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  Polarization Sums in QCD for the calculation of parton model splitting functions

+ 4 like - 0 dislike

Before i state the actual problem, here's a premise. In the case of a Spin 1 massive particle it's possible to demonstrate that $$\sum_{\lambda=0,\pm1}\epsilon_{\lambda}^{* \ \mu}\epsilon_{\lambda}^{\nu}=-g_{\mu\nu}+\frac{q^\mu q^\nu}{q^2}$$ for a massless particle it will be $$\sum_{\lambda=\pm1}\epsilon_{\lambda}^{* \ \mu}\epsilon_{\lambda}^{\nu}=-g_{\mu\nu}+\frac{q^\mu n^\nu+ q^\nu n^\mu}{q \cdot n}-\frac{q^\mu q^\nu}{(q\cdot n)^2}$$
where $n^\mu=(1,0,0,0)$ and $$n\cdot \epsilon=0\\
q\cdot \epsilon=0\\
q\cdot n=q^{0}$$
Ok, now to my understeanding in QED due to the gauge invariance of the theory under $U(1)$  follows the Ward identity: $$q_\mu \mathcal{M^\mu}=0$$ which implies that for all practical purposes we can drop all the terms except for $-g_{\mu\nu}$

My problem lies in the calculation of a QCD process (all particles are assumed massless ) $g \ (gluon)\rightarrow q \ \bar{q}$ needed to compute the splitting function $P_{qg}$ (the probability that a gluon converts into a quark wich carries a fraction of the impulse of the gluon) which is paramtrized in the following way $$K_A (gluon)=(p^0,0,p) \\ K_B(q)=(zp+\frac{p_{\perp}^2}{2 \ zp},p_{\perp},zp) \\
K_C(\bar{q})=((1-z)p+\frac{p_{\perp}^2}{2 \ (1-z)p},-p_{\perp},(1-z)p)$$ such that $$K_A=K_B+K_C$$ provided that $$p^0=p+\frac{p_{\perp}^2}{2 \ z(1-z)p}$$ which gives the gluon a small virtuality. Up to $O(p_{\perp}^4)$ the following identities are true: $$\tag 1 K_A\cdot K_B=\frac{K^2_A}{2}\\
K_A\cdot K_C=\frac{K^2_A}{2}\\
K_B\cdot K_C=\frac{K^2_A}{2}\\
K^2_A=\frac{p_{\perp}^2}{z(1-z)}$$now the authors of the article state that is important to consider: $$\sum_{\lambda=\pm1}\epsilon_{\lambda}^{* \mu}\epsilon_{\lambda}^{\nu}=-g_{\mu\nu}+\frac{K_A^\mu n^\nu+ K_A^\nu n^\mu}{q \cdot n}-\frac{K_A^\mu K_A^\nu}{(K_A\cdot n)^2}$$ because the middle term $$\frac{K_A^\mu n^\nu+ K_A^\nu n^\mu}{q \cdot n}$$ when plugged in the trace (which comes from the sum over polarizations of $\mathcal{ M}(g\rightarrow q \ \bar{q})$) $$ tr({K}\!\!\!/_C  \gamma^\mu   {K}\!\!\!/_B \gamma^\nu)$$ gives a non zero contribution.

I have two problems: 

the first is a conceptual one, why in a QCD calculation i have to consider all the polarization sum terms unlike in QED? Is it due to the fact that QCD is non- abelian? If so, where it comes from mathematically speaking?

The second problem is a practical one: the product $$\frac{K_A^\mu n^\nu+ K_A^\nu n^\mu}{q \cdot n} \cdot tr({K}\!\!\!/_C  \gamma^\mu {K}\!\!\!/_B \gamma^\nu)=-8 p_{\perp}^2+O(p_{\perp}^4)$$ but if i actually do the calculation it yelds me zero since $$tr(\gamma^\alpha \gamma^\mu \gamma^\beta \gamma^\nu)=4(g_{\alpha\mu}g_{\beta\nu}-g_{\alpha\beta}g_{\mu\nu}+g_{\alpha\nu}g_{\beta\mu})$$ then we have form the product : $$\frac{1}{K_A \cdot n}[tr(K\!\!\!/_C K\!\!\!/_A K\!\!\!/_B n\!\!\!/)]+tr(K\!\!\!/_C n\!\!\!/ K\!\!\!/_B K\!\!\!/_A)]$$ which should become $$\frac{8}{K_A \cdot n}[(K_C \cdot K_A)(K_B \cdot n)-(K_C \cdot K_B)(K_A \cdot n)+(K_B \cdot K_A)(K_C \cdot n)]$$ but from eq. $(1)$ we know that becomes: $$\frac{8}{K_A \cdot n }\cdot \left( \frac{K_A^2}{2} \right)[(K_B+K_C-K_A)\cdot n]$$ which should be zero for the conservation of energy! what i am doing wrong?  thank you for any help.

asked Sep 2, 2015 in Theoretical Physics by fra (155 points) [ revision history ]
edited Sep 4, 2015 by fra

1 Answer

+ 2 like - 0 dislike

For the practical problem: When i wrote the parametrization for the four momentums of the particles involved in the scattering i did wrongly assume that the gluon's momentum $K_A$ should have been corrected in order to account for it's virtuality. What did evade me was that the virtuality of the gluon was already codified in the quark and antiquark momentums $K_B$ and $K_C$ via the term $p^2_\perp$. Infact let's write $K_A=(p,0,0,p)$ (as if it was real) and the other two as in the question above. Then from conservation of energy and by taking the square we have: $$K^2_A=(K_B+K_C)^2=\frac{p^2_\perp}{2z(1-z)}+O(p^4_\perp)$$ so, since $K^2_A\neq0 $ we see that we have already accounted for the gluon having a small virtuality which comes from the parametrization of the quarks four momenta. There is no need then to further modify $K_A$ in order to account for it's vituality. Furthermore the substitution i made in the question effectively cancelled the virtuality of the gluon! (like i added it in $K_B$ and $K_C$ and subtracted it in $K_A$)

 If we try to do the calculation of: (the prevoius relationships of the momenta are mostly false now!) $$\frac{8}{K_A \cdot n}[(K_C \cdot K_A)(K_B \cdot n)-(K_C \cdot K_B)(K_A \cdot n)+(K_B \cdot K_A)(K_C \cdot n)]$$ 

now we get, knowing that:

$$K_A \cdot n=p\\K_B \cdot n=zp+\frac{p^2_\perp}{2zp}\\K_C \cdot n=(1-z)p+\frac{p^2_\perp}{2(1-z)p}\\K_A \cdot K_B=\frac{p^2_\perp}{2z}\\K_A \cdot K_C=\frac{p^2_\perp}{2(1-z)}\\K_B \cdot K_C=p^2_\perp+\frac{z \ p^2_\perp}{2(1-z)}
+\frac{(1-z) \ p^2_\perp}{2z}$$
exactly the result it should have come:  $$\frac{8}{K_A \cdot n}[(K_C \cdot K_A)(K_B \cdot n)-(K_C \cdot K_B)(K_A \cdot n)+(K_B \cdot K_A)(K_C \cdot n)]=-8p^2_\perp+O(p^4_\perp)$$

To the conceptual side: In non abelian gauge theory the non physical degrees of freedom of the gluon do not cancel themeselves when calculating scattering amplitudes as in QED. A way to say this is: it depends upon the fact that the underyling gauge symmetries are different and that causes a modification to the generating functional and so to the ward identities. Indeed that's why there are the ghost fields that have the exact role of eliminating the non physical degrees of freedom. 

Now then in order to obtain the correct physical amplitude we should account for the ghost contributions to the process. This isn't very convenient when calculating such an easy process as a tree level amplitude, it's much much easier to manually cut the non physical degrees of freedom by replacing the full polarization sum with only the transverse one.

answered Sep 15, 2015 by fra (155 points) [ no revision ]

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