Quantum Field Theory Spring 2019 Problem sheet 3 (Part I)

Transcription

1 Quantum Field Theory Spring 2019 Problem sheet 3 (Part I) Part I is based on material that has come up in class, you can do it at home. Go straight to Part II. 1. This question will be part of a take-home assignment for extra credit. You are not allowed to communicate with anyone other than Ana, Guadalupe or Valeri. You are encouraged to check your ideas and get feedback before you hand in the assignment. Make two examination-type questions (with answers) for a closed book exam. One should be a short question, something that can be answered in a few lines. The other, a longer question or problem, requiring calculations, maybe with several parts. Your questions don t necessarily have to be original, but they should be relevant, clearly formulated and fair. Something you would like to see in the final exam (perhaps you will!). Make sure you quote all your sources. For the rest of the problems you can work as usual with others. 2. Consider plane-wave solutions to the Dirac equation (conventions as in Peskin and Schroeder chapter 3). Check that the plane-wave solutions with positive frequency, Ψ = u(p)e ipx are given by ( ) p σξ u(p) = p σξ where ξ is an arbitrary two-component spinor, whereas the ones with negative frequency Ψ = v(p)e +ipx are given by ( ) p ση v(p) = p ση and η is a two-component spinor unrelated to ξ. Notice that there are two independent choices for the spinor ξ and two independent choices for η. We can call these ξ 1, ξ 2 and take them mutually orthogonal, (ξ 1 ) ξ 2 = 0 (or (ξ r ) ξ s = δ rs ), and η 1, η 2, also mutually orthogonal, (η r ) η s = δ rs. When you have finished, check that you agree with the normalization conditions in eqs. (3.55) (3.65) of Peskin Schroeder. An intermediate step is to show that the equation satisfied by u(p) is ( p/ m)u(p) = 0 whereas v(p) has to satisfy ( p/ + m)v(p) = Check the spin sum rules (conventions as in Peskin Schroeder section 3.3), u s (p)ū s (p) = p/ + m s v s (p) v s (p) = p/ m s

2 for two choices ( of ) the spinors ( ) ξ s : 1 0 a) ξ 1 =, ξ 0 2 = 1 b) ξ 1 = ξ, ξ 2 = ξ The spinors in part b) have spin up and down (parallel and antiparallel) along an axis with polar coordinates (θ, φ), see Peskin Schroeder.. Dirac adjoints. a) If ( i / + m)ψ(x) = 0 find the equation satisfied by Ψ = Ψ γ 0. b) Check also that ( p/ m)u(p) = 0 ū( p/ m) = 0 c) Check that the current J µ = e Ψγ µ Ψ is conserved and its time component (the charge density) is positive definite (unlike in the Klein-Gordon case!).

3 Quantum Field Theory Spring 2019 Problem sheet 3 (Part II) 1. Convince yourself of the following property of Mandelstam s variables for scattering of two particles into two particles: s + t + u = i (m i ) 2 It may be easier to consider a scattering diagram with four particles with masses m 1 and incoming momenta p 1,...p so that momentum conservation gives p 1 + p 2 + p 3 + p = 0. Then s = (p 1 + p 2 ) 2, t = (p 1 + p 3 ) 2, u = (p 1 + p ) a) Convince yourself that the two-body invariant phase element is given in the centerof-mass frame by dπ 2 = dω 1 32π 2 2 p 1 s (center of mass) where p 1 is the momentum of one of the final particles, that scatters into a solid angle Ω. [The definition of the integral dπ 2 includes the four-momentum conservation factor (2π) δ(e 1 +E 2 s)δ 3 ( p 1 + p 2 )]. The last factor tends to 1 at high energy (do you agree?) b) Insert this expression in the differential cross section and convince yourself that if all particles have equal masses (in particular, in the massless limit at very high energies) the center of mass cross section is ( ) dσ = 1 M 2 dω com 6π 2 (all four masses identical; distinguishable particles) s 3. Show that the Mandelstam variables for scattering of two massless particles into two massless particles with scattering angle θ in the center of mass frame are t = s sin 2 (θ/2), u = s cos 2 (θ/2). Check that this is consistent with the property s + t + u = i m2 i that you proved in problem 1. These expressions should be approximately correct in the high energy limit, when masses can be neglected. Notice that t 0 at forward scattering (θ = 0) whereas u 0 for backward scattering (θ = π). In the next three problems we practice computing Feynman diagrams for QED processes.. Go to page 131 of Peskin and Schroeder, where you can find the diagram for the lowest order contribution to the process e + e µ + µ.

4 a) Copy the diagram carefully. Go to Appendix A.1 and look at the Feynman rules. Why is this the lowest order contribution to the process? (is there a lower order diagram allowed for this process?). b) Convince yourself that there are no other diagrams contributing to the perturbation series at this lowest order. Notice that squaring the amplitude obtained from this diagram will give you a probability which is O(α 2 ) (α is the fine structure constant). c) Write down the value of the diagram (the amplitude) using the Feynman rules (notice the implicit µ label on the vertex in the Feynman rules, because of the γ µ ). Remember to impose momentum conservation at each vertex and to integrate over all undetermined momenta if there are any left. Check for symmetry factors. d) check your answer by going back to page 131. e) Write down two diagrams which contribute to the same process at next-to-leading order. You don t have to calculate them. f) Write down two other diagrams that contribute at higher order in the perturbation series.. 5. Write down the diagram for the lowest order contribution to electron-muon scattering, e µ e µ and repeat a) to f). The answer to c) is on page 153 of Peskin and Schroeder. Notice that this diagram looks like the previous diagram turned on its side. This leads to a relation between the probabilities for both processes (after squaring the amplitudes and summing over polarizations to get the unpolarized cross sections) which is known as crossing symmetry. 6. Compute the value of the diagram on eq. (7.15) of Peskin and Schroeder. Some comments: Notice that the fermion propagator is a matrix, so you have to follow the arrows in the fermion propagators to get the matrices in the right order (actually you have to follow them backwards, since S F ab has the outgoing state on the left, i.e. the row index). On the other hand the photon propagator can be anywhere in the integral. Also, because of the arrow, there is no symmetry factor to worry about. The answer looks like this (subscript 2 because the diagram is second order): iσ 2 (p) = ( ie) 2 d k i( k/ + m) (2π) γµ k 2 m 2 + iɛ γ µ i (p k) 2 + iɛ 7. Now consider the electron two-point function (also for QED), < Ω T ψ(x) ψ(y) Ω >.

5 a) Draw the first two terms in the perturbation expansion. It is a sum of (connected) diagrams with external points x, y that starts with the Feynman propagator, d p i( p/ + m) e ip x eip y (2π) p 2 m 2 + iɛ The next order term is a diagram like the one you just calculated with the external points added. Write it down, your answer should be: where A is the matrix d p i( p/ + m) e ip x (2π) p 2 m 2 + iɛ [ iσ i( p/ + m) 2] eip y p 2 m 2 + iɛ = d p (2π) e ip (x y) i( p/ + m) p 2 m 2 + iɛ A i( p/ + m) A = [ iσ 2 ] p 2 m 2 + iɛ b) The next order has several terms. Draw one where a second photon is emitted before the first one is absorbed; another where the second photon is emitted after the first one has been absorbed (there are others, can you think of one?). Notice that this last one looks like d p i( p/ + m) e ip x (2π) p 2 m 2 + iɛ [ iσ i( p/ + m) 2] p 2 m 2 + iɛ [ iσ i( p/ + m) 2] eip y p 2 m 2 = + iɛ d p e ip (x y) i( p/ + m) (2π) p 2 m 2 + iɛ AA (Recall the exponentiation of vacuum bubbles; do you see any pattern emerging? You could continue to sum a whole class of diagrams in which photons are emitted and absorbed one after another. Here there are no (1/n!) factors so the series is not an exponential, it is geometric. More on this later.) 8. a) Convince yourself that you agree with the calculation of the unpolarized cross section for e + e µ + µ at lowest order O(α 2 ). In the center-of-mass frame, and ignoring the electron mass, dσ dω = α2 1 m2 µ [( m 2 µ ) ( m 2 µ ) cos 2 s E 2 E 2 E 2 θ ] b) Why can you ignore the electron mass in this computation? c) What is the total cross section for a center-of-mass energy of 100 MeV? (Hint: take a look at fig. 5.1 on page 137 of Peskin and Schroeder. What is the mass of the muon?)

6 d) Show that in the massless (high energy) limit, 1 spins M 2 128π 2 α2 s 2 [( ) 2 t + 2 ( ) 2 ] u 2 9. a) Go to page 9 in Peskin and Schroeder. The claim is that all diagrams in figure 1. contribute to the O(α 3 ) term in the e + e µ + µ cross section. Do you agree? (notice that some of those diagrams have three vertices and some have four vertices) b) Is it consistent to ignore the electron mass at tree level in this case?