Physics 523, Quantum Field Theory II Midterm Examination

Transcription

1 Physics 53, Quantum Fiel Theory II Miterm Examination Due Monay, 9 th March 004 Jacob Lewis Bourjaily University of Michigan, Department of Physics, Ann Arbor, MI

2 PHYSICS 53: QUANTUM FIELD THEORY II MIDTERM EXAMINATION. Loop Integrals in Dimensional Regularization We are to verify the ientity q π) n)q m q m ) n+ = 0. Noting the results of homework 6 an the elementary properties of the Γ-function, we may procee irectly. q π) n)q m q m ) n+ = n) )n i 4π) / = )n i 4π) / Γn + ) Γ ) ) n Γn + ) m ) n / m )n+ i n + 4π) / Γn + ) [ / n) Γ ) )] n Γ n + + m m, ) n / m ) n+ / m ) n+ /, = )n i [ n /)Γn /) + Γn + /)], 4π) / Γn + ) m ) n / = )n i [ Γn + /) + Γn + /)], 4π) / Γn + ) m ) n / = 0. q n)q m π) q m ) n+ = 0..a) Let us now evaluate the following loop integral, Ip, m, m ) = ie q π) q + p/) m + iɛ)q p/) m + iɛ). óπɛ ɛδɛι δɛ ιξαι To evaluate this integral lucily, let us first introuce the change of variables k q +p/. Introucing the Feynman parameter x, the integral becomes, k x π) [xk p) m + iɛ) + x)k m +. iɛ)] Introucing the variables, 0 l k xp an xx )p + xm + x)m, we see that Ip, m, m l ) = x 0 π) [l + iɛ], [ )] i Γ = x, 0 4π) / / i [ ] 4 4π) x 0 ɛ log γ E + log4π) + Oɛ) Ip, m, m i [ ] ) 4 4π) x 0 ɛ + log xx )p + xm + γ E + log4π)..b) x)m óπɛ ɛδɛι δɛ ιξαι

3 JACOB LEWIS BOURJAILY The One-Loop Structure of Quantum Electroynamics While stuying the superficial ivergences of quantum electroynamics, we note that gauge invariance an hence the War ientity mae several superficially ivergent iagrams either converge or vanish. We are to verify these claims explicitly. Superficially, the one-point function of the photon has a cubic ivergence. Let us emonstrate that in fact, to the one-loop orer, the one-point function of the photon vanishes. To one-loop orer, we see that = k The amplitue for the above iagram is given by q im = )ɛ [ ] k q) π) Tr i k + m e ) k m e + iɛ) ieγ ), = ɛ k Tr kγ + mγ ) q)e 4π) k m e + iɛ), = ɛ k k q)4e 4π) k m e + iɛ), = 0. Therefore, to one-loop orer, = 0. óπɛ ɛδɛι δɛ ιξαι Similarly, we argue that although the photon three-point function has a superficial, linear ivergence, its amplitue shoul also vanish. Let us now emonstrate this fact. To one-loop orer, we see that k + k = k k3 k k 3 Note that the secon iagram has been labele the same as the first iagram but with relative minus signs on the momenta fa k. This is because the Feynman propagator has the property that kf= i k + m e) k m whereas fa kf= i k + m e) e + iɛ) k m e + iɛ). Let us consier the evaluation of the first iagram. Its amplitue is proportional to integration over Tr [γ k + m e )γ k + m e )γ k 3 + m e )]. Because only those traces over an even number of γ-matrices are non-vanishing, this is equal to Tr [γ k γ k γ k 3 ] + m e Tr [γ k γ γ ] + Tr [γ γ k γ ] Tr [γ γ γ k 3 ]). Notice that the only remaining traces involve an o number of momenta k. Similarly, we see that the amplitue of the secon iagram is proportional to integration over Tr [ k 3 + m e )γ k + m e )γ k + m e )γ ] = Tr [ k 3 γ k γ k γ ] m e Tr [ k 3 γ γ γ ] + Tr [γ k γ γ ] Tr [γ γ k γ ]). But, noting ientity 5.7) of Peskin an Schroeer, the traces of each expression are equal. Therefore, the negative contribution from the secon iagram cancels the contribution from the first. k k k 3 k = k k 3. Therefore, to one-loop orer, = 0. óπɛ ɛδɛι δɛ ιξαι

4 PHYSICS 53: QUANTUM FIELD THEORY II MIDTERM EXAMINATION 3 Lastly, our analysis showe that the photon four-point function has a logarithmic, superficial ivergence, but by gauge invariance this amplitue is convergent. We are to emonstrate that the photon four-point function oes not iverge to the one-loop orer in perturbation theory. To one-loop orer, we see that I II III IV V VI = σ σ) σ ) σ ) σ ) σ ) σ ) Because it is our task to emonstrate that the above amplitue converges rather than actually compute the amplitue we may make several helpful simplifications. To illustrate the first major simplification, let us analyze the first iagram, I). p 3 k p 3 p 4 k k p 3 p 4 σ p k p p = e 4 k π) Tr [ k p + m e )γ k + m e )γ k p 3 + m e )γ k p 3 p 4 + m e )γ σ ] k p ) m e)k m e)k p 3 ) m e)k p 3 p 4 ) m e), = e 4 k π) Tr [ kγ kγ kγ kγ σ ] k m e) 4 + finite terms. Therefore, we see that the ivergent part of each iagram is a function of only the orer of γ-matrices in the trace. Now, we claim that the ivergence of iagram I) is the same as II), III) IV), an V) VI). First, note that the relative change of sign for the loop momentum k between each pair will not change the ivergence of the iagram because each involves only k 4 = k) 4. Seconly, the orering of the vertices are precisely reverse for each pair an so by ientity 5.7) of Peskin an Schroeer they are equal. Therefore the total ivergence of these six iagrams will be twice that of I), III), an V) alone. Let us continue to compute the ivergence of iagram I) before illustrating the sum of all six iagrams. Because, as we will show, the sum of the iagrams will converge, we will continue without imensional regularization. In our calculation below, we will repeately make use of γ-matrix algebra prove in homework incluing that of semester I). Also, note our use of ientity A.4) from Peskin an Schroeer. Let us begin to evaluate the ivergence of iagram I). The integran is proportional to Tr [ kγ kγ kγ kγ σ ] = k α k β k γ k δ Tr [ γ α γ γ β γ γ γ γ γ δ γ σ], + ) k ) g αβ g γδ + g αγ g βδ + g αδ g βγ ) Tr [ γ α γ γ β γ γ γ γ γ δ γ σ], Tr[γγ γ γ γγ γ γ σ ] + Tr[γγ γγ γγ γ γ σ ] + Tr[γγ γγ γγ γ γ σ ], = Tr[ γ )γ γ )γ σ ] + Tr[ )γ γγ γ γ γ σ ] + Tr[γγ γ )γ γ γ σ ], = 4Tr[γ γ γ γ σ ] Tr[γ 4g γ σ ] Tr[ γ γ γ γ σ ], = 8Tr[γ γ γ γ σ ] 8g Tr[γ γ σ ], = 3 g σ g g σ g + g σ g ) 3g g σ, g g σ g g σ + g σ g ). Therefore, when we evaluate the amplitue for all six iagrams, the ivergent integral will be over a term proportional to g g σ g g σ + g σ g ) together with the analogous terms uner the other two istinct permutations. Therefore, the amplitue s ivergence will be proportional to, g g σ g g σ + g σ g ) + g g σ g g σ + g σ g ) + g g σ g σ g + g g σ ) = 0. Therefore, the photon s four-point function is convergent to loop-orer in QED. óπɛ ɛδɛι δɛ ιξαι It is easier for our purposes to work with = 4 trace-algebra. Because the total ivergence will vanish in = 4, it must also vanish in general imensional regularization.

5 4 JACOB LEWIS BOURJAILY The β-function of Quantum Chromoynamics We are given that, at one-loop orer in perturbation theory, the ivergent parts of the counter terms of quantum chromoynamics are δ = 7 g Λ log 4π) M, δ = g Λ log 4π) M, an δ 3 = 5 ) g 3 n Λ f log 4π) M, where the δ i are efine in analogy to quantum electroynamics. We see that these irectly imply that B g = 7 g 4π), A f = g 4π), an A gl = 5 ) g 4π), where A f correspons to fermion self-energy an A gl correspons to gluon self-energy. Let us now compute the β-function for the strong coupling g. This correspons to the iagram, δ δ δ δ 3 Therefore, because β g = gb g ga F ga gl, we see that β g = ) g 3 6π. In homework 0, we compute the general running coupling constat associate with quantum chromoynamics. To relate that result with our work here, we shoul set the unetermine constant β to ). So from our results of homework 0, we see that the square of the running coupling g is g = g 3.a) + g 8π 3 n ). 3.b) f logp/m) We see that the coupling constant will be asymptotically free if > /3n f. This is because asymptotic freeom is irectly a result of a negative β-function. It is clear that β g < 0 only if n f < 33/ = 6.5. Also, again by the results of homework 0, we see that at large energy p/m ), the square of the coupling constant can be approximate by g 8π p ) M. 3.c) logp/m)

6 PHYSICS 53: QUANTUM FIELD THEORY II MIDTERM EXAMINATION 5 ± W ± e ± e