Final Exam Solutions

Transcription

1 Final Exam Solutions. Pair production of electrons from two photons. a) I refer to the initial four-momentum of the cosmic ray photon by q µ and the photon in the background q µ. The requirement for the pair production is q + q ) = q q > 4m e. We choose our coordinate system such that q is along the z- axis, q µ = E,,, ). According to the assumption in the problem, photons in the background have four-momenta q µ = E, sin θ cos φ, sin θ sin φ, cos θ) with E =3 4 ev. q q =E E cos θ), which is maximized when cos θ =. Therefore the minimum energy is determined by q q ) max =4E E > 4m e, i.e., E >m e /E =.5 MeV) /3 4 ev) = GeV = 87 TeV. b) I believe these are obvious. c) You always have the following common factors. The amplitude has two coupling constants e, the phase space β f 8π and the flux factor. The kinematical factors s in the amplitude are different between two cases. i) Close to the threshold, the wave functions of electron, positron are roughly m,andthet- oru- channel electron propagators p + m)/p m ) /m. The photon polarization vectors are O). Therefore, the amplitude behaves as M e. The cross section then is σ β f 8π 8m e ) = πα β f. ii) Well above the threshold, the electron, positron 4m spinors are E, while the electron propagator /E. Therefore the amplitude is roughly M e. There is, however, an angular dependence of the amplitude, which is actually M e sinθ/)/ cos θ) ore cosθ/)/ + cos θ) which gives us a logarithmic enhancement factor. We ll put this back in at the very end. Setting the angular dependence aside, we obtain σ 8π s e ) = 4πα.Nowwe s put the logarithmic enhancement factor in, and the estimate is σ 4πα log s s m. d) The amplitude is im = ie) ūp)γ µ i k m γν v p)ɛ µ q )ɛ ν q ) ) +ūp)γ µ i k m γν v p)ɛ µ q )ɛ ν q )) )

2 Here, k = p q and k = p q. The relative sign between two amplitudes is plus, because they are related by Bose symmetry, q q. e) We calculated the spin-summed squared amplitude for the Compton scattering in the class. For initial four-momenta p i k i ) of electron photon), and similarly for final ones, we found M helicities =8e 4 p i k f + p i k i +m p i k i p i k f ) + m 4 ) p i k i p i k f p i k i p i k f ). Using crossing, we can obtain the spin-summed squared amplitude for γγ e + e by substituting k i q, k f q, p i p, p f p, and changing the overall sign. M helicities =8e 4 p q + p q +m p q p q + ) m 4 + ) p q p q p q p q.3) The symmetry between q q in the result is apparent. What is less apparent is the symmetry p p: p q = p + q p q ) = m + q p) =p q. This symmetry holds thanks to the charge conjugation invariance of the QED which interchanges electron momentum p) and positron momentum p) in the final states, but does not affect the initial state. f) The total cross section is given by σ = ) β f 8π s dω 4π helicities M 4) First, one needs to write out the inner products of four-momenta explicitly. Let us use the center-of-momentum frame to simplify it. Remember the total cross section is Lorentz-invariant.) Use q µ = E,,, ), q µ = E,,, ), p µ = E,psin θ,,pcos θ), p µ =E, p sin θ,, p cos θ), with p = Eβ = E m. Here, we have chosen the rotational invariance to fix the p direction to lie in the xz plane. Then, p q = E + Epcos θ, p q = E Epcos θ. The spin-summed

3 squared amplitude becomes helicities { E M = 8e 4 Epcos θ E + Epcos θ + E + Epcos θ E Epcos θ ) +m E + Epcos θ + E Epcos θ ) m 4 E + Epcos θ + E Epcos θ. 5) An integral of the terms inside the curly brackets over cos θ = togives 4 E 3 p 4 Em p +8E 4 +8E m 4 m 4 ) arctanh p E ) E 3 p 6) Hey, I obtained a lot messier expression before; this is a lot simpler!) Therefore, ) β f d cos θ σ = M 8π s helicities = β f { 4 E 3 p 4 Em p +8E 4 +8E m 4 m 4 ) arctanh p ) } 4 8π s 8e4 E E 3 p { = e4 β f 4 E 3 p 4 Em p +8E 4 +8E m 4 m 4 ) arctanh p ) } E 7) 6πs E 3 p We show a plot of the cross section in Fig.. g) The terms in the curly brackets have a limit 4 when p, E m. Therefore, σ = e4 β f β f 6πs 4=πα. 8) m The estimate in b) was a factor of four off. Not too bad. h) Let us use an approximate order of magnitude for the cross section, πα /m.the mean free path due to a scattering with the background photons is roughly l πα m ζ3) π T 3 =. 33 MeV 9) and using hc = 97 MeV fm this is a useful constant to remember!) l 9.7 cm = 3.5 kpc ) 3

4 . σ [b]. s [MeV ] Figure : The cross section of γγ e + e in the unit of barn b = 4 cm,asa function of s = E CM. This is somewhat shorter than the distance between the Earth and our galactic center 8 kpc, but we cannot say it is definitely shorter because we made only a crude estimate. To be more precise, we need to integrate over the Planckian distribution of photon energies and also directions). However, this is not important anyway because there are many other and more important processes which can scatter incoming photons inside the galactic disk. The background photons are important when there is basically nothing between the source and the Earth, which is the case for an extragalactic source of high energy photons. The calculated mean free path is much shorter than an intergalactic distance. Therefore, ultra-high-energy gamma rays from extrargalactic sources are all converted to electron-positron pairs due to scattering with microwave background photons. Their fate is to cause showers due to electromagnetic cascades. The produced eletron-positron pairs further lose energies by emitting breamsstrahlung photons, or cause further pair production by scattering with microwave background photons, and result in a larger number of electrons, positrons, and photons in a narrow cone. They are called showers. Such high energy photons, therefore, can still be detected in a form of showers even though the primary photon itself won t reach the Earth directly. But this tells us that looking for photons traditional astronomy) wouldn t be so useful at very high energies. To the best of my knowledge, whether there exist infrared radiations which may potentially scatter the high-energy photons is not understood. 4

5 Note Many of you asked what the heck ζ3) is. This is a very standard outcome in statistical mechanics of relativistic particles, but maybe you are not familiar with it. The number density of a boson is given by n = g d 3 p π h) 3 e βe µ) As usual, β =/kt. The factor g = counts the spin degrees of freedom. Photons do not have a chemical potential, and E = cp. Therefore, n = g 4πp dp π h) 3 e βcp Now you expand the integrand in the following manner, n = g π h 3 p dp e nβcp 3) Integrating each terms in the sum gives the Γ-function, and we obtain n= n= n = g π h 3 Γ3) nβc) = g kt)3 3 π hc) 3 n= ) ) n 3 4) The last factor is ζ3) =.6... which is an irrational number like e or π. In general, ζs) n= n s Finally, we have n = g ζ3) π ) 3 kt 5) hc I have further taken the natural unit, k = h = c = in the problem. If you have a thermal gas of relativistic fermions, the only change in the above calculation is that you have a sum over series with alternating signs, e βcp + = ) n e nβcp 6) n= The result of the phase space integral also has a sum over alternating signs, n = g π hc) 3 5 ) n 7) n 3 n=

6 The summation can be done using the following trick. n= ) n n 3 = n= n 3 + ) = ζ3) 3 ζ3) = 3 ζ3) 8) 3 4 Therefore, n = g 3 ) 3 ζ3) kt 9) 4 π hc for relativistic fermions. You can also calculate the energy density in a similar way. It helps to know that ζ) = π /6. You obtain Stefan Boltzman law of the energy density of the black body radiation in this manner. 6

7 . beta-function in φ 4 theory. a) The Feynman rule is that we put a factor of iλ for each four-point scalar vertex. The one-loop amplitude is i λ d 4 p i ) π) 4 p m The reason for / is the following. From the definition of the two-point function in the interaction picture, the term at Oλ) is given by the correlation function, Tφx)φy) d 4 z i) λ 4! φz)4 ) There are four choices on which φz) is contracted with φx), and three more choices on which of the remaining three φz) is contracted with φy). There still remains two φz) s, which are contracted with each other to give the above oneloop amplitude. Therefore, the overall factor is i λ 4! 4 3= iλ ) It has an additional factor of / compared to a naive expectation. In general, a loop amplitude which involves a real scalar by itself has an additional multiplicative factor of /. The result does not depend on the four-momentum of the external line. Therefore, it cannot contribute to the coefficient of q term in the PI two-point function, and hence not to the wave-function renormalization. The wave function renormalization appears only at the two-loop level in this theory. On the other hand, the above one-loop amplitude renormalizes the mass of the scalar boson. From the power counting, it is easy to see that it is quadratically divergent! This makes φ 4 theory very sensitive to physics at the ultraviolet cutoff. Not very nice. This is called the naturalness problem and is relevant to the Higgs boson in the standard model. b) Call the initial four-momenta to be p and p, and final ones p 3 and p 4. one-loop amplitude then is given by im = d 4 q i i iλ) 4π) q p p ) m q m The +p + p ) p p 3 )) + p + p ) p p 4 )) 3) 7

8 Here again we have a multiplicative factor of /. The reason is the same as the above case. We need to calculate the four-point correlation function at Oλ ). It comes from the following correlation function, Tφx )φx )φx 3 )φx 4 ) d 4 z i) λ! 4! φz ) 4 d 4 z i) λ 4! φz ) 4 4) Here, the reason for /! is because it is the second-order term in the Taylor expansion of Te i d 4 zh int. But this is not the reason for the final /. First of all, there are three different types of Wick contractions. One is to contract φx )and φx )withthesamez, sayφz ) 4,andφx 3 )andφx 4 )withφz ) 4 s-channel). The other is to contract φx )andφx 3 )withthesamez t-channel), the last φx )andφx 4 )withthesamez u-channel). This is why we have a sum of three different terms. Now let us look at one particular contraction, where φx ) and φx ) are contracted with the same z s-channel diagram). First of all, there are two choices, whether they are contracted with z or z. Both of them give the same amplitude, and they are added together. This factor of two cancels the factor of /! in the Taylor expansion. We pick z and drop the factor of /! hereafter. Now there are four choices on which φz ) to contract with φx ), and three more on which remaining three φz ) to contract with φx ). At this point, we have a factor of iλ/4! 4 3 = iλ/. Exactly the same situation occurs with the contraction of φx 3 ), φx 4 )withφz ) 4. At this point, two φz ) and φz ) remain uncontracted. They are contracted with each other, with two possibilities. Therefore, the overall coupling constant factor is iλ/) = iλ) /. You can check that exactly the same factor appears also for t-channel and u-channel amplitudes. c) After setting all external momenta to vanish, we obtain im = 3 d 4 ) p λ 5) π) 4 p m because s-, t-, and u-channel amplitudes become the same. One way to evaluate the amplitude is to use Pauli Villars regulator with a mass M, and do the following. im = 3 λ = 3 λ d 4 p π) 4 p m d 4 p M dµ ) π) 4 m 8 ) p M ) ) 3 6) p µ

9 Now we perform the p integration first using the by-now-familiar formula, im = 3 M λ dµ ) )3 i Γ) m 4π) Γ3) µ = 3 λ ) i Γ) 4π) Γ3) log M m λ = i 3 4π) log M 7) m Another method is an explicit cutoff. Going back to Eq. 5), we first perfom the Wick rotation and cutoff the momentum integration with p E <M, im = 3 d 4 ) p E λ i π) 4 p E + 8) m = i 3 M π p ) λ E dp E π) 4 p E + 9) m The integration can be done in parts, im = i 3 λ p M M 4π) E p E + m dp E p E + m = i 3 λ +logp 4π) E + m ) ) = i 3 λ +log M ) 3) 4π) m The results based on two different methods disagree on the constant term, but the logarithmically divergent piece have the same coefficient. d) The calculated im above is added to the bare couping iλ at the lowest order, and hence, λ = λ 3 λ 4π) log M 3) m According to the definition of the β-function in the problem, βλ) = M dλ dm = 3 λ 3) λ 4π) Here, we used the fact that the λ in the one-loop term can be replaced by λ up to higher order corrections. 9

10 Note In this problem, a couple of you asked me what is the momentum scale Q here analogous to that in the effective QED coupling constant. The answer is somewhat complicated. The amplitude here depends on three kinematical variables, s, t, andu. This is actually the one-particle-irreducible PI) four-point function Γ 4) p,p,p 3,p 4 ), or equivalently, Γ 4) s, t, u). How you define an effective coupling constant as a function of a single momentum scale Q is up to you. One choice made in the textbook is s = t = u = Q, which is in the unphysical region, i.e., such a kinematics never arises for an S-matrix element. But this is a popular definition. Another choice is to use dimensional regularization and the so-called MS renormalization scheme, which is also popular. Whatever your definition is, the following is always true, which does not require you to define the effective coupling constant as a function of a single kinematical variable. It is the repetition of the argument we did in the class. The PI four-point function is a function of external momenta, the bare coupling constant, and the ultraviolet cutoff parameter M. To be more explicit, Γ 4) = Γ 4) s, t, u, λ,m ). Now we study the same function with a scaling of the fourmomenta, Γ 4) e α s, e α t, e α u, λ,m ). In other words, we study the same PI fourpoint function with all kinematical variables scaled by the same factor e α.ifweneglect the mass m, a dimensional analysis tells you that Γ 4) e α s, e α t, e α u, λ,m )= Γ 4) s, t, u, λ,e α M ). Therefore, s s + t t + u ) u Γ 4) s, t, u, λ,m ) = α Γ4) e α s, e α t, e α u, λ,m ) α= = α Γ4) s, t, u, λ,e α M ) α= = M M Γ4) s, t, u, λ,m ) 33) And at least at the one-loop level) we find the last expression to be independent of the kinematics s, t, oru. Therefore, what you have calculated is useful independent of the precise definition of the effective coupling constant λq ). e) I am aware that this question was confusing.) In dimensional regularization method, the bare coupling constant λ has a mass dimension of ɛ, whereitake the space-time dimension to be D = 4 ɛ. The one-loop amplitude in Eq. 5) is now imµ ɛ = 3 d D ) p λ = 3 i Γɛ) π) D p m λ 4π) ɛ Γ) m ) ɛ 34)

11 using the formulae which we discussed in the class. The factor µ ɛ is introduced to make the amplitude M dimensionless, and µ is an arbitrary parameter with unit mass dimension. By differentiating the total amplitude with tree-level term) by µ with λ fixed, λ λ = M = λ µ ɛ 3 4π) ɛ µ ɛ Γɛ)m ) ɛ 35) λ β = µ dλ dµ = λ ɛ)µ ɛ 3 4π) ɛ ɛ)µ ɛ Γɛ)m ) ɛ 3 36) 4π) In the last part, we took the limit ɛ, and used a perturbative expansion λ = λµ ɛ + Oλ ). Note Usually we expand the amplitude in power series in ɛ, λ λ = M = λ µ ɛ 3 λ µ ɛ ) 4π) ɛ λ µ ɛ 3 λ µ ɛ ) 4π) ɛ γ + Oɛ ) m µ ) ɛ m γ +log4π) log ɛ µ + Oɛ) ) 37) If you define the renormalized coupling constant λµ ) by absorbing the pole and γ +log4π) pieces, λµ )=λ µ ɛ 3 λ µ ɛ ) ) 4π) ɛ γ +log4π) 38) it is called the MS coupling constant. This is scheme is very commonly used in QCD. Note This excercise is a sloppy way of calculating the β-funcion in dimensional regularization, which works only at one-loop. I learnt this method from the book by Ryder. Let me describe what the correct method is. As usual, we focus on the PI four-point function, Γ 4) s, t, u, λ ) which is calculated with the bare coupling λ. Note that in the bare theory, Γ 4) has a mass dimension of ɛ. What we are interested is the momentum dependence of the four-point function. We scale all four-momenta by an overall common factor e α, and use the dimensional analysis, Γ 4) e α s, e α t, e α u, λ )=e ɛα Γ 4) s, t, u, e ɛα λ ) 39)

12 Now we define the renormalized four-point function by Γ 4) R s, t, u, λ) =µ ɛ Z / ) 4 Γ 4) s, t, u, λ ) 4) which is dimension-less. Note that we regard the renormalized four-point function as a function of λ, notλ. Z is the wave-function renormalization factor. In our case, it is unity at the one-loop level. Using the definition of Γ R and also the dimensional analysis of Γ, we find Γ 4) R e α s, e α t, e α u, λ, µ )=Γ 4) R s, t, u, λ, e α µ ) 4) Therefore, the correct way is to fix λ, differentiate it with µ and change the sign. However, it is equivalent to fixing λ and not changing the sign because the amplitude is λ µ ɛ m ) ɛ = λ µ ɛ m ) ɛ 4) at this order. f) By writing down the differential equation, in a slightly different form, both sides can be easily integrated and we obtain Q dλ dq = 3 4π) λ 43) dλ λ = 3 4π) d log Q, 44) λq ) + λq ) = 3 4π) log Q, 45) or, λq λq )= ) 46) 3 λq ) log Q 4π) Q Therefore, the coupling constant grows for larger energies, or descreases for smaller energies: infrared-free. The situation is very similar to that in the QED. However, if we have such a theory with a relatively large size of λ at energy scale directly observable by experiments, the coupling constant may diverge not too far above such energy scales. If you require, for instance, that the coupling constant Q

13 stays finite up to the Planck scale, you obtain an upper bound on the size of λm Z ), for instance. Actually, the Higgs boson in the Standard Model has φ 4 -type coupling, and the mass squared of the Higgs boson is proportional to coupling constant λ. Therefore, an upper bound on λ translates into an upper bound on the Higgs boson mass. By requiring that the coupling constant remains perturbative up to the Planck scale, we obtain m H < 5 GeV or so. 3

14 3. Electron Electric Dipole Moment. The main point of this calculation is that the contribution vanishes if only φ L or φ R ) propagates inside the vertex diagram because it would give a combination ūp ) +γ 5 γ 5 up), 47) andevenifiσ µν q ν is obtained inside,itcommuteswithγ 5 and the end result vanishes because +γ 5 γ 5 =. Therefore, we need a diagram which starts with φ L and ends with φ R or vice versa. This is why we need to use vertex iam or ia m in the diagram. If the diagram starts with φ L and ends with φ R, the structure of the amplitude becomes ūp ) γ 5 γ 5 up), 48) and iσ µν q ν term in would survive. Given this point, and using the Feynman rule for the scalar QED given in the book p. 3), we find the following amplitude when φ L is emitted from the electron, converts to φ R on the way and is absorbed by the electron: im LR = [ d 4 k π) 4 ūp ) i e) γ 5 i k µ i e) γ 5 up) i ia i m) ie)p + p k) µ i p k) ML p k) ML i ie)p + p k) µ i ia i m) p k) MR p k) ML p k) M R + p k) MR First note that the photino propagator simplifies to ]. 49) γ 5 i γ 5 k µ = γ 5 i k + µ) γ 5 k µ = iµ γ 5 k µ 5) because { k, γ 5 } =. We also assume M L = M R,andthen im LR =e 3 A mµūp ) γ 5 d 4 k up) π) 4 [ p + p k) µ [p k) M ] [p k) M ] + p + p k) µ [p k) M ][p k) M ] ] k µ. 5) 4

15 We use Feynman parameter integral to simplify the integrand as usual. By using the identity abc = d 3 zδ z z z 3 ) [z a + z b + z 3 c], 5) 3 we act / b on both sides and find ab c =6 d 3 z zδ z z z 3 ) [z a + z b + z 3 c]. 53) 4 Since the integrand we have has the form ab c + abc,weuse ab c + abc =6 d 3 z + z 3 zδ z z z 3 ) [z a + z b + z 3 c]. 54) 4 d 4 [ k p + p k) µ π) 4 [p k) M ] [p k) M ] + p + p k) µ ] [p k) M ][p k) M ] k µ d 4 k = π) 6 d 3 z + z 3 )p + p k) µ zδ z 4 z z 3 ) 55) [k z p k z 3 p k M ] 4 where we ignored p ) = p = m M in the denominator and set µ = M. We shift k k +z p +z 3 p, and we ignore p p =m q M in the denominator = d 4 k π) 6 4 d 3 zδ z z z 3 ) z ) z 3 )p + z )p k) µ [k M ] 4. 56) d 4 k integration can be done using the general formula, and we find = d 3 i z ) z 3 )p + z )p ) µ zδ z z z 3 ). 57) 4π) M ) Now the integration volume is symmetric under z z 3, and we can replace z 3 ) by [ z 3 )+ z )]/ = z z 3 = z. Therefore, = z i z )z p + p ) µ dz dz = i 4π) M ) 4π) M p + 4 p ) µ. 58) By combining with the prefactors and setting µ = M), we find im LR =e 3 A mm i 4π) M 4 ūp ) γ 5 up)p + p ) µ. 59) 5

16 Now we use Gordon decomposition technique. As done in the class, we know that Therefore, ūp ) γ 5 up)p + p ) µ =ūp ) γ 5 mγ µ up) = p µ up) iσ µν p ν up), ūp)mγ µ = ūp)p µ +ūp)iσ µν p ν. mγ µ + iσ µν p ν )up)+ūp )mγ µ iσ µν p ν ) γ 5 up) = γ µ piece + ūp ) γ 5 iσ µν q ν )up). 6) We used q = p p. The last piece is of our interest. We now find im LR =e 3 A mm i 4π) M 4 ūp ) γ 5 iσ µν q ν )up). 6) Another amplitude of creating φ R first and absorbing φ L is obtained by simple replacements, γ 5 +γ 5, A m Am. Therefore, the total is given by im total = e 3 m i 4π) M 4 ūp ) iσ µν q ν )ReA + iγ 5 ImA)up) = i α m e ) iσ µν q 4π M 3 ν )ReA + iγ 5 ImA)up). 6) 6ūp Now comes the interpretation of this amplitude as the electric dipole moment of the electron. First notice that the amplitudes can be regarded as the matrix element of an effective Feynman rule ih eff.wefocusontheσ µν γ 5 piece. The effective Hamiltonian can be read off as p H eff p = α m e )iσ µν q 4π M 3 ν )iγ 5 ImA)up)A µ q). 63) 6ūp Here, A µ q) is the vector potential in the momentum space. Since we are applying an electric field, let us take A and other components zero. Then the matrix we need to understand is σ i γ 5. Taking Pauli Dirac representation suitable for non-relativistic limit, we find σ σ i γ 5 = iγ γ i γ 5 = iα i i γ 5 = i σ i 6 ) ) = iσ i iσ i ). 64)

17 Non-relativistic electron states occupy only the upper two components, and we keep only iσ i in our expression. Then p H eff p = α 4π m M 3 e 6 ImA ūp )σ i up) iq i A q). 65) Since iq i A = iq i A = i A = E i, the matrix element is then [ α p H eff p =ūp ) 4π m e M 3 6 ] ImA σ E up). 66) Clearly this is an electric dipole moment d e = α m e ImA. 67) 4π M 3 6 Taking A = Me iφ,wecanwriteima = M sin φ. d e = α 4π m e M 6 sin φ = e cm sin φ ) TeV. 68) M From Gene s result d e =.8 ±. ±.) 6 e cm, we combine statistical and systematic errors in quadrature.) +.) =.6. Taking 95% confidence level i.e., two sigma), we find d e <.5 6 e cm. Comparing this to the expression found from supersymmetry above, we conclude M>44 GeVsin φ) /. 69) If the phase of A is order unity sin φ. This is actually one of the most stringent constraints on supersymmetry. 7