Spin and Polarisation Terms

Recall the spinor/polarisation completeness relations:

${\sum }_s{u}_{\alpha }(p,s){\bar{u}}_{\beta }(p,s)=(/p+m{)}_{\alpha \beta }$

${\sum }_s{v}_{\alpha }(p,s){\bar{v}}_{\beta }(p,s)=(/p-m{)}_{\alpha \beta }$

${\sum }_{\lambda }{ϵ}_{\mu }^a(p,\lambda ){ϵ}_{\nu }^{\ast a}(p,\lambda )=-g_{\mu \nu }+\frac{p_{\mu }{n}_{\nu }+p_{\nu }{n}_{\mu }}{(p.n)}$

Here, s labels the spin states of the fermions, λ labels the two helicities of the gluon and n is an arbitrary reference vector.

In QED, we could use the replacement, ${\sum }_{\lambda }{ϵ}_{\mu }{ϵ}_{\nu }^{\ast }\to -g_{\mu \nu }$ to simplify our calculations.

But in QCD, this does not give the correct result!

QED Case

Consider the scattering $q\bar{q}\to \gamma \gamma$.

The amplitude may be written as:

$M_{q\bar{q}\to \gamma \gamma }=M_{q\bar{q}\to \gamma \gamma }^{{\mu }_1{\mu }_2}{ϵ}_{{\mu }_1}^{\ast }(k_1,{\lambda }_1){ϵ}_{{\mu }_2}^{\ast }(k_2,{\lambda }_2)$

$M_{q\bar{q}\to \gamma \gamma }^{{\mu }_1{\mu }_2}=M_{q\bar{q}\to \gamma \gamma }^{(1),{\mu }_1{\mu }_2}+M_{q\bar{q}\to \gamma \gamma }^{(2),{\mu }_1{\mu }_2}$

Charge conservation (or Ward identity) tells us that the QED amplitude vanishes when we replace the photon polarisation with its momentum, as

QED: $k_{1,{\mu }_1}{M}_{q\bar{q}\to \gamma \gamma }^{{\mu }_1{\mu }_2}=k_{2,{\mu }_2}{M}_{q\bar{q}\to \gamma \gamma }^{{\mu }_1{\mu }_2}=0$

This justifies the polarisation sum replacement rule, as the terms in the sum polarisation proportional to the photon momentum do not contribute to the squared amplitude.

QCD Case

Analogous to QED, considering the scattering $q\bar{q}\to gg$.

(Re-labelling needed) The amplitude can be written as:

$M_{q\bar{q}\to gg}=M_{q\bar{q}\to gg}^{{\mu }_1{\mu }_2}{ϵ}_{{\mu }_1}^{\ast }(k_1,{\lambda }_1){ϵ}_{{\mu }_2}^{\ast }(k_2,{\lambda }_2)$

Considering the polarisation stripped amplitude $M_{q\bar{q}\to gg}^{{\mu }_1{\mu }_2}$ in QCD, we obtain,

$M_{q\bar{q}\to gg}^{{\mu }_1{\mu }_2}=(t^{a_2}{t}^{a_1})M_{{\mu }_1{\mu }_2}^{(1)}+(t^{a_1}{t}^{a_2})M_{{\mu }_1{\mu }_2}^{(2)}+i{f}^{a_1{a}_2{a}_3}{t}^{a_3}{M}_{{\mu }_1{\mu }_2}^{(3)}$

$=(t^{a_1}{t}^{a_2})\underset{=QED}{\underbrace{[M_{{\mu }_1{\mu }_2}^{(1)}+M_{{\mu }_1{\mu }_2}^{(2)}]}}+i{f}^{a_1{a}_2{a}_3}{t}^{a_3}[-M_{{\mu }_1{\mu }_2}^{(1)}+M_{{\mu }_1{\mu }_2}^{(3)}]$

Hence, in QCD, $k_{1,{\mu }_1}{M}_{q\bar{q}\to \gamma \gamma }^{{\mu }_1{\mu }_2}\ne k_{2,{\mu }_2}{M}_{q\bar{q}\to \gamma \gamma }^{{\mu }_1{\mu }_2}\ne 0$. Instead, the gluons are physical, i.e., if their polarisations are transverse, ${ϵ}_{{\mu }_1}{k}_1^{{\mu }_1}={ϵ}_{{\mu }_2}{k}_2^{{\mu }_2}=0$

then, $\text{QCD: }{k}_{1,{\mu }_1}{M}_{q\bar{q}\to \gamma \gamma }^{{\mu }_1{\mu }_2}{\epsilon }_{{\mu }_2}^{\ast }(k_2,{\lambda }_2)=k_{2,{\mu }_2}{M}_{q\bar{q}\to \gamma \gamma }^{{\mu }_1{\mu }_2}{\epsilon }_{{\mu }_1}^{\ast }(k_1,{\lambda }_1)=0$.