.38)

First try at quantisation, for operators at the same times:

(3) [Ai(x), πj(y)] = iδijδ (x − y) (4.39) If we try and use the projection operator on this we see that the RHS of the above equation will change. But if we utilise that P 2 = P we gain the invariant expression:

∇ ∇ [A (x), π (y)] = i δ − i j δ(3)(x − y) (4.40) i j ij ∇2 Following from this one can ﬁnd the creation and annihilation operators:

Z ↔ 3 −ip·x ar(p) = iri (p) d xe ∂0Ai(x) (4.41) Z ↔ a†(p) = i∗ (p) d3xeip·x∂ A (x) r ri 0 i where we have used the short hand notation:

↔ f∂0g ≡ f∂0g − (∂0f)g (4.42)

Which gives us the following commutation relations:

† 3 [ar(p), a (q)] = (2π) δrs2Epδ(p − q) s (4.43) † † [ar(p), as(q)] = [ar(p), as(q)] = 0 Which can be seen to be the same as the Klein-Gordon equation except with the added information from the polarisation states. Thus a† and a can be interpreted as creation and annihilation operators for the photon. As A is a real ﬁeld, photons are their own “antiparticle”. Lastly we turn towards the Hamiltonian in order to asses the number generator and propagator. The Hamiltonian from classical physics is known: Z 1 H = d3x (E2 + B2) − A0(∇~ · ~π) 2 | {z } =0 (4.44) X Z d3p = E a†(p)a (p) + ∞ constant (2π)3 p r r r Which is equal to the sum over all states (Number of photons × energy). (The number operator for ˆ † photons with polarisation r and momentum p is unsurprisingly N = ar(p)ar(p))

4.5 Propagators in electromagnetism Time ordered ﬁeld product:

43 Z 3 3 d pd q 1 X ip·x ∗ −ip·x † iq·x∗ −iq·x † ∆ij = 0 e ar + e a r e as + e a s 0 (2π)6 4E E r r s s p q r,s

Z 3 3 d pd q 1 X † †¨¨ ∗ i(px−qy) = 0 [ar, a ] +¨a ar 0 se (2π)6 4E E s s r p q r,s

Z d3pd3q 1 X = 2E δ(p − q)δ ∗ ei(px−qy) (4.46) (2π)3 4E E p rs r s p q r,s

Z d3p 1 X = ∗(p) (p)eip(x−y) (2π)3 2E r r p r

Z 3 d p 1 pipj ip(x−y) = 3 δij − 2 e (2π) 2Ep p An explanation: 1. First we can remove two terms as a† acting on the left is zero and a acting on the right is zero, halving our workload. 2. We make a commutator by adding and subtracting a term because we know how to work with commentators. We can then lose the extra term as a is acting on the right. 3. Next we lose more terms by integrating over q. 4. Finally summing over r to give ourselves something nice and familiar. Which is exactly like the propagator for the Klein-Gordon equation except we now have the term from the projection operator. We can express this in a complete result as an integration over all space-time with an added imaginary term allowing for a soluble contour integral:

Z d4p p p 1 ∆ = eip(x−y) δ − i j (4.47) ij (2π)4 ij p2 p2 − i ∇ ∇ ∆ = −i δ − i j δ(4)(x − y) (4.48) ij ij ∇2

We also note that ∆i0 and ∆00 are zero. So this is a projected form of the Green’s function and acting with P on the propagator leaves it invariant.

4.6 Example: complex scalar ﬁelds Consider the Lagrangian of a complex scalar ﬁeld:

L = −∂φ†∂φ − m2φ†φ (4.49) We calculate the corresponding Noether currents from the translations:

φ → eiΛφ φ† → e−iΛφ (4.50) to be:

J = i φ†∂φ − ∂φ†φ (4.51) Let us suppose that our scalar ﬁeld Λ isn’t constant:

φ → e+ieΛφ φ† → e−ieΛφ† (4.52)

44 Our Lagrangian is no longer invariant if our Λ is position dependent but we can make it invariant with the transformation:

∂a → Da = (∂a − ieAa) (4.53)

Where Aa is just the Maxwell ﬁeld. You can look at this as a covariant derivative with connection coeﬃcient ∼ Aa. This allows us to re-express our Lagrangian as:

L = −Dφ†Dφ† − φ†φ (4.54) which is invariant as long as

Aa → Aa + ∂aΛ (4.55) and we have the following interaction terms:

∂φ†∂φ, Aφ∂φ, A2φ2 (4.56)

45 5 Feynman path integrals

To start with we will look at normal quantum mechanics and proceed to generalise this to QFT. We begin with a classical Lagrangian to describe a dynamical system with n degrees of freedom qi where i = 1, ..., n L(qi, q˙i) (5.1) We can then move to Hamiltonian formalism to place qi andq ˙i on more symmetric footing. To do this we replaceq ˙i with pi which is the conjugate momenta i.e. ∂L pi = (5.2) ∂q˙i As usual the Hamiltonian is then

H(qi, pi) =q ˙ipi − L (5.3) We can then proceed to quantise this in a similar way to section 2 by exchanging the coordinate and momenta terms for operators that satisfy the commutation relations:

i i [ˆq (t), pˆj(t)] = i~δjI (5.4) where the commutation relations are evaluated at equal times. This is easily generalised to scalar ﬁeld theory:

˙ (3) [φˆ(x, t), φˆ(y, t)] = i~δ (x − y)I (5.5) Equal time evaluations is one of the problems with the canonical quantisation approach to QFT. It means that there is a preferred frame in which to measure time and therefore loses relativistic invariance. It is also non-covariant and Lorentz invariance is not manifest meaning that perturbation expansions are hard to derive in terms of Feynman rules. The Path Integral approach is equivalent to the standard operator approach and is manifestly symmetric between space and time. It does this by removing the idea from classical physics that a particle travels a single unique trajectory and replaces it with a sum over all possible trajectories.

5.1 Derivation of the Feynman path Integral for a single particle Starting with the operatorsp ˆ andq ˆ and the classical Hamiltonian

p2 H(q, p) = + V (q) (5.6) 2m which corresponds to an operator in QM and is found by changing p and q into the corresponding operators that obey the commutation relation [ˆq, pˆ] = iI. We ﬁnd the quantum mechanical Hamiltonian to be: Hˆ = H(ˆq, pˆ) (5.7) In the Schrodinger picture the Hamiltonian is independent of time and the states evolve. The formal solution to the Schrodinger equation is given by

|ψ(t)i = exp(−iHtˆ ) |ψ(0)i (5.8)

We can also use position eigenstates as usual:

qˆ(t) |q, ti = q |q, ti (5.9) where q is any real number and for convenience we use normalisation as hq0, t| |q, ti = δ(q0 − q). Since the states {|qi} are not dependent on time, they form a basis for any state. This allows us to deﬁne a wave function

ψ(q, t) = hq | ψ(t)i (5.10)

46 The path integral approach expresses time evolution of states in terms of possible trajectories of particles. Firstly we can write the wave function as

This has now changed the Schrodinger equation into an expression with an integral with the evolution kernel K(q, q0; t) = hq| exp(−iHtˆ ) |q0i (5.14) The next step is to imagine that the wave function is evolving over some time period T and then to divide this period up into small intervals that are not required to be equally spaced. This gives us 0 = t0 < t1 < t2 < ... < tn+1 = T . This allows us to write the exponential as a product of these diﬀerent splits.

exp(−iHTˆ ) = exp(−iHˆ (tn+1 − tn)) exp(−iHˆ (tn − tn−1))... exp(−iHtˆ 1) (5.15)

We can then introduce a series of complete states for each tr were r = 1, ..., n into the expression for K to get

n Z Y K(q, q0; t) = dqr hqr+1| exp(−iHˆ (tr+1 − tr) |qri hq1| exp(−iHtˆ 1) |q0i (5.16) r=1

We can see from this that the integration goes over all values of q for each time segment t1, ..., tn as the value for q evolves from q0 at t = 0 to q at t = T and demonstrates the integration over all possible trajectories. There is no easy way to ﬁnd explicit solutions with an arbitrary potential so we approximate the factors of:

hqr+1| exp(−iHδtˆ ) |qri (5.17) where the δt arises from making the diﬀerence between the time segments small i.e. δt = tr+1 − tr.

Now we look explicity at free theory so that we can set V = 0, giving the result ! pˆ2 K (q, q0; t) ≡ hq| exp −i t |q0i (5.18) 0 2m This is valid for any value of t and can be shown by using a comeplete set of eigenstates with eigenvalues p when acted on by the momentum operator such that Z dp I = |pi hp| (5.19) 2π A basic result from QM is that

0 hq|pi = eip·q, hp|q0i = e−ip·q (5.20) The changes the kernel to:

Z ˆ2 Z ˆ2 0 dp p 0 dp −i p t ip·(q−q0) K (q, q ; t) = hq| exp(−i t) |pi hp|q i = e 2m e (5.21) 0 2π 2m 2π

0 0 m(q−q ) Now we make the substitution p = p − t to get

47 0 2 Z 0 02 0 2 r 0 im(q−q ) dp −i p t im(q−q ) m K (q, q ; t) = e 2t e 2m = e 2t (5.22) 0 2π 2πit 0 −i π where in the last step we rotated the integrand by p = e 4 r to give

Z 0 Z r dp −i λ p2 −i π dr − λ r2 1 −i π 2π e 2 = e 4 e 2 = e 4 (5.23) 2π 2π 2π λ 0 0 Note that there is a limiting case t → 0, we get K0(q, q ; t) → δ(q − q ) as required. We now return to ﬁnding an approximation for exp(−iHδtˆ ). In order to do this we must note that

eAˆ+Bˆ 6= eAêBˆ (5.24) where Aˆ and Bˆ are operators. One can find an expression using the Baker-Campbell-Hausdorff formula:

1 exp(Aˆ) exp(Bˆ) = exp Aˆ + Bˆ + [A,ˆ Bˆ] + ... (5.25) 2 and therefore we can use a small parameter to say:

exp((Aˆ + Bˆ)) = exp(Aˆ) exp(Bˆ)(1 + O(2)) (5.26) which implies:

exp(Aˆ + Bˆ) = lim (exp(A/nˆ ) exp(B/nˆ ))n (5.27) n→+∞ Hence, as δt is small, we can write: ! pˆ2 exp(−iHδtˆ ) = exp −i δt exp(−iV (ˆq)δt)(1 + O(δt2)) (5.28) 2m and now we can take δt → 0 to get: ! pˆ2 hq | exp(−iHδtˆ ) |q i = hq | exp −i δt exp(−iV (ˆq)δt) |q i (5.29) r+1 r r+1 2m r

We now return to calculating the kernel and take tr+1 − tr = δt so time is divided into equal increments:

1 (n+1) Z n n 2 ! ! m 2 Y X m qr+1 − qr K(q, q ; T ) = dq exp i − V (q ) δt (5.32) 0 2πiδt r 2 δt r r=1 r=0 The path integral is obtained by considering the limit n → ∞ and δt → 0. Therefore in the exponent

n 2 ! Z T Z T X m qr+q − qr 1 − V (q ) δt → dt mq˙2 − V (q) → dtL(q, q˙) = S[q] (5.33) 2 δt r 2 r=0 0 0 where L is the classical lagrangian and the notation S[q] denotes a dependence on the function q(t) at a given interval. We also note that this function has the property q(0) = q0 and q(T ) = q. We also use another deﬁnition,

48 r n r m Y m · dq → d[q] (5.34) 2πiδt 2πiδt r r=1 This allows us to ﬁnally write the integral kernel as: Z iS[q] K(q, q0; T ) = hq| exp(−iHTˆ |q0i = d[q]e (5.35)

iS This is the path integral and is a sum over all paths between q and q0 weighted by e . We can deﬁne d[q] such that our potential is zero and we obtain:

r 2 m im(qf − qi) K(qf , qi; T ) = exp (5.36) V =0 2πiT 2T

5.2 Example: simple harmonic oscillator We will now look at the Harmonic Oscillator. Firstly consider the potential, 1 V (q) = mω2q2 (5.37) 2 The harmonic oscillator together with the Hydrogen atom are the only two known solvable problems using the path integral. Every path q(t) has the the property

q(0) = qi, q(T ) = qf (5.38) as usual. We now consider the classical path qc(t) and expand q(t) around it. The classical path is deﬁned as:

2 q¨c + ω qc = 0, qc(0) = qi, qc(T ) = qf (5.39) It would be normal to solve this by writing down a linear combination of sin(ωt) and cos(ωt) and then work out the prefactors but in this case it is easier to choose sin(ωt) and sin ω(T − t). This is obviously equivalent to having cos(ωt) if ωT = nπ for some integer n. The solution can then be written down by inspection as: 1 q (t) = (q sin ωt + q sin ω(T − t)) (5.40) c sin ωT f i The action for a particular path is given by:

Z T m ˙2 2 2 S[qc] = dt(qc − w qc ) (5.41) 2 0 Integrating by parts gives:

Z T m T m 2 S[qc] = [qcq˙c]0 − dtqc(q ¨c + ω qc) (5.42) 2 2 0

The remaining integral term then vanishes as qc obeys the equations of motion. We can therefore say: m S[q ] = (q q˙ (T ) − q q˙ (0)) (5.43) c 2 f c i c

Now using the explicit solution for qc we get:

2 2 −2qf qi + (q + q )cosωT S[q ] = mω f i (5.44) c 2 sin ωT Note for consistency that as ω → 0:

m (q − q )2 S[q ] = f i (5.45) c 2 T

49 We can now write the general path as:

q(t) = qc(t) + f(t), f(0) = f(T ) = 0 (5.46) and the general action quadratic in q then

S[q] = S[qc] + S[f] (5.47)

This is exact since the action is stationary at the classical path qc. There are no linear terms in f. In this we have assumed that the functions are small, meaning that the integral will only have contributions from paths close to the classical one. We can also assume that

d[q] = d[f] (5.48) This is true for ordinary integrals, namely d(x + a) = dx. We can then write the integral kernal as Z Z iS[q] iS[qc] iS[f] K(qf , qi; T ) = d[q]e = e d[f]e (5.49)

Note that this is independent of the initial and ﬁnal points q0 and q; all this information is contained within the prefactor. We now need to calculate the second factor and we will use a method that will prove to be useful later on. Expanding f(t) in terms of a complete set of Fourier sine series gives

∞ r X 2 nπt f(t) = a sin (5.50) n T T n=1 The sine functions form an orthonormal basis for functions vanishing at t = 0 and t = T . We can then use integration by parts and noting that f(0) = f(T ) = 0 to write:

m Z T m X n2π2 S[f] = − dtf(f¨+ ω2f) = a2 − ω2 (5.51) 2 2 n T 2 0 n where we have used:

Z T r 2 r 2 dt sin(nπt) sin(mπt) = δnm. (5.52) 0 T T We assume the relation:

∞ Y d[f] = C dan (5.53) n=1 where C is a normalisation constant. We can now express the integral as:

∞ Z Z m 2 n2π2 2 iS[f] Y i 2 an 2 −ω d[f]e = C dane T (5.54) n=1 We can now do this integral by rotating the contour:

Z +∞ r i λy2 2πi dye 2 = (5.55) −∞ λ By absorbing constants we now get Z ∞ r iS[f] Y 1 m d[f]e = C0 ,C0 = (5.56) q ω2T 2 2πiT n=1 1 − n2π2 Now we can say that the inﬁnite product is

50 ∞ Y ω2T 2 sin ωT 1 − = (5.57) n2π2 ωT n=1 Our ﬁnal result is then:

Z r mω d[f]eiS[f] = (5.58) 2πi sin ωT This gives us the integral kernal as:

r 2 2 ! mω (q + q ) cos ωT − 2qf qi K(q , q ; T ) = exp imω f i (5.59) f i 2πi sin ωT 2 sin ωT

We can check this result by looking at the eigenfunctions of |ni of Hˆ with:

1 X E = (n + )ω, I = |ni hn| (5.60) n 2 n Using standard quantum mechanics we have a formula:

X 1 ∗ −i(n+ 2 )ωT K(qf , qi; T ) = ψn(qf )ψn(qi)e (5.61) n Now take the special situation where T = −iτ, τ → ∞,

2i sin ωT → eωτ , 2 cos ωT → eωτ We can then see that the result obtained using the path integral is consistent with quantum mechanics and given by, r mω − 1 ωτ − 1 mω(q2 +q2) ∗ − 1 ωτ K(q , q ; −iτ) ≈ e 2 e 2 f i = ψ (q)ψ (q )e 2 (5.62) F i π 0 0 0

2 Integrals of the form R dxeiλx do not converge and are therefore ill-deﬁned. We would rather consider 2 integrals of the form R dxe−λx for λ > 0. It is the same for path integrals,

Z T 1 S[q] = dt mq˙2 − V (q) (5.63) 0 2 This is normally a real quantity but we use analytic continuation of time to obtain well deﬁned integrals:

t → −iτ, T → −iτ1 (5.64) so that:

2 2 ! dq Z τ1 1 dq q˙2 → − , iS[q] → − dτ m + V (q) (5.65) dτ 0 2 dτ meaning the path integral becomes

" 2# Z Z τ1 1 dq Z τ1 hq| exp(−Hτˆ 1) |q0i = d[q] exp − dτ m exp − dτV (q) (5.66) 0 2 dτ 0 The first exponential in this term defines a measure on paths and can be thought of as a probability measure. This can then be defined for a wide range of potentials V subject to the requirement that V is bounded from below. These types of integrals (with analytic continuation) first appeared in Brownian motion.

51 5.3 Gaussian integrals and free ﬁelds Let us consider a set of coordinates represented by a column vector,

T n ~x = (x1, ..., xn) ∈ R (5.67) and also deﬁne the scalar product to be

n 0 T 0 X 0 ~x · ~x = ~x · ~x = xixi (5.68) i=1 To start we let A~ be a symmetric n × n positive deﬁnite matrix where a complex A~ is Hermitian. We can then consider the Gaussian integral:

Z 1 ~ n − 2 ~x·Ax ZA~ = d xe (5.69) which is essentially equivalent to free ﬁeld theory which will become clear. If we then deﬁne an orthogonal matrix U~ where   λ1 0 0 ~ ~ ~ T ~ UAU = D =  0 ... 0  , λi > 0 (5.70) 0 0 λn We can now compute the integral by making a change of variables to let,

~x0 = U~x,~ dnx = dnx0 (5.71) Therefore we get Z n Z 1 0 ~ 0 Y 1 02 m 0 − 2 ~x ·Dx 0 − 2 λixi ZA~ = d x e = dxie (5.72) i=1 Where we have used the factorisation

n ~0 ~ 0 X 02 x · D~x = λixi (5.73) i=1 and therefore we get the answer

n r n Y 2π (2π) 2 Z = = (5.74) A~ λ p i=1 i det A~ By analytic continuation into the complex plane we can see that adding i into the exponent will give

n 1 Z 2 − n − i ~x·Ax~ 2π p 2 Z = d xe 2 = det A~ (5.75) A~ i If we now consider ~b to be an n dimensional vector too we can have an extension to the linear term in the exponent

Z 1 ~ ~ n − 2 ~x·Ax+b·~x ZA,~ ~b = d xe (5.76) We can reduce this to the previous case by noticing that we can write 1 1 1 ~x · Ax~ −~b~x = ~x0 · A~x~ 0 − ~b · A~−1b, x~0 = ~x − A−~1b (5.77) 2 2 2 We therefore get

n 1~ −~1 (2π) 2 2 b·A b ZA,~ ~b = e p (5.78) det A~

52 We will now look at expectation values Z 1 n − 1 ~x·Ax~ hf(~x)i = d xf(~x)e 2 , h1i = 1 (5.79) ZA,~ ~0 We are going to use a trick here that ∂ ~b·~x f(~x) = f e |~ (5.80) ∂~b b=0 We can now evaluate the expectation, Z 1 n ∂ − 1 ~x·Ax~ +~b·~x hf(~x)i = d xf e 2 | ~ ~b=0 ZA,~ ~0 ∂b 1 ∂ = f Z | (5.81) ~ A,~ ~b ~b=0 ZA,~ ~0 ∂b ∂ 1~b·A−~1b = f e 2 |~ ∂~b b=0 Now let us consider some simple cases,

hxii = 0,

∂ 1~ −~1 (5.82) ~−1~ 2 b·A b −1 hxixji = (A b)je |~b=0 = Aij ∂bi

-1 i A ij j

Figure 2: Feynman Diagram corresponding to equation 5.82

We can also see that

hxi1 xi2 ...xin i = 0 for odd n, −1 −1 −1 −1 −1 −1 (5.83) hxixjxkxli = Aij Akl + Aik Ajl + Ail Ajk

-1 -1 -1 -1 -1 -1 A ij A kl A ik A jl A il A jk i j i j i j

+ +

k l k l k l

Figure 3: Feynman Diagram corresponding to equation 5.83

53 5.4 Non-Gaussian integrals and perturbation expansions Non-Gaussian integrals correspond to interacting ﬁeld theory, unlike the previous Gaussian ones. The most basic integral is given by, Z 1 n − 1 ~x·Ax~ +~b·~x−V (~x) Z = d xe 2 (5.84) ZA,~ ~0 where we require that V (~x is bounded from below V (~0) = 0. Now we use the same trick as before to get Z 1 −V ( ∂ ) n − 1 ~x·Ax~ +~b·~x Z = e ∂~b d xe 2 (5.85) ZA,~ ~0 Subbing in we can then get

1 −V ( ∂ ) 1~b·A−~1b Z = e ∂~b e 2 (5.86) ZA,~ ~0 We would then evaluate this by expanding out the ﬁrst exponential term but this would give rise to a perturbation expansion. We will explore an alternative way of doing this that might make the manipulations easier

Lemma ∂ ~ ∂ ~x·~b G F (b) = F G(~x)e |~x=0 (5.87) ∂~b ∂~x Proof We will consider the special case G(~x) = e~x·~α, F (~x)eβ~·~x then the LHS becomes ∂ ∂ ·~α β~·(~b+~α) G F (~b) = e ∂~b F (~b) = F (~b + ~α) = e (5.88) ∂~b Now the RHS can be evaluated ∂ ~x·~b β~· ∂ ~x·(~α+~b) (~x+β~)·(~α+~b) β~·(~b+~α) F G(~x)e | = e ∂~x e | = e | = e (5.89) ∂~x ~x=0 ~x=0 ~x=0 We can then say that this result is true for any F and G because they can be expressed as Fourier series. Now let us apply this to our expression for Z,

1 ∂ A~−1 ∂ −V (~x+~b·~x Z = e 2 ∂~x ∂~x e |~x=0 (5.90) Setting ~b = 0 for simplicity and expanding both exponentials we get

1 −1 1 −1 −1 1 −1 −1 1 −1 −1 Z =1 − Aij Vij − Aij Akl Vijkl + Aij VijAkl Vkl + VijAik Ajl Vkl 2 8 8 4 (5.91) 1 1 + V A−1A−1A−1 V + V A−1A−1 A−1V + ... 8 ijk ij kl mn lmn 12 ijk il jm kn lmn where we have used the notation ∂ ∂ ∂ Vi1i2...ik = ... V (~x)|~x=0 (5.92) ∂xi1 ∂xi2 ∂xik and assumed that V (~0) = Vi(~0) = 0. Expression 5.91 is extremely messy and make it hard to see what is going on. It is much better represented using vacuum diagrams as in Figure 5. It is useful to know that the Euler formula holds for all connected diagrams and is given by

L = I − V + 1 (5.93) where L is the number of closed loops, I is the number of internal lines and V is the number of vertices. The Feynman rules for diagrammatic expressions of integrals of this type are given by,

−1 • Lines, with end points labelled by i and j, represent Aij

54 • Vertices represent −Vi1...ik • Contract all indices

1 • Symmetry factor S where S is the dimension of the symmetry group of the graph

n -1 5 ... i A ij j

4 Vi1...Vin

2 1

−1 Figure 4: Diagramatic representation for equation 5.91 is done by associating Aij with lines and −Vi1...in with vertices.

-1 1 1 - + 2 8 8

1 1 1 - + + 4 8 12

Figure 5: Feynman Diagram corresponding to equation 5.91

55 6 Feynman path integrals for scalar ﬁelds 6.1 A further introduction

The path integral in quantum mechanics makes use of the fact that qi(0) = ~q0 and qi(T ) = ~qf .

Z Z T ! −iHt Y X h~qf | e |~q0i = dqi(t)dpi(t) exp i piq˙i − H dt (6.1) i 0

where pi is the moment conjugate to qi. We then assume that this applies to ﬁeld theory and consider a real scalar ﬁeld:

φ = φ(x, t) π = π(x, t) (6.2) where φ(x) is the generalised coordinate and π(x) is the momentum conjugate to the generalised coordinate. We can then say that the degrees of freedom are thus φ and π at each point in space with the values φi at time ti and φf at time tf . The path integral then becomes: Z Z iH(tf −ti) Y X ˙ hφf | e |φii = Dφ(x)Dp(x) exp i pφ − H dt x (6.3) Z Y Z tf Z = Dφ(x, t)Dp(x, t) exp i dt dxpφ˙ − H x ti Let us now look at a Lagrangian density for a scalar field with a potential term 1 1 L = − (∂φ)2 − m2φ2 − V (φ) (6.4) 2 2 where the potential term V (φ) allows particles to interact. In practice we have λ V (φ) = φ4, λ > 0 (6.5) 4! this is an important example for the potential, because φ4 is a renormalisable interaction i.e. the infinities are controllable. If λ < 0 this still leads to a potential that is unbounded from below and thus is unstable. We can also have interactions with potential: g V (φ) = φ3, g > 0 (6.6) 3! but due to the fact that this potential does not have a minimum, it is always unstable so we will not consider this. It is a nice toy model though. In general we have: λ V (φ) = φp (6.7) p! but p > 4 gives you unrenomalisable results i.e. uncontrollable infinities. We will therefore focus on φ4 interactions. The Hamiltonian is positive-semidefinite and given by Z 1 1 1 H = d3x p2 + m2φ2 + (∇φ)2 + V (φ) (6.8) 2 2 2 therefore the path integral becomes

Z Z tf Z Y 1 1 1 2 hφ | eiH(tf −ti) |φ i = Dφ(x, t)Dp(x, t) exp i dt d3xpφ˙ − p2 − m2φ2 − (∇φ) − V (φ) f i 2 2 2 x,t ti (6.9) We now wish to do the p integral which runs from −∞ to +∞. To do this we look at a single p integral Z i pφ˙− 1 p2 Dpe ( 2 ) (6.10)

56 This integral occurs at each point in space and is convergent. We can evaluate is by rotating the contour −i π of integration p = e 4 z. The single p integral therefore becomes: Z −i π −i π 1 2 e 4 dz exp e 4 zφ˙ − z 2 Z 2 −i π 1 −i π i 2 (6.11) ⇒e 4 dz exp − z − e 4 φ˙ + φ˙ 2 2 ˙ −i π i φ 1 ⇒e 4 e 2 (2π) 2 therefore we get:

Z Z tf Z N iH(t −t ) Y 3 1 2 1 2 1 2 2 h −i π 1 i hφ | e f i |φ i = Dφ(x, t) exp i dt d x φ˙ − (∇φ) − m φ − V (φ) e 4 (2π) 2 f i 2 2 2 x,t ti (6.12) where N is the number of points in spacetime. This deﬁnes a quantum ﬁeld theory with an expression that puts space and time on equal footing.

Z Z tf iH(tf −ti) Y 4 hφf | e |φii = η Dφ(x, t) exp i Ld x (6.13) x,t ti

6.2 Non-interacting systems We now want to introduce the idea of a source term J(x) for any ﬁeld and set V (φ) = 0

Z Y Z 1 1 Z[J] = η Dφ(x) exp i d4x − (∂φ)2 − m2φ2 + Jφ (6.14) 2 2 x

where the x now denotes all points in spacetime and the limits were taken such that tf → ∞ and ti → −∞. We also take φf , φi → 0 with these limits. Thus, the initial and ﬁnal states are the vacuum and we can integrate the Lagrangian by parts to get

Z Y Z 1 Z[J] = η Dφ(x) exp i d4x φOφ + Jφ (6.15) 2 x 2 where O is the Klein-Gordon operator O = − m . We therefore can say that Z Y Z 1 1 Z[J] = η Dφ(x) exp i d4x (φ + JO−1)O(φ + O−1J) − JO−1J (6.16) 2 2 x O−1 is the Green’s function for this operator:

2 (−x + m )∆F (x, y) = −iδ(x − y) −1 ⇒ O = −i∆F (x, y) (6.17) Z ⇒ (O−1f)(x) = dyO−1(x, y)f(y)

Therefore we get

Z Y Z 1 i Z Z[J] = η Dφ(x) exp i d4x (φ + JO−1)O(φ + O−1J) exp − d4xd4yJ(x)(−i)∆ (x, y)J(y) 2 2 F x Z Y Z 1 1 Z = η Dφ0(x) exp i d4x φ0Oφ0 exp − d4xd4yJ(x)∆ (x, y)J(y) 2 2 F x (6.18) where φ0 = φ + JO−1. We observe that only the last factor in the integral depends on J. We can thus take the ratio with Z[0] to give:

57 Z[J] i Z = exp − d4xd4yJ(x)(−i)∆ (x, y)J(y) (6.19) Z[0] 2 F Now we take the functional derivative with respect to the source term and put J equal to zero at the end since the initial and ﬁnal state are the vacuum:

Z Z δ Z[J] 4 i 4 4 = − d y∆F (x1 − y)J(y) exp − d xd yJ(x)(−i)∆F (x, y)J(y) δJ(x1) Z[0] 2

Z δ δ Z[J] i 4 4 − = +∆F (x1 − x2) exp − d xd yJ(x)(−i)∆F (x, y)J(y) + terms that vanish at J = 0 δJ(x1) δJ(x2) Z[0] 2

= ∆F (x1 − x2)

= h0| T φ(x1)φ(x2) |0i (6.20) This gives directly the probability amplitude for creating a particle at x1 and letting it be annihilated at x2. This simplest Feynman diagram is used to symbolise this and is shown in ﬁgure 6

Figure 6: Feynman Diagram corresponding to equation 6.20

Wick’s theorem Wicks theorem states that this process can be done n times to get n δ Z(J) X (−1) |J=0 = ∆F (xi − xj)...∆F (...) δJ(x2n) Z(0) all pairs (6.21)

= h0| T φ(x1)...φ(x2n) |0i where the sum of all paired terms in the ﬁrst line signiﬁes Wick’s Theorem. Note that the amount of derivatives taken must be an even number. If this number were odd, all terms depend on J and go to zero. As an example let us put n = 2:

h0| T φ(x1)...φ(x4) |0i = ∆F (x1 − x2)∆F (x3 − x4)

+ ∆F (x1 − x3)∆F (x2 − x4) (6.22)

+ ∆F (x1 − x4)∆F (x2 − x3) and the Feynman diagrams are given in Figure 7. If there are 2n objects, then the number of pairs is (2n)! given by 2nn! .

58 Figure 7: Feynman Diagram corresponding to equation 6.22

6.3 Interacting systems Let us now add an interaction term into the path integral to give

Z Y Z 1 1 λφ4 Z[J] = η Dφ(x) exp i − (∂φ)2 − m2φ2 + Jφ − d4x (6.23) 2 2 4! x We can then perform a power series expansion in positive powers of λ to get

" ∞ r r# Z Y Z 1 1 X λ (−i)r Z Z[J] = η Dφ(x) exp i − (∂φ)2 − m2φ2 + Jφ d4x φ4(z)d4z 2 2 4! r! x r=0 (6.24) This turns out to be an asymptotic series in λ for λ > 0, which is manageable for small orders in λ.

∞ r r X λ (−i)r Z h0| T φ4(z)d4z |0i (6.25) 4! r! r=0

If r is ﬁxed, we can use Wick’s theorem to re-express (6.25) in terms of ∆F . We now proceed as before and notice that diﬀerentiation with respect to J does not notice the interaction term because it is independent of J so we get

∞ r r δ Z(J) X λ (−i)r Z (−1)n = h0| T φ(x )...φ(x ) φ4(z)d4z |0i (6.26) δJ(x ) Z(0) 1 2n 4! r! 2n r=0 We can then use Wick’s theorem on terms that look like φ2n+4r. Thus, this series can be completed order by order in n and r. The number of terms needed is then

(2(2n + 4r))! (6.27) 22n+4r(2n + 4r)!

6.3.1 n = 1, r = 0 This is the most basic example and is just given by

h0| T φ(x1)φ(x2) |0i = ∆F (x1 − x2) (6.28)

6.3.2 n = 1, r = 1 It gets slightly more complicated as there are now 15 terms from Wick’s theorem

59 λ Z −iλ Z −i h0| T φ(x )φ(x ) d4zφ4(z) |0i = ∆ (x − x ) · 3∆2 (z − z)d4z + 4! 1 2 4! F 1 2 F (6.29) −iλ Z 12∆ (x − z)∆ (x − z)∆ (z − z)d4z 4! F 1 F 2 F where the corresponding diagrams are given in Figure 8.

Figure 8: Feynman Diagrams for the 2-point function when n = 1 and r = 1 corresponding to (6.29). The “ﬁgure of 8” is a disconnected particle which is not of much use and the loop is a virtual particle.

6.3.3 n = 1, r = 2 This ﬁnal example has 945 terms according to Wick’s theorem so is extremely long!

λ2 1 Z Z − h0| T φ(x )φ(x ) d4zφ4(z) d4wφ4(w) |0i = (4!)2 2! 1 2 λ2 1 Z Z − d4z d4w9∆ (x − x )∆2 (z − z)∆2 (w − w)+ (4!)2 2! F 1 2 F F 4 12∆F (x1 − x2)∆F (z − w)+ 2 (6.30) 6∆F (x1 − x2)∆F (z − z)∆F (w − w)∆F (z − w)+ 2 288∆F (x1 − z)∆F (x2 − z)∆F (z − w)∆F (w − w)+ 2 72∆F (x1 − z)∆F (x2 − z)∆F (z − z)∆F (w − w)+ 3 192∆F (x1 − z)∆F (x2 − w)∆F (z − w)+ 128∆F (x1 − z)∆F (x2 − w)∆F (z − w)∆F (z − z)∆F (w − w) The corresponding Feynman diagrams are given in Figure 9.

+ + +

Figure 9: Feynman Diagrams for the 2-point function when n = 1 and r = 2 corresponding to equation (6.30). We see that compared to Figure 8 a very quick development in complexity of the diagrams as r goes from 1 to 2.

60 6.3.4 4-point function The 4-point function is given by

h0| T φ(x1)φ(x2)φ(x3)φ(x4) |0i (6.31)

+ +

(a) Terms for λ0

+ +

+ + + ...

(b) Terms for λ1

+ ... + + ... + + ...

Figure 10: Feynman Diagrams for the four-point function for varying orders in λn

You can then evaluate this order by order. As an example we show the ﬁrst term of λ1 and using Wick’s Theorem 1 Z (−i) λ d4z [24∆ (x − z)∆ (x − z)∆ (x − z)∆ (x − z) + ...] (6.32) 4! F 1 F 2 F 3 F 4

6.3.5 Zero-point function As another example we give the zero-point function

−iλ −λ2 h0|0i + h0| φ4(z) |0i + h0| φ4(z)φ4(w) |0i + ... (6.33) 4! 2(4!)2 The ﬁrst term here means a theory with no interactions h0|0i 6= 1 in the interacting theory. The vacuum of the interacting theory is modiﬁed by this process We can use this fact in our interacting theory:

61 + + + ...

Figure 11: Feynman Diagrams for the zero-point function.

I h0| T φ(x1)...φ(xn) |0iI = h0| T φ(x1)...φ(xn) |0i (6.34) I h0|0iI where the RHS is restricted to only connected diagrams. By modifying the vacuum of the interacting theory, we have rid ourselves of all the disconnected diagrams:

+ + + + ...

(a) Two-point function

+ + ... + + ... + + ...

(b) Four-point function

Figure 12: Feynman Diagrams for expressions with the zero-point function terms removed.

6.4 Solving the path integral 6.4.1 Feynman rules in coordinate space If we consider the time ordered interaction to order λp given by,

h0| T φ(x1)...φ(xn) |0i (6.35) the Feynman rules for solving the path integral are given by

• Draw all diagrams that contain p-vertices and n extended legs • For each vertex you get a factor of (−iλ)p

• Each internal line contributes ∆F (End point 1 − End point 2)

• Each external line ∆F (x1 − ...) • Divide by the symmetry factor S

R 4 4 • Integrate over the position of each vertex d z1...d zp

62 3 4

2 v2 v3

The symmetry factor S is the dimension of the automorphism group of the diagram i.e. the number of diﬀerent ways you can draw the diagram without changing its shape. This is equivalent to calculating the number of pairings froim Wick’s theorem.

6.4.2 Feynman rules in momentum space In practice, use of momentum space is a lot easier. Each vertex will now contain the following factor Z 4 (1) (2) (3) (4) d z∆F (z − ...)∆F (z − ...)∆F (z − ...)∆F (z − ...) (6.36)

i·(z−...) These contain the e for each∆F therefore

d4zei(p1+p2+p3+p4)z (6.37)

Now doing the z integral gives a delta function for p1 + p2 + p3 + p4. We therefore have momentum conservation.

p3 p4

k -p1-p1-k

p1 p2

Figure 13: Feynman diagram for a process where n = 2, r = 2. Particles are labelled by their momenta.

The Feynman rules in momentum space are as follows,

• Draw all diagrams to the appropriate order

63 • Label each line by it’s momentum

−i • For each line there is a factor of p2+m2−i

R d4k • Integrate over all undetermined momenta with weight (2π)4 • Multiply by a factor of (−iλ)p where p is the number of vertices • Divide by the symmetry factor S

For the diagram in Figure 13, we have the following contribution:

1 Z d4k 1 1 1 1 (−λ)2 2 (2π)4 p2 + m2 − i p2 + m2 − i p2 + m2 − i p2 + m2 − i 3 4 2 1 (6.38) 1 1 × 2 2 2 2 2 2 k + m − i (p1 + p2 + k ) + m − i The symmetry factor of this process is 2.

6.5 Scattering amplitudes In experiments we can calculate the scattering cross section so we need to calculate a value in order to test our quantum ﬁeld theory. This scattering cross section is classically given by: Number of particles scattered σ = (6.39) T × velocity of beam × Number density of particles in the beam where the denominator on the RHS is the ﬂux of the beam of particles.

Initial State Final State

E1, v1 E2, v2 Energies:

E1,E2, ... EN Momentum:

p1,p2, ... pN

(a) (b)

Figure 14: Diagrammatic representation of how we should look at scattering interactions.

We now introduce the concept of an S-matrix that interacts with initial and ﬁnal states to produce the scattering

hf| S |ii S = I + iT (6.40) where T = (2π)4δ(4) (P momenta) M and M is the amplitude of the process in question. M is related to the amplitude of h0| T φ...φ |0i but without factors for the external legs. The Feynman rules for calculating M are as follows

• Draw all connected Feynman diagrams to the relevant order • Label each line with an appropriate momentum subject to momentum conservation at each vertex • Each vertex gets a factor of (−iλ)

−i • Each internal line gets a factor of p2+m2−i

64 d4k • Integrate over all undetermined momenta with weight (2π)4 • Divide by a symmetry factor • Add up all contributions

For the scattering in Figure 15, we are in the frame in which energies are E1, E2 and v1, v2. This gives us a diﬀerential scattering cross section as follows,

3 ! 1 1 Y d pi 1 2 dσ = 3 |M | (6.41) 2E1 · 2E2 |v1 − v2| (2π) 2Ei ﬁnal state momenta In φ4 theory we calculate the lowest order scattering in λ for four-particle scattering (Figure 15) in the Center-of-Mass frame:

Z 3 3 1 1 2 4 d p3d p4 (3) dσ = λ (2π) 6 δ ( p1 +p2 +p3 + p4)δ(E1 + E2 − E3 − E4) (6.42) 2E12E2 |v1 − v2| (2π) 4E3E4 | {z } COM-frame where we have

~p1 = (0, 0, p)

~p2 = (0, 0, −p) (6.43) ~p3 = (p sin θ cos φ, p sin θ sin φ, p cos φ)

~p4 = ~p3 where we have used that the particles were scattered in some arbitrary direction labelled with polar coordinates θ and φ. We now do the p4 integral to get

p3 p4

p1 p2

Figure 15: Lowest order scattering in λ

Z 3 1 1 2 4 d p3 dσ = λ (2π) 6 δ(E1 + E2 − E3 − E4) (6.44) 2E12E2 |v1 − v2| (2π) 4E3E4 We can now use the fact that R d3p = R p2dp sin θdθdφ to ﬁne the diﬀerential scattering cross section as

dσ 1 1 λ2 Z ∞ p2dp = 2 δ(E1 + E2 − E3 − E4) (6.45) dφ 2E12E2 |v1 − v2| 4π 0 4E3E4 Since we are in the centre of mass frame all momenta have the same modulus,

|v1| = |v2|,E1 = E2 = E3 = E4 = E (6.46)

2 E2−m2 We also know that v = E2 from E = γm. The diﬀerential scattering cross section then becomes

65 2 ∞ 2 dσ 1 1 λ Z p dp 1 2 2 2 = 2 2 δ(2E − 2(p + m ) ) dφ 2E12E2 |v1 − v2| 2π 0 4E

λ2 1 1 p2 = √ 2 3 1 1 128π E 2 2 2 2 2 E − m |2 · 2 (p + m ) · 2p| (6.47) 2 2 2 1 λ 1 1 1 (p + m ) 2 √ 2£ = 2 3 p 128π E E2 − m2 2 p¡

λ2 1 λ2 1 = 2 2 = 2 2 256π E 64π ECOM where in the last step we have used the fact that 2E = ECOM .

66 7 Feynman path integrals for the Dirac ﬁeld

We can build an expression for the path integral while considering the dirac equation from a comparison to the case with a scalar ﬁeld: Z Z Y 4 a Z[η, η¯] = DΨDΨ¯ exp i d xΨ(¯ iγ ∂a − m)Ψ +η ¯Ψ + Ψ¯ η (7.1)

However, we have a diﬃculty. Ψ and Ψ¯ obey anti-commutation rules and so there exists a point such that { Ψ(x), Ψ(¯ x0) } = 0 and so before we start to work with this expression we need to deﬁne a new algebra.

7.1 Grassmann calculus 7.1.1 Grassmann numbers

“Numbers” which anti-commute are known as Grassmann numbers. Let us look at C:

z = a + ib i2 = −1 for complex numbers. (7.2) The generalisation to the Grassmann case is: Ψ2 = 0. We deﬁne a Grassmann number as:

f = a + Ψb, a, b ∈ R, C (7.3) and Ψ is a Grassmann number. For commuting f we have:

f = a + Ψb = a − bΨ (7.4) as Ψ2 = 0. This is all one has if there is only a single element to the Grassmann numbers. We can begin to understand Ψ by noticing that it acts a little like a 1-form.

7.1.2 Grassmann variables

When we have many Grassmann numbers we obtain Grassmann variables: Ψi i = 1,...,N and [N/2] { Ψi, Ψj } = 0. Note: this should remind you of a Cliﬀord algebra, of dimension z . A general function of N Grassmann numbers can then be written using an analog of a Taylor series for Grassmann variables as: 1 1 f(Ψ) = a + biΨi + cij ΨiΨj + ... + dij...N ΨiΨj ... ΨN (7.5) 2 |{z} N! | {z } i↔j Totally antisym antysym

As dij...N is totally antisymmetric we can say: dij...N = dij...N , where d is now just a constant. Moving back to the simple expression: f = a + Ψb for a commuting f ⇒ a commuting and b anticommuting. In other words (maths words): aΨ = Ψa, b2 = 0, bΨ + Ψb = 0, ab = ba,[bΨ, a] = 0. This gives Z2 grading so each quantity is either odd, or even: Odd × Odd = Even Odd × Even = Odd Even × Even = Even

Lets now move on to how we can being to use our old calculus tools on these new numbers (remember, we need a crash course so that we can do the path integral!)

67 7.1.3 Differentiation Let us first get to grips with what happens when we differentiate a Grassmann variable. So we start with:

f = a + Ψb. (7.6) Differentiate with respect to Ψ: ∂f = b (7.7) ∂Ψ However due to the anti-commutation we need to be careful about which side we differentiate from. Here, and in general, we will only differential from the left, which is explicitly denoted as: −→ ∂ (Ψb) = b (7.8) ∂Ψ It would be just as correct to differentiate from the right though, quickly outlined: ←− ←− ∂ ∂ (Ψb) = (−bΨ) = −b (7.9) ∂Ψ ∂Ψ

7.1.4 Integration So now we’ve diﬀerentiated lets learn how to undo this and get to a description for integration. We want to impose:

• Integration is linear • Integration is invariant under translations.

In other words, (maths):

∞ ∞ Z Z dxcf(x) = c dxf(x) (7.10a)

−∞ −∞ ∞ ∞ Z Z dxf(x + a) = dxf(x) (7.10b)

−∞ −∞

The only deﬁnition of integration for Grassmann variables that has these two properties is:

Z dΨ 1 = 0 (7.11a) Z dΨ Ψ = 1 (7.11b) Z dΨ(a + bΨ) = b (7.11c) noting that Ψ2 = 0 and Ψ−1 seems fairly meaningless. Hence, diﬀerentiation and integration are the same operation. Now suppose we have N Grassmann variables: Ψi, i = 1,...,N, we return to the previous expression: 1 1 f(Ψ) = a + b Ψ + c Ψ Ψ + ... + d Ψ Ψ ... Ψ (7.12) i i 2 ij i j N! ij...N i j N Let us look at how we could integrate this general function:

68 Z dΨN dΨN−1 . . . dΨ2dΨ1 f(Ψ). (7.13) | {z } Arbitrary ordering gives ⇒ d

Note: the ordering of the dΨi is completely arbitrary but our choice simplifies our choice giving us only d. A different order would give us a horridly messy prefactor, which doesn’t sound like fun at all. The rule for a single integration is thus: Z dΨiΨj = δij (7.14) and so is again the “same” as differentiation.

7.1.5 A change of variables This is easiest to describe with a bunch of algebra:

0 0 Ψi → Ψi Ψi = JijΨj (7.15)

0 1 0 0 1 0 0 f(Ψ) = a + bi(JijΨ ) + (JikJjlΨ Ψ )cij + ... + (Ji j Ψ ) ... (Ji j Ψ )di ...i (7.16) j 2 k l N! 1 1 j1 N N jN 1 N Z Z −1 0 0 d = dΨN . . . dΨ1f(Ψ) → (det J) dΨN . . . dΨ1f(Ψ) (7.17)

So one ﬁnds a “Jacobian” factor det J has the opposite sign of the exponent to the case of regular commuting variables.

7.2 Road to solving the path integral 1 T Let us look at the kind of things that we ﬁnd in path integrals. They contain exp( 2 Ψ MΨ) so suppose that there is only Ψ1 and Ψ2: 1 T 1 0 m Ψ1 Ψ MΨ = (Ψ1Ψ2) 2 2 −m 0 Ψ2 1 (7.18) = (Ψ mΨ − Ψ mΨ ) 2 1 2 2 1 = mΨ1Ψ2 The taylor expansion of an exponential is very easy with Grassmann numbers as only the term linear in Ψ survives: 1 exp(Ψ) = 1 + Ψ + Ψ2 + ... 2 (7.19) = 1 + Ψ and so: 1 exp( ΨT MΨ) = 1 + ΨT MΨ = 1 + mΨ Ψ . (7.20) 2 1 2 giving us the simpliﬁcation: Z 1 Z dΨ dΨ exp( ΨT MΨ) = dΨ dΨ (1 + mΨ Ψ ) = m (7.21) 2 1 2 2 1 1 2 Note: the positive power of m. What happens when we move to 2N Grassmann variables? For our 1 T exp( 2 Ψ MΨ), M is a 2n × 2n antisymmetric matrix and Ψ = (Ψi,..., Ψ2N ) In general, M = U T MUˆ for a unitary matrix U and Mˆ in it’s canonical form:

69   0 m1 −m1 0    ˆ  0 m2  M =   (7.22)  −m2 0   .  ..

! Or we can say: Mˆ is multiples of the symplectic form with eigenvalues mI · mI > 0. Turning back to the integral again:

Z N 1 T 1 Y 2 dΨ . . . dΨ exp( Ψ MΨ) = (det Mˆ ) 2 det Mˆ = ( m ) (7.23) 2N 1 2 i Here we have chosen the positive square root.

7.2.1 Aside: complex Grassmann numbers Now let us move to complex Grassmann numbers as it is these which we ﬁnd inside the Dirac equation. These are generalised much in the same way as with the reals using Cayley Dickson constructions. 1 1 χ = √ (Ψ1 + iΨ2)χ ¯ = √ (Ψ1 − iΨ2) (7.24) 2 2 Complex Grassmann integration has measures:

dχdχ¯ = −idΨ2dΨ1 (7.25) giving the basic integral for complex Grassmann variables: Z DΨDΨˆ exp(iΨ¯ OΨ) Z (7.26) DΨDΨˆ exp(iΨ¯ MΨ) = det M

Note: if we were dealing with a scalar ﬁeld we would have: Z − 1 DΦ exp(ΦMΦ) = (det M) 2 (7.27)

7.3 Solution of the path integral Resulting in the path integral that we need to solve:   Z Z ¯ 4 ¯ a ¯ Z0[η, η¯] = DΨDΨ exp i d xΨ(iγ ∂a − m)Ψ +η ¯Ψ + ηΨ  (7.28) | {z } | {z } for each point Source terms in spacetime

We want h0|T Ψ(x1) ... Ψ(¯ y1) ... |0i to be given by the path integral and so we look at: Z Z 4 a DΨDΨΨ(ˆ x1) ... Ψ(ˆ y1) ... exp i d xΨ(¯ iγ ∂a − m)Ψ (7.29)

with the following in the pre-factor of the path integral:

δ Z0(η, η¯) → −iΨ(¯ x1)Z0(...) (7.30a) δη(x1) δ Z0(η, η¯) → iΨ(x1)Z0(...) (7.30b) δη¯(x1)

70 So to solve this Grassmann integral we look at a standard solution: Z DχDχ¯ exp χ†Mχ + η†χ + χ†η = (det M) exp(−η†M −1η) (7.31) and apply this to the Dirac ﬁeld to obtain: Z Z 4 a Z0(J, J¯) = DΨDΨ¯ exp i d xΨ(¯ iγ ∂a − m)Ψ +η ¯Ψ + ηΨ¯ Z (7.32) = (Inﬁnite Factor) exp d4xd4yη¯(x)G(x, y)η(y)

where η (¯η) are (conjugate) spinors and G(x, y) is the greens function for Dirac algebra:

Z d4p 1m − γap G(x, y) = eip(x−y) a (7.33) (2π)4 p2 + m2 − i where we have two implicit spinor indices. Now we diﬀerentiate with respect to η, η¯ which is the same as putting Ψ, Ψ¯ into the integrand. From this we obtain an analogue of Wick’s theorem: X h0|T Ψ ... |0i = (Greens functions) (7.34) pairs Note: the left hand side disappears when the number of Ψ and Ψ¯ are equal.

7.4 Overview of solution rules 7.4.1 Propagators

Fermions

Photons

Scalars

Figure 16: The three diﬀerent lines we use when constructing Feynman diagrams.

Scalar Propagator −i (7.35) p2 + m2 − i Massless Vector (Photon) Propagator

papb −i(ηab − 2 ) p (7.36) p2 − i Dirac ﬁeld Propagator

−i(γap − m) a (7.37) p2 + m2 − i

71 7.4.2 Allowed interactions Self Interactions

gΦ3 λΦ4 (7.38) 6 4! Associated vertex terms: −ig and −iλ

Scalar-Dirac Interactions cΨΨΦ¯ , cΨ¯ γ5ΨΦ, cΨΨΦ¯ 2 (7.39) | {z } Non-renormalizable Associated vertex terms: −ic and −icγ5

Minimal coupling of EM ﬁeld to scalars The interactions are gauge invariant, giving the couplings:

−iqΛ Aa → A − a + ∂aΛ, Φ → e Φ (7.40) resulting in the interactions:

∗ a 2 a qΦ ∂aΦA , q Φ ∗ ΦAaA (7.41) a 2 2 Associated vertex terms: −iq(ip2 − ip1) and −iq

Dirac-EM ﬁeld coupling

∂a → ∂a − iqAa (7.42) gives gauge invariant couplings:

iqΛ A − a → Aa + ∂aΛ, Ψ → ie Ψ (7.43) adding two more interactions:

a a 5 qΨ¯ γ ΨAa, qΨ¯ γ γ ΨAa (7.44) Associated vertex terms: −iqγa and −iqγaγ5

The 8 rules for calculating the matrix elements:

hf|ii = i(2π)4δ(4) (momentum) T

1. Draw all Feynman diagrams and label each line with it’s momentum. Do this up to whatever order in the coupling you might want for a given process. 2. Associate to each vertex the appropriate coupling. 3. Add the propergators for each internal line. 4. External line factors: • Scalars: 1 ∗ • Photons: these need a polarisation vector. Ingoing (outgoing) has a factor s(p)(s(p))

72 -ig -iλ (a)

-ic -iγ5c (b)

a a -iq(-ip -ip ) 2 2 1 -iq (c)

-iqγa -iqγaγ5 (d)

Figure 17: Feynman diagrams corresponding to the diﬀerent interactions: (a) self interactions, (b) scalar-Dirac interactions, (c) scalar-EM interactions, (d) Dirac-EM interactions. Above each vertex is the corresponding constant which relates to it.

• Fermions: × Ingoing Particle ↑ us(p) Outgoing Particle ↑ u¯s(p) × × Ingoing Anti-particle ↓ v¯s(p) Outgoing Anti-particle ↓ vs(p) × where × denotes the external part of the Feynman Diagram. 5. Integrate over all undetermined momenta with weight: Z d4k (2π)4

6. Closed Fermion loop 1 for each loop. 7. Divide by the symmetry factor. 8. Add up all the diagrams → verify gauge invariance.

73 7.5 Worked examples 7.5.1 Fermions coupled to a scalar field e+e− → 2Φ, interaction: gΨΨΦ¯ (7.45) we obtain two different tree diagrams from this expression, shown in in figure 18.

p1 p3 p1 p4 e- Φ e- Φ

p - p p - p 1 3 + 1 4

p2 p4 p2 p3 e+ Φ e+ Φ

Figure 18: Feynman Diagrams for the two possible cases of e−e+ → 2φ.

The amplitudes are constructed using the above rules to obtain for the left diagram in ﬁgure 18:

a −i(i (p1 − p3)aγ − m) 2 Tleft = ∆s2 (−p2) 2 2 us1 (p1)(−ig) ) (7.46) (p1 − p3) + m − i

and for the right hand diagram we simply swap p3 with p4 (which i will explicitly write out because I can copy/paste):

a −i(i (p1 − p4)aγ − m) 2 Tright = ∆s2 (−p2) 2 2 us1 (p1)(−ig) ) (7.47) (p1 − p4) + m − i Following from this we obtain the scattering cross-section in the centre of mass (COM) frame as:

dσ 1 2 = 2 |Matrix Element| (7.48) dΩ 64π ECOM

7.5.2 Fermions coupled to the electromagnetic ﬁeld Due to the nature of the rules being universal the method for calculating the matrix elements and cross-section is nearly identical. We start by drawing the two tree diagrams seen in ﬁgure 19.

− − − − a e + e → e + e interaction: eΨ¯ γ ΨAa (7.49)

p1 p3 p1 p4

p - p γ p - p γ 1 3 + 1 4

p2 p4 p2 p3

Figure 19: Feynman Diagrams for the two possible cases of electron scattering.

From this we do exactly the same thing, just applying the above rules. For the ﬁrst diagram we obtain:

74 a b 2 Tleft = ∆ab(p1 − p3)¯us3 (p3)γ us1 (p1)¯us4 (p4)γ us2 (p2)(−ie) (7.50) where the only internal line (and hence only propagator) is the photon propagator. Making the most of copy/paste we swap around p3 and p4 again to get the amplitude for the second diagram:

a b 2 Tright = ∆ab(p1 − p4)¯us4 (p4)γ us1 (p1)¯us3 (p3)γ us2 (p2)(−ie) (7.51) Looking at the scattering cross-section:

dσ 1 2 = 2 |Matrix Element| (7.52) dΩ 64π ECOM | {z } Various s speciﬁed Often one does not measure the polarisation states of the particles, we then can average over the appropriate spin-states: ! dσ 1 X 1 = |T |2 (7.53) dΩ 2 64π2E s COM ﬁnially simplifying our result with the expressions derived from 3.64.

75 8 A little beyond perturbation theory

Throughout the course we have been studying the solutions of Quantum Field Theory using only perturbative methods, expanding using Feynman diagrams. However if we turn our attention back to Scalar ﬁelds we see that in special cases we are able to work exactly and produce two quite exceptional theoretical results: the Soliton and the Higgs particle.

8.1 Real ﬁelds: the soliton 1 1 L = − (∂Φ)2 − λ(Φ2 − v2)2, (8.1) 2 8 where v is a real constant. When looking at the potential term (V) we can notice that it is quartic in Φ and so would usually be ignored while solving with perturbation theory. What happens if we keep it in? Looking at the turning points in the potential we see a maximum at Φ = 0 and two minima at V = 0 for Φ± = ±v. Looking at the equations of motion for the Lagrangian we obtain:

0 Φ = V (Φ) (8.2) where we see that the two solutions are the turning points of the potential. The maximum, Φ = 0, satisﬁes the classical E.O.M but is an unstable solution. (Think of the particle coming to rest on the top of a pin-head). At V (Φ) = 0 we obtain the classically stable state - however, how do we deal with this degeneracy of two possible vacuum states, both with zero energy? Let us decompose our potential:

Φ = Φ± + ΦQuantum, (8.3) looking at derivatives of the potential: λ V (Φ) = (Φ2 − v2)2 8 λ V 0(Φ) = (Φ2 − v2) · Φ (8.4) 2 λ λ V 00(Φ) = (Φ2 − v2)Φ + Φ · 2Φ 2 2 we obtain the resultant Lagrangian: 1 L = − (∂Φ )2 − λv2Φ2 + O(Φ3, Φ4,...) (8.5) 2 q q q q

where we now see the eﬀective mass of the quantum excitations around the Φ± vacua is: √ m = 2λv. (8.6)

So diﬀerent parts of the universe might be in diﬀerent vacua (the two states Φ± separated by a domain wall in the x-direction) independent of the y/z-directions. There are true solutions to the classical equations of motion such that:

x → ∞ Φ → Φ− (8.7) x → −∞ Φ → Φ+ when the wavefunction is of the form: √ ! λ Φ = v tanh (x − x ) (8.8) 8 0 with resulting energy density: 1 E = (∇Φ)2 + V (Φ) (8.9) 2

76 The only obviously measurable elements are m, (the eﬀective mass of ﬁeld Φq) and λ. The energy per m3 unit area of the domain wall is proportional to λ . We never would have been able to see this phenomena using perturbative solutions as it will only ever give us an expansion of positive powers of λ. The object we have described above is a Soliton, which can be thought of as a self-reinforcing solitary wave that maintains its shape while it traveling at a constant speed (thanks wikipedia). These types of phenomena are universal as long as there exist multiple discrete vacua.

8.2 Complex ﬁelds: the Higgs 1 L = −∂Φ∗∂Φ + m2Φ∗Ψ − λΦ2Φ∗2 (8.10) 4 Looking at the Noether symmetry:

Φ → Φiα (8.11) We see that Φ = 0 is a maximum. We have two minimum values at:

2m2 |Φ|2 = (8.12) λ where we see there is no explicit phase dependance, leading to the famous Mexican hat potential arriba arriba arriba!

Figure 20: The Mexican hat potential. We see complete spherical symmetry due to the potential having no phase dependence.

So how should we choose our vacuum states? r 2m2 Φ = eiθ, θ = arg(Φ ) Classical λ Classical ΦComplete = ΦClassical + ΦQuantum (8.13) r ! 2m2 σ(x) iπ(x) = + √ exp λ 2 fπ where σ(x) and π(x) are variations of the modulus and argument. We follow previous methods and look at the Lagrangian:

r !2 1 2m2 σ 1 m4 1√ 1 2 √ 2 2 2 3 4 L = (∂σ) = + 2 (∂π) − − + m σ + λmσ + λσ (8.14) 2 λ 2 fπ λ 2 16

From this we see that σ(x) is a massive scalar ﬁeld with some interesting interactions, whereas π(x) is massless.

77 8.2.1 Goldstone’s theorem If we break a continuous symmetry there is always a massless particle. Here we broke the symmetry by speciﬁng a classical solution. We can couple the above Lagrangian to the EM ﬁeld:

∂a → ∂a − ieAa (8.15) 1 ab where we have given our Φ an electric charge. So we add in a factor − 4 FabF so we now have Maxwell and complex scalars. Let us look at the lowest orders of quadratic terms in A: 1 L = − F F ab − λA Aa + O(A4) (8.16) 4 ab a from which we can deduce that there must be a term of the form: e2A2Φ2. Now we’ve added a massive spin-1 particle with mass: 2em m = √ (8.17) λ We see that the massless ﬁeld π has disappeared and A has become massive. This massive term is known as the Higgs term.

End of Lecture Series