Name: Vladyslav Shtabovenko Date: 08/22/17-05:02:29 PM Z


Am 21.08.2017 um 16:04 schrieb Maksym:
> Thank you for the responses.
>
> I don’t understand why I can’t sum to zero when calculating the matrix elements using QED-like axial-vector theory. The only thing which is relevant for the unitarity is the internal structure of the gauge group, which in my toy-like case is just the direct product of the abelian gauge groups. In my initial question, the 1- GA5 coupling from the toy example corresponds to Z-boson which longitudinal degree of freedom decouples for low energies, so I again can make the replacement of the sum over the polarizations by -g_\mu\nu (in my question, the sum over the polarizations appears in the Z propagator).

The replacement of the sum over polarizations by -g_\mu\nu corresponds
to summing over 4 polarizations, including timelike and longitudinal
components. A massless gauge field (e.g. a photon) has only 2 physical
degrees of freedom. A massive gauge field only three. In both cases the
sum over the polarizations of such particles is not just -g_\mu\nu.

Pure QED is really a special case where this works. There is a reason,
why in Eq. 5.55 Peskin puts an arrow and not an equality sign.

If one includes unphysical polarizations to the matrix elements squared
then there must be additional contributions that cancel those, e.g.
ghosts in QCD. Chapter 16.1 in Peskin explains that quite well,
including the issues with unitarity if unphysical contributions are not
removed.

Furthermore, you mentioned that there is a second diagram. I would look
at the squared sum of the full amplitude, not just different pieces.

Anyhow, you can do whatever you want as long as you know that it is
correct. If the result does not make any sense, then probably something
is not correct, right?

>
> Also, this doesn’t tell nothing about why the unitarity is restored just under changing of the sign of the mp term in the propagators nominator of my two examples, without any other changes.
>

To me this looks like an accidental effect. Feel free to repeat the
calculation with FORM or any other tool of your choice and let me know
if you get a different answer.

Cheers,
Vladyslav