0
$\begingroup$

The following amplitudes of the 3-gluon vertex are obtained only from momentum conservation (shown in Quantum Field Theory and the Standard Model, Schwartz):

$$M^{+++}=C^{abc}\frac{1}{\langle12\rangle\langle23\rangle\langle31\rangle}$$

$$M^{++-}=\frac{[12]^3}{[13][32]}$$

But by using the Feynman rule for 3 gluons and then applying spinor helicity formalism and choosing some reference momenta, it is not possible to obtain these amplitudes.

For example, given the amplitude for the 3-gluon diagram

$$M=-gf^{abc}((p_1-p_2).\epsilon_3 (\epsilon_1.\epsilon_2) + (p_2-p_3).\epsilon_1 (\epsilon_2.\epsilon_3) + (p_3-p_1).\epsilon_2 (\epsilon_1.\epsilon_3))$$

for the following choice of reference momenta

$$r_{\epsilon_1(p_1)}=p_3, r_{\epsilon_2(p_2)}=p_3 , r_{\epsilon_3(p_3)}=p_1$$

I get the final result $$M^{+++}=-gf^{abc}(\frac{[12][32]}{\langle31\rangle} + \frac{[12][31]}{\langle32\rangle})$$

or some other different choice of reference momenta

$$r_{\epsilon_1(p_1)}=p_3, r_{\epsilon_2(p_2)}=p_1 , r_{\epsilon_3(p_3)}=p_2$$

I get the final result $$M^{+++}=-gf^{abc} (\frac{[12][32]}{\langle31\rangle} + \frac{[12][31]}{\langle32\rangle} + \frac{[32][31]}{\langle12\rangle})$$

I don't understand why these amplitudes that I get using Feynman rules are different from the ones in the book and why different reference momenta lead to different final amplitudes.

$\endgroup$
3
  • $\begingroup$ I don’t see a problem with two different methods yielding the same correct result. $\endgroup$ Jun 4 at 12:25
  • $\begingroup$ I have a different result using Feynman rules. $\endgroup$
    – Ana
    Jun 4 at 14:59
  • 1
    $\begingroup$ In that case, you should include your detailed computation in the question. $\endgroup$ Jun 4 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.