November 30, 2016

Theoretical Exercises 2016

Thomas Becher

Albert Einstein Center for Fundamental Physics Institute for Theoretical Physics University of Bern

Contents

1 Gaussian integrals and Wick’s theorem 1 1.1 Gaussian integrals ...... 1 1.2 Generating function, Wick’s theorem ...... 2

2 Integrals over Grassmann variables 4

3 Dirac Algebra 6 3.1 Levi-Civita symbol ...... 6 3.2 Dirac Algebra ...... 8

4 Klein-Gordon, Maxwell and Proca equations 11 4.1 Klein Gordon equation ...... 11 4.2 Electromagnetic field in relativistic notation ...... 11 4.3 Proca equation ...... 12 4.4 Gauge invariance ...... 13

5 Spinors and Dirac equation 15 5.1 Lorentz transformation of spinor fields ...... 15 5.2 Dirac Lagrangian and Dirac equation ...... 19 5.3 Plane-wave solutions ...... 19

6 Non-relativistic limit of the Dirac equation 22 6.1 Dirac equation in the presence of an electromagnetic field ...... 22 6.2 Magnetic moment of the electron ...... 23 7 Interacting fields in perturbation theory 25 7.1 Relativistically invariant interaction Lagrangians ...... 25 7.2 Perturbation theory ...... 27 7.3 Wick’s theorem ...... 28 7.4 Feynman rules ...... 30 7.4.1 φ4-theory ...... 30 7.4.2 QED ...... 31 7.4.3 QED with an extra neutral scalar S ...... 32

8 Computation of scattering cross sections 33 8.1 Reduction formula and S-matrix ...... 33 8.2 Cross section and decay rate ...... 35

9 Diphoton production and a scalar resonance 37 9.1 Decays of the scalar S ...... 38 9.2 Diphoton production ...... 39 9.3 Finite-width effects ...... 41 9.4 Data analysis ...... 42

2 1 Gaussian integrals and Wick’s theorem

1.1 Gaussian integrals Exercise 1.1. Compute the basic, one-dimensional Gaussian integral and show that it is given by

2 π ∞ dx e λx = , λ > 0. (1) − λ Z−∞ r The standard trick to obtain this result is to take the square and use polar coordinates to evaluate the integral. Notice that λ can be complex, as long as Re(λ) > 0.

Exercise 1.2. Show that

2 2 π a ∞ dx e λx +ax = exp , λ > 0. (2) − λ 4λ Z−∞ r  

Next, we turn to n-dimensional Gaussian integrals. We consider an element x of Rn and a symmetric n n matrix M and define a quadratic form xT Mx (we view x as a column matrix × to avoid writing vector arrows). Integrating over Rn, we obtain the Gaussian integral

n T 1 d x exp x Mx = − (3) − N Z
Exercise 1.3. Show that det M = . (4) n N r π What are the conditions on the matrix M for the integral (3) to exist?

Exercise 1.4. Similarly, show that for a Rn ∈

n T T 1 T 1 d x exp x Mx + a x = exp a M − a . (5) N − 4 Z  

Gaussian integrals appear in many areas of physics, in particular in statistical physics and in path-integral quantization. In Feynman’s path-integral formulation of quantum mechanics, one needs to compute an integral of the form

x , t x , t = x(t) eiS[x(t)] (6) h f f | 0 i D Z to obtain the probability amplitude that a particle which was at point x0 at time t0 can be found at point x1 at time t1, where S[x(t)] is the action associated with the path x(t) and the symbol x(t) indicates that one should sum over all paths x(t) starting at x(t0) = x0 and ending atDx(t ) = x . R f f 1 As it stands, it is not clear what (6) means. To define expression (6), one discretizes time t = i ∆ such than t = t . After which the integral over all paths takes the form i · n+1 f n x(t) dx = dnx (7) D −→ i i=1 Z Y Z Z where xi = x(ti) is the position after i time steps. We see that we encounter integrals over Rn as in (3). To evaluate the oscillatory expression (6), one first computes it for imaginary (Euclidean) time t = iτ with τ R. Performing this so-called Wick rotation leads to − ∈ iS[x(t)] SE[x(τ)] so that one can work with a real exponent as in (3). After performing the integrals→ − and taking after taking ∆ 0 one then analytically continues result back to physical time-values . Doing so, the path→ integral for a harmonic oscillator (or a free particle) boils down to the evaluation of Gaussian integrals as in (3). All these steps will be explained and carried out in the Advanced Concepts course, in these exercises we focus on the computation of the Gaussian integrals. The path integral formulation plays an important role in QFT, where it is also the basis for numerical computations. In field theories one integrates over all field configurations φ(t, ~x) instead of the path x(t). In discretized form, one then integrates over the field values at points φi φ(ti, ~xi). The discrete set of points (ti, ~xi) is called a lattice and the to obtain the continuum≡ result one takes the limit where the distance between lattice points (the lattice spacing) goes to zero.

1.2 Generating function, Wick’s theorem If we include the normalization factor , we can view the integrand in (3) N ρ(x) = exp xT Mx , (8) N − as a probability distribution in Rn since it is normalized and
strictly positive as long as M is a real, symmetric and positive1 matrix. We can then compute expectation values

A(x) = dnx ρ(x) A(X) . (9) h i Z The m-point correlation functions x . . . x , (10) h i1 im i play an important role when one computes path integrals and we will now analyze them in detail. The result is the Wick theorem, which provides the basis for perturbative computations in QFT. To obtain the expectation values (10), let us consider

1 (b) = ebixi = dnx ρ(x)ebixi = b . . . b dnx ρ(x) x . . . x . (11) Z h i m! i1 im i1 im m 0 Z X≥ Z 1All its eigenvalues are strictly positive.

2 The quantity (b) is the generating function of the moments of the probability distribution ρ(x): Z 1 (b) = b . . . b x . . . x . (12) Z n! i1 im h i1 im i m 0 X≥ The inverse relation can be written ∂ ∂ x . . . x = ... (b) . (13) h i1 im i ∂b ∂b Z  i1 im b=0 In general, for an arbitrary probability density ρ(x), (b) cannot be exactly calculated. But it can easily be evaluated for the Gaussian probabilityZ density. According to result (5), the generating function of the moments of the Gaussian distribution is 1 (b) = dnx exp xT Mx + bT x Z N −2 Z   = exp(bT x) (14) h i 1 = exp bT M 1b . 2 −   The inverse relation is

∂ ∂ 1 1 x . . . x = ... exp b (M − ) b . (15) h i1 im i ∂b ∂b 2 i ij j  i1 im  b=0

Exercise 1.5. Prove that: ∂ 1 exp b (M 1) b = (M 1) b exp 1 b (M 1) b , ∂b 2 i − ij j − i1j j 2 i − ij j i1  
∂ ∂ 1 1 1 1 1 exp bi(M − )ijbj = (M − )i1i2 + (M − )i1jbj(M − )i2kbk ∂bi1 ∂bi2 2   h i 1 1 exp b (M − ) b . · 2 j jk k
∂ ∂ 1 1 Exercise 1.6. Find a general expression for ... exp bi(M − )ijbj . Looking at the ∂bi1 ∂bin 2 explicit expressions for the lowest few derivatives, you should observe a simple pattern, which
can then be established using induction.

Exercise 1.7. Use the previous results at b = 0, as in the inverse relation (15), to verify that 1 x x = (M − ) , h i1 i2 i i1i2 1 1 1 1 x x x x = (M − ) (M − ) + (M − ) (M − ) (16) h i1 i2 i3 i4 i i1i2 i3i4 i1i3 i2i4 1 1 +(M − )i1i4 (M − )i2i3 ,

3 and the correlators of an odd number of points vanish,

x . . . x = 0 . h i1 i2m+1 i

Exercise 1.8. Derive Wick’s theorem:

xi1 . . . xi2m = xk1 xk2 ... xk2m 1 xk2m , (17) h i h i h − i P X where the sum is over all pairings P , i.e. all possible ways to group the indices the indices i1, i2, . . . , i2m into m pairs (k1, k2),..., (k2m 1, k2m). The theorem states for the Gaussian integral all higher-point correlators reduce− to products of the 2-point correlation function, which is given by the inverse of the matrix in the exponent of the Gaussian integral

1 x x = (M − ) (18) h i1 i2 i i1i2 as we have shown above.

(As usual in mathematics, the proof of the final theorem is trivial after you have prepared it with a highly nontrivial proof of a lemma, in our case the general form of the derivative in Exercise 1.6.)

2 Integrals over Grassmann variables

To construct a consistent quantum theory of fermionic fields, the field operators must fulfil anti-commutation relations

(3) ψ (x), ψ† (y) = δ (~x ~y)δ , (19) α β − αβ  ψα(x), ψβ(y) = ψα† (x), ψβ† (y) = 0 , at equal times x0 = y0. To construct a path-integral  representation of a such a theory, one uses anti-commuting variables, which are also called Grassmann numbers. They are introduced and explored in this part of the exercises. We consider a set of n such numbers gi, i = 1 . . . n, which fulfil the Grassmann algebra ηi, ηj = 0 , (20) 2 implying that ηi = 0, which makes the algebra extremely simple. The most general function of two Grassmann variables is

f(η1, η2) = f00 + f10 η1 + f01 η2 + f11 η1η2 , (21) since η2η1 = η1η2 and all higher-order terms vanish. The expansion coefficients are ordinary numbers. The− Taylor series expansion of Grassmann functions is thus always finite and exact. For the rest of this section, roman letters refer to real or complex numbers, greek letters denote

4 the Grassmann variables. Exercise 2.1. The Grassmann algebra can be implemented using anti-commuting matrices. Provide a matrix representation for the case n = 2.

Exercise 2.2. Equation (21) shows that the n = 2 algebra is four-dimensional. What is the dimension for general n?

The integral over a Grassmann function f(η) = a + b η is defined as

dη f(η) = b . (22) Z Exercise 2.3. Show that, up to a normalization factor, this definition follows from the requirements of linearity and invariance of the integral under a shift η η + θ. → One can also define the derivative of a Grassmann function ∂ ∂ f(η) = (a + b η) = b , (23) ∂η ∂η which happens to be the same as the integral! For multiple integrals and derivatives one needs to adopt a sign convention. We define ∂ ∂ dη dη η η = η η = +1 , (24) 2 1 1 2 ∂η ∂η 1 2 Z Z 2 1 i.e. we perform the innermost integral (or derivative) first. Exercise 2.4. Compute the Gaussian integrals

ηT Aη I(A) = dηn dηn 1 ... dη1 e− , (25) − Z Z Z where η = (η1, . . . , ηn) for n = 2, 3, 4. Hints: Taylor expand and note that only the terms proportional to η1η2 . . . ηn contribute, which involve exactly one power of each variable. Note further, that the matrix A can be chosen anti-symmetric since the symmetric part does not contribute.

Under a variable change θ = aη, we have

1 = dθ θ = d(aη) aη (26) Z Z 1 d(aη) = dη . → a Note that this is exactly the opposite of the behavior of regular (bosonic) integrals, where d(ax) = a dx. The Grassmann integral behaves like a derivative under variable transforma- tions, which is maybe not surprising, since it is the same as the derivative.

5 Exercise 2.5. For a general variable transformation ξi = Bijηj, prove that

1 dξn . . . dξ1 = (det B)− dηn . . . dη1 . (27)

Hint: proceed as in (26), using ηi1 . . . ηin = i1...in η1 . . . ηn.

To work with the complex valued Dirac field, one introduces complex Grassmann variables 1 1 η = (η1 + iη2) , η∗ = (η1 iη2) . √2 √2 −

One can treat η and η∗ as independent variables and defines

dη∗dη η η∗ = 1 . (28) Z Exercise 2.6. Derive the following identites for Gaussian integrals with complex Grassmann variables

n η†Aη dηi∗ dηi e− = det A, (29) i=1  Y Z Z  n 1 η†Aη+η†θ+θ†η +θ†A− θ dηi∗ dηi e− = det A e , (30) i=1  Y Z Z  for a hermitian matrix A. Hint: Derive (29) by performing a change of variables which diag- onalizes A and complete the square to obtain (30).

Except for the normalization, the only difference to regular (bosonic) Gaussian integrals is the 1 appearance of det A instead of (det A)− . As in the bosonic case, all moment integrals can be obtained by taking derivatives of the result (30) with respect to θi and θj∗. Exercise 2.7. Use this technique to compute the integral

n η†Aη dηk∗ dηk e− ηiηj∗ . (31)  kY=1 Z Z  The Wick theorem for Grassmann integrals thus has exactly the same form as in the bosonic case, but one needs to keep track of the minus signs which arise when variables are reordered.

3 Dirac Algebra

3.1 Levi-Civita symbol Vectors or tensors under Lorentz transformations carry greek indices µ, ν, τ, . . . . In d = 4 space-time dimensions, the indices take the values 0, 1, 2, 3. The coordinates of a space-time

6 point are written as xµ with x0 = t (we set c = 1) and ~x = (x1, x2, x3). We use roman indices to denote the space components xi with i = 1, 2, 3. We write the metric in four-dimensional Minkowski space in the form 1 0 0 0 0 1 0 0 g =  −  . (32) µν 0 0 1 0  −   0 0 0 1     −  Sometimes also the notation ηµν is used for the metric in flat Minkowski space. The inverse µν metric g is numerically equal to gµν and verifies, by definition,

µν µ µν µ g gνρ = δρ , g gµν = δµ = d = 4. Throughout, we use the Einstein convention and sum over repeated indices. The metric gµν is an invariant tensor, i.e. the components in (32) are the same in every frame. There is a second invariant tensor, the fully antisymmetric Levi-Civita symbol µνρσ and we define 0123 = 1 (33) All other components follow from antisymmetry.

Exercise 3.1. Show that the contravariant tensor fulfils

0123 = 1 . (34) − Raising all four indices using the inverse metric induces a minus sign.

For both physics and formal reasons, it is sometimes interesting to work in space-times with d = 4. In d dimensions, we can define a Levi-Civita symbol with d indices µ1µ2...µd . This tensor fulfills6 the relation

δµ1 . . . δµ1 ν1 νd µ1µ2...µ d 1 . .  d  = ( 1) − . . . (35) ν1ν2...νd . . − δµd . . . δµd ν1 νd

The symbol ... denotes the determinant of the d d matrix. Products of -tensors can thus always be eliminated| | in favor of metric tensors. A× related, useful relation is the following

 Aν1 Aν2 ...Aνd = det(A)  . (36) ν1ν2...νd µ1 µ2 µd µ1µ2...µd By choosing a proper Lorentz transformation Λ for the matrix A one also immediately sees that  is indeed an invariant tensor.

Exercise 3.2. Write out and derive relation (35) in d = 2 space-time dimensions.

7 Exercise 3.3. Derive relation (35) for d space-time dimensions.

Exercise 3.4. Derive (36).

For d = 4, one obtains several useful identities from contracting indices in (35)

µνρσ = δν δρδσ + δν δρδσ + δνδρ δσ δν δρ δσ δν δρδσ δνδρδσ , (37) µαβγ − α β γ β γ α γ α β − β α γ − α γ β − γ β α µνρσ = 2 δρ δσ δρδσ ,
(38) µναβ − α β − β α µνρσ = 6 δσ,
(39) µνρα − α µνρσ = 24. (40) µνρσ − Exercise 3.5. Derive two of these four relations.

3.2 Dirac Algebra The Dirac equation is formulated using complex n n matrices γµ (called Dirac matrices or simply γ-matrices) which fulfill ×

γµ, γν = γµγν + γνγµ = 2 gµν 1 , (41) { } where 1 is the n n identity matrix. The algebra generated by these matrices (i.e. sums of products of matrices× with complex coefficients), defines the Clifford (or Dirac) algebra. The reason that the Dirac matrices are useful, is that they allow one to obtain easily obtain representations of the Lorentz group for particles with half-integer spin. Indeed, the matrix i Sµν = [γµ, γν] (42) 4 fulfils the commutation relations of the Lorentz algebra, which read

[J µν,J ρσ] = i (gνρJ µσ gµρJ νσ gνσJ µρ + gµσJ νρ) , (43) − − where J µν are the generators of the Lorentz transformations. Due to antisymmetry under µ ν, there are six generators J µν, which contain the three boosts J 0i and three rotations J ij↔.

Exercise 3.6. Show that J µν = Sµν fulfills the commutations relations (167). To see this, derive first that [Sµν, γρ] = iγµgνρ iγνgµρ (44) − and then use the Jacobi identity

[A, [B,C]] + [B, [C,A]] + [C, [A, B]] = 0 (45)

8 with A = Sµν, B = γρ, C = γσ.

The Dirac algebra can be introduced in different space-time dimensions and both for a Minkowski metric (32) or for an ordinary Euclidean metric gij = δij. Indeed, for three Eu- clidean dimensions, the Pauli matrices generate the Clifford algebra

σi, σj = 2 δij 1 , (46) { } for n = 2. For d = 4 space-time dimensions 2 2 matrices are not sufficient. The reason is × that an arbitrary 2 2 matrix A can be written as a linear combination A = c01+ci σi so that it is not possible to× find a fourth anti-commuting matrix. The lowest dimensional matrices which can realize the d = 4 algebra have n = 4.

Exercise 3.7. Find an explicit representation of the algebra for n = 4. Consider matrices which have blocks of σi, 1 or 0 as 2 2 submatrices. (Obviously, solutions are in every field- theory book, but it should be fun to× find a representation yourself.)

µ µ µ 1 Exercise 3.8. Show that if γ is a representation, then alsoγ ˜ = Sγ S− is one, for an arbitrary invertible n n matrix S. × Also the opposite statement is true (but more difficult to show): any representationsγ ˜µ and γµ can be related with a suitable matrix S. The representations are thus equivalent up to a change of basis in the vector space on which they act. Many computations, in particular all exercises in the rest of this section, can be performed without relying on an explicit represen- tation of the γ matrices.

Exercise 3.9. Using only the properties of the algebra (41), derive the following expressions

µ γ γµ = d 1 (47) γµγνγ = (d 2) γν, (48) µ − − γµγνγργ = (d 2)γνγρ + 2γργν (49) µ − γµγνγργσγ = 2γσγργν + (4 d)γνγργσ (50) µ − − Simplify the third relation for d = 4 dimensions.

In many QFT computations one needs to compute traces of products of γ-matrices. These can be evaluated using an explicit representation, but it is more elegant to compute them in a representation independent way using (41). Let us first compute the trace of a single matrix. Consider (2 d)tr(γν) = tr(γµγνγ ) = tr(γµγ γν) = d tr(γν) (51) − µ µ It follows that the trace must vanish for d = 1. Similarly, one can show than the trace of any odd number of Dirac matrices vanishes. 6

9 Exercise 3.10. Show that tr(γνγργσ) = 0.

Next, consider the trace of two matrices 1 tr(γµγν) = tr(γµγν + γνγµ) = gµν tr(1) . (52) 2 There is a simple algorithm to compute any trace along these lines. Consider the trace of the product of m matrices and use the cyclicity of the trace to rewrite

µ1 µ2 µm µ1 µ2 µm µm µ1 µ2 µm 1 2 tr(γ γ . . . γ ) = tr(γ γ . . . γ ) + tr(γ γ γ . . . γ − ) (53)

In a second step one now uses (41) to anti-commute γµm in the second term back to the right- most position. One obtains commutator terms containing fewer γ matrices and finally the term tr(γµ1 γµ2 . . . γµm ), but with a minus sign since one had to anticommute (m 1) times. Because of the sign, this term cancels the first term on the right-hand side of (53).− The end result is that the trace of m matrices gets reduced to sum of traces with (m 2) matrices. − Exercise 3.11. Use this algorithm to show that tr(γµγνγργσ) = (gµνgρσ gµρgνσ + gµσgνρ) tr(1) (54) − tr(γµγνγργσγαγβ) = gµνtr(γργσγαγβ) gµρtr(γνγσγαγβ) + gµσtr(γνγργαγβ) − gµαtr(γνγργσγβ) + gµβtr(γνγργσγα). (55) −

Exercise 3.12. Voluntary: write a small recursive code, e.g. in Mathematica, which com- putes the trace of an arbitrary number of Dirac matrices.

Let us now consider d = 4 dimensions. For this case, one can then define an additional matrix γ = iγ0γ1γ2γ3 = iγ γ γ γ . (56) 5 − 0 1 2 3 Exercise 3.13. Show that this matrix fulfils i γ , γµ = 0, (γ )2 = 1 γ =  γµγνγργσ . (57) { 5 } 5 5 −24 µνρσ

Exercise 3.14. Verify the following trace relations

tr(γ5) = 0 , (58) µ ν tr(γ γ γ5) = 0 , (59) tr(γµγνγργσγ ) = 4iµνρσ . (60) 5 −

10 4 Klein-Gordon, Maxwell and Proca equations

4.1 Klein Gordon equation The relativistic wave equation for a free spin zero field has the form m2 ϕ(x) = 0 (61) − − where ϕ(x) is a real or complex scalar field with
mass m. This is called the the Klein-Gordon equation.

Exercise 4.1. Derive the equality d3k f(~k) = d4k δ(k2 m2) θ(k0) f(~k) , (62) 2ω − Z k Z 2 2 where ωk = √m + k . Show that the integration measure on the right-hand side is invariant under proper Lorentz transformations. The same is thus true for the one on the left-hand side.

Exercise 4.2. Show that the following general superposition of plane waves

3 d k ikx ikx ϕ(x) = a(~k)e− + b(~k)e , (63) 3 (2π) 2ωk Z h i 0 with k = ωk, solves the Klein-Gordon equation.

~ ~ Exercise 4.3. For a real field ϕ(x) = ϕ∗(x) the expansion coefficients a(k) and b(k) are related. What is the relation?

4.2 Electromagnetic field in relativistic notation The Maxwell equations (in Lorentz-Heaviside units) are

E~ = ρ , B~ ∂ E~ = ~j , (64) ∇ · ∇ × − t B~ = 0 , E~ + ∂ B~ = 0 . (65) ∇ · ∇ × t Exercise 4.4. Show that the homogeneous equations (65) are automatically satisfied if the fields are written in terms of the 4-vector potential Aµ = (φ, A~), with

B~ = A~ and E~ = ∂ A~ φ . (66) ∇ × − t − ∇

Exercise 4.5. Show that the inhomogeneous equations (64) can be written in terms of the field strength tensor F µν = ∂µAν ∂νAµ (67) − 11 and current vector jµ = (ρ,~j) as

µν ν ∂µF = j . (68) Show that Ei = F 0i and Bi = 1 ijkF . − − 2 jk

4.3 Proca equation

Exercise 4.6. Derive the equations of motion for the field Aµ from the Lagrangian density 1 m2 (A) = F µν F + A Aµ jµA . (69) L −4 µν 2 µ − µ and show that they reduce to the Maxwell equations (68) for m = 0. The equation with m = 0 is called the Proca equation and describes the propagation of a massive spin 1 field. 6

2 µ Exercise 4.7. Show that the Proca equation implies the consistency relation m ∂µA = µ µ ∂µj . For a conserved current ∂µj = 0 and nonzero mass m, this imposes the condition µ µ ∂µA = 0. This condition implies that not all four components of the field A are independent.

µ Exercise 4.8. Show that using ∂µA = 0 the Proca equation simplifies to m2 A (x) = j . (70) − − µ µ In other words, each component of the Proca
field fulfils the Klein-Gordon equation.

As in (117), we can write the solution as a superposition of plane waves. For a real field it is usually written in the form

4 3 d k ikx ikx A (x) =  (~k, λ)a(~k, λ)e− + ∗ (~k, λ)a∗(~k, λ)e , (71) µ 3 µ µ (2π) 2ωk Xλ=0 Z h i ~ where the four auxiliary polarization vectors µ(k, λ) can be chosen such that they form an orthonormal basis of Minkowski space

~ µ ~ µ∗ (k, λ) (k, λ0) = gλλ0 . (72) In principle, these vectors can be chosen real, but sometimes it is convenient to work with complex ones (circular polarizations). The first two of the vectors (the transverse polarizations) are usually chosen to have the form µ(~k, 1) = (0,~(~k, 1)) µ(~k, 2) = (0,~(~k, 2)) (73) ~ ~ ~ ~ with k (k, 1) = k (k, 2) = 0 and ~∗(k, i) ~(k, j) = δ . The third, space-like polarization · · · ij vector is chosen to have its three-vector components along ~k. This longitudinal polarization vector has the form µ(~k, 3) = (A, B ~k) , (74)

12 where the parameters A and B are chosen such that it is orthogonal to k (~k, 3) = 0 and normalized. The final polarization vector is timelike ·

µ(~k, 0) = C kµ , (75) along the direction of kµ.

Exercise 4.9. Determine the coefficients A, B and C explicitly and show that the con- µ ~ sistency condition ∂µA = 0 implies that a(k, 0) = 0. The massive field thus contains three independent solutions (“polarizations”) for a given momentum.

Exercise 4.10. Show that the four polarization vectors introduced above fulfil the com- pleteness relation 3 ~ ~ gλλ0 µ∗ (k, λ)ν(k, λ) = gµν . (76) λ,λX0=0

Exercise 4.11. Derive from this that the three physical polarization vectors fulfil the completeness relation 3 kµkν ∗ (~k, λ) (~k, λ) = g + , (77) µ ν − µν m2 Xλ=1 which follows immediately from rewriting the previous relation (76).

Please note the factors of 1/m in the polarization vectors and in (76), which make the limit m 0 nontrivial. → 4.4 Gauge invariance

Exercise 4.12. Show that for m = 0 the Lagrangian Aµ is invariant under gauge transfor- mations A (x) A (x) + ∂ α(x) . (78) µ → µ µ These transformations leave F µν and therefore also the electric and magnetic fields unchanged.

Exercise 4.13. Show that for any given field Aµ one can always make a gauge transfor- µ mation such that the condition ∂µA = 0 is fulfilled. This is called Lorenz gauge (since the imposed condition is Lorentz invariant. . . ).

µ Exercise 4.14. The condition ∂µA = 0 eliminates one unphysical degree of freedom. µ However, even imposing ∂µA = 0 does not fix the gauge freedom completely, since we can still perform gauge transformations which fulfil the scalar wave equation

 α(x) = 0 . (79)

13 Such transformations correspond to another unphysical degree of freedom so that we conclude that the photon field only contains only two physical degrees of freedom.

The Proca action for a massive spin one field is not gauge invariant, but the Stueckelberg action (see http://www.stuckelberg.org)

1 m2 (A, φ) = F µν F + (∂ φ + A )(∂µφ + Aµ) jµA , (80) L −4 µν 2 µ µ − µ which involves an additional real scalar field φ is invariant under the simultaneous gauge transformation

A (x) A (x) + ∂ α(x) . φ(x) φ(x) α(x) . (81) µ → µ µ → − Choosing the gauge condition φ(x) = 0, the Stueckelberg action reduces to the Proca action. One can thus view a massive spin one field as a combination of a gauge invariant field Aµ(x), with two degrees of freedom, and an additional scalar field φ, which provides the third degree of freedom (the longitudinal polarization). An explicit realization of this is the Higgs mechanism, where additional scalar fields provide a mass term for the gauge fields of the weak interactions.

14 5 Spinors and Dirac equation

5.1 Lorentz transformation of spinor fields

To construct theory for a field φα(x), one first writes down an action. To get relativistic equations, this action must be Lorentz invariant. To construct such an action for a given field, it is obviously important to know how the field transforms φα(x) under Lorentz transformations

µ µ µ ν x x0 = Λ x (82) → ν

We know how Lorentz transformations act on a scalar field ϕ(x) and a vector field Aµ(x)

1 ϕ(x) ϕ0(x) = ϕ(Λ− x) , (83) → µ µ µ ν 1 A (x) A0 (x) = Λ A (Λ− x) . (84) → ν Exercise 5.15. Show that the term m2 ∆ (x) = A (x)Aµ(x) L 2 µ 1 in the Proca Lagrangian density transforms as a scalar, i.e. ∆ 0(x) = ∆ (Λ− x). Remember that the metric tensor fulfills the identity L L

µ ν ρσ µν Λρ Λσ g = g . (85)

Exercise 5.16. Show that the Lagrangian of a free massless scalar m2 ∆ (x) = ∂ ϕ(x)∂µϕ(x) L 2 µ transforms as a scalar.

The general transformation law for a field φα(x) under Lorentz transformations is

β 1 φ (x) φ0 (x) = D (Λ) φ (Λ− x) , (86) α → α α β where the matrices D(Λ) are a representation of the Lorentz group, i. e.

D(Λ2) D(Λ1) = D(Λ2Λ1) , (87) and D(1) = 1. To find different Lorentz invariant theories, one should now classify all possible representations of the Lorentz group. In the following, we will construct a representation for 1 spin- 2 particles. This is the most fundamental representation since higher spin representa- 1 tions can be obtained from products of spin- 2 representations. Rather than analyzing full transformations, it is convenient to look at infinitesimal Lorentz transformations. One writes

µ µ µ Λν = δν + Ων (88)

15 µ where Ων is an infinitesimally small matrix.

µ Exercise 5.17. Show that the condition (85) implies that Ων is an antisymmetric matrix.

A general anti-symmetric matrix has six independent entries and can therefore be param- eterized as i Ωµ = ω (J αβ)µ , (89) ν −2 αβ ν where the six antisymmetric matrices J αβ correspond to the six different, independent Lorentz- transformations (3 rotations and 3 boosts) and the six epsilotic parameters ωαβ determine the angles and velocities of the boosts. The matrices J αβ are called the generators of the Lorentz transformations and have the form

(J αβ)µ = i(gµα δβ gµβδα) . (90) ν ν − ν They fulfill the following commutation relations

J αβ,J ρσ = i (gβρJ ασ gαρJ βσ gβσJ αρ + gασJ βρ) , (91) − − These commutation relations of the generators encode the Lorentz group in the same way that the commutation relations J i,J j = iijkJ k (92) describe the the rotation group. It is convenient  to analyze groups using the algebra of their generators and one can later reconstruct the finite transformations by exponentiation. A general Lorentz transformation can be written as i Λ = exp( ω J αβ) . (93) −2 αβ

Exercise 5.18. Consider ω12 = ω12 = θ and all other components of ωαβ zero. Show that µ − the resulting transformation Λν describes an infinitesimal rotation along the z-axis.

Exercise 5.19. Consider ω01 = ω10 = β and all other components zero. Show that the µ − resulting transformation Λν describes an infinitesimal boost along the x-axis.

In Part 3 of the exercises we considered the Dirac matrices and we have shown that the six matrices i Sαβ = γα, γβ (94) 4 fulfill the commutation relations (91) of the Lorentz group. To get an explicit form of these matrices, we need an explicit form of the Dirac matrices. We will use the chiral representation. Writing the 4 4 matrices in 2 2 blocks, the matrices have the form × × 0 1 0 σi γ0 = , γi = , (95) 1 0 σi 0 ! − !

16 which is also called the Weyl representation. Using the matrices (94), we can now construct the spinor representation of the Lorentz group. We consider a field ψ with four complex components, called a Dirac spinor, which transforms as 1 ψ(x) ψ0(x) = D(Λ)ψ(Λ− x) (96) → where i Λ = exp( ω J αβ) , (97) −2 αβ i D(Λ) = exp( ω Sαβ) . (98) −2 αβ

Exercise 5.20. Consider a general rotation and write the rotation parameters as ωij =  ϕ . Show that the transformation takes the form − ijk k exp(iϕ σi/2) 0 D(Λ) = i . (99) i 0 exp(iϕiσ /2)!

Note that the boost matrices S(Λ) are not unitary. Because of this, ψ†(x)ψ(x) does not trans- form as a scalar.

Exercise 5.21. Consider a rotation along the z-axis and show that after a rotation with ω = ϕ = 2π one obtains the remarkable result 12 − 3 ψ(x) ψ0(x) = ψ(x) . (100) → − Spinors pick up sign under a 2π rotation (while vectors rotate onto themselves)!

Exercise 5.22. Consider a boost and write the boost parameters as ω0i = βi. Show that in the chiral representation the boost matrix takes the form

exp(+β σi) 0 D(Λ) = i . (101) 0 exp( β σi) − i ! It is interesting to note that our representation matrices block-diagonal, see (99) and (101). This means that the spinor representation is reducible. One can split the spinor into two component spinors ψ (x) ψ(x) = L (102) ψR(x)! which independently transform. The left-handed and right-handed component spinors ψL,R(x) are called Weyl spinors can be extracted in a general representation of the γ matrices using the projection operators

ψL(x) = PL ψ(x) ψR(x) = PR ψ(x) , (103)

17 where 1 1 P = (1 + γ ) ψ(x) P (x) = (1 γ ) , (104) L 2 5 R 2 − 5 where, in the chiral representation, 1 0 γ5 = − (105) 0 1! The spinors are called left and right handed, because they have positive and negative helicity (projection of the spin on the momentum) in the massless case. The representations of the Weyl spinors are irreducible. Note that the boost matrices D(Λ) in (101) are not unitary. Because of this, ψ†(x)ψ(x) does not transform as a scalar. To find an invariant, note that the Dirac matrices in the chiral representation have the property 0 µ 0 µ γ γ γ = (γ )† (106) which can easily be verified.

Exercise 5.23. Show that this implies

αβ 0 αβ 0 (S )† = γ S γ . (107) −

¯ 0 Exercise 5.24. After defining the adjoint spinor ψ(x) ψ†(x)γ one finds with (107) that it transforms as ≡ 1 1 ψ¯(x) ψ¯0(x) = ψ¯(Λ− x) D(Λ)− . (108) → It is easiest to show this using an infinitesimal transformation. Note that this implies that the product ψ¯(x) ψ(x) transforms as a Lorentz scalar.

Exercise 5.25. Show that 1 µ µ ν D(Λ)− γ D(Λ) = Λν γ (109) It is easiest to verify this using an infinitesimal transformation and the relation

Sαβ, γµ = iγαgβµ iγβgαµ = (J αβ)µγν , (110) − ν which you have derived in (44). 

Exercise 5.26. Show that the relation (109) immediately implies that ψ¯(x)γµψ(x) trans- forms as a Lorentz vector and ψ¯(x)γµγνψ(x) as a tensor.

18 5.2 Dirac Lagrangian and Dirac equation From the considerations above it is clear that the action

S = d4x ψ¯(x)(iγµ∂ m) ψ(x) (111) µ − Z is a Lorentz scalar. Up to terms involving higher derivatives or γ5, this action is unique. The factor i in the derivative term makes the action real.

Exercise 5.27. Show that S is real.

To get the equations of motion of motion one can consider independent variations of the real and imaginary parts of the field ψ(x). A shortcut leading to the same result is to consider independent variations of ψ and ψ¯. The variation of ψ¯ leads to the Dirac equation

(iγµ∂ m) ψ(x) = 0 (112) µ −

µ Exercise 5.28. Multiply the Diract equation by (iγ ∂µ + m) from the left and show that each component of the spinor field fulfils the Klein-Gordon equation

m2 ψ(x) = 0 . (113) − −

5.3 Plane-wave solutions Similar to the Klein-Gordon equation, we now find plane-wave solutions to this equation. Let us first consider a plane wave of the form

ikx ϕ(x) = u(~k)e− (114)

0 with k = ωk so that it fulfills the Klein-Gordon equation. The plane wave is a solution of the Dirac equation if the spinor u(~k) fulfils the matrix equation

m k σ (k/ m) u(~k) = − · u(~k) = 0 (115) − k σ¯ m · − ! where we have inserted the chiral representation of the γ matrices and have introduced three common short-hand notations

k/ = γµk σµ = (1, σi) , σ¯µ = (1, σi) . (116) µ − The first one is called the Feynman slash.

19 As for the case of the Klein-Gordon equation, we also have negative frequency solutions ϕ(x) = v(~k)e+ikx . (117) Exercise 5.29. Show that for these (which describe anti-particles) the spinor v(~k) fufils the equation +m k σ (k/ + m) v(~k) = · v(~k) = 0 (118) k σ¯ +m · !

Consider the positive frequency solutions with vanishing three-momentum ~k = 0 and show that the spinor ξ u(0) = √m (119) ξ! fullfills the equation for an arbitrary two-component spinor ξ. According to (99) the two- compoment spinor transforms under rotations as ξ exp(iϕ σi/2) ξ (120) → i This is the familiar transfromation law of quantum mechanical two-component spinors and 1 the spinor indeed describes two spin states of a spin- 2 particle. The spinor ξ = (1, 0) fulfils σ3 ξ = +1 ξ and describes a particle with spin along the positive z direction, while ξ = (0, 1) has the spin in the negative z direction. − Our work on the Lorentz transformations of the spinors now pays off. The solutions for particles with momentum ~p can be obtained by performing a boost of the solution for the particle at rest. We have derived the explicit form of the boost in (101) and will now apply to the spinor (119).

Exercise 5.30. Consider a boost along the three direction ω03 = ω30 = β. What value of β does is needed to arrive at a boost of the form −

µ µ µ µ 3 k = (m, 0, 0, 0) k0 = Λ k = (E, 0, 0, k ) (121) → ν Note that the explicit form of the boost matrix was given in (93). (Since the other components do not play a role, it is convenient to write the vectors in the form kµ = (E, k3) to keep the notation compact.)

Exercise 5.31. Use the value of the boost parameter to boost the spinor u(0) for the case ξ = (1, 0) so that it has three momentum ~k = (0, 0, k3). The relevant boost matrix is (101).

Exercise 5.32. Same exercise for the case ξ = (0, 1).

It turns out that the general boosted solution can be written in the elegant form

√k σ ξ u(~k) = · , (122) √k σ¯ ξ · ! 20 where the square root of the matrix is obtained by first diagonalizing, taking the square root of the eigenvalues and transforming back.

Exercise 5.33. Derive the matrix relation

k σ k σ¯ = k2 = m2 (123) · · and use it to verify that (122) indeed solves (115).

Analogously, the general boosted solution for the anti-particle spinors can be written

√k σ η v(~k) = · . (124) √k σ¯ η − · !

It is convenient to introduce orthonormal basis spinors ξs (e.g. ξ1 = (1, 0) and ξ1 = (1, 0)) which fulfil

ξr†ξs = δrs (125) ~ and then denote the solution for the corresponding spinor by us(k) where s = 1, 2. The two compoment spinors then fulfil the completeness relation

1 0 ξ ξ† = . (126) s s 0 1 s=1,2 ! X Exercise 5.34. Using the explicit result (122), show that the spinors are normalized as

~ ~ us†(k) us(k) = 2ωk , (127) ~ ~ u¯s(k) us(k) = 2m . (128)

Exercise 5.35. Using the results (122), (126) and (123) show that

~ ~ us(k)u ¯s(k) = k/ + m . (129) s=1,2 X With a similar computation (which you are not asked to do), one finds

v (~k)v ¯ (~k) = k/ m . (130) s s − s=1,2 X

21 6 Non-relativistic limit of the Dirac equation

6.1 Dirac equation in the presence of an electromagnetic field The Schr¨odingerequation for an electron in a background electromagnetic field can be obtained from the equation of the free particle (i~ )2 i∂ χ = ∇ χ (131) t 2m by performing the substitutions i∂ iD = i∂ eφ , i~ iD~ = i~ + eA.~ (132) t → t t − ∇ → ∇ This gives the correct equation, except for the term describing the coupling of the electron spin to the magnetic field, which has the form ∆H = ~µ B~ (133) − · where the magnetic moment is e ~σ ~µ = gµ S~ = g . (134) − B − 2m 2

The quantity µB = e/(2m) is called the Bohr magneton. The extra term in the Hamiltonian ∆H, as well as the value of the so called gyromagnetic factor g, which is equal to 2 to very good approximation, will be derived from the Dirac equation in the next subsection. Historically, this factor of g = 2 came as a surprise, since the energy associated with orbital angular ~ ~ ~ momentum L of an electron in a magnetic field is ∆H = µBL B without an extra factor. In relativistic notation (remember that ∂µ = (∂ , ~ )) , the· substitution (132) reads t −∇ i∂ iD = i∂ eA , (135) µ → µ µ − µ and the quantity Dµ is called the covariant derivative. This suggests that the Dirac equation for an electron in an electric field is (iD/ m) ψ(x) = [γ (i∂µ eAµ) m] ψ(x) = 0 (136) − µ − − The deeper reason for the substitution (135) is that it leads to gauge invariance, i.e. invariance under the local phase symmetry ψ(x) eiα(x)ψ(x) , (137) → 1 A (x) A (x) ∂ α(x) . (138) µ → µ − e µ Where α(x) is an arbitrary, space-time dependent function. In the non-relativistic case, it is the field χ(x), which acquires the phase factor χ(x) eiα(x)χ(x). Gauge symmetry plays a key role in particle physics and lies at the heart of the→ Standard Model.

Exercise 6.36. Show that after the substitution (132) the Schr¨odinger(131) and the Dirac equation (136) are gauge invariant.

22 6.2 Magnetic moment of the electron We now consider the Dirac equation for a nonrelativistic electron, for which the components of the three momentum are much smaller than the mass ~p m. The energy | |  ~p2 p0 = m2 + ~p2 = m + + ..., 2m on the other hand, is large becausep it includes the rest mass. To expand the Dirac equation in the nonrelativistic limit, we first remove the large phase in the Dirac field, which arises due to the mass, and define χ(x) ψ(x) = e imt , (139) − η(x)   where χ(x) and η(x) are two-component fields. Having removed this factor, all momentum components of the fields χ and η are small for a solution which describes an electron near 2 2 its mass-shell p = m . In other words Dµχ(x) m χ(x) and Dµη(x) m η(x). For our discussion, it is furthermore convenient to use the so-called standard representation for the Dirac matrices 1 0 0 σ γ0 = , γi = i . (140) 0 1 σ 0 − ! − i ! This differs from chiral representation (95) we used earlier. Both representations are com- monly used. For massless particles, the chiral representation is especially convenient, while the standard representation is useful to study the non-relativistic limit. As stressed earlier, all representations are equivalent, so that the choice is a matter of convenience.

Exercise 6.37. Write out the coupled equations for the fields χ(x) and η(x). To keep the expressions compact, write everything in terms of covariant derivatives. Drop suppressed terms in the lower-component equation and show that it implies that the field η is suppressed with respect to χ. For this reason η is called the ”small” component of the Dirac field (this is true for particle solutions, for anti-particles the upper components are suppressed).

Exercise 6.38. Show that (~σ ~a)(~σ ~b) = ~a ~b + i~σ (~a ~b) . (141) · · · · ×

Exercise 6.39. Solve the approximated lower component equation for η and plug it into the upper-component equation for χ. This leads to the Pauli equation for an electron in an electromagnetic field ~ 2 (iD) e ~ iDtχ = ~σ B χ . (142) " 2m − 2m · # and one can now read off that g = 2.

23 The value of g changes because of quantum corrections associated with the electromagnetic α field. Schwinger computed the first order correction in 1948 and found g = 2 + π 2(1 + e2 ≈ 0.00116), where α = 4π 1/137. This was in good agreement with the deviation from 2, which had been discovered≈ experimentally shortly before. This agreement was a first triumph of the then new (and still poorly understood) theory of quantum electrodynamics. g 2 In modern experiments the anomalous magnetic moment a = −2 of electrons and muons are measured with exquisite accuracy. The muon and electron magnetic moments differ be- cause the quantum corrections depend on the mass of the lepton. (The muon has the same electric charge as the electron but is about 200 times heavier.) The current experimental results are

exp 12 a = (1159652180.73 0.28) 10− (143) e ± × exp 12 a = (1165920890 630) 10− (144) µ ± × The first, extremely precise number is used to obtain the value of the electric charge α. The scond number is compared to theoretical predictions and provides a test of the Standard Model of particle physics. The Standard Model theoretical prediction obtained by computing all quantum corrections up to order α4 and even the most important of order α5 (an enormous task!) is SM 12 a = (1165917900 662) 10− , (145) µ ± × and deviates from the experimental value.

Exercise 6.40. By how many standard deviations does the predicted theoretical value deviate from the measurement? What is the probability that this deviation is a statistical fluctuation, assuming that the distributions of both the experimental and theoretical values are Gaussian?

This deviation has been around since the experimental measurement in the year 2001. Despite a lot of scrutiny of the theory result (and some small mistakes which were identified), the discrepancy has persisted. This year, a new more precise measurement has started, see http://muon-g-2.fnal.gov. It will be interesting to see what happens!

24 7 Interacting fields in perturbation theory

7.1 Relativistically invariant interaction Lagrangians Three examples of Lagrangian densities describing interacting fields are (φ is a real scalar, Φ a complex scalar field)

2 1 µ m 2 λ 4 4 = ∂ φ(x)∂ φ(x) φ(x) φ (x) , (146) Lφ 2 µ − 2 − 4! 1 = F µν F + ψ¯(x)(iD/ m) ψ(x) , (147) LQED −4 µν −

1 µν µ 2 λ 2 = F F + (D Φ)∗(D Φ) + µ Φ∗Φ (Φ∗Φ) , (148) LHiggs −4 µν µ − 2 where Dµ = ∂µ + ieAµ. The first Lagrangian describes a scalar field with self interaction, the second one is Quantum Electro-Dynamics (QED) and describes a fermion interacting with the photon field. The third Lagrangian describes a electrically charged, complex scalar field which has both electromagnetic and self interactions. It was used by Higgs in Phys. Rev. Lett. 13 (1964) 508 to illustrate how one can use a scalar field to generate a massive photon. Higgs’ paper was initially rejected, because “we know” that the photon is massless, but the same mechanism now generates the masses of the Z and W bosons in the Standard Model. All of these are local theories since the Lagrangian is obtained by an integral of a local function: L(t) = d3x (x) L Z For a given field content of the theory one can easily write down many additional local, Lorentz invariant terms in the Lagrangian. At first sight, it looks like there is infinite freedom in constructing local field theories, even for given particle content. However, this freedom is strongly reduced by imposing renormalizability which dictates that only terms in which the mass dimension D 4 are allowed in the Lagrangian. This condition arises once one considers quantum corrections,≤ as will be explained in the QFT lectures. While renormal- izability restricts the the operators in fundamental theories, higher-dimensionional operators are allowed in low-energy effective theories. An effective theory is a quantum field theory which breaks down above a certain energy scale Λ (e.g. because there are additional particles with masses M Λ not included in the effective theory). In effective theories one includes higher-dimensional∼ operators, but their contributions are suppressed by powers of Λ. In the following we will therefore also consider higher-dimensional operators.

Exercise 7.1. In natural units ~ = c = 1, the action S is a dimensionless quantity. Use this fact to derive the mass dimensions of the fields in the above Lagrangians.

Exercise 7.2. Consider a theory with only a single real scalar field φ(x) with a symmetry φ(x) φ(x). What is the most general Lagrangian density if terms up to dimension D = 6 are included?→ − Which ones remain for D 4? ≤

25 Exercise 7.3. Consider QED, a gauge invariant theory of a single fermion field ψ coupled to the gauge field Aµ. Again, write down all terms up to mass dimension D = 6. Which ones remain for D 4? All terms can be constructed from covariant derivatives Dµ and fermion fields ψ(x). The≤ QED Lagrangian is invariant under parity and one therefore does not need ¯ to write down pseudo-scalar terms such as ψγ5ψ, which change sign under parity.

Exercise 7.4. Show that the Higgs Lagrangian is invariant under a gauge transformation 1 Φ(x) eiα(x) Φ(x) ,A (x) A (x) ∂ α(x) . (149) → µ → µ − e µ

Exercise 7.5. Note that the mass term in Higgs has the ”wrong” sign. The Higgs potential 2 λ 2 L V = µ Φ∗Φ + (Φ∗Φ) has a minimum for a nonzero value of the field: − 2 v µ2 1/2 Φ(x) = φ0 = = . (150) | | √2 λ   The quantity v is called the vacuum expectation value. Parameterize the complex Higgs field by two real fields β(x) and h(x) as 1 Φ(x) = eiβ(x) (v + h(x)) , (151) √2 i.e. by a phase-field β(x) and the deviation h(x) from the vacuum value. Since the theory is invariant under the gauge transformation (149), we can perform a transformation which eliminates the field β(x). This is called ”unitary gauge”. Write down the Lagrangian LHiggs in unitary gauge. Show that the photon field Aµ picks up a mass term of the form 1 ∆ = m2 A Aµ . (152) L 2 A µ

Derive the mass mA of the photon.

Early this year, LHC data for the production of two photons showed tantalizing hints for the existence of an additional scalar particle S with a mass of MS = 750 GeV. With more data, it became clear that the hints were simply a statistical fluctuation, but it is an interesting exercise to try to write down a Lagrangian for a new scalar particle S. In the following we assume that the world is described by QED (with a single massive fermion, the electron e) and we will add terms to the Lagrangian which describe the interaction of the photon and the fermion with the new scalar S. Take into account the following statements: Since the new particle arises in the decay S γγ, it is electrically neutral and transforms • trivially under a gauge transformation. →

+ Instead of a hadron collider such as the LHC, we consider e e− collisions which then • produce the scalar S.

26 For the moment we assume that S(x) is a scalar field (as opposed to a pseudoscalar). • We also assume that our theory is parity invariant.

The theory, and therefore also the new interactions with the scalar S, must be gauge • invariant.

Exercise 7.6. Write down the lowest dimensional interaction terms of the fermion ψ with the new scalar S. Such terms should be present, otherwise S would not be directly produced + in the e e− collisions.

Exercise 7.7. Write down the lowest dimensional interaction terms of the scalar S with the photon field Aµ.

Exercise 7.8. Assume that S is a pseudoscalar, but that the theory is parity invariant. Write the lowest dimensional interactions in this case.

7.2 Perturbation theory All the information on a quantum field theory can be extracted from the correlation functions of the fields. Of particular interest are time-ordered correlation functions

Ω T φ(x ) . . . φ(x ) Ω . (153) h | { 1 n } | i The time-ordering operator T orders the fields such that the fields with later time arguments are to the left of earlier ones. From the two-point function, one can obtain information about the spectrum of the theory and the higher-point functions yield the amplitudes for scattering particles. Interacting theories are in general too complicated to allow for an exact computation of these correlation functions, but we can simplify the problem by treating the interaction as a perturbation. For φ4-theory, for example, we write the Lagrangian as

λ 4 = 0 + int = Klein Gordon φ (x) (154) L L L L − − 4! and expand in the coupling constant λ. The prescription to compute correlation functions in perturbation theory is quite simple. One computes

1 Ω T φ(x ) . . . φ(x ) Ω = 0 T φ(x ) . . . φ(x ) exp i d4z (z) 0 , (155) h | { 1 n } | i Z h | 1 n LI | i   Z  where the correlation function on the right-hand side is computed in the free theory. The normalization factor Z = 0 T exp i d4z (z) 0 (156) h | LI | i   Z 

27 arises because the vaccuum of the free theory 0 is not the same as the vacuum of the interacting theory Ω . The formula (155) will be| derivedi in the QFT lectures (the derivation is trivial in the path-integral| i formalism, see below). To do perturbation theory, the factor

4 4 1 4 4 exp i d z (z) = 1 + i d z (z) d z0 d z (z) (z0) + ... (157) LI LI − 2 LI LI  Z  Z Z Z is expanded and the higher-order terms are suppressed by higher powers of the coupling constant. The interaction terms in the theory are polynomials in the fields and so the entire computation reduces to the evaluation of correlation functions in the free theory

0 T φ(x ) . . . φ(x ) 0 , (158) h | { 1 n } | i where some of the fields arise from expanding the interactions in the coupling constant using (157).

7.3 Wick’s theorem An important result for the evaluation of correlation functions in the free theory is the Wick theorem. It states that

0 T φ(x1) . . . φ(xn) 0 = ∆F (xi1 xi2 )∆F (xi2 xi3 ) ... ∆F (xin 1 xin ) , (159) h | { } | i − − − − pairings X where the Feynman propagator

∆ (x y) = 0 T φ(x)φ(y) 0 . (160) F − h | { } | i For a free theory, all higher-point correlators reduce to products of two-point correlation functions. The Feynman propagator has the Fourier representation

4 d k i ik(x y) ∆ (x y) = e− − (161) F − (2π)4 k2 m2 + iε Z − There are different ways to derive Wick’s theorem. One way is to use the path integral formalism, where a free field theory is represented by a set of high-dimensional Gaussian integrals for which we derived Wick’s theorem earlier in Section 1. Let us briefly sketch how this is done. After adding a source term b(x)φ(x) to the Lagrangian, the Euclidean path integral for a scalar field has the form

SE [φ]+R d4x b(x)φ(x) (b) = φ(x) e− 0 . (162) Z D Z Time-ordered products of fields are obtained as 1 S0[φ(x)] 0 T φ(x ) . . . φ(x ) 0 = φ(x) φ(x ) . . . φ(x ) e− . (163) h | { 1 n } | i (0) D 1 n Z Z

28 One can get all these correlation functions by expanding in the source terms in the generating functional (b) as in (11). Z To make sense of (162) one discretizes the theory on a lattice with points xn. The field variables φ φ(x ) and corresponding source values b b(x ) at the discrete points play n ≡ n n ≡ n the same role as the coordinates xn and the coefficients bn in Section 1. If the lattice is chosen to have finite volume and lattice spacing, it contains a finite number of points and associated field variables φn. The discretized Euclidean action with the source term then takes the form ˆ ˆ ˆ ˆ SE = φnMnmφm + bnφn , (164) n,m n X X ˆ ˆ where we have rescaled the fields φn and bn to make them dimensionless. To obtain the matrix E Mnm explicitly one has to discretize the derivatives in the action S0 . Computing (b) now ˆ Z reduces to evaluating (11) (with xn φn) and we can immediately use the Wick theorem (17) derived for these integrals. After→ taking the continuum and infinite volume limit one then obtains the theorem (159). To derive the theorem (159) in canonical quantization, one expresses the field operators in terms of creation and annihilation operators

+ φ(x) = φ (x) + φ−(x) , (165) where

3 3 + d p 1 ipx d p 1 ipx φ (x) = a e− and φ−(x) = a†e . (166) (2π)3 2ω p (2π)3 2ω p Z p Z p p Then one commutes all annihilation operators ap to the right and all creation operators ap† to the left such that they annihilate the vacuum. This implies that the matrix elements (159) vanish up to commutator terms which arise during the rearrangement. The necessary commutator [φ+(x), φ (y)] for x0 > y0 ∆ (x y) = − (167) F + 0 0 − ([φ (y), φ−(x)] for y > x is equal to the Feynman propagator (161). To demonstrate that the commutator terms have indeed the form (159), one can perform an inductive proof, as is done e.g. in Peskin’s QFT book.

Exercise 7.9. Insert the decomposition (165) into the two point function

0 T φ(x)φ(y) 0 (168) h | { } | i and express it in terms of the commutator (167).

Exercise 7.10. Derive the Fourier representation (161) for the commutator (167) using the commutation relations

3 (3) [ap, a† ] = (2π) (2ωp)δ (~p ~p0)[ap, ap ] = 0 . (169) p0 − 0 29 for the creation and annihilation operators. To see that (161) is equivalent to the expression obtained using the commutation relations it is simplest to first integrate over k0 in (161) using the residue theorem.

7.4 Feynman rules 7.4.1 φ4-theory Exercise 7.11. Compute the four-point correlation function

Ω T φ(x )φ(x )φ(x )φ(x ) Ω (170) h | { 1 2 3 4 } | i at first order in λ in φ4-theory. The number of contractions is quite large, but many of them are equivalent because the fields in the interaction Lagrangian live at the same point. It is good enough to write down the expression, you do not need to perform the integration over the position of the interaction vertex. Represent the contractions graphically by drawing the propagator ∆F (x y) as a line from x to y. Draw the diagrams for both the numerator in (155) and also the− normalization factor Z in (156). Which contributions cancel against the normalization factor?

Exercise 7.12. Compute the Fourier transform of the connected part of the correlator (170). The connected piece is the most relevant, since the scattering amplitude is contained in it.

Instead of using Wick’s theorem, the results for the correlation functions can easily be obtained using Feynman rules. For the case of φ4-theory, the momentum-space rules for the connected part of the n-point correlation function at m-th order in perturbation theory are as follows: 1. Draw all connected diagrams with n external legs and m interaction vertices. Then convert each diagram into the mathematical expression for the corresponding contraction of fields using the rules which follow. 2. Each line represents a propagator i = (171) p2 m2 + i p −

3. At each vertex impose momentum conservation and insert a factor

= iλ (172) −

d4p 4. Integrate over each momentum p flowing inside a closed loop: (2π)4 .

4 5. Include an overall momentum conservation delta-function, (2π)Rδ(Ptot), where Ptot is the incoming minus the outgoing momentum.

30 6. For loop diagrams: divide by the appropriate symmetry factor.

λ 4 Let us comment on the symmetry factor in the last point. The interaction Lagrangian 4! φ (x) involves a factor 4! which is absent in rule 3 because there are usually 4! possibilities to− contract other fields to the vertex. Indeed, at tree level, the factor of 4! is always cancelled, but in loop diagrams, there are sometimes fewer possible contractions. At the loop level one has to count the number of equivalent contractions and multiply by a correction (or ”symmetry”) factor if necessary.

7.4.2 QED The Feynman rules for QED are quite similar to the scalar case. The Wick theorem ap- plies also for vector and fermion fields (except that one has to be careful about signs when anti-commuting fermion fields), but we now need expressions for the fermion and photon propagators. The photon propgator reads

4 µν µ ν d k i ik(x y) D (x y) = 0 T A (x) A (y) 0 = ( g ) e− − . (173) F − h | | i (2π)4 k2 + i − µν Z   The numerator gµν arises from the sum over polarizations and includes unphysical degrees of − µν freedom since ∂µDF = 0, but one can show that the unphysical polarizations do not contribute to physical quantities6 in QED. The photon propagator is not unique. The form (174) only holds in Feynman gauge. The fermion propagator reads

4 ik γµ µν d k µ αβ ik(x y) S (x y) = 0 T ψ (x) ψ¯ (y) 0 = e− − . (174) F − h | α β | i (2π)4 k2 + i Z   µ Typically, the abbreviation k/ = kµγ is used for the numerator and often one does not explic- itly write out the Dirac indices α and β. Armed with these expressions and the Wick theorem, we can now also evaluate correlation functions in QED.

Exercise 7.13. Compute the three-point function

Ω T ψ (x )A (x )ψ¯ (x ) Ω . (175) h | α 1 µ 2 β 3 | i Represent the result graphically, using a line with an arrow to represent the fermion propaga- tor and a wiggly line for the photon. Read off the Feynman rule for the QED vertex.

The QED Feynman rules in momentum space have the same structure as the ones in the scalar case, but the propagators and vertices now carry Dirac indices α, β and Lorentz indices µ, ν.

1. Draw all connected diagrams with n external legs and m interaction vertices. Then convert each diagram into the mathematical expression for the corresponding contraction of fields using the rules which follow.

31 Figure 1: Interaction vertices in . The double line denotes the scalar particle S. LQED+S

2. Each line represents a propagator

i(/p + m) = βα (176) p2 m2 + i p −

ig = − µν (177) p p2 + i 3. The vertex is

= ie(γµ) (178) − βα

d4p 4. Integrate over each momentum p flowing inside a closed loop: (2π)4 .

R4 (4) 5. Include an overall momentum conservation delta-function, (2π) δ (Ptot), where Ptot is the incoming minus the outgoing momentum. 6. For loop diagrams: divide by the appropriate symmetry factor and include a factor of ( 1) for each closed fermion loop. − 7.4.3 QED with an extra neutral scalar S The lowest dimensional interaction terms of a neutral scalar S with a fermion and with photons are = g S(x)ψ¯(x) ψ(x) g S(x)F (x)F µν(x) (179) Lint ψ − γγ µν Exercise 7.14. Read off the Feynman rules associated with the two vertices shown in Figure 1. To get the Feynman rules in momentum space, it is easiest to write the action S = d4x in momentum space by Fourier transforming the fields using int Lint 4 R d k ikx S(x) = S˜(k) e− . (180) (2π)4 Z and similarly for the other fields. Performing the integral over x yields the momentum conser- vation δ-function at the vertex and the remaining expression is the Feynman rules. To make

32 the Feynman rule for the Sγγ vertex user friendly, one should symmetrize it in the two photon fields.

8 Computation of scattering cross sections

8.1 Reduction formula and S-matrix In Section 7.2 we have discussed carefully how to obtain vacuum correlation functions of fields in perturbation theory. The scattering amplitudes can be extracted from these using the Lehmann Symanzik Zimmerman (LSZ) reduction, but a thorough derivation of this relation is beyond the scope of these exercises. The corresponding formalism is discussed in every field theory book. Let us briefly sketch how the reduction formula is obtained by considering the four-point correlation function. To extract the contribution of an outgoing particle generated with field φ(x1) one considers the Fourier transform with respect to x1,

F (p , x , x , x ) = d4x eip1x1 Ω T φ(x )φ(x )φ(x )φ(x ) Ω . (181) 1 2 3 4 1 h | { 1 2 3 4 } | i Z One can show (and this is the subtle part) that if there is a single particle state with mass m, then for p0 > 0 the function F develops a pole at p2 = m2, which describes the propagation of an outgoing particle at very large times i F (p , x , x , x ) = Ω φ(0) ~p ; out ~p ; out T φ(x )φ(x )φ(x ) Ω + ... (182) 1 2 3 4 p2 m2 h | | 1 i h 1 | { 2 3 4 } | i 1 − The dots represent contributions from other states, which do not give rise to a pole. While we don’t want to go into the technical details, it is easy to see that the Feynman diagrams in momentum space do involve poles coming from the external propagators. The expectation value 1 Ω φ(0) ~p = Z 2 = 1 + O(λ) (183) h | | 1i is called the on-shell wave-function renormalization constant and can be computed from the two-point function for which i iZ F (p , x ) = Ω φ(0) ~p ~p φ(x ) Ω + = eip1 x2 . (184) 1 2 p2 m2 h | | 1i h 1| 2 | i ··· p2 m2 1 − 1 − We can repeat the procedure for the other three fields to get

4 ip1x1 4 ip2x2 4 iq1x3 4 iq2x4 d x e d x e d x e− d x e− Ω T φ(x )φ(x )φ(x )φ(x ) Ω 1 2 3 4 h | { 1 2 3 4 } | i Z Z Z Z i√Z i√Z i√Z i√Z = ~p , ~p ; out ~q , ~q ; in . (185) h 1 2 | 1 2 i p2 m2 p2 m2 q2 m2 q2 m2 1 − 2 − 1 − 2 − 33 To get the 2 2 scattering amplitude one thus computes the Fourier transform of the four- point function→ and divides by the external propagators. A Green’s function for which the external propagators have been divided away is called an amputated Green’s function. The quantity S = ~p , ~p ; out ~q , ~q ; in (186) fi h 1 2 | 1 2 i is called the S-matrix. The nontrivial part of the S-matrix, in which the incoming particles interact, arises from the connected diagrams and is called the scattering matrix . It is defined as M (2π)4δ(4)(q + q p p ) i (q , q p , p ) = ~p , ~p ; out ~q , ~q ; in . (187) 1 2 − 1 − 2 M 1 2 → 1 2 h 1 2 | 1 2 i connected In practical terms, the computation of is quite simple. One uses the momentum-space

Feynman rules for the connected Green’sM functions, removes the external propagators and multiplies by the appropriate Z-factor (at lowest order Z = 1). For φ4-theory one obtains for the 2 2 amplitude at lowest order → = λ . (188) M −

For external fermions and photons, the reduction formula has the same structure, but one needs the matrix elements 1 2 s Ω ψ (0) e−(~p,s) = Z u (~p) , h | α | i ψ α 1 Ω ψ¯ (0) e+(~p,s) = Z 2 v¯s (~p) , (189) h | α | i ψ α 1 Ω A (0) γ(~p,r) = Z 2 r (~p) . h | µ | i A µ and their complex conjugate.

Exercise 8.1. Verify the matrix elements (189) using the expressions for the free field operators. The expressions can be found in Jean-Pierre Derendinger’s Quantum Field Theory, I and II Notations and conventions and they are obtained by replacing the Fourier expansion coefficients by creation and annihilation operators. The relevant expression for a vector field is (71). The Z-factors for free fields are equal to 1.

One can formulate the above prescription as Feynman rules for the scattering amplitude . Suppressing the Z-factors, which for our leading-order computations are equal to 1, Mone first computes the amputated amplitude using the Feynman rules given in the previous Section. Then, for the external lines one multiplies with 1. A trivial factor 1 for in- or outgoing scalar fields, 2. us(~p) for an incoming fermion with spin s andu ¯s(~p) for an outgoing fermion, 3.¯vs(~p) for an incoming anti-fermion and vs(~p) for an outgoing anti-fermion,

r r 4.  (~p) for an incoming photon of polarization r and ∗ (~p) for an outgoing photon.

34 8.2 Cross section and decay rate With the scattering matrix at hand, one can compute the probability P that two particles A and B scatter into a givenM final state. This probability is not just a property of the particles and their interactions. It is proportional the probability densities of the incoming particles and on their relative velocity. The cross section σ is defined by dividing out these factors: dP = ~v ~v ρ (x)ρ (x) σ , (190) dt d3x | A − B| A B z where it is assumed that the incoming particles both fly along the z-axis and vi = pi /Ei. The 2 n cross section is given by → 1 n d3p n dσ = i (p , p p ) 2(2π)4δ(4) p + p p , 2E 2E v v (2π)32E |M A B → { f } | A B − i A B| A − B| i=1 i i=1 ! Y X (191) This relation will be derived in class. To obtain it, one computes the probability P for scattering normalized Gaussian wave packets of the form d3k 1 φ(~p,L) φ(~k, ~p,L) ψ(~k) (192) | i ≡ (2π)3 √2E | i Z k with (~p ~k)2L2 φ(~k, ~p,L) = e− − (193) N and ~p 1/L. The normalization factor is chosen such that φ(~p,L) φ(~p,L) = 1. The particle| | density ρ(x) = φ(x) 2 is obtained from the Fourier transformh i| i | | 3 d k ikx φ(x) = φ(~k, ~p,L)e− . (194) (2π)3 Z For the derivation one uses the fact that the wave packets are sharply peaked at ~k ~p, so that one can replace the momenta ~k in the amplitudes by ~p, the typical momentum≈ of the wave packet, up to corrections of order ~p 1/L. We will also compute the decay rate| | for  the heavy scalar. In the limit where the decay width Γ is much smaller than the particle mass M, its decay into a particular final state can be computed as

1 n d3p n dΓ = i (p p ) 2(2π)4δ(4) p p . (195) 2M (2π)32E |M → { f } | − i i=1 i i=1 ! Y X The total decay rate Γtot of an unstable particle is the sum of the rates into all possible decay channels and the lifetime is τ = 1/Γtot. Both the scattering amplitude and the particle decay require that one evaluates phase-space integrals, i.e. integrals of the form

n n d3p n dΦ (q p ) = i δ(4)(q p ) . (196) n → i (2π)32E − i i=1 i=1 i i=1 X Y X 35 To compute those integrals one eliminates some integrals using the momentum conservation δ-functions. The remaining one, need to be parameterized in terms of suitable variables, for example energies and angles.

Exercise 8.2. Show that the two-body phase space takes the form

~p1 dΦ2(q p1 + p2) = dcos θ dφ | | . (197) → 4√s0 The phase-space is Lorentz invariant, but the angles and the three momentum refer to the rest frame of the decaying particle. The angles θ and φ are relative to some fixed axis in the rest frame of q. For a scattering process q = q1 + q2 and the rest frame of q is the center-of-mass frame and θ is chosen as the angle between ~q1 and ~p1. Show that the three momentum is given by

1 2 2 ~q1 = ~q2 = λ(s , p , p ) , (198) | | | | 2√s 0 1 2 0 q 2 + 2 2 2 where λ(a, b, c) = (a b c) 4bc. For the decay S e e−, we have q1 = q2 = me, but derive the formula for− the− general− case of unequal masses.→

Exercise 8.3. Using this result, show that for 2 2 scattering the cross section in the center-of-mass frame (191) simplifies to → dσ 1 ~p = | 1| (p , p p , p ) 2 , (199) dΩ 2E 2E v v (2π)24E |M A B → 1 2 | c.m. A B| A − B| cm where Ωc.m. is the solid angle of particle 1. Since the cross section does not depend on the azimuthal angle dΩc.m. = 2π sin θc.m.dθc.m., where θc.m. is the scattering angle in the center of mass frame.

Exercise 8.4. Compute the total cross section for 2 2 scattering in φ4 theory in the center-of-mass frame at a given center-of-mass energy. →

+ + As a final warm-up exercise, let’s compute muon production, e e− µ µ−, the simplest leading-order QED process. We evaluate the cross section in the center-of-mass→ frame in the high-energy limit, where one can neglect the electron and muon masses. The Feynman rules for the muon are exactly the same as for the electron. We work in the center-of-mass frame and choose to parameterize the momenta as p = E(1, 0, 0, 1) , p = E(1, 0, 0, 1) , (200) 1 2 − q = E(1, sin θ, 0, cos θ) , q = E(1, sin θ, 0, cos θ) . (201) 1 2 − − + + Exercise 8.5. Compute the scattering amplitude (e−(p1)e (p2) µ−(q1)µ (q2)). Only a single diagram contributes to this process. M →

36 Interpreting the 750 GeV digamma excess: a review

ALESSANDRO STRUMIA CERN, INFN and Dipartimento di Fisica, Universit`adi Pisa

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I(JP )=?(0?) z (750000) ����� �� ��� = J needs confirmation � �� ��� OMITTED FROM SUMMARY TABLE Needs confirmation. z MASS VALUE(GeV) EVTS DOCUMENTID TECN COMMENT 750 30 OUR AVERAGE ATLAS, CMS pp z ± We do not use the following data for average, fits, limits, etc. �� �� •• • ••• �� ��� �� ��� / / z WIDTH VALUE(GeV) CL% DOCUMENTID TECN COMMENT <100 95 ATLAS, CMS pp z We do not use the following data for average, fits, limits, etc.

������ ������ •• • ••• z DECAY MODES Mode Fraction (i/) 1 seen 2 Z, ZZ, jj expected � �

We summarise the main experimental, phenomenological and theoretical issues related to the 750 GeV digamma excess. ��� ��� ��� ���� ���� ��� ��� ��� ���� ����

�γγ �� ��� �γγ �� ��� Contents

1 Data 2 Figure 1 – Left: spectra measured by ATLASFigure (blue) and 2: CMSRight: (red) An at p entrys =8TeV for(lighter thesupposed colors) and 13 new TeV diphoton particle F (we call it S in these (darker). Right: (warning: adult content) spectrumexercises) obtained for summing the Particle ATLAS and Data CMS Group counts2 Widths at (PDG)ps =13TeV prepared. by an overexcited theoretical physicist 3 [5]. Left: A naive combination of3E the↵ 2015ective data Lagrangian from ALTAS [1, 3] and CMS data [2, 4] from 4 2 Widths [5]. 4 The everybody’s model 7 The cross section for single production of a boson with spin J can be written in the narrow- z arXiv:1605.09401v1 [hep-ph] 30 May 2016 5 Composite diphoton 9 width approximation in terms of its decayExercise widths 8.6. into partonsCompute}, } the= unpolarized(z }), as amplitude squared, which is obtained after summing ! over the final state spins r1, r2 and averaging over the initial spins s1, s2 2J +1 } 6 What next? 11 (pp z)= C} . (2) ! s M } z 1 7 Connection2 with1 Dark Matter, axions, vacuum stability, baryogenesis... 13 X = † (202) 4 |M| 4 M M A resonance with spin J = 1 is excluded because it cannot decay into s(Lee-Yang,s ,r ,r theorem).s ,s ,r ,r 1 8X2 Who1 2 ordered that?1 X2 1 2 14 The luminosity factors C} of the main partons are [10] The conjugate Dirac structure follows9 Conclusions from the relations (µ = 1 ... 4) 14 ps Cb¯b Ccc¯ Css¯ Cdd¯ Cuu¯ Cgg 0 µ 0 µ 5 5 8 TeV 1.07 2.77.2 89 158 174γ γ .†γ = γ , γ † = γ . (203) 13 TeV 15.3 36 83 627 1054 2137 The sum over spins can be simplified using the completeness relations QCD corrections enhance the cross section by K 1.5 and K 1.2[10,245]. Some authors gg ' qq¯ ' also consider SM vectors as partons, e.g. C 10 (60) at ps = 8 (13) TeV [141,29,10,178,281]. ⇠ us(p)¯us(p) = /p + m vs(p)¯vs(p) = /p m . (204) The diagonal lines in fig. 2 shows the ratio of cross-section 13/8 predicted by the various − s s partons: data disfavor the partons that give the smallestX enhancement (light quarks and SM X vectors), favouring z production from(We heavy are quarks neglecting or gluons the —mass and in an this even exercise.) larger 13/8 enhancement would give a better fit.d dσ dThis can be achieved if pp collisions produceExercise some heavier 8.7. particleCompute (e.g. a heavy the vector muon with production mass 1.5TeV) cross sections and the total cross section σ. ⇠ dΩ that decays into z and something invisible [7, 10, 46, 69, 62, 297, 315], with a phase space almost closed in order to reproduce the lack of extra particles and transverse momentum in the excess events. Such kinematics can be used to fake a large z width [107]. A large z width can be faked in other ways: by having two or more nearby narrow resonances (for example the scalar and pseudo-scalar components of a SU(2)L doublet splitted by 2 9 Diphoton production and a scalar resonance v /M 40 GeV, where v is the Higgs vev) [10, 174], or by assuming that z decays into pairs of light particles z ⇠ with mass m < GeV, that decay into two or more photons collimated within an angle ✓ m/M ,suchthatthey ⇠ z appear in the⇠ detector as a single [42,56,72,This189,222 year,,297]. there This latter was possibility big excitement allows to get about a large a tree-level possible new boson (with even spin, so most likely ( like) and can be tested by better studying the events (multiple traveling in the material before z spin zero) being observed in the scattering cross section pp γ(q1)γ(q2) + X at the LHC, ! + the electromagnetic calorimeter give more e e conversions than a single photon [204]); furthermore it can → !i.e. the scattering of protons into two energetic photons plus any other particles, collectively lead to a displaced vertex, and to no decays into other electroweak vectors.

37 6 6 Selected data sample and interpretation of the results

mentum. This allows to test the transverse energy range up to roughly 150(100) GeV in the barrel (endcap) region. For the barrel region, the energy scale corrections are found to be stable within 0.4% in the probed range. A stability better than 0.8% is observed for photon candidates in the endcaps.

6 Selected8 Results data sample and interpretation of the results A total of 6284 (2791) photon pairs are selected in the EBEB (EBEE) category. Out of these, 461 (800)8.1 pairs Compatibility have an invariant with the mass background-only above 500 GeV. The hypothesis invariant mass distribution of the selected events is shown in Fig. 2. A parametrisation of the spectrum of the form f (mgg)= a+b logFigure(mgg) 3 shows the diphoton invariant mass distribution together with the background-only fit, for events mgg · , obtained through an unbinned maximum likelihood fit to the selected events, is shown.selected This parametric in the 2015 form and incorresponds 2016 datasets. to the one chosen to model the background in the hypothesis tests, as detailed in Section 8.

-1 -1 Preliminary4 Preliminary4 CMS10 ATLAS Preliminary 12.9 fb (13 TeV) CMS 10 ATLAS Preliminary 12.9 fb (13 TeV) Data Data 3 Data Data 10 3 3 10 EBEB Fit modelBackground-only fit 10EBEE Fit model Background-only fit ± Spin-01 s.d. Selection 102 ± 1 s.d.Spin-0 Selection

Events / 20 GeV 2 Events / 20 GeV 2 2 10 ± 2 s.d. 10 ± 2 s.d. 10 s = 13 TeV, 2015, 3.2 fb-1 s = 13 TeV, 2016, 12.2 fb-1

10 10 10 Events / 20 GeV 10 Events / 20 GeV 1 1 1 1 10−1 10−1

500 1000 1500 2000 2500 500 1000 1500 2000 2500 15 15

stat 2 10 stat 2 10 σ 5 σ 5 0 0 0 0 −5 −5 -2 -2 (data-fit)/ −10 (data-fit)/ −10 400Data - fitted background 600 800 100012001400160018002000 400 Data - fitted background 600 800 100012001400160018002000 500 1000 1500 2000 2500 500 1000 1500 2000 2500 m [GeV] m [GeV] mγγ (GeV) γγ mγγ (GeV) γγ Figure 2: Observed invariant mass(a) spectra for the EBEB (left) and EBEE (right). No(b) event with mgg > 2000 GeV is selected in the analysis. The results of a likelihood fit to the background- only hypothesis are also shown. The shaded regions show the 1 and 2 standard deviation FigureFigure 3: Invariant-mass 3: The 2016 distribution data ofdoes the selected not diphoton show any candidates, excess with for them background-onlyγγ = 750 GeV. fit overlaid, for uncertainty(a) 2015 bands data associated and (b) 2016 with data. the The fit, di and↵erence reflect between the statistical the data and uncertainty this fit is shownof the indata. the Thebottom panel. The lower panelsarrow shown show in the the difference lower panel between indicates the a values data and outside fit, dividedthe range by with the more statistical than one uncertainty standard deviation. There denotedin the datais by no points. dataX. event Both with largem > 2500 LHC GeV. experiments have observed an excess of photon pairs in a 2 2 regionThe results where ofm theγγ search= (q are1 + interpretedq2) 750 in the GeV. framework Figure of 2 a shows composite the statistical number hypoth- of diphoton events The 2015 data have been reanalyzed≈ with improved photon reconstruction algorithms. The significance measuredesis test. by A simultaneous ATLAS and fit CMS to the and invariant the excess mass spectra is clearly of the visible. EBEB and The EBEE results event shown cat- in the plots of the largest excess above the background-only hypothesis decreased from 3.9 standard deviations of wereegories based is used on about to study 3 thefb compatibility1 of luminosity of the per data experiment, with the background-only collected in and 2015. the Thesig- bump visible Ref. [1] to 3.4 standard− deviations. The corresponding signal mass and width also changed, from a mass innal+background Figure 2 could hypotheses. be explained as the process pp S and the subsequent decay S γγ. of 750 GeV and a relative width of 6% to a mass of 730 GeV→ and a relative width of 8%. These di↵erences→ AboutThe test 500are statistics mainly (!) papers due used to in two on the events the hypothesis subject being a tests↵ected were are by basedwritten the new on reconstruction the in profilethe first likelihood algorithms six months ratio: used of for this the reprocessing year, mostly containingof the explanations 2015 data. In one of theevent excess selected in both terms by this of analysis various and new the onephysics of Ref. models. [1], one of Fortunately, the two photon this ˆ candidates is at ⌘ = 1.53, where the improvedL(µ calibrationS + B qµ) of photon candidates near the transition region year the experiments| | have collectedq(µ)= much2log more· data| and in the summer they released results of the electromagnetic calorimeter1 leads toL( aµˆ decreaseS + B qˆ) of the diphoton invariant mass from 757 GeV to based on more than 12 fb− . The result is shown· | in Figure 3. There is no excess present in the larger722 data GeV. Inset, a second so that event the that earlier passed thesignal selection was in just Ref. a [1 statistical], one track previously fluctuation. associated to one of where theS and twoB selectedare theprobability photon candidates, density reconstructed functions for as the a convertedresonant diphoton photon, is production not considered pro- as originating Whilefrom the the excitement photon conversion. has faded, The photon the candidate excess thus provides fails the forpiso a< nice0.05 requirement set of exercises. and the event In the T + following,does we not assume pass the selection. that a similar excess was discovered in the process e (p1)e−(p2) γ(q )γ(q ). For simplicity, we will assume that the world is described by QED with a single→ 1 In2 the 2016 data set, no significant deviation from the background-only hypothesis is observed at the flavor ofvalue fermions, of the mass electrons corresponding of mass to theme most, and significant a heavy excess scalar in 2015 particle data. TheS of compatibility mass M, between the 2016 data and the best fit signal associated to the largest excess in the 2015 data is investigated by 1 M 2 = + (∂ S)(∂µS) S2 + g Sψ¯ ψ g SF F µν . (205) LQED+S LQED 2 µ − 2 ψ − γγ µν You have derived the Feynman rules for this theory11 in Section 7.4.3 in we will use it for all of the following exercises. We will compute diphoton production in this model and see how the resonance S can be observed in the cross section.

9.1 Decays of the scalar S The heavy particle S can decay into ligher ones and we now analyze the decays which arise at tree level in . LQED+S

38 (1) (2) (3)

+ Figure 4: Tree-level diagrams for e−e γγ in the presence of a scalar S. →

+ Exercise 9.8. Compute the decay width for the process S e−(p )e (p ). → 1 2 Exercise 9.9. Compute the decay width Γ(S γ(q )γ(q )). → 1 2

Exercise 9.10. The decay S γγ arises even when gγγ = 0, but only at the one-loop level. Draw the relevant Feynman one-loop→ diagrams for this case.

Exercise 9.11. Voluntary bonus exercise: Provide a rough estimate (factors of the cou- plings and dimensional analysis) under which conditions the tree-level decay mediated by gγγ dominates over the loop level process. In the following, we assume that these conditions are fulfilled and neglect loop processes.

9.2 Diphoton production

+ In the following, we will compute the cross section for the process e−(p1)e (p2) γ(q1)γ(q2) in the presence of a scalar S. Usually this process is called electron positron→ annihilation and it is the classic example which demonstrates how in quantum field theories particles are created and annihilated. We will work in the center-of-mass frame and will parameterize

p = E(1, 0, 0, β) , p = E(1, 0, 0, β) , (206) 1 2 − q = E(1, sin θ, 0, cos θ) , q = E(1, sin θ, 0, cos θ) , (207) 1 2 − − where β = ~p /E = 1 m2/E2. | | − e p Exercise 9.12. Compute the amplitudes , and for Feynman diagrams shown in M1 M2 M3 Figure 4. The first two diagrams capture the contributon of QED, QED = 1 + 2, while the last diagram is due to the new particle S. M M M

We now compute the squared amplitude, averaged over the spins of the incoming fermions and summed over the polarizations, i.e. 1 2 (208) 4 |M| s1,sX2,λ1,λ2 39 To get the sum over polatizations, one can use the substitution

∗ (k, λ) (k, λ) g . (209) µ ν → − µν Xλ Note that this amounts to summing both over both physical and unphysical photons and so the substitution looks problematic. However, the contribution from the unphysical photon states cancels out in QED cross sections as will be shown in the QFT lecture. The reason is the Ward identity, the statement that the amplitudes vanish when one substitutes µ(k, λ) kµ for any of the polarization vectors involved in a given process. → The squared amplitude has the form

2 = † + † + † + † (210) |M| MQEDMQED M3M3 MQEDM3 M3MQED   2 In the context of the search for a new resonance, the first contribution QED is called the background. The second one is the signal and the third contribution is the|M interference| of sig- nal and background. After performing the spin sums using (204), each contribution involves a trace of γ matrices. Having three diagrams, we end up with nine traces which need to be evaluated. To make the following computations less tedious we now set me to zero. Since we work at high energies, the effect of the electron mass on the result should be very minor.

Exercise 9.13. Show that the interference term vanishes in the limit m 0 so that e → 2 = 2 + 2 + (m ) . (211) |M| |MQED| |M3| O e Hint: remember that traces odd numbers of γ matrices vanish.

Exercise 9.14. Compute the signal process 2. |M3| Exercise 9.15. Compute the QED result 2. |MQED| dσ Exercise 9.16. Based on these results, provide the differential cross section dθ .

Exercise 9.17. How does the QED cross section behave in the limit θ 0 ? How would it behave if we had not neglected the electron mass in the propagator denominators?→

The detectors in realistic experiments cannot cover the entire solid angle. The reason is that there are beams of particles entering and exiting the detector, and the particle detectors cannot be placed too close to the beams. For this reason, particles with scattering angles below a certain value θ cannot be detected. At the LHC the minimal angle is θ 2◦. min min ≈ Exercise 9.18. Compute the integrated cross section σ after imposing a cut that the angle of the scattered particles with respect to the beams has to be larger than θmin.

40 Figure 5: Loop corrections which lead to a shift of the pole in the propagator of the scalar S.

9.3 Finite-width effects

2 2 2 If the center of mass energy of the collision, s = (p1 + p2) = 4E , is equal to MS our result for the cross section explodes, because the propagator of the scalar S hits the mass shell. This is obviously nonsensical. Near this pole, loop corrections can significantly alter the result, despite the fact that they are suppressed by higher powers of the coupling constants. Indeed, the loop effects shift the pole into the complex plane because they involve imaginary parts which arise because the particle S can decay. The relevant diagrams are shown in Figure 5. The computation of these loop diagrams is beyond the scope of these exercise but the net effect is to shift the propagator 1 1 + ... (212) p2 M 2 → p2 M 2 + iΓ M − S − S tot S where the dots indicate non-pole terms. The quantity Γtot is the total decay rate. If S only + decays to e e and γγ,Γ = Γ + + Γ , but even in our simple model, other decays, such − tot e e− γγ + + as S e e e e also occur. (You have computed the decay rates Γ + and Γ in Section − − e e− γγ 9.1.) → Exercise 9.19. Show that (212) amounts to replacing 1 1 + ... (213) (s M 2)2 → (s M 2)2 + M 2 Γ2 − S − S S tot in the cross section. This is called the Breit-Wigner shape.

2 Instead of a pole, the cross section now displays a peak at s = MS with height proportional 2 2 to 1/(MS Γtot). This peak structure, on top of a smooth background, is what is shown by the red curve in Figure 2.

Exercise 9.20. Show that in the narrow width limit Γ/M 0 → 1 π δ(s M 2) . (214) (s M 2)2 + M 2 Γ2 → Γ M − − This replacement of the propagator with a δ-function is called the narrow width approximation.

41 100

80

60 Bin - counts 40

20

0 200 400 600 800 1000

Ecms [GeV]

+ Figure 6: Fake e e− γγ data from nonexistent exeriment. →

Exercise 9.21. Show that the signal cross section in the narrow width approximation is obtained as

+ + σ(e−(p )e (p ) γ(q )γ(q )) = σ(e−(p )e (p ) S(q)) Br(S(q) γ(q )γ(q )) (215) 1 2 → 1 2 1 2 → × → 1 2 where the branching ratio for the decay into two photons is defined as 1 Br(S γ(q1)γ(q2)) = Γ(S γ(q1)γ(q2)) (216) → Γtot →

Physicswise, (215) states that the full process can be viewed as the cross section for producing the particle S, times the probability that it decays into two photons. Note that the lifetime τ of a particle is τ = 1/Γtot. A narrow width therefore corresponds to a long lifetime. For a short- lived particle the approximation (215) does not work (among other things due to interference effects), but for a long-lived particle we can view the entire process as a production of an S, which much later later decays into γγ.

9.4 Data analysis Figure 6 shows how an experimental signal for our scalar particle could look in data. The (fake) data shows an excess of events around s = m2 (770 GeV)2. This excess could be γγ ≈ due to a scalar resonance with mS 770 GeV, but also due to a statistical fluctuation. Note that the data shows fluctuations at≈ lower energies.

Exercise 9.22. What is the probability that the peak near 770 GeV is due to a statisti- cal fluctuation? More precisely, what is the probability that a fluctuation of the background

42 occurs which is as large as the observed signal? Would the observed excess count as a 5σ discovery of the new particle?

For a large number N of events in a bin, one expects fluctuations of √N around the mean value. A criterion to assess a theoretical model is to compute the quantity n 1 χ2 = (N N th)2 , (217) N th i − i i=1 i X th where Ni is the expected number of events in the bin, according to the theoretical model. For a typical fluctuations, one expects that for n bins χ2 n. The χ2 distribution for n independent Gaussian random variables is known. For n random≈ values, the probability density at a value x = χ2 reads

1 x/2 n 1 p (x) = e− x 2 − . (218) n n/2 n 2 Γ 2

Exercise 9.23. Show that pn(x) is properly normalized and that the average is indeed n.

Exercise 9.24. Derive pn(x), which is obtained from the n-dimensional Gaussian integral

n/2 n 2 1 P y2 p (x) = (2π)− d y δ x y e− 2 i i . (219) n − i i Z  X  Hint: use n-dimensional spherical coordinates. Remember that you already checked the nor- malization in the previous exercise!

To solve Problem 9.4, assume that there is no resonance present. Then the appropriate the- th ory is QED and the number of events at Ecms = √s should behave like N (s) σ(s) = k/s, where k is a constant. The bins and bin counts in Figure 6 can be found on the∼ course web page. Compute χ2 as a function of k and then choose k to minimize the χ2 value. This minimum value will be much larger than n because of the peak near 770 GeV. What is the probability p of a statistical fluctuation which leads to such an equal or larger χ2 value? If we compare to a one dimensional Gaussian distribution, how many σ do we have to be away from the mean to have the same small probability?

Exercise 9.25. Consider now only the four data bins in the energy region 740 GeV to 820 GeV and perform the same test, with the k value obtained previously. What is the p-value and the σ-value in this case? This second version is called the local p-value and the associated σ significance is larger because we have zoomed in on the excess. In the case of the diphoton resonance, the local excess was around 4σ, while the global one was merely 2σ.

Exercise 9.26. Bonus exercise: What is the mass mS, width Γ and the coupling combina- tion gψgγγ extracted from the data?

43 References

[1] ATLAS collaboration, ATLAS-CONF-2015-081.

[2] CMS Collaboration, CMS-PAS-EXO-15-004.

[3] ATLAS collaboration, ATLAS-CONF-2016-018.

[4] CMS Collaboration, CMS-PAS-EXO-16-018.

[5] A. Strumia, arXiv:1605.09401 [hep-ph].

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