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which, in the limit e —» 0, gives k = /30, where (30, defined in Eq. (17.25), determines how the coupling constant "runs". If k is positive, the theory is asymptotically free.

The key to the demonstration is therefore the calculation of Zg. From Eq. (16.121), there are three equivalent forms for Zg:

where we ignore the possibility of using Z4 (why?). The diagrams which contribute to the six renormalization constants which enter (17.32) are shown in Fig. 17.5. We will choose the last combination in (17.32) which is comparatively simple to calculate.

We begin with the lowest order calculation of Z3, the renormalization constant for the ghost propagator, labeled in Fig. 17.6. Using the Feynman gauge (a = 1) and the Feynman rules for QCD, summarized in Appendix B, we obtain

This integral can be shown to be zero in dimensional regularization (see Prob. 17.1), but we will find it convenient to transform it and add it to the other gluonic contributions. To this end, shift k —> k + and multiply and divide by A-. Then f ddk I f ddk A- _ f ddk -k2-\g2 J (2n)d A+ J (2tr)<* J (2n)d A+A- ' 1 ' '

With this transformation, (17.40) may be added to the other purely gluonic contributions to II (from graphs A and B), giving

/rldk l x {2 {fqu - <r<?2) + (d- 2) [<T (k2 + \q2) - 2 k»kv] }.

We now evaluate this by combining denominators and shifting k —> k + I<7(1 - 2x). The new denominator becomes

and using the replacement k^k" —► k2g^u/d, we obtain n^(gluon) = -i92RN8ab J\x x{(îY-r«2) [2- (f -1) (i -2x)2]

First look at the gauge violating g^v term. We can show that it is identically zero by direct integration:

Finally, to complete the calculation, we must evaluate the gcc vertex corrections. These come from the two diagrams shown in Fig. 17.8. The first of these diagrams is

{2ir)d A+A-Ao x [<T (fc - fg)" - gl/p2kfl + g^ (k + §q)"} , (17.51)

where A0 = —(p — k)2. We need to calculate only the divergent part, which comes from the terms proportional to two powers of k in the numerator. The shift in k which we will eventually make will not change these terms, and using faef fcfdfbde = |N facb = —^Nfabc

(which is proved in Appendix D), we obtain immediately

2 J (2n)d A+A-AQ Using (17.52), the second diagram (B) gives

N f ddk

9RKbc(B) - 93Rfabc 2 J {2iv)d j4+j4_j4o ddk 1 (2n)d A+A^Ao

where we have again kept only the divergent part (terms proportional to two powers of k in the numerator). Adding diagrams A and B together gives

9r~KI

abc n r ddk

-gRJabc

Now, combining the denominators and shifting k, as we did in the evaluation of Eqs. (16.44M 16.48), give

where

Fig. 17.9 The square of the e+e~ annihilation amplitude to second order in the strong coupling constant g consists of 5 contributions. The leading term, of order is the square of the e+ + e~ —► q + q amplitude and is shown in (A). The corrections of order g2 come from the square of each e+ + e~ —> q + q + g amplitude, diagrams (B) and (C), and the interference of the two amplitudes, diagrams (D) and (E). The vertical dashed line indicates that the intermediate particles are on their mass-shell in these diagrams.

Fig. 17.9 The square of the e+e~ annihilation amplitude to second order in the strong coupling constant g consists of 5 contributions. The leading term, of order is the square of the e+ + e~ —► q + q amplitude and is shown in (A). The corrections of order g2 come from the square of each e+ + e~ —> q + q + g amplitude, diagrams (B) and (C), and the interference of the two amplitudes, diagrams (D) and (E). The vertical dashed line indicates that the intermediate particles are on their mass-shell in these diagrams.

As we discussed in Chapter 10, the total hadronic e+e~ cross section depends on the square of the matrix elements given in Figs. 10.7 and 10.8. The square of these matrix elements are shown in Fig. 17.9. In these figures the vertical dotted line which cuts the diagram means that all particles cut by this line are on their mass-shell, and from the general unitarity principle discussed in Sec. 11.7, we know that these contributions are just the imaginary parts of the corresponding diagrams with vacuum polarization insertions. In fact, because the external e+e~

Fig. 17.10 The correction to the ratio R is equal to the ratio of the imaginary part of the fourth order self-energy to the imaginary part of the second order self-energy.

Fig. 17.11 The diagram with overlapping divergences has two cuts, giving the contributions of both Fig. 17.9D and E. Hence Fig. 17.10 is consistent with Fig. 17.9.

couplings are the same in all graphs, the ratio R is just the ratio of the imaginary parts of the vacuum polarization diagrams themselves, as illustrated in Fig. 17.10. Here there are only three fourth order graphs, while there are four contributions shown in Fig. 17.9, but the imaginary part of the overlapping diagram has two cuts, as illustrated in Fig. 17.11, so the counting is correct.

In Chapter 16 we calculated the leading fourth order contributions to the vacuum polarization. From Eq. (16.114), the finite part which dominates at high q2 is a a

This was a purely electromagnetic result but can be modified to apply to the problem under consideration if we multiply each fourth order graph by the correct color factors, which are simply tr (§A0 iAa) = 4 .

The lowest order graph has a color factor of tr 1 = 3. Hence the result (17.63) is modified for QCD to

where as = g2/4n is the strong fine structure constant, which can be expected to run as shown in Eq. (17.62). The imaginary part of the log for large positive q2 is simply

Im log

and hence for each flavor of quark the hadronic production cross section is proportional to

L 7T

Multiplying this by the square of the charge of each quark flavor, Qf, summing over all flavors produced, and dividing by the same factor for /i+/x", which is simply —a, give

We have obtained the QCD correction factor reported in Chapter 10.

This calculation was very simple because the difficult work of obtaining Eq. (16.114) had already been carried out. The calculation illustrates how QED calculations may, in some cases, be extended to QCD.

PROBLEM

17.1 In Eq. (17.41) we transformed the integral

Evaluate the right-hand side of this equation by combining denominators and shifting k, and thereby show that it is identically zero. Use the same method to prove that for any a > 0. Can you find an argument to justify this? (See Muta (1987). Sec. 2.5.5.) ]

[ In dimensional regularization, it is often assumed that

With this definition, the spinors satisfy the following normalization and orthogonality relations:

u(p,s)u(p,s') = 2m6as> v (p, s) u (p, a') = 0 (A 19)

The completeness relations are expressed in terms of the positive and negative energy projection operators u (p, s) u (p, s) =j> +m = 2mA+ (p)

The u and v spinors are related by charge conjugation:

Relativistic Quantum Mechanics and Field Theory

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APPENDIX B

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