1
$\begingroup$

In order to keep the correct degrees of freedom (which are 2) for massless gauge fields one imposes,

$$p^\mu \epsilon_\mu = 0 \tag1$$

Together with the gauge redundacy/equivalence relation,

$$\epsilon_\mu \sim \epsilon'_\mu = \epsilon_\mu + F(p)p_\mu \tag2$$

Because $\epsilon$ and $\epsilon'$ satisfy Eq. (1).

Now, the amplitude of any process with 2 gauge fields in the final state, $\epsilon_\mu\epsilon_\nu M_{\mu\nu}$, has to be gauge invariant and due to Eq. (2), that implies

$$p^\mu·M_{\mu\nu} = 0 \tag3$$

Nevertheless, for non-abelian theories our massless gauge fields transform for small parameter $\omega = \omega(x)$ and $g$ the coupling constant as

$$A_\mu \rightarrow A_\mu + \partial_\mu\omega - ig[A_\mu, \omega] \tag4$$

From here you can say that the gauge redudancy for non-abelian fields should be:

$$\epsilon_\mu \sim \epsilon'_\mu = \epsilon_\mu + F(p)p_\mu + G(p)\epsilon_\mu \tag5$$

But now, from Eq. (5) you get that Eq. (2) is changed, in order to preserve gauge invariance, by

$$(p^\mu + \epsilon^\mu)M_{\mu\nu} = 0 \tag6$$

But in many book appears that in QCD we hold Eq. (2) which should imply null amplitude accordingly to Eq. (6). What am I misunderstanding?

$\endgroup$

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.