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Chiral Perturbation Theory : pi0 pi0 Z vertex?

by Hepth
Tags: chiral, perturbation, theory, vertex
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Hepth
#1
Dec3-12, 11:59 AM
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Not sure if anyone has any experience with chiral perturbation theory, but I'm trying to see what all of the vertices are for interactions with a single Z boson. I've looked at the lagrangian up to order p^4 so far, and it seems that the Z only interacts with charged pions/kaons.
I'm using the external field method. Is there a reason why these don't appear?
I imagine that I have the generic Z current:
[tex]
\sum_{q = u,d,s} \bar{q} \left( c_L \gamma^{\mu} P_L + c_R \gamma^{\mu} P_R\right) q \cdot Z_{\mu}
[/tex]
where the constants are calculable, but it should be noted that due to weak isospin and charge, the down and strange currents are the same constant, with the up being different.

I imagine this then correlates to chiral perturbation theory as an external field matrix where these constants lie along the diagonal for the left and right handed external fields, which then go into the covariant derivative.

So just naively I would think that if I'm looking for say, K0 in, Z and K0 out:
[tex]
\langle K0(p_i - p_Z) | \bar{s} \left( c_L \gamma^{\mu} P_L + c_R \gamma^{\mu} P_R\right) s | K0(p_i) \rangle \eta_{p_Z}^{*}
[/tex]

As the Vector parts of the current are -1 parity, the axial +1, and the Kaons -1, We have
(-1)(c_v (-1)) (-1) = -1 c_v
so a vector, such as the momenta of the kaons or the Z.

The axial parts
(-1) c_a (+1) (-1) = +1 c_a, we have no way to construct a +1 axial vector (polarization is already factorized out)

so only the vector component remains, which is a combination of the left and right components. (I guess I could have just started with that form).

So its momenta dependent, and we have two degrees of freedom, the pZ and the pK.

Now, due to indices we can have things of the form
[tex]
p_Z \cdot \eta_{p_Z} \rightarrow 0\\
p_K \cdot \eta_{p_Z}
[/tex]

So I expect my vertex to be of the form of the second product.

Now when looking for where this contribution comes from in the chiral lagrangian, at O(p^2) we would expect the kinetic term:
[tex]
L_{kin} = -\frac{f^2}{4} tr\left(D^{\mu} U D_{\mu} U^{\dagger}\right)
[/tex]
to generate these, and it does for the K+-, pi+- vertices, but if you work it out (and I think I'm doing it right) the neutral contributions all disappear.

Perhaps its due to the d/s being the same mass in this case,(though the pions/eta also disappear) and I need to break that symmetry to get this term, but then when I do the lagrangians at p^4 with mass included, I still don't get a term with it...

Any ideas or suggestions? http://arxiv.org/abs/hep-ph/9406283 , section 21( page 44) has both the p^2 and p^4 lagrangians.

I'm just trying to see which order I have to go to in order to get this vertex, as it MUST exist right? I mean, theres a definite cross section for e+ e- -> pi0 pi0, and I believe it occurs through the Z -> (ssbar)-> glu/photon off leg -> (ddbar) and the opposite diagram.

Any suggestions, even if only reading materials would be greatly appreciated!

-Hepth

ps Its possible I'm doing the matrix algebra incorrectly, but I'm fairly sure I'm doing it right. (95% sure).
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